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s,),
Proof. We assume that the assumption of a) holds and assume indirectly that X has a wellordered subset W of type w,, say W = {w,,Iv < w,), where the order of the w, corresponds to the order of their indices v. If there exists an ordinal T < w, such that all w, with T 5 v < W, begin with a sequence F, we say: Nearly all w, begin with F. Let K O be the set of all 0components of the elements of W . Then there follows: (1) K O has a greatest element ao.
For otherwise KO would be cofinal in W, which has type w,, and then also KO, and a fortiori So,had to contain a subset of type w,, which yields a contradiction. So we have obtained: For p = 1 the following statement (*) holds: (*) p is an ordinal 5 A, and there exists a sequence (a,),<, with a, E S, for v < p, such that nearly all w, begin with (a,),<,. Let now T be an ordinal 5 X such that (*) holds for all p with O
+
Case 1. T is a successor ordinal p 1. Then, according to our induction hypothesis, there exists a sequence (a,),<,, with a, E S,, for v < p, such that nearly all w, begin with it. Then, quite analogous to the existence of ao,there follows the existence of an element a,, E Spsuch that nearly all w, have a,, as p  component. And then also nearly all w, begin with (a,),<,. So (*) also holds for T (instead of p). Case 2. r is a limit ordinal. Then for very p < T there exists an ordinal v, < w, and a sequence (a,),<, such that all w, with v, v < w, begin with (a,),<,. Since w, is regular and because of 171 < N, the supremum a of {v,lp < r ) is still < w,. And then all w, with a v < w, begin with (a,),<,. Again (*) holds for T (instead of p). Using transfinite induction now there follows that (*) also holds for X (instead of p). But this yields a contradiction, because then nearly all w, must begin with a sequence (a,),<x. And that means that they must be equal which gives a contradiction. For the proof of b) we can use a). Therefore we consider the sets S, with the inverse orders >,of the 5, and consider the set X, but now with the lexicographic order 51with respect to the orders 2, .This
< <
sL
sL)
If in 2.2 all (S,, 5,) are the same linearly ordered set (S, <) we obtain as a corollary to 2.2:
2.3 Theorem. Let w, be a regular initial ordinal, X < w, an ordinal, and S a linearly ordered set. Then there holds: a) If S is w,  free, then the lexicographic product S((X)) is also w,  free. b) If S is w:  free, then S((X)) is also w2y  free.
4.2. PROPERTIES OF LEXICOGRAPHIC PRODUCTS
93
2.4 Remark. The Theorems 2.2 and 2.3 are no longer valid if we don't require that w, is regular: Let S be a linearly ordered set wl wo). In 2.2 we of order type C{tp(w,)lv E (Z 5 0)) (= take X = a := wo. Then X := S((wo)) has a subset of type wwo (= w, . . Iv < wO).Indeed, in X we have subsets Xu = wo w l . {Y) x W, x (1) x (1) x  . for v < WO, and for p < v < wo each element of XCLis < each element of Xu so that the union u{X,IV < wo) is a subset of X of type wwo.But S has no subset of type wwosince every wellordered subset of S intersects only finitely many of the summands of one of the types w,.
+
+ + + +
Now we consider the special case where X and S are ordinals: 2.5 Theorem. Let p and X be ordinals. Then there holds: a) For every wellordered set T p((X)) we have cf(T) 5 sup{p, A). b) For every inversely wellordered subset S p((X)) there holds coin S 5 A*.
Proof. If ,u or X is 0, this is trivial. So let both be # 0. a) If T has a last element, we have cf(T) = 1, and the statement a) is trivial, also if T is empty. In the other case cf(T) is a regular ordinal w,, and p((X)) is not w,  free. Then u , or X is W, because of 2.3 a), and this yields sup{p, A) w, = cf(T). b) Let S be an inversely wellordered subset of p((X)). If S is empty or has a first element, then our statement is clear. In the other case we have coin S = w;, where wp is regular. Then p((X)) is not w;  free. According to 2.3 b) now p is not wz  free or X wp. But every wellordered set, and then also p, is already w:  free. So X must be wp and thus W; = coin S A*.
>
>
>
<
>
From 2.3 we can derive, and this now also for singular w, : 2.6 Theorem. Let w, be an arbitrary initial ordinal, n E N , X < w,. Then there holds: a) n((X)) has no subset of type w, and no subset of type WE. b) n((w,)) has no subset of type w,+l and no subset of type w;+~.
Proof. If w, is regular, a) follows immediately from 2.3. If now w, is singular, and then a a limit ordinal, then from X < w, there also follows X < wp for an ordinal P < a, where P is a successor number so that wp is regular. Because of 2.3, then n((X)) has no subsets of one of the types wp, wi, a fortiori no subset of one of the types w,, WE.
b) is now a consequence of a) by replacing X by w,, and w, by w,+l. Until now we investigated which ordinals (resp. their inverses) do not occur as subtypes of linearly ordered sets. In the other direction we have a trivial statement: 2.7 Theorem. Let p be a limit ordinal, X an ordkaal P((X)) has a cofinal subset of type p. Y
> 0. Then
Proof. Every set {avlv < p) of elements a, E p((X)), where a, has as 0component, is cofinal in p((X)).
Many posets, e.g. Z,Q and R, have the property to be symmetric in the following sense: 2.8 Definition. A linear order on a set S is called symmetric and (S,5 ) a symmetrically ordered set, if there exists a bijective mapping f : S 4 S such that for x, y E S we have x 5 y & f (x) 2 f (y). Then f is said to be a symmetry of (S, 5 ) . An order type is said to be symmetric if it has a symmetric realization. The symmetry of the factors of a lexicographic product of linearly ordered sets is transferred also to their product:
2.9 Theorem. Let X be an ordinal, and (S,, 5,) a symmetrically linearly ordered set for every v < A. Then also the lexicographic product X := X,<xS, is symmetrically ordered.
Proof. For every Y < X there exists a symmetry g, : S, + S,. For a = (a,),<x E X we put f (a) := (g,(a,)),<x. If now b = (b,),<x E X, we have: a 5 b a a, 5 bv for v < X a gv(bv) 5 g,(a,) for v < X ++ f (b) 5 f ( a ) , and thus we have obtained a symmetry f of (X, 5 ) because f is also bijective. Our next investigations concern the existence of infima and suprema of subsets of lexicographic products. Together with this we can describe which lexicographic products are dense (resp. continuous). 2.10 Theorem. Let X be an ordinal, and for v < X let S, be a linearly ordered set without gaps, which has a first element e, and a last element z,, so that S, is completely ordered. Then for every nonempty subset T of the lexicographic product X = X,<xSv there exist inf T and sup T . So the set X has no gaps, but X has a first and B last element, and thus it is a complete lattice.
4.2. PROPERTIES OF LEXICOGRAPHIC PRODUCTS
95
Proof. Let T be a nonempty subset of X. Let a0 E S o be the infimum of the set of 0components of the elements of T . We make the following induction hypothesis: Let T be an ordinal < A, for which already elements a , E S , have been defined for all v < 7.Let T T be the set of all sequences of T that have (a,),<, as r  segment. If TT = 0 we define a, := z,. If TT # 0 we define a, to be the infimum of the set of T  components of the elements of T,. Recursively we have so defined a sequence a = (a,),<x E X . It will follow that a = inf T . First there holds: ( 1 ) a is a lower bound of T . If there would exist a t = (t,),<x E T with t < a , let r be the first index where the components t, and a, are different, i. e. where t, < a, holds. Then t has (a,),<, as T  segment. But then an element a, 5 t, had to be defined, which gives a contradiction. Also a is the greatest lower bound of T . For assume indirectly that there exists an element b = (b,),<x E X with a < b 5 T. Let a be the first ordinal for which a, < b,. If there is no element in T which begins with (a,),<,, then by definition a, = z, b, would follow with contradiction. And if an element of T begins with (a,),<,, then a, is the infimum of the set of a  components of the elements of T,, and b, must be all a  components of the elements of T,, hence b, 5 a, with contradiction. The first element of X is (e,),<x and the last (z,),<x. And so X is infcomplete and complete. If finally ( A ,B) is a proper cut in X,it is no gap because there exists inf B , and this is the least element of B or the greatest of A.
>
<
Ordinals are linearly ordered sets without gaps, and so 2.10 yields the following statement:
2.11 Theorem. Let p be a successor ordinal and X an arbitrary ordinal. Then p ( ( X ) ) has no gaps. If also X is > 0, then p ( ( X ) ) has a first and a last element and is therefore a complete lattice. We supplement 2.11 by:
2.12 Theorem. Let p be a limit ordinal and X 5 w. Then p ( ( X ) ) has no gaps. It is no complete lattice because it has no greatest element. But every nonempty subset has an infimum. Proof. If X < w holds, then p ( ( X ) ) is wellordered according to 1.13. Assume now X = w , and let T be a nonempty subset of p ( ( w ) ) .
We define elements ao, a l , . . . , a,, . . . for v < w as in the proof of 2.10. Then evidently (a,) ,<, = inf T. If now ( A ,B ) is a proper cut in p ( ( w ) ) ,then there exists inf B, and it is the first element of B or the last element of A, so that ( A ,B ) is no gap. We cannot sharpen 2.12, so that also each nonempty subset T of p ( ( w ) ) has a minimum. E.g. the set T of all sequences E p ( ( w ) ) , in which exactly one component is = 1, and all others = 0, is a counterexample. For the sets p ( ( X ) ) there is only one case left which is not yet settled by 2.11 and 2.12, namely: 2.13 Theorem. Let p be a limit ordinal and X an ordinal Then ,u((X))has gaps.
> w.
Proof. Let a = (avlv < W ) be an w  sequence of ordinals a, < p. And let A be the set of all sequences x = (x,lv < A), where xu = a, for v < W , xw < p, xu = 0 for w < v < A. Then A has no last element since p is a limit ordinal. The element m = (mvlv < X),where mu = a, for v < w , m , = O f o r w < v < A, is =minA. Let now b = (b,(v < A) be an upper bound of A, and thus > x for all x E A since ,u has no greatest element. Then we also have b > m. Let 6 be the least ordinal with bs > ms. Now 6 w is impossible. So 6 < w holds, and then b cannot be a least upper bound of A. The cut formed by the initial segment I S ( A ) and the final segment of upper bounds of A is thus a gap.
>
Finally we investigate what can be said about the existence of neighboring elements in the sets p ( ( v ) ) . 2.14 Theorem. Let ,u and X be ordinals > 0. 1) If p and X are lirnit numbers, then p ( ( X ) ) is dense. There are no neighboring elements. 2) If p is a limit number, X a successor number p 1, then there exist i n p ( ( X ) ) neighboring elements. Two sequences a = (a,),<x and b = (bu)u<x of p ( ( X ) ) with a < b are neighboring, i# a, = b, for all v < p, and for their last components a p , bp there holds b p = a p 1. 3) If p is a successor number p 1, X arbitrary, then there exist of neighboring elements i n p ( ( X ) ) . Two elements (a,),<x < p ( ( X ) ) are neighboring i f for the first index r where they difer there
+
+
+
4.3. UNIVERSALLY ORDERED SETS
+
97
holds: b, = a, 1, and a, = p, b, = 0 for all v with ordinals v of course don't exist if if = T 1 holds.)
+
T
< v < A. (Such
Proof. 1) If a = (a,),<x < (ZI,),<~ = b holds for two elements a, b E p((X)), let 7 be the index where a and b differ for the first time, so that a, < b, holds. Every x E p((X)), that has (a,lv 5 r ) as initial segment, and as ( r 1) component the ordinal a,+l 1, is then strictly between a and b. And thus p((X)) is dense. 2) Assume that a < bare neighboring elements, and let r be the index where a and b differ for the first time, hence a, < b,. If also a, 1 < b, would hold, then a and b would evidently not be neighboring. So we assume a, 1 = 13,. If now r 1 < X would hold, then one could find as in 1) an element z strictly between a and b. Therefore we have r 1 = X and thus T = p. So the condition of 2) is necessary. Clearly it is also sufficient. 3) is immediately clear.
+
+
+
+
+
+
Now we can give an answer to the question, for which ordinals p, X the set p((X)) is continuously ordered, i. e. without steps and gaps: 2.15 Theorem. p((X)) is continuously ordered i# p is a limit number and X = wo.
Proof. p((X)) can only be continuous, if it is dense, and then p and
X have to be limit numbers due to 2.14. Because of 2.13, X must be wo. So the condition is necessary, and 2.14 and 2.12 show that it is also
sufficient.
4.3
Universally ordered sets and the sets Ha of normal type rl,
In this section we study N,  universally ordered sets and in particular the sets of type h, (which is the same as Hausdorff's normal type q,). These sets have many nice properties, and they are important to bring a good oversight into the system of linearly ordered sets. 3.1 Definition. A linearly ordered set (S,5 ) is said to be N, universal, if there holds: Every linearly ordered set M of cardinality N, is isomorphic to a subset of S (which is equipped with the induced order), the same: M is embeddable in S.
<
The statement of the following example can easily be verified, but it follows also later in a more general context: 3.2 Example. The set Q of rational numbers with their usual order is an No  universal set of cardinality No.
Of course it is an easy thing to construct a linearly ordered set which is N,  universal, if a is a given ordinal. But if we seek su.ch a set which has the minimal possible cardinality the situation changes. Using the GCH we shall see that there always exists an N,  universal set of cardinality N,. For the theory of universally ordered sets the following sets H, are of fundamental importance. They were introduced by Sierpinski in 1949 ([164]) subsequently to previous work of Hausdorff ([81],Chapter VI) : 3.3 Definition. A dyadic sequence is a sequence (a,lv < p), where p is an ordinal, and where each a, is 0 or 1. Let a be an ordinal. Then we define H, to be the set of all dyadic sequences cp = (cp,lv < w,) E 2((w,)) that have a last digit 1. So for every cp E H, there exists an ordinal (which depends on cp) p < w, with cp, = 1 and cp, = 0 for p < v < w,. We always consider H, as equipped with the lexicographic order of first differences. It can easily be verified that Ho is isomorphic to the set Q with its usual order, and so Ho is No  universal. This makes the conjecture plausible that generally H, is N,  universal.This is indeed the case. It was proved by Sierpinski [l64] for regular cardinals N, and by Gillman [53] also for singular cardinals N,. In 1958 Mendelson [I181 gave a very short and elegant proof which we now present: 3.4 Theorem. For every ordinal a the set H, is N,  universal.
Proof. Let (S,5 ) be a linearly ordered set with IS1 = N,. Then we can represent S in the form S = {u,Iv < w,), where for v < p < w, we have u, # up. Now we construct a <  preserving mapping f : S +H, recursively. The image sequence f (up) of up shall be denoted by up = (ap,Iv < w,). Let f (uo) be the sequence (1,0,O, . . .) of length w,, which has the digit 1 at the first place and at the other components only zeros. Let now p be an ordinal < w,, and for all v < p let f (u,) already be defined. Then f (up) is defined as the sequence a, = (a,, Iv < w,), for which a,, = 1, a,, = 0 for p < v < w,, and which for v < p satisfies
4.3. UNIVERSALLY ORDERED SETS (1) a,, = 0 +=+ up < u,, and (then of course) (1') a,, = 1 e uV < up. Evidently the a,, p < w,, are elements of Ha. We now verify, that f maps the set S isomorphically onto the set A := {aplp < w,). Let v < p < w,. Then we distinguish two cases: Case 1. u, < u,. Here we have up, = 1 because of (1')) and further (2) If a,, = 1 holds for a K < v, then also a,, = 1. Indeed, from the assumption and (1') there now follows u, < u, (< up), and then again with (1') a,, = 1. Then (2) has as a consequence, that at the place, where a, and a, differ for the first time, the corresponding component of a, must be 0 and that of a, must be 1, so that a, < a, follows. Case 2. u, < u,. Then (1) yields a,, = 0 < a,, = 1. If further we have a,, = 1 for a K < v, then (1') yields u, < u, (< u,), hence u, < u,, and again because of (1') this results in a,, = 1. Because of a,, = 0 < 1 = a,, the least ordinal K for which a,, # a,, holds is I v, and then we have a,, < a,,. So finally a, < a, follows. Next we wish to determine the cardinality of the sets H a . But before we introduce the cardinals t,, which often occur in set theory (partly in other notation): 3.5 Definition. Let a be an ordinal. We put to := No. For successor numbers a 1 we put t,+l := 2N". And if a is a limit ordinal we define t, := x u < , 2Nv.Then we have
+
<
3.6 Theorem. 1) For every ordinal a there holds N, %. 2) If we assume the GCH, there holds N, = t, for all ordinals a. 3) Also for the nonlimitordinals a > 0 we have f& = x u < , 2'~.
Proof. 1) For a = 0 , l ) holds by definition. Let now a be an ordinal > 0, and 1) already be proved for all ordinals v < a. If a is no limit ordinal and thus of the form a = ,6' 1, then there follows: ~ tB+1= ta. (1) Na = NB+1 5 2 N = If a is a limit number, then we have: Nu (2) = x u < , Nu I x u < , Nv+1 I x u < , 2  LSo 1) is also valid for a, and then by transfinite induction it holds in general.
+
2) Applying the GCH the signs 1 in ( 1 ) and ( 2 ) can be sharpened to =, and 2) follows. 3 ) L e t a = P + 1 . T h e n t , = t a + l = 2 N @ = Cv
Proof. First we define: If H is a set of (transfinite) sequences s = (s,),<~ of zeros and ones, where X is an ordinal and where K is an ordinal 1 A, we define (Hlr;) to be the set of all those sequences of H , in which all digits 1 occur at positions < r;. Then we have Ho = UnEN(Holn).Then I (Hoin)1 1 2n and lHol 5 CnEN 2n = No. Since also 1 Hal > N o holds the statement is valid for a = 0. Let now a be an ordinal > 0 and the statement already be proved for all ordinals P < a. Then H, = U{(H,lv)lv < w,). If a is a successor number p 1, then for every v < wp+l we have I (Halv)I 1 214 1 2N@,hence IHg+11 5 Np+1 .2" = 2N@= tS+1. Also IHp+ll tp+l holds because Hp+l has a subset which is equipotent to 2 ( ( w p ) ) where , the last set has cardinality 2N@. Let now a be a limit ordinal. Then H, = u,<,(H~w,). So there follows [ H aI IHvI = Xu<, t, = t,, further I H, I I HvI = t, for all v < a, and then also I H, I t, (= C,.,, t,) .
+
>
< z,<,
>
>
In the rest of this section a always denotes an ordinal and w, = cf(w,) Now we shall prove that H, is an %  set, and thus an q, is regular. First we show:
 set if w,
3.8 Theorem. Let a be an element of Ha. Then ( H a < a ) has a cofinal subset of type w,, and ( H a > a ) has a coinitial subset of type w:. Thus every element of H, has the elementcharacter (w,, w;). And so H, has only symmetric elements.
Proof. Let a := (a,),<,,. There is a T with a, = 1, and a, = 0 for T < v < w,. For the ordinals p with T < p < w, let f,, be that sequence of Ha, which has (a,),<, as initial segment, and which satisfies f , ( ~ ) = 0, f,(v) = 1 for T < v 1 p, and f,(v) = 0 for p < v < w,. The set {f,Ir < p < w,) then has the order type w, and is cofinal in (Ha < a ) .
4.3. UNIVERSALLY ORDERED SETS
101
On the other hand we define for the ordinals p with T < p < w, a sequence gp E Ha as follows: gp has (a,),sT as initial segment, gp(p) = 1 and gp(v) = 0 for the ordinals v # p that satisfy T < v < w,. Then {gplr < )(I < w,) has the order type w: and is coinitial in (Ha > a). For the gaps of H, we have only a "onesided" analog to 3.8:
3.9 Theorem. To every gap (A, B) of H, there exists a subset of type w, which is cofinal in A, or a subset of type w: which is coinitial in B. Every gap of H, has a character (wy,w;) or (up, w;), where wp is a regular ordinal 5 w,.
Proof. Let (A, B) be a gap of H,.We consider A as a subset of in 2((w,)). Then according to 2.11, A has a supremum s = (s,),<,, 2((w,)). This cannot belong to A or to B, because in the first case s would be the greatest element of A, in the second the least element of B. Both is impossible because (A, B ) is a gap. Thus we have s E 2((w,)) \Ha. Of course s has N, zeros or N, ones, precisely: There exists a subset T E w, of type w,, so that s, = 0 for v E T, or there exists a subset U C w, of type w,, so that s, = 1 for v E LT.In the first case we define for every T E T a sequence f T E H, by: (*) f T has ( s ~ ) as ~ initial < ~ segment, f T ( r ) = 1, and fT(v) = 0 for r
3.10 Theorem. H, is an q, qa
 set.
If w, is regular, then H, is an
 set.
3.11 Definition. We denote the order type of H, by h,. If w, is regular, we shall see in the following that h, is nothing else but what Hausdorff called the normal type qa.
The types h, have many features in common with the order type of rational numbers, because they are the generalizations of this to higher cardinalities. Now we establish a characterization theorem for the types h, with regular w,, which has no connection to the special realizing sets Ha 3.12 Theorem [64] (Characterization Theorem). Let w, be regular and S a linearly ordered set. Then S has the ordertype h, exactly i f the following two conditions are satisfied: 1) S is an q,  set. 2) S is a union of N, many subsets S,, v < w,, where every S, is free from w, and w.:
Proof. H, is an q,  set, and we have H, = u{TpIp < w,), where Tp is the set of all sequences (avlv < w,) of Ha, in which the last digit 1 is already in the segment (a,),
f;'
(*) A"
> S, and Bv _> Tv, where all A,
w,  free and w:
and all Bv with v
< w,
are
 free.
+
If T is a successor number a 1, then one can construct, in analogy to the construction of Fo,an isomorphic mapping F, : A, + B,, which satisfies (*) for v = T and which extends F,. If T is a limit ordinal, let f, be the limitmapping of the Fv,v < T. It maps UV<,Av isomorphically onto Uv
4.3. UNIVERSALLY ORDERED SETS
103
from w, and w i because of 3.3.8, and so one can extend f, by a twofold application of 3.3.6 to an isomorphic mapping FT: A, + BT,where the following holds: S A, U{Aul~< 7) U STand H, BT U{BuI~< 7) U T,, where AT and BT are free from w, and w i . Then for p < v < w, the mapping Fu extends Fp,and the limit mapping of the F u , v < w,, is an isomorphic mapping of S onto Ha.
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The characterization theorem has a lot of applications. There holds the following minimal property, whose substance is due to Hausdorff [81], p. 183:
3.13 Theorem. Every q,
 set contains a subset of type ha.
Proof. Let E be an q,  set. If w, is regular, then H, is a union of N , many subsets which are free from w, and from w i , and thus H, is by 3.3.9 embeddable in E. So H, is isomorphic to a subset T C E of type ha. If w, is singular, then E is also an qa+l  set because of 3.3.5. Then it has a subset of type according to the already proved regular case. This yields a fortiori our statement. 3.14 Corollary. A n q,  set has a cardinality 1 t,. If here w, is singular, then it has a cardinality 2 t,+l. For a regular ordinal w, there exists an rl,  set of cardinality N, iff t, = N,. Hausdorff ([$I], p.182) proved:
3.15 Theorem. If two q,  sets both have the cardinality N,, then they are isomorphic. (Indeed, each of them is isomorphic to Ha.) Proof. Let E be an q,  set of cardinality N,. Then E is a union of N , oneelement sets,  which of course are free from w, and w:. Then E is isomorphic to H, because of 3.12. 3.16 Remark. 3.15 contains the following statement: If % = N, holds, then every two q,  sets of cardinality t, are isomorphic. Here the assumption that t, = N , holds, is not superfluous: Gillinan [52] proved: If 2'" # N,+l holds, then there exist two nonisomorphic qa+l  sets of cardinality 2'". And further: If for ordinals S, ,6 we have S 2 ,6 2 and N6 2 2'p, then there exist two nonisomorphic qp+1  sets of cardinality N g .
+
Concerning the existence of q,  sets of cardinality N,, Gillman [52] proved: If a is a limit number, then there exists an q,  set of cardinality N, iff N , is regular and 2Nfl 5 N , is valid for every P < a. The latter holds for the socalled strongly inaccessable numbers. The characterization theorem gives very short proofs in some cases:
3.17 Theorem. If w, is regular and E a linearly ordered set of type h,, and i f T is a subset of E, which is again an q,  set, then T has the order type h,. Proof. T satisfies the conditions of 3.12 a fortiori because Ha does this. The last theorem immediately yields the next one, which is related to a result of Padmavally [135],p. 67:
3.18 Theorem. If w, is regular and E a linearly ordered set of type ha, and i f D C E is dense in E, then D has also the type h,. Proof. According to 3.3.10 D is an q,  set, and so 3.17 proves the statement. 3.19 Theorem. Let w, be a regular ordinal. Then every open interval (a,b) of H, has again the order type h,. And for every a E H, the sets ( H , < a ) and (Ha > a ) have the order type h,. Proof. Let a < b be elements of Ha. Then the open interval (a,b) of H, and the sets ( H a < a) and ( H a > a ) are 7, sets by 3.3.11 and the union of N , many subsets which are free from w, and w:. Thus 3.12 proves our assertion. In 4.6.3' the first part of the last theorem is generalized to arbitrary initial ordinals w,. Also the next theorem will be generalized later to arbitrary w, in 4.4.5.
3.20 Theorem. For regular w, there holds h,. h, = ha. Proof. We realize the type ha . h, by the lexicographic product
H, x Ha. We prove first that the latter is an q,  set. Let A and B be nonempty subsets of H, x H, with /A/< N , > IBI, which satisfy a < b for all a E A and all b E B. We assume indirectly that A and B are neighboring. Case 1. There exists an element z E H, such that A and B contain elements which have z as first component. We put A' := {x E
4.3. UNIVERSALLY ORDERED SETS
105
H,l(z,x) E A ) and B' := { y E H,l(z, y ) E B ) . Then A' and B' are neighboring in H , because A and B are neighboring in H, x Ha. But this and IA'I 5 IAl < N , > IBI 2 lB1l contradict the fact that H , is an r], set. Case 2. There is no element z which satisfies the assumption of Case 1. Then we put A" := { a E Ha 13% E H, with (a,x ) E A ) and B" := {b E H,13x E H, with ( b , x ) E B ) . By assumption we have A" n B" = 0 and then a < b for all a E A" and all b E B". Now A" and B" must be neighboring in H , since A and B are neighboring in H, x Ha. This leads to a contradiction as in Case 1. If C is cofinal (resp. coinitial) in H , x H, then the set of its first components is cofinal (resp. coinitial) in H , and thus of a cardinality 2 N,. The latter then also holds for the set C. We now define the sets which were introduced by Hausdorff as carrier sets for his normal types 7, :
3.21 Definition. 3((w,)) is by 1.9 the set of all transfinite sequences of length w, of digits 0,1,2. We order it lexicographically. Let then Hk be the subset of those sequences (avlv < w,) E 3((w,)) for which there exists an ordinal p < w, such that a, 1 for all v with p 5 v < w,.

Then there holds:
3.22 Theorem. For regular w, the set Hk has the order type h,. Proof. Quite analogously to the proof of 3.10 it follows that Hk is an qa  set. And it is a union of N , subsets which are free from w, and WE; this follows from 2.3. With the last theorem we immediately see that h, is a symmetric order type (see 2.8):
3.23 Theorem. For every ordinal a the lexicographically ordered set Hk is symmetric.

Proof. We define a bijective mapping f : HA + Hk by: If (a,(v < w,) is an element of HA we put f (a,lv < w,) := (2  a,(v < w,). This mapping interchanges 0 with 2, and we have for x,y E H& : x5Y f ( 42 f ( y ) . The next theorem is related to a result of Padmavally [134],p.260. In 4.6 it will be generalized on arbitrary w, :
3.24 Theorem. Let w, be a regular initial number, S a linearly ordered set, and S = u{S,IV < w,}, where every S , is embeddable i n Ha. Then also S is embeddable in Ha. Proof. By 3.12 every S , is a fortiori representable in the form S , = u{S,,I,U < w,} , where each S,, is free from w, and w; .Then S = Uv<w, (Up<w,Svp) = U K ~ (U{SV,l~ , < 7,p < 7). Here every summand US,, is a union of 1712 < N, sets which are free from w, and w.: Therefore itself is also free from w, and w;T, by 3.3.8. According to 3.3.9 then S is embeddable in every q,  set and thus also in Ha.
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In 1.9.6 we had introduced the relations and < for order types. The next two theorems describe properties of the types h,. 3.25 Theorem. Let E be an q,  set, where w, is regular. Then there holds h, = tp(E) or h, < tp(E). Proof. If E is isomorphic to a subset T of H,, then according to 3.17, T and with it E, has the type ha. If E is not isomorphic to a subset of H, we obtain, because of h, 5 t p E , ha < tp(E). 3.26 Theorem. Let w, be a regular ordinal, T 5 Ha and t p ( T ) < ha. Then tp(H, \ T ) = ha. Proof. No interval of H, is a subset of T because of 3.19. Then H, \ T is dense in H,, and therefore it has the type h, because of 3.18.
In 3.1.17 we had introduced the final order in a cartesian product of linearly ordered sets. We now present an application of this concept for sequences of real numbers. If (an)nENand are converging sequences of real numbers with limits a resp. b, then we have the following elementary fact: If a < b holds, then there exists an index no E N such that a, < b, holds for all n 2 no. This simple observation gives rise to the following definition: 3.27 Definition. Let S be the set of all sequences (an)nENof real numbers. For (a,) and (b,) E S we put (a,) < (bn) U 3no E N such that a, < bn for all n 2 no. Then < is a transitive relation, and if we put (a,) (b,) (a,) < (bn) or (a,) = (b,), then is an order in S, which of course is not linear.
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In this context Hausdorff ([8l], p. 191) proved:
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4.3. UNIVERSALLY ORDERED SETS
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3.28 Theorem. Let A be a countable subset of S. Then there exists an element in S which is < all elements of A. And symmetrically: There exists an element in S which is > all elements of A. Let A, B be countable subsets of S, such that a < b holds for all a E A, b E B, then there exists an element x E S with a < x < b for all ~ E A~ , E B . Proof. Let A = {anln E N ) be a countable subset of S, where ~ x1 < a l l , 2 2 < min{a12, aa2), . . . , a, = ( u ~for~n E) N.~We~choose xn < min{aln, . . . ,an,), . . . . Then x := ( x , ) , ~ ~is evidently < a for every a E A. Let now B := {b,ln E N) be another countable subset of S such that a < b holds for all a E A, b E B, and bn = (bni)iENfor n E N. We say: ak is less than bl from i on, if ak, < bl, for all v i. Now for every pair k, 1 E N there exists a least number i(k, 1) such that ak is less than bl from i(k, I) on. For every n E N we put in := max{i(k, l)lk, 1 n). n. Then there also Then we have ak < bl from in on for all k,l exists a strictly increasing sequence X1 < X2 <    < An < .  . of natural numbers such that for every n E N there holds: ak < bl from An on for all k, 1 5 n. Every j E N that is Xlsatisfies An j < X,+l for exactly one n E N. Then we choose an element x j E R satisfying max{akjIk n ) < X j < min{bkjlk n). Let x be the sequence (xj)jEN. Then we have a < x < b for all a E A and b E B.
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Now Hausdorff's observation ([811,p. 192) follows:
3.29 Theorem. Let M be a maximal chain of the set S of all sequences of real numbers, equipped with the final order. Then M is an 71  set. M has cardinality 2N0 = IRI. If the Continuum. Hypothesis is assumed, then M has the order type hl (= normal type ql). Proof. The first statement follows immediately from 3.28. And IMI 1 ~ 1 =~ 02 N =~ el. If CH is assumed, then by 3.13 we have el then el = Nlholds, and the rest follows from 3.15.
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Hausdorff ([81], p.193) had mentioned that several other order relations for sets of sequences of real numbers can be defined (and were inves
tigated in the literature), which have some resemblance with the notion which we have treated. In particular the following order for sequences with nonvanishing elements was studied: < (bn)nEN e lim = 0. The variations of Definition 3.27 have a similar behaviour as the order which we studied, and so we don't treat them in more detail.
4.4
Generalizations to the case of a singular w,
In the last section we established several theorems for the sets H, (and types h,) where w,was regular. We now generalize most of them to the case where w, is singular, which implies that (L, is a limit ordinal. 4.1 Lemma. Let S be a linearly ordered set of type h,, where w, is regular. Let further S* E S be a set which is free from w, and from WE.Let (A, B) be a cut (see 1.10.15) i n the set S* (where S* is equipped with the order induced from S). Then the set T of all x E S , that satisfy A < x < B, has order type h,.
Proof. For the set A there holds: A is empty, or A has a last element, or A has a cofinal subset whose type is an ordinal < w,. For B the corresponding properties hold. In every possible case it follows that A and B are not neighboring in the q,  set S. Thus T is not empty. Now A and T are neighboring in S, and therefore T must be coinitial with WE. Analogously T is cofinal with w,. Since T is also a segment of S, it follows easily that T is an q,  set, and then it has the type ha became of 3.17. 4.2 Definition. If M is a set of transfinite sequences (s,),<, of length T where the components s, are 0 or 1, then for p < T we define the subset M[p]to be the set of all those s E M that have a last digit 1, and this at an s, with v < p.
Now we can supplement our characterization theorem 3.12 by
4.3 Theorem. Let w, be a singular initial ordinal. Then a linearly ordered set S has the order type h, ifl S is representable i n the form S = U,<,S,+l,where S,+l has the order type h,+l, and where there holds S,+l C_ S,+l for p < v < a.
4.4. THE CASE OF A SINGULAR w,
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Proof. a is a limit ordinal, because w, is singular, and then we have H, = U,<,H,[W,+~]. Here H,[W,+~]has the type hV+l,and the summands increase with v. The above condition is thus necessary. Let now S satisfy the given condition. Then we first define Sx := U,<xS,+l for the limit ordinals X which are < a . Then there exists an isomorphic mapping of S1onto H,[wl] because both sets have the type hl. Applying Lemma 4.1 this mapping can be extended to an isomorphic H,[wl]. Indeed, the sets S1 and mapping of S2 S1 onto Ha[w2] H, [wl] are free from w2 and w;, and S2and H,[w2] are sets of type ha. If in general T is an ordinal with 1 < T < a and if for every v with 0 < v < T an isomorphic mapping cp, : S, +H,[w,] is already defined such that for v > p > 0 the mapping cp, extends the mapping cpp,we define cp, inductively as follows: If T is a limit ordinal then cp, shall be the limit of the mappings cp,, 0 < v < T . Then cp, is an isomorphic mapping of STonto H,[w,]. If T is a successor number, then again using Lemma 4.1 we extend the mapping cpTl to an isomorphic mapping of STonto H,[w,]. By transfinite induction we so obtain an isomorphic mapping of S onto H, as the limit of the mappings cp,, v < a .
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For later use we establish the following lemma, which can be proved in a very similar way as 4.3 using 4.1: 4.4 Lemma. If w, is singular, S = U,<,S, a linearly ordered set, where every summand S,, v < a, is free from w,+l and w;+~, then S is embeddable i n H,.
We can now generalize 3.20 to the following 4.5 Theorem. h,
 h,
= h, for every ordinal a.
Proof. Let w, be singular, the regular case was already settled in 3.20. For every ordinal v 5 a let P, be the set of all pairs of elements of H,[w,] (see Definition 4.2!), ordered according to the principle of first differences. Then P, = U,<,P,+l, t p P, = h, h,, and tp(P,+l) = h,+l  h,+l = h,+l because of 3.20. The sets P,+l increase with v, and so Theorem 4.3 yields our statement. Next we generalize Theorem 3.24: 4.6 Theorem. Let S = U{S,Iv < w,) be a linearly ordered set, where every S,, v < w,, is embeddable i n Ha. Then also S is embeddable i n Ha.
Proof. The case of a regular w, was already settled in 3.24. Let therefore w, be singular and thus a! a limit ordinal. Then we have w, = U,<,w,. Since S, is embeddable in H, = Up
4.7 Theorem. Let a < b be elements of Ha. Then the open interval (a, b) has a subset of type h,, and moreover a segment of type ha.
Proof. For regular w, this is already contained in 3.19. The set H, is dense, and so there exists a z E H, with a < x < b. For u < w, let a,, b,, z, be the v  component of a, b, z respectively, and let 6 be the first ordinal, for which the 6  components as and b6 differ, so that as = 0 and bs = 1. If zs = 0, then, due to a < z, there is a p > 6 with ap = 0 and zp = 1. Then all sequences (x,lv < w,) of H,, for which xu = z, for v 5 p holds, are in (a, b). If 4 6 ) = 1,there is a p > 6 where zp = 0 and b, = 1. Then all sequences (xvlv < w,) of Ha, for which xu = xu holds for v p, are in (a, b). In both cases we have a dyadic sequence s of length p + 1 < w,, such that all prolongations of s to a sequence E H, are in (a, b). If now we take the set of all sequences that arise from s by adhering to s the sequences of Ha we obtain a subset of (a, b), which is isomorphic to Ha.
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4.5 The method of succesively adjoining cuts In this section we deal with a principle of how one can obtain from a given linear order type a greater one by "adjoining cuts". Dedekind had already applied a similar method in order to give an exact construction of the irrational numbers. The more general method which is treated here was applied by Cuesta Dutari [16], [17]. Subsequently the author rediscovered his method and some of his results and extended some of
4.5. THE METHOD OF SUCCESIVELY ADJOINING CUTS
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them. Later the method of successively adjoining cuts was also used by Conway [14]in the context of the construction of surreal numbers. For Alling [4]. details see e.g. the book of Our construction principle yields interesting insights into the hierarchy of order types. In particular we obtain a new characterization of the order types ha, namely one which does not make use of representations by lexicographic orderings. We recall the former definition (1.10.15)of a cut (A,B ) of a linearly ordered set ( T ,<) and extend it:
orm man
5.1 Definition. Let (T,<) be a linearly ordered set. A cut of T is a pair (A,B ) , where A is an initial segment of T and where B = T\A. Here A and B , also T , are allowed to be empty ! We denote the set of all cuts of T by S ( T ) . If T and S ( T )are disjoint we now introduce a linear order in the set T U S ( T ) as follows : In this set the elements of T shall have the same order to each other as in (T,5 ) .If ( A ,B ) and (C,D) are cuts of T , we put ( A ,B ) (C,D ) e A C_ C. If t E T and ( A ,B ) E S ( T ) then we put t < A,B) e t E A, and ( A , B ) < t e t E B. It can easily be verified that the relation 5 which is so defined is a linear order. (T,<) is an ordered subset of it.
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The following is easily seen: Every element x of T has an immediate predecessor and an immediate successor in T U S ( T ) ,which are not in T , namely the cuts (T < x, T 3 x) and (T 5 x, T > x).
5.2 Definition. If TI and T2are subsets of a linearly ordered set, we call TI strictly dense (resp. dense) in T2,if for every two elements a < b of T2 there exists an element c E TI with a < c < b (resp. a c b).
< <
Then we verify: 5.3 Lemma. If T is linearly ordered, then T is strictly dense i n S ( T ) ,and S ( T ) is strictly dense i n T . Then further each of the sets T and S ( T ) is dense i n T U S ( T ) .
Proof. Let ( A ,B ) and (C,D) be cuts of T with (A,B ) < (C,D ) . Then C\A # 0, and there exists an element x E C\A which now fulfills ( A ,B ) < x < (C,D). If a < b are elements of T , the cut ( { x E Tlx 5 a ) , { x E Tlx > a ) ) is strictly between a and b. The rest is evident. An easy consequence of Lemma 5.3 is:
5.4 Lemma. If T is linearly ordered and i f T uS(T) has a subset L of order type A (resp.A*), where X is a limit ordinal, then already T has a subset L' of type A (resp. A*), which generates the same initial (resp. final) segment of T as L does.
Proof. Let A = {a,lv < A) be a wellordered subset of T U S(T) with a, < ap for v < p < A. Because of 5.3 there now exist elements t, E T with a, 5 t, 5 a,+l. The set of all t,, v < A, then evidently contains a wellordered set of type A which fulfills our statement. The case of A* is of course symmetric to that of A*. The next lemma stands in analogy to the completeness theorem of Dedekind, which states  roughly speaking  that if one adjoins to a linearly ordered set the cuts that are gaps, one obtains a linearly ordered set without gaps. 5.5 Lemma. If T is linearly ordered, then the set T U S(T) has a first element (0, T ) and a last element (T, 0) and has no gaps. For this one also says: It is ordertheoretically closed.
Proof. We assume indirectly that there exists a gap (A, B ) in T U S(T). Then A has a wellordered cofinal subset A' of order type XI, and B has a coinitial inversely wellordered subset B' of order type A2, where A1 and A:! are limit ordinals. According to 5.4 there now exist subsets A" and B" of T of types A1 resp. A; such that A1'and B" are neighboring in T. So they define a cut of T which is a gap. This cut had to be an element of S(T). It would be greater than all elements of A and smaller than all elements of B. This contradicts the fact that (A, B ) is a cut in T U S(T). The rest is clear. Now we derive a relation between the order types of T and S(T), which can be proved using a method of Padmavally [133]:
5.6 Lemma. Let T be a linearly ordered set. Then there holds tp(T) < tp(S(T)), and then of course also tp(T) < tp(T U S(T)).
Proof. The relation tp(T) 5 tp(S(T)) is trivial: If we map t ( ~ T ) 4 (A, B), where A = {x E Tlx 5 t) and B = T\A, this yields an isomorphic mapping of T in S(T). We now assume indirectly that there exists a <  preserving mapping f : S(T) + T. Let a0 be the first element (this is (0,T)) of S(T). It fulfills evidently a0 < f (ao). Let then a1 be the first element of S(T) which is > f (ao) (if such an element still exists). In general we proceed
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as follows: Let a, < f (a,) hold for an ordinal v. Then there exists the least cut in T which is > f (a,),namely ( T 5 f (a,), T > f (a,)) =: a,+l. We now have a, < a,+l, and since f is <  preserving, f (a,) < f (a,+l). This entails that f (a,+l) is in the second component of the cut a,+l and thus > a,+l. So we can resume: From a, < f (a,) there follows a,+l < f (a,+l). If for a limit ordinal A we have already defined elements a, E S(T) for all v < A, such that a, < f (a,) holds for all v > A, and such that the a, increase with v, we define ax to be the cut (A, B), where A is the initial segment of T which is generated by the elements f (a,), v < A, and B = S(T)\A. Then ax is greater than all elements f (a,), v < A, and then also greater than all a,, v < A. By transfinite induction we could so construct for every ordinal v an element a, E S(T), such that the a, strictly increase with v. This is impossible of course, and thus our statement is proved. Now we introduce the method of successively adjoining cuts: 5.7 Definition. Let To be the empty set 0. In this set, which we consider as linearly ordered, exists exactly one cut, namely (a,@).We denote the set which contains this pair and nothing else by TI, so that TI = To U S(To) holds. Then we define T2 := Tl U S(Tl) and so on. For instance, TI has one element, T2 has three, in general for n E N the set Tn+1 has 2 . ITn] 1 elements. Generally our construction runs as follows: Let T be an ordinal such that for all v < T a linearly ordered set Tu is already defined so that for p < v < T the set Tp is an ordered subset of Tu. (The sets T,, v < T, thus form a tower of linearly ordered sets.) If now T is a successor number, so that T  1 exists, we put T, := TTl U S(T,l) and order it according to 5.1. If T is a limit number, we put T, := UU<,Tu. And we order T, in the obvious way: We put a < b for elements of T, iff a < b holds in at least one Tu with v < T. Instead of Tu we also write T(v). The above definition generalizes at the same time the construction of von Neumann of the system of ordinal numbers and the construction of the Dedekind cut. Now e.g. T(wo) has the order type of the set of rational numbers, ordered by magnitude. In general we have the following theorem of Cuesta Dutari [16]:
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5.8 Theorem. T(w,) is an q,
 set, if
w, is regular.
Proof. Let w, be regular, and let A and B be neighboring subsets of T (w,). We assume indirectly that both have a cardinality < N,. Then also /A U BI < N , holds. We have T(w,) = U { T u I ~< w,) , and then the set of indices v, for which Tu has elements in common with A U B cannot be cofinal with the regular w,. So there exists an index /3 < w, for which A U B C Tp holds. But then our construction would furnish an element x E Tp+l5 T(w,), which is greater than all elements of A and smaller than all elements of B, but this contradicts the assumption that A and B are neighboring. In a similar way it follows that T(w,) is cofinal with w, and coinitial with w:. 5.9 Definition. We denote the order type of T, by t, or by t ( v ) . According to 5.6 we then have t p < t, for p < v. The statement of 5.8 can be sharpened essentially to the following:
5.10 Theorem ([64], Satz 14). For every ordinal a the set T(w,) has the type h,.
Proof. Let first w, be regular, and let v be an ordinal < w,. We prove: ( 1 ) T, has no subset of type w, (and analogously no subset of type w:>If (1) would be false there would exist a least ordinal X for which TA has a subset of type w,. According to 5.4 then X must be a limit number. If now W is a wellordered subset of TAof cardinality N,, then we have W = u,
4.6 Special properties of the sets TAfor indecomposable A. The sets TAwith an indecomposable X have some remarkable properties. We recall this concept:
4.6. PROPERTIES OF TAFOR INDECOMPOSABLE X
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6.1 Definition. An ordinal X is said to be indecomposable, if in every representation X =. a ,B as sum of ordinals a,,B there follows a=Xorp=X. A wellknown theorem of set theory states that the indecomposable ordinals are exactly the powers wT, where T is an ordinal. In particular the initial ordinals w, are indecomposable. Further we define: A linearly ordered set S is said to be homogeneous, if for each two elements a, b E S there is an isomorphism of S onto S, which maps a onto b. If S is a linearly ordered set, T C S, (A, B) a cut in T , x an element of S, then we say: x is within the cut (A,B) iff A < x < B holds. For every x E Tx there exists a smallest index 6 5 X for which x E T6. Then 6 is called the degree of x.
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Now there holds: 6.2 Lemma. Let X be an indecomposable ordinal, and let x < y be elements of Tx, t an element of Tx n (x, y) of least degree 6. Then S is a successor number, and t is uniquely defined.
Proof. TAis dense, and so there exists an element t E TAn(x, y) ,and then also one with lowest degree 6. If S would be a limit ordinal we would have T6 = Uv
Proof. First we prove as the main part: If a < b are elements of TA, then the open interval ( a ,b) of Tx has again the type t x . To this purpose we construct an isomorphic mapping fx of TAonto (a, b). Let a be the least ordinal such that a and b are both in T,. We map the oneelement set TI into (a, b), where T: := f [TI] contains only the (by 6.2 uniquely defined) element of least degree S +1 which is in (a, b). This degree is
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indeed a successor number. And we have 6 1 a 1 because a and b are in T,, so that there exist elements of T,+l between them. Now for v = 1 the following induction hypothesis is satisfied: (1) f, : Tv + T,* is an isomorphic mapping with (a, b) nT6+, 5 T,* 5 (a, b) n T,+,+l. Suppose now that (1) is valid for a fixed ordinal v < A. Then we define f,+l (x) := f,(x) for all x E T,. If x E Tv+1\Tv holds, then x is a cut (A, B) E TV+l of T,. If here A and B are nonempty, then there is exactly one element x*of (a, b) n T6+,+1 between A* := f,[A] and B* := f,[B]. For if there would be two, there would by 5.3 exist an element of (a, b) n Ts+, between them, and this contradicts the fact that A* U B* 1 f,[T,] = T,* (a, b) n Tb+,. Then we put f,+l(x) := x*. If x = ( A ,B ) is an improper cut, where A or B is empty, we have to discuss the cuts (0, T,) and (T,, 0). These cases are symmetric, and so we restrict ourselves to the first case. Here there is an element x*of Ta+,+2 between a and T,* because of (1) and a E T, C_ T,+,+l. If there is also such an element which is in Ts+,+1 (and then this is uniquely determined by 6.2) we choose this as x*. In any case we map x onto x* by f,+l. We treat (T,+l, 0) analogously. So we obtain an isomorphic mapping f,+l : TV+l+ T;+l := f,+l [T,+l] which satisfies (1) for v 1. If p is a limit ordinal 5 A such that (1) holds for all v < p, we define f p : Tp := u{T,Iv < p) + u{T; Iv < p) =: T; to be the limit mapping of the f,, v < p. Then f p satisfies (1) for p, and so (1) holds for all v < A. The limit mapping fx of the f,, v < A, is then an isomorphism of Tx onto (a, b). The rest now follows easily: Let Sl and S2be segments as introduced in the statement. Then we choose elements do E Sl and eo E S2 and further a sequence do < dl <   . < d, < . . 1 T < w, which is cofinal in S1 and a sequence eo < el < . . . < e, < . . . I T < wy which is cofinal in S2. According to that what we have proved already '[d,, d,+l] is isomorphic to [e,, e,+l] for every *r < wy. Together this yields an isomorphic mapping of {x E Sllz 2 do) onto {x E S21x L: eO). The counterpart of this follows symmetrically, and therefore S1 is isomorphic to S2. If s, is the smallest and g, the greatest element of T,+l for v < A, then {svlv < A) is coinitial in Tx, and {g, lv < A) is cofinal in Tx. Thus Tx and all open intervals of Tx are cofinal with A and coinitial with A*, and so everything is proved.
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By 5.10 we have Ha E T(w,), and so 6.3 implies the corollary, which generalizes 3.19 to arbitrary initial ordinals w,: 6.3' Corollary. For every ordinal a holds: Each open interval of Ha is isomorphic to Ha.
Our last theorem implies that TA is a homogeneous chain if X is indecomposable : 6.4 Theorem. Let X be indecomposable. Then for every two elements a,b of Tx there exists an isomorphic mapping of Tx onto itself (a socalled automorphism) which maps a onto b.
Proof. By 6.3 the initial segments (TA < a ) and (TA < b) are isomorphic, say by a mapping 9, also (TA> a) and (TA> b) are isomorphic, say by a mapping $. Then the mapping which maps a onto b and the other elements x of Tx onto ~ ( x (resp. ) $(x)) fulfills our assertion. The question arises whether the statement of 6.3 can be generalized to more general ordinals. But this is not the case. To see this we first mention: 6.5 Lemma. If a and P are ordinals, and if C is a cut i n T,, then the set of all elements of T,+p that are within the cut C, has the type tp.
The proof follows immediately from the construction of the sets T,. 6.6 Remark. Let X be a decomposable limit ordinal.Then there exist open intervals of Tx which are not isomorphic. Indeed, there exists a representation X = a+P, where a and p are ordinals between 0 and A. Then also there holds X = ( a 1) P since the limit number P satisfies 1 p = P. In T,+l there exist neighboring elements x, y and because of 6.5 the set of elements of T,+p, that are between x and y, is an open interval of Tx of type t g . On the other hand there exist open intervals of Tx of type t x . If e.g. a is the least and b the greatest element of T2\ TI, then the open interval (a, b) of Tx evidently has the type t x . If X is a successor ordinal 2 2 there trivially exist open intervals of Tx which are not isomorphic. For then there exist neighboring elements in Tx, and the open interval with these endpoints is empty. And also there exist nonempty open intervals.
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Next we study the idempotency property of certain types t x . To this purpose we define:
6.7 Definition. For every ordinal a we denote T a x T, by P ( a ) . We consider P ( a ) as linearly ordered by first differences. An easy consequence is then: 6.8 Lemma. P ( a ) is isomorphic to a subset of T,+,. Proof. P ( a ) has the order type t,  t,. If now we have a cut in T,, then there exists because of 6.5 a subset M of elements of To+,, which lie within this cut, such that M has order type t,. If then a E T,, let s(a) denote the cut ( A ,B ) in T,, where A contains all elements 5 a. Let S(a) be a subset of T,+, of type t, , whose elements lie within the cut s(a).Then the set S = U{S(a)laE T,) has the type t , . t , = t p ( P ( a ) ) , and so we have an isomorphic mapping of P ( a ) onto S E T,+,. We shall prove that tx . t x = t x holds for indecomposable ordinals X and prepare this with some lemmas.
6.9 Lemma. Let a and /3 be ordinals, C = ( A ,B ) a cut in P ( a ) , M the set of all elements of P ( a + p) that lie within C. Then M has an order type > tp. Proof. Case 1. There exists an element x E T, such that A and B have elements which have x as first component. Then { y ( ( xy,) E A ) and {yl(x,y ) E B ) are neighboring in T,, and so they define a cut C, in T,. By 6.5 the set U of elements of T,+@that lie within C, has the order type tp. Then the elements of { x ) x U lie within the cut C, and we have { x ) x U E P(a+P) so that M > {x) x U holds.The last set has order type t p U = tp, and so Case 1 is settled. Case 2. No element of T, is the first component of an element of A and of an element of B. Then the set Alof the first components of the elements of A and the set B1 of the first components of the elements of B are disjoint and constitute a cut C1 = (A1,B1) in T,. By 6.5 the set V of all elements of To+@that lie within Cl has order type t p . Then all elements of (V x T,+p) belong to P ( a P), and then also to M because they lie within C.The order type of their set is tpV = tp.
+
6.10 Lemma. Let a and
P
>
be ordinals with wV 5 a , @< wV+'(= 2 satisfying
wV  w ) . Then we can choose natural numbers n , m (1) a ( n  1)
> p  2, and
>
4.6. PROPERTIES OF TAFOR INDECOMPOSABLE X
119
a) Let f : T ,  + P(P) be a <  preserving mapping. Then there T,., and a <  preserving surjective mapping f * : exists a set S, S, + P(P) which extends f. b) There exists a <  preserving mapping f' : T,., + P(P . m ) , which extends f *, and then also f.
Proof. First one can choose n so great that (1) holds because of a 2 wV and p . 2 < wV+l.Then in (2) m can be chosen so great that (2) holds because of a  n < wv+land ,B 2 wV. a) Let C = ( C l ,C2) be a cut in T,. The set M ( C ) of elements of T,., that lie within C has the order type t ( a  (n  1)). For we have T ( a n) = T ( a a . (n I ) ) , and then this follows from 6.5 by putting P = a ( n  1). Let further C' := ( f ( C l ) ,f ( C 2 ) )be the corresponding cut in f (T,) P(P). Then the set M' of elements of P(P) that lie within the cut C' has (trivially) an order type 5 tp P(P) which is 5 tg+g by 6.8. Now we can define, due to ( I ) , an isomorphic mapping between a subset of M and the set MI. We perform this action with all cuts C of T , and combine this with f. Then there results a mapping f * which satisfies a). Here S, is T, U u { M ( C )IC is a cut of T,). b) Let D = ( D l ,D2) be a cut in S, T,., and D' := ( f * ( D l )f, * ( D 2 ) )the corresponding cut in f * (S,) C P(P). The set E of elements of T,., that lie within D has trivially an order type 5 t,.,. And the set E' of elements of P(P m ) that lie within D' has (according to 6.9 by taking there P . (m 1) for P and P for a ) an order type 2 tp.(ml).And so by (2) there exists an isomorphic mapping fg of E onto a subset of E'. Then f * , together with the mappings f & furnishes a mapping f' which satisfies b).
+
z
In a special form of 6.10, namely by replacing a by wV . ni, ,B by wv . mi, f by fi, and f' by fi+1we obtain:
6.10'. Lemma. Suppose we have a <  preserving mapping fi : T(wvni)+ P(wvmi),where ni and mi are natural numbers. Then there ~ + ~ ) , exists a <  preserving mapping fi+1 : T ( w ~ . ~+~P+( ~w )~  ~ where ni+l and mi+l are natural numbers with ni+l > ni and mi+l > mi, which extends f i , and which satisfies fi+l(T(wV ni+1))2 P(wV . mi). 6.11 Theorem [64]. t x t x = t x holds for every indecomposable ordinal A, which means for the ordinals X = wVwhere v is an ordinal.
Proof. For v = 0 we have w0 = 1,and to = 1 = 1  1. In the case v = 1 we have X = w1 = wo and we must show that T(wo) x T(wo) is isomorphic to T(wo). But these sets have order types ho ho and ho which are equal by 4.5. By the way: Every set T(w,), where a is an arbitrary initial ordinal has by 5.10 the order type h,, and for this there holds h, . ha = h, by 4.5. Now we assume that the theorem is already proved for a fixed ordinal v, so that we have an isomorphic mapping fo : T(wV)+ P(wV).Then for i = 0, no = 1 = mo the following condition is satisfied: (1) fi : T(wV. ni) + P ( w V  mi) is a <  preserving mapping. Let now i be a fixed natural number for which (1) holds. Then by 6.10' there exists a <  preserving mapping fi+l : T(wV ni+1) + P(wV.mi+l) with ni+l > ni and mi+l > mi which extends fi and which satisfies: (2) fi+l(T(wV. ni+l)) 2 P(wV. mi). By induction on i we can so define a sequence of <  preserving mappings fi : T(wV. ni) + P ( w V mi) for all i E N,where the ni, i E N, and the mi, i E N, form strictly ascending sequences of natural numbers, and where for i < j the function f j is an extension of fi with fj(T(wV . ni+i)) 2 P(wV. mi). The limit mapping of the fi, i E N,(which extends all of these), maps uiENT(wV . ni) = T(wV. W) = T(w'+') isomorphically onto UiEN P(wV = P(wV+l). Therefore our statement also holds for v 1. mi) If now p is a limit number such that for all ordinals v < p the sets T(wV)and P(wV)have the same order type, so that there exists an isomorphic mapping f V : T(wV)+ P(wV),the limit mapping which extends all f V with v < p is an isomorphism of T(wp) onto P(wp).
+
The property which was expressed in 6.11 is restricted to the indecomposable ordinals. Namely there holds: 6.12 Theorem [64]. Let 6 be a decomposable ordinal. Then there holds t(6) t(6) > t(6).
Proof. For finite ordinals 6 the assertion is trivial. So we assume 6 > wo. The decomposable 6 has a representation 6 = a P, where a and p are ordinals < 6. In the set P(6) (see Definition 6.7), whose order type is t(6) .t(6), exists a system Jp of disjoint segments, such that each of them contains a subset of type t(b),and such that also Jp has the
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4.6. PROPERTIES OF TAFOR INDECOMPOSABLE X
121
order type t(S) with respect to the natural order of Jp, which is given by Il 5 I2 for 11,I2 E JP e x 5 y for all x E Il and all y E 12. Now it suffices to prove: (1) In T(S) there is no such system of disjoint segments. Let JT be a system of disjoint segments of T(S) such that each of them contains a subset of type t(6). Then we have: (2) Every S E JT contains an element of T(a). For otherwise there would exist a segment S E JT such that all elements of S would lie within the same cut of T(a). But then S would not contain a subset of type t(S) because of 6.5, which states that the set of all elements of T(S) that lie within a cut of T ( a ) has the order type t(P) < t(6). So (2) holds. Now all sets I' := I rlT(a), I E JT,are nonempty, and so to the set of segments I E JT there corresponds a set GT of nonempty disjoint segments I' of T(a) which has the (induced) order type tpGT = tpJT. So finally by 5.6 we have tpJT = tpGT 5 tpT(a) < t p T ( a 1) 5 tp T(S), so that tpJT < tpT(S) follows, and (1) is proved.
+
6.13 Definition.We call an order type T intervalhomogeneous, if all nonempty open intervals of a realization of T are isomorphic. We call an order type T unattainable (by summation), iff there holds: If A and Bi, i E A, are linearly ordered sets with order types t p A < T and t p Bi < T for every i E A, then also the ordered sum CiEAtpBi is
< 7. This definition entails that for unattainable order types T there holds: If a! and /3 are order types < 7, then also a . P < T holds. A sufficient condition for being unattainable is presented in: 6.14 Theorem. Let T be an intervalhomogeneous and idempotent linear order type. Then T is unattainable.
Proof. Let A and Bi, i E A, be linearly ordered sets with types tp(A) < T and tp(Bi) < T for a11 i E A. Then there holds T ~ where , T is a set of type T and every T, = 7. The CiEAtpBi 5 CxET latter sum is T . T = 7, SO that CiEAtpBi 1 T holds. We now assume indirectly that T 5 CiEAtpBi holds. Then there Bi, exists a <  preserving mapping f of T into the ordered sum XiEA which is the set of all pairs (a, b) with a E A, b E B,. Let A' be the set
of all elements a that occur as the first component of such a pair (a, b) which is ascribed by f to an element of T. Because of T > tpA 2 tpA' there exist two elements t' < t" in T, whose f  images have the same first component, say a'. Then a' is also the first component of the f  images of all t E T that fulfill t' < t < t". This set {t E Tlt' < t < t") =: T* has again the type T because of the intervalhomogenity of T. Due to the injectivity off the seccnd components of the images of the elements of T* must be different and their set, which is contained in Bat , must have the type T. But this contradicts t p Bat < T. An immediate consequence of the last theorem, 6.3 and 6.11 is now: 6.15 Theorem. If X is an indecomposable ordinal, t(X) is an unattainable order type.
4.7
Relations between the order types of lexicographic products
As we have seen in the previous sections the ordered sets p((v)), where p and v are ordinals, and certain subsets of them like the sets H, are rather important for the theory. They are comparatively easy to describe because the ordinals p, v which determine them, are wellordered sets. They give a good oversight over the structure of the class of linearly ordered sets. In this section we investigate, following [71], how the sets p((v)) and the types h, are related.
7.1 Definition. If T is a linearly ordered set and a an ordinal, let T((a)) denote the lexicographically ordered set of all transfinite sequences of length a of elements of T. Its order type shall be denoted by da),where T is the order type of T. The property to be an q,  set is productive in the following sense: 7.2 Theorem. Let T be an ordinal > 0, and let Ev,v < T, be q, sets. Then the lexicographic product L := LV<,Ev is again an q,  set.
Proof. Let (A, B) be a proper cut in L. We denote the set of the 0components of the elements of A (resp. B) by A. (resp. Bo). If A. n Bo = 0, then (Ao,Bo) defines a proper cut in Eo,and then A. has a cofinal subset of type w,, or Bo has a coinitial subset of type w.: In the first case also A has a cofinal subset of type w,, in the second also B has a coinitial subset of type WE.
4.7. ORDER TYPES OF LEXICOGRAPHIC PRODUCTS
123
Now we assume that A. nBo is nonempty. Then it contains exactly one element xo E Eo. Generally we now consider all sequences (x,),<~ of elements xu E Ev for ordinals p 5 T, that are initial segments of elements of A and of elements of B. Let a be the supremum of the ordinals p which here occur. Then we have a sequence x = (xu),<, of which all the before mentioned sequences are initial segments. We distinguish two cases: Case 1. x is an initial segment of elements of A and of B. Then a is < T because of A fl B = 0.And according to the definition of a there is no sequence of length a 1 (which prolongs x with an additional component) which is still a common initial segment of elements of A and of B. Let A' (resp.B') be the set of the a  components of those elements of A (resp. B) that have x as initial segment. Then A' n B' = 0, and thus (A', B') defines a cut in E,. Again it follows as before, that A is cofinal with w,, or B is coinitial with w:. Case 2. x is only an initial segment of elements of A, not of B. (The case with B instead of A is symmetric.) Then A contains all sequences of L, that have x as initial segment and an arbitrary element of E, in the a  component. Then, of course, A is cofinal with w, since E, is that also. So finally we have obtained that there are no nonempty neighboring subsets of L which both have a cardinality < N,. It is trivial that L has a cofinal subset of type w, and a coinitial subset of type w: because this also holds for the set Eo. And therefore L is an q,  set.
+
7.3 Remark. By the way, the property to be an q,  set, is also summative in the following sense: If E is an q,  set, and if Ei is an q,  set for every i E I, then the ordered sum Ei is also an q,  set. This can easily be verified. 7.4 Lemma. Let w, be a regular initial ordinal and a an ordinal < w,. Then the lexicographically ordered set L := H,((a)) has type ha.
Proof. Let s = (s,),<, be an element of L, hence s, E H, for v < a. Each of the sequences s, has a last digit 1, say at the position ~ ( v. The ) set { ~ ( vlv) < a) has a cardinality 5 la1 < N, and is therefore not cofinal in the regular ordinal w,. So there exists an ordinal K < w, such that all n(v), v < a, are < K, and then we have s E (H, [K])((a)) (see Definition 4.2), and H,((a)) = u{(H, [ ~ ] ) ( ( IKa )< ) w,) is a union of N, subsets which are free from w, and w.: Indeed, Ha [K] is isomorphic
to a subset of 2 ( ( ~ )which ) is free from w, and w: by 2.2. The latter also holds for ( H a[ K ] )( ( a ) )because of 2.2. The rest follows with 7.2 and 3.12. More generally we now obtain: 7.5 Theorem. For every a order type h,.
< w,
= cf(w,) the set H,((a)) has the
Proof. If w, is regular we are done because of 7.4. So we assume that w, is singular. Then we have H, = U,<,H,[W,+~] and further: ( 1 ) H a ( ( 0 ) )= Uy
7.6 Theorem. For w, = cf(w,) the set S := H a ( ( w y ) )is embeddable in 2((w,)). Proof. Let c be a fixed element of Ha. We consider the sequences s = (sVIv< w,) E H,((w,)) for which there exists a first ordinal P such that s, = c for all v with C,? 5 v < w,. We denote their set by S c ( P )and put Sc := U{Sc(P)IP< w,). Since H, is dense it is easily seen that also Sc is dense in S. For every p < w, now Sc(P) is isomorphic to a subset of of H a ( @ ) ) , and so its type is 5 tpH,((P)) = h, because of 7.5. Then Sc is the union of N , 5 N , subsets Sc(P), each of which has a type 5 ha, and then Sc itself has a type 5 h, by 4.6. So we have obtained: S , is dense in S and embeddable in Ha, say by a mapping f. Now 2((w,)) has no gaps and contains Ha. Then f can be extended to an isomorphic mapping of S in 2((w,)). Indeed, every x
4.7. ORDER TYPES OF LEXICOGRAPHIC PRODUCTS
125
E S\S, determines a gap (S, < x, S, > x) in S,. Then there is an element x' E 2((w,)) which satisfies f (a) < x' < f (b) for all a E (S, < x) and b E (S, > x). We ascribe this x' to x and obtain the desired extension.
Also the counterpart of 7.6 can be proved: 7.7 Theorem. For w, = cf(w,) the set 2((w,)) is embeddable in Ha((wy))
Proof. For the regular ordinals w, our assertion is trivial since then wy = W, holds. So we presuppose that w, is singular. Then there exists a strictly ascending sequence of initial ordinals w,,, v < w,,with C{w,, Iv < w,} = w,. This entails 2((w,,)) x 2((w,,)) x  . x 2((w,,)) x . . . Iv < wy, where 2((w,)) the last product is ordered by the principle of first differences. Here every factor 2((w,,)) is isomorphic to a subset of Ha, and so 2((w,)) is isomorphic to a subset of H, ((w,)).

7.8 Remark. We have seen that from the order types h?) and 2(Wa)each is 5 the other. But these order types are not equal. This is an immediate consequence of the fact that Ha((wy)) is dense, whereas 2((w,)) is not dense: Every sequence (s,Iv < w,) of it which has a last digit 1, say at the position p, has an immediate predecessor, namely the element (t,), for which t, = s, for v < p, tp = 0, t, = 1 for p < v < w,.
7.9 Definition. If S is a linearly ordered set, then a subset I C_ S is called isolated, if for every x E I there exist two elements ai, bi in S with ai < x 5 bi such that the intervals (ai, bi], i E I, are pairwise disjoint. In this connection the following holds: 7.10 Theorem (Padmavally [133]). Let S be a linearly ordered set which has no gaps, but a first element xo and a last element. And let I be an isolated subset of S. Then there is no <  preserving mapping of S in I.
Proof. Suppose indirectly that f : S + I is a <  preserving mapping. The first element xo of S is not in the isolated set I and so xo < f (xo) =: xlholds. From xo < XI we obtain (xl =) f (xo) < f (xl) =: x2 and so on. Precisely: Let T be an ordinal such that x, E I is already defined for 1 5 v < r such that the x, strictly increase with
v and that xu < f (xu) holds for all v < T. If T is a successor ordinal g + 1, then xu < f (xu) =: x, satisfies x, = f (x,) < f (x,). If T is a limit ordinal then x, := sups{x,~v < T} exists because of 2.10. And x, is not in I because I is isolated. Now f (x,) is 2 f (xu) > xu for all v < T, and then also f (x,) 2 sups{x,~v < T) = x, holds. But f (x,) = x, is impossible since f (x,) is in I, but x, is not. Thus also x, < f (x,) holds. Applying transfinite induction this process of defining elements xu could be performed over the whole class of ordinals v, which of course yields a contradiction. For the lexicographic products we obtain as a corollary:
7.11 Theorem. For every ordinal a the set 2((w,)) is not embeddable in Ha. Proof. The set H, x {e), where e is an arbitrary element of Ha, is an isolated subset of H, x Ha. For let el, ez be elements of H, with el < e < ea. Then the intervals [(x,el), (x, ez)], x E Ha, of H, x H, contain (x, e) and are pairwise disjoint. Now H, x Ha is by 4.5 isomorphic to H, and then embeddable in 2((w,)), say by a mapping f. The set I := f [{H, x {e}] is then an isolated subset of 2((w,)), and it follows from 7.10 that there is no < preserving mapping of 2((w,)) in I, and I is isomorphic to Ha. By the way, using the GCH our assertion would be trivial since then 2Na = 12((w,))/ > N, = ]Halholds.
7.12 Theorem. The set S := w,((w,)) (where W, = cf(w,)).
is not embeddable in Ha
Proof. Our assertion would follow immediately if the GCH is adopted, > N, = [Hal.But we don't for then we would have lw,((w,))l = need the GCH. If w, is regular, we have w, = w,, and then our assertion is an immediate consequence of 7.11. So we assume that w, is singular. And then we define ordinals w, as in the proof of 7.7. Further for v < w, we define Hkv as the set of all those sequences of H, that have their last digit 1 at a position /I < w,, . We assume indirectly that there exists a <  preserving mapping f of S in Ha. If (T,),<~ is a sequence of length X w, of ordinals < w,, we define S[T,/U < A] to be the set of all elements of S that have (T,),<~ as initial segment. First we consider the sets S[~],whereT < w,. There
N?
<
4.7. ORDER TYPES O F LEXICOGRAPHIC PRODUCTS
127
n H;, = 8 because H;, has no must exist a TO < w, such that f [S[ro]] subset of type w,. Suppose now that we have already defined ordinals r, < w, for v < A, where X is an ordinal with 0 < X < wy, such that f [S[rv(v < p]]f~H& = ~8 holds for all p < A. Then we conclude as before that there exists an ordinal TA < w,,such that f [S[~,lv 5 A]] n H;, = 0, because H;, has no subset of type w,. By transfinite induction we so obtain a sequence of length w7 of ordinals r, < w,, i. e. an element of S, for which f [S[rvlv < wr]]n u{H;" Iv < w7) = 8 which is impossible since the last union is Ha. 7.13 Theorem (Hausdorff [82],1908). Let S be a linearly ordered set which has no gaps, but a first element a and a last element x > a. Let p be an arbitrary ordinal. Then there is no <  preserving mapping of S((P + 1)) i n S W . Proof. S((p)) is, of course, isomorphic to the set S* of all x E S((p 1)) that have z as last component. Now S*is an isolated subset of S((p + 1)). For let x = ( x , ) , < ~ + ~be an element of S*, and thus xp = 2. Then x E (y, x] := Ix,where y = (y,),
+
+
As a corollary to 7.13 we mention:
7.14 Theorem. For every ordinal p the set 2((p morphic to a subset of 2((p)).
+ 1)) is not iso
Now we supplement the statement of 7.6. This theorem has the following immediate consequence:
7.15 Remark. For wy = cf(w,)) the set w,((w,)) is embeddable i n 2((wa))* This statement is best possible with regard to the ordinal w7. Namely there holds: 7.16 Theorem. For w7 = cf(w,) the set M := w,((w7 embeddable i n 2((w,)).
+ 1))is not
Proof. If in 7.13 we take w, instead of p, we obtain that 2((w,+ I)), and then a fortiori w, ((w,+l)) is not embeddable in 2((w,)). Thus we are done, if w, is regular.
Let now w, be singular. Then there exists a (with v) strictly increasing sequence of inital ordinals w,,, v < w,, for which C{wffvIv < w,) = w, holds. We assume indirectly that there exists a <  preserving mapping f : w,((w, + 1)) + 2((w,)). If t = (r,lv < p) is a sequence of ordinals < w, and p < w,, we define S(t) to be the set of all sequences of w,((w, I)), that begin with t, i. e. which have t as initial segment. And if d = (6,lv < K) is a dyadic sequence, so that 6, E {O,1) for v < K, then we denote the set of all sequences of 2((w,)), that begin with d, by D(d). First we prove: (I) There is an ordinal T < w, and a dyadic sequence d = (6,lv < w,,) such that f [S(r)]E D(d) holds. There are three consecutive ordinals K, K 1,K 2 < w, such that the f  image sets f [S(T)],f [S(T I)], f [S(T 2)] intersect the same segment D(d) with a d = (6,lv < w,,) of 2((w,)). For if this would be false, then each D(6,lv < w,,) would intersect at most two of the sets f [S(t)], where t is an ordinal < w,. And then the set { f [S(t)]lt < w,) would have, relative to the natural order of 1.8.1, an order type 5 C{2ili E 2((w,,))), where each 2i is the order type 2. But it has the order type w,, which is not embeddable in the last ordered sum. If now we put T := K 1 the set f [S(T)]fulfills (I). Suppose now that for an ordinal p < w, we have already defined for all v < p ordinals 7, < w, and for all v < w,, ordinals 6, E {O,1) such that
+
+
+
+
+
+
(11) f [S(~ulv< P)] E D(6uIv E C Wlv~<, 11) holds. Then, in the same way as before, we can find an ordinal rpand 5 p)] E D(6, lv E a dyadic sequence of length w,, such that f [S(T,~U C w, Iv 5 p). If X is a limit ordinal for which (11) holds for all p < A, then also (11) is valid for A. So, finally, we have by transfinite induction obtained a sequence t = (r,lv < w,) of ordinals Tu < w, such that all 1))that begin with t, are mapped by f on one sequences of w,((w, element of 2((w,)). This is a contradiction because there are still N, sequences in w,((w, 1)) which begin with t.
+ +
We remark that in [108] it was indicated that w,((w,)) is embeddable in 2((w,)) for all w,. But this is by 7.16 only true for the regular w,. The citation in the footnote of 1701, p. 245 should be corrected adequately. In analogy to H, we define sets G, as follows:
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129
7.17 Definition. Let a be an ordinal. Then the set of all sequences of 2((w,)) that have a last digit 0 is denoted by G,. The following statement is easily verified:
7.18 Theorem. Every element s E G , has an immediate successor t i n 2((w,)), and every element t E H , has an immediate predecessor s E G,. All elements of 2((w,)) that have an immediate successor (resp. immediate predecessor) are i n G , (resp. i n H a ) . The mapping which ascribes to every s E G , its immediate successor i n H , constitutes an isomorphic mapping of G , onto Ha. Thus also G , has the order type h,. Proof. Let s = (s,Iv < w,) E G , have its last digit 0 at the position T . We put t, = s, for z/ < T , t, = 1, i, = 0 for T < v < w,. Then the sequence t = (t,lv < w,) is evidently the immediate successor of s. The rest follows easily. Later we wish to prove a theorem of Kurepa/Papic [108],according to which all lexicographic products (v l)((w,)) with 1 5 v L. w, have the same order type. We first establish:
+
+
7.19 Lemma. Let w, be an initial ordinal and S := ( v l)((w,)) where v is an ordinal with 0 < v < w,. W e define D to be the set of all those sequences a E S satisfying
(*) a = ( a p l p < w,), and there exists a last component a, # v . (For v = 1 the set D is = G,.) Then D has the order type h,. Proof. First we assume that w, is regular. Then there holds: ( 1 ) Every a E D has the elementcharacter (w,,wG). Let a satisfy (*). For K < p < W , let b, be that sequence of D which arises from a by exchanging the p  component of a by 0. Then {b, I K < p < w,) is cofinal in ( D < a ) . For K < p < W , we define sequences cp = (cP,1r < w,) by c,, = a, for T < K , cp, = a , 1, c,, = 0 for K < T 5 p cpT = v f o r p < r < w , . Then {c,lp < w,) is coinitial in (D > a ) . And thus ( 1 ) is proved.
+
For s,,
=v
< W, let (s,,) for K < p < w,.
K
be the sequence where s,, = 0 for p 5 K and Then {s,~K < w,) is a coinitial subset of D of
type WE. For K < W, we define t, = (t,,lp < w,), where t,, = 0 and t,, = v for all p # K. Then { t , l ~ < w,) is a cofinal subset of D of type w,. Let now (A,B ) be a gap of D. According to 2.10 then there exists the supremum s = (s,lp < w,) of A in S. Because of s E S\D and s # ( v , v , . . .) we have two possible cases: Case 1. The set of indices p , for which s, is # v , is cofinal in w,. If Case 1 does not hold, then the set of indices with s, # v is not cofinal in w,, and then we have: Case 2. There exists a least ordinal X < w,, such that s, = v for p < w,. T h i ~X must be a limit ordinal, for otherwise s all p with X would at position X  1 have a last component # v , and then s would belong to D with contradiction. And due to s 4 D the set of ordinals K < A, for which s, # v holds, is cofinal in A. In Case 1 let { p p l p < w,) be the set of all indices p with s, # v. For p < w, we define bp to be that sequence which arises from s by replacing all components s, with p > p p by v. Then bp has a last component # v at the position p p and is thus in D. Now {bplp < w,) is coinitial in ( S > s), and then also coinitial in B. In Case 2 we define d, for p < w, as follows: d, arises from s by exchanging the ( A p )  component sx+, by 0. Then d, E D, and the set {d,lp < w,) is cofinal in ( S < s), and then also cofinal in A. So we have obtained that every gap (A,B ) has character (wp, w:) in Case 1 (resp. character (w,,wE) in Case 2 ) where wp is an initial ordinal. Thus D is an 7,  set. Now D is the union of N, sets D,, p < w,, where D, consists of all those sequences of D that have their last component # v at a position < p. Then D, is embeddable in ( v 1 )( ( p ) ) and thus free from w, and wTy by 2.2. Due to 3.12 our lemma now is proved for regular w,. Let w, be singular. Then we have a representation w, = '&,w,+l. We define D,+l to be the set of all sequences of D, for which the last component which is # v , is at a position < w,+l. Then it follows from the first part of this proof that D,+l has the order type h,+l. Because our assertion now follows from 4.3. of D = U,<,D,+l
<
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7.20 Lemma. Under the assumptions of 7.19, D is dense i n S = (u
+ l)((wa)).
4.7. ORDER TYPES OF LEXICOGRAPHIC PRODUCTS
131
Proof. Let a = (apIp < w,) and b = (b,lp < w,) be sequences of S with a < b, T the first ordinal where their components a, and b, are different, so that a, < b, holds. Let then x = (x,lp < w,) be defined by x, = a, for p 5 T , xp = v for T < p < w,. Then we have x E D, for its last component # v is x, = a,, and this is (< b,, and a fortiori) < v. Further we have a 5 x < b.
D is not strictly dense in S since S has elements which are neighboring. Now we can prove:
7.21. Theorem (KurepaIPapic [108]) Let w, be an initial ordinal and 1 5 v < w,. Then the lexicographically ordered set S := (v + 1)((w,)) is isomorphic to 2((w,)). Proof. With the concepts of 7.19 we have an isomorphic mapping f : D +=G,. Let D' be the set of immediate successors of elements of D. Then we extend f by mapping every x E D' onto the immediate successor (in 2((w,)) ) of f (x), which is in Ha.Further we map the first (resp. last) element of (v l)((w,)) onto the first (resp. last) element of 2((w,)). Let now x be an element of S\(D U D'), which is neither first nor last element of S. Then ((D < x), (D > x)) determines a gap in D of type (oo, oo) (see Definition 1.10.15), and (f [(D < x)], f [(D > x)] ) determines the corresponding gap in f [Dl 2((w,)). Since this latter set has no gaps there exists exactly one element y E 2((w,)) with f [(D < x)] < y < f [(D > x)]. We map x onto y. By extending f in this way it can be seen that we so obtain an isomorphic mapping of S onto 2((w,)).
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A fortiori we also have: 7.22 Theorem. Let v be an ordinal dable in 2((w,)).
< w,.
Then ~((w,))is embed
For even (v + 1)((w,)) is embeddable in 2((w,)).
7.23 Remark. Theorem 7.21 cannot be sharpened so that also ~((w,))would be isomorphic to 2((w,)). Indeed, if X is a limit ordinal < w,, the set X((w,)) is cofinal with X and with cf(X). In particular, it has no last element. Moreover, for different ordinals A, e.g. for different initial ordinals, the cofinalitytypes cf(X) can be different.
At the end of this paragraph we compile several results. We can give a complete discussion of which lexicographic products p ( ( v ) ) are embeddable in Ha resp. in 2((w,)).
7.24 holds: 1) If 2 ) If 3 ) If Ha 4 ) If 5 ) If
Theorem. Let p, v be ordinals 2 2, wy = cf(w,). Then there p 2 w,+l, then p ( ( v ) ) is not embeddable i n Ha. w, 5 p < w,+l and v < wy, then p ( ( v ) ) is embeddable in Ha. w, 5 p < w,+l and v 2 wy, then p ( ( v ) ) is not embeddable i n p p
< w, and v < w,, then p ( ( v ) ) is embeddable i n Ha. < W , and v 2 w,, then p ( ( v ) ) is not embeddable i n Ha.
Proof. 1 ) follows from the fact that H , has no subset of type w,+l, due to 2.6, b). 2) p is embeddable in Ha and then p ( ( v ) ) in H,((v)), which has the type h, by 7.5. 3 ) w,((w,)) is not embeddable in Ha by 7.12. A fortiori this holds for ~ ( ( 4 . 4 ) If w, is regular, then p ( ( v ) ) is embeddable in H , ( ( v ) ) , and this in H a by 7.4. If w, is singular, then there exists a regular initial ordinal w, < w,, for which p and v are < up,and then p ( ( v ) ) is embeddable in (w, l ) ( ( ~ , + ~ the ) )latter , in 2 ( ( ~ ~ +by~ 7.22, ) ) and this in Ha. 5 ) 2((w,)) is not embeddable in H , by 7.11. So we are done because ofp>2andv2wa.
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In the cases 2) and 4) of 7.24 the set p ( ( v ) ) is a fortiori also embeddable in 2((w,)), so we have only to discuss the cases 1),3),5) in the next theorem:
7.25 Theorem. Let p, v be ordinals 2 2, wy = cf(w,). Then there holds: 1) If p 2 w,+l, then p ( ( v ) ) is not embeddable i n 2((w,)). 3 ) a) If w, 5 p < w,+l and v = wy, then p ( ( v ) ) is embeddable i n 2((wa)). 3 ) b) If w, 5 p < w,+l and v > wy, then p ( ( v ) ) is not embeddable in Ww,)). 5 ) a) If p < W , and v = w,, then p ( ( v ) ) is embeddable i n 2((w,)). 5 ) b) If p < W a and v > w,, then p ( ( v ) ) is not embeddable i n 2((wa)).
4.7. ORDER TYPES OF LEXICOGRAPHIC PRODUCTS
133
Proof. 1) 2((w,)) has no subset of type w,+l by 2.6, b). 3) a) p((v)) is embeddable in H,((wy)), and this in 2((w,)) by 7.6. 1))is not embeddable in 2((w,)) by 7.16. A fortiori 3) b) w,((w, this holds for p ((v)). 5 ) a) p((v)) is embeddable in 2((w,)) by 7.22. 5) b) 2((w, + 1))is not embeddable in 2((w,)) by 7.14. A fortiori this holds for p((v)).
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We supplement our investigations by considering which well (resp. inversely wellordered) subsets the lexicographically ordered sets p((v)), where p and v are ordinals, can have: 7.26 Theorem. Let p and v be ordinals < w,, which are not 0. Then the lexicographically ordered set p((v)) has no subset of type w,.
Proof. For a = 0 the statement is trivial. So let a be > 0. From p < w, we conclude 1p1 5 Np for a ,O < a, and therefore p is embeddable in H p , and further p((v)) in Hp((v)). The set Hp is w,  free. If now w, is regular we obtain from 2.2, a) that Hp((v)) is w,  free, and this a fortiori holds for p((v)). If w, is singular, a is a limit ordinal, and then there exists a regular initial ordinal wp < w, which is greater than p and v. Now by the first case p((v)) has no subset of type wp, and a fortiori no subset of type w, . The Theorem 7.26 is tight:
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7.27 Remark. If p is w, and v = 1, then p((v)) is isomorphic with w,. And if p is 2 and v w,, then already 2((w,)) H, contains a subset of type w,.
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7.28 Theorem. Let p be an arbitrary ordinal, v < w,. Then p((v)) has no subset of type w.: And this statement is tight.
Proof. p has no subset of type w:, a fortiori no subset of type w.: If w, is regular, then it follows from 2.2, b) that p((v)) is w2,  free. Let now w, be singular, so that a is a limit ordinal. Then there exists a regular initial ordinal wp < w, which is > v. Then by the above case p((v)) has no subset of type w;, and a fortiori no subset of type w; . 7.28 is tight: If p is 2 and v 2 w, then p((v)) contains 2((w,)) and then also a subset of type w.:
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7.28' Corollary. If w, is singular and w, = cf w,, then w,((w,)) has no subset of type ~ l ; + and ~ , a fortiori no subset of type w:. Now we can supplement 7.15 by:
7.29 Theorem. If w, is singular and w, = cf w,, then 2((w,)) is not embeddable in w,((w7)). Proof. By 7.28' w,((w7)) has no subset of type w:, but 2((w,)) does. In the following we introduce certain subsets of 2((w,)), that constitute higherdimensional generalizations of the set R of real numbers. The set 2((w,)) has jumps: There are elements a, b in 2((w,)), where a is an immediate predecessor of b. Namely we have, recalling a previous result:
7.30 Remark. Let a and b be elements of 2((w,)). Then the sequence a = (a,lv < w,) is an immediate predecessor of the sequence b = (b,lu < w,) i f a has a last digit 0 , say a, = 0 , b, = a, for p < v , bu = 1, and b,, = O f o r v < p < w,. In order to obtain a subset of 2((w,)) which is dense, we eliminate from 2((w,) the elements of G, (of 7.17) and define:
7.31 Definition. For an ordinal a we put C , := 2((w,))\(G, {a, I ) ) , where o is the first and I the last element of 2((w,)).
U
For these sets we now have:
7.32 Theorem. Let cl, be an ordinal. Then C , is dense, and its subset H , is strictly dense i n C,. Each open interval of C , has a segment which is isomorphic lo C,. Further C , is continuously ordered, that means it has no gaps, and C , has no first and no last element. The set Co is isomorphic to the set R of real numbers. Proof. Let a = (a,lv < w,), b = (b,lv < w,) E C , satisfy a < b. At the first index p, for which a , # b, holds, we have a, = 0 and b, = 1. Since a is not in G, there is an ordinal r > p with a, = 0 , and then the element z = (z,lv < w,), which satisfies z, = a, for v < r, z, = 1, and z, = 0 for r < v < w,, is in Ha and satisfies a < z < b. Let S be the set of all x = (x,lv < w,) E C , that satisfy xu = a, for v < T and x , = 1. These fulfill a < x < b and form a segment C ( a ,b). This is isomorphic to the set F of final segments (x,~T< v < w,) of the
4.7. ORDER TYPES OF LEXICOGRAPHIC PRODUCTS
135
x E S, and F is isomorphic to one of the sets C,, {o) U C,, C, U {I), {o) U C, U {I). If we omit the first (resp. last) element of S, if it has one, we obtain a segment C S C (a, b), which is isomorphic to C,. If A is a nonempty initial segment of C,, and if the complementary final segment B := C,\A is also nonempty, then A has a last or B a first element. For A has, due to 2.10, as a subset of 2((w,)) a supremum s in 2((w,)). If s E C,, then s is also the supremum of A in C,. If s $! C,, then s E G, has the form (. . . , O , 1 , 1,1,.. .) and then the upper neighbor (. . . ,1,0,0,O, . . .) of s in 2((w,)) is the supremum of A in C,. B has an infimum i in 2((w,)). This is not in G,. For if i would be = (. . .O, 1,1,1,.. .) with a last component 0, then its upper neighbor (. . . l , 0,0,O, . . .) would be a lower bound of B, and i could not be = inf B in 2((w,)). So i is also the infimum of B in C,. If now s is the last element of A, we are done. If s $! A, then i must be the first element of B. Indeed, we have s 5 i, and s < i is impossible since C, is dense. Now s = i E C, = A U B, and so s = i belongs to A or B. The rest is clear. The order type of the set C, is a subtype of that of 2((w,)), but the converse also holds: 7.33 Theorem. For each ordinal a the set 2((w,)) is embeddable in C,. And IC, I = 2Na.
Proof. If 6 = (6,lv < w,) is a dyadic sequence E 2((w,)), we ascribe to 6 an element f (6) E C, as follows: We enlarge the sequence 6 by putting an additional digit 0 directly behind each digit 1 which occurs in the sequence 6. The sequences which so arise are not in G,, and thus in C, U {o). If then 6 < e are sequences E 2((w,)), their image sequences satisfy f (6) < f ( E ) , SO that 2((w,)) is isomorphic to a subset of C, U (0). Let now g be the mapping which ascribes to every 6 E 2((w,)) that sequence E 2((w,)), which arises from 6 by putting the digit 1 before 6. Then the composition f o g : 2((w,)) + C, is <  preserving, and this proves our assertion. Since H, is dense in C, we can state that, using GCH, there is a ,,l in which a subset of cardinality linearly ordered set of cardinality N+ N, is dense. The question arises, what can be said if we don't presuppose the GCH. In this context we can now apply the previous theorems for a short proof of a theorem of Sierpinski [163], with which he gave a positive
answer to questions of Knaster (resp. Kuratowski). We formulate it in a more general form, which was already indicated in [163]:
7.34 Theorem [163]. Let m be an infinite cardinal. Then there is a linearly ordered set S of cardinality > m, which has a subset D of cardinality 5 m, which is strictly dense i n S.
Proof. For m = No the sets D := Q and S = R satify the assertion. So we suppose m > No. There exist cardinals k 5 m with 2k > m, (e.g. k = m). Let N, be the least cardinal k with this property, so that we have:
<
me (1) N, (2) 2N0 > m. (3) 2k 5 m for all cardinals k < N,. Now S := C, (of 7.31) satisfies the assertion; C, has by 7.33 and (2) the cardinality 2N0 > m, and D := H, is strictly dense in C,. So we are done if we have proved /Hal 5 m. We distinguish two cases: Case 1. a is a successor number p 1. Then I H, I = t, = 2 N 5~ m by (3). Case 2, a is a limit ordinal. Then IH,I = t, = '&,2 N ~ m  la1 by (3) because of Nu < N, for v < a . For v < a we have v 5 w, and thus Ivl 5 Iw,l = Nu and Ivl < 2Nu5 m by (3). This yields v < w(m) for all v < a . Then a = sup{vlv < a) is 5 w(m). This entails la1 5 m, so that [Hal me m = m follows.
+
<
<
As a consequence of 7.34 we also obtain a positive answer to the following question of Kuratowski: Does there exist a chain of subsets of (R,C) which has a cardinality > 2N0? First we prove in general:
7.35 Theorem [163]. Let m be an infinite cardinal, IMI = m. Then ( p ( M ) ,C) has a chain of cardinality > m.
Proof. For the cardinal m we determine sets S and D as in 7.34, so that ID1 5 IMI = m holds. The mapping which assigns to each x E S the initial segment (D < x) of D, is injective since D is strictly dense in S. And the set of all initial segments (D < x), x E S, ordered by C, is a chain of (p(D), 2)of cardinality IS1 > m. Since (V(D),C) is isomorphic to a subset of ($(M), E) our assertion follows. If in the last theorem we take m := 2N0and M := R, the above question is answered.
4.8. CANTOR'S NORMAL FORM
137
4.8 Cantor's normal form. Indecomposable ordinals In the following we need some concepts and theorems concerning the exponentiation of ordinals. We extract here only what we use in the sequel. 8.1 Definition. For the least infinite ordinal w we define w0 := 1 and w1 := w. If we have already defined wV for an ordinal v we put wV+l := wV w (= W" wV  . IW many summands). If for a limit ordinal X the power wV is already introduced for all v < X we put wX := U{wVlv < A) (= sup{wUlv< A)). By transfinite induction we so have defined the powers wVof w for all exponents v which are ordinals. (This concept must not be confused with the exponentiation of cardinal has cardinality numbers. So wW has only cardinality No, whereas > No.) An ordinal 5 is said to be indecomposable, if it is not the sum of two smaller ordinals: = a + @ a = 5 (and t h e n p = 0) or p = 5 (and then a < 5).
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*
8.2 Theorem. Every power wV is indecomposable.
Proof. For v = 0 or 1 this is trivial. Let v be an ordinal for v. Suppose indirectly which the statement is true for all ordinals W ~ +=l a p, where a and /3 are < wV+l. Then there exist numbers a , b € N o w i t h a < w V . a a n d p < w V . b . For i f a would be > w V  n f o r all n < w, a would also be U{wV  nln E N) = wV+l,and similar for p. N o w a + ~ w o u l d b e < w V ~ a + w V ~ b = w V ~ (
<
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8.3 Theorem. Let p, v be ordinals with p < v. Then there holds < w". Further we have wp > p for all ordinals p. Proof. wp+' > wp follows immediately from the definition; also for limit ordinals X we have wX> wp for all p < A. And wp > p follows wp
easily by induction. Here we remark without proof that it can happen that wp = p holds.
8.4 Lemma. Let p be an ordinal > 0. Then there exists a greatest ordinal a for which w" is 5 p. And then further there exists a greatest number n E No for which w" n 5 p.
Proof. Let S be the set of all ordinals v which satisfy wV 5 p. This set contains with an ordinal v also all ordinals 7 < v. Then the supremum a of S is an ordinal for which still w" 5 p holds, and then a is the greatest exponent with this property. Now we consider the sequence w" . 0, w" . 1,. . . . . Here the first member is 5 p. But not all are 5 p, because in this case we would have that also w"+l = u { w a mlm E No) would be 5 p. So the number n exists. Now we can construct Cantor's normal form: 8.5 Theorem and Definition. For every ordinal p > 0 we have wan . a,, where n is a a unique representation p = wffO. a0 nonnegative integer, a 0 > . . . > a, is a strictly decreasing sequence of ordinals, and ao, . . . ,a, are natural numbers. This is called Cantor's normal form (or representation) of p.
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Proof. First we prove the existence. Suppose that for all ordinals that are # 0 we have already proved that they have a uniquely determined representation of the above kind. Then by 8.4 there exists a greatest ordinal a 0 with w"0 5 p and further a greatest number no E N with wffO  a0 5 p. Then the ordinal p, which is uniquely determined by p = w"O  a0 + p, is 0 or has, due to the induction hypothesis, a represen+wan . a, of the kind under consideration, where tation p = w"l  a1 in particular a1 < a 0 holds. So the existence is proved by induction. Uniqueness follows so: Let this be proved for all ordinals < p. In every representation of p in the above form a 0 is the greatest ordinal for which w"0 is 5 p, and so a 0 is uniquely determined. The same holds for ao, due to 8.4. Since by induction hypothesis also p is uniquely determined, the uniqueness of the representation now easily follows by induction on p.
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Using Cantor's normal form a possibility arises to introduce an addition in the class of ordinals which is different from the usual addition, but has nice properties: 8.6 Definition. Let a and P be ordinals > 0 with normal form +wan . a,, p = wS0 . bo +wpm bm. (1) a = waO a0 We can choose here a common system of powers of w by putting several coefficients a, resp. bv equal 0. So we can assume w.r.0.g. that (1) holds with m = n and a, = P, for v = 0,. . . ,n, but where now the a, and bv are only 0.
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4.8. CANTOR'S NORMAL FORM
139
And then we define the Hessenberg natural sum of a and P to be the ordinal a CBp := w a O . (ao bo) ... +wan  (a, b,). For the sake of completeness we further put a @ P := a P, if a or p is 0.
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+
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One can immediately see that this operation @ is commutative and associative. In the following, sets of ordinals are always considered as ordered, (in the obvious sense) by magnitude. We prove some properties of indecomposable ordinals: 8.7 Lemma. Let a be an ordinal > 0, X < w", L E w" a subset of type A. Then the set wa\L has the order type w".
Proof. For a = 1 the statement is trivial. We make the assumption that it is satisfied for a fixed a > 0. Let then X < wa+l and L a subset of wa+lof type A. We have w"+l = w" w" .  . (w many summands), so that wa+' is partitioned in segments Sl,S2,. . . , which each have type w", and where all elements of a segment S, are less than all elements of a segment Sn if m < n holds. The set F of those n, for which the set SnnL has type w", is finite because otherwise L would have type wa+'. For the n of the infinite set N\F then S,\L has by induction hypothesis type w". These sets together form a subset of w"+l of type wa+'.
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As a reformulation of 8.7 and a generalization of 8.2 we have: 8.8 Lemma. If
W"
= AU B, then
A or B has the type
W"
8.9 Definition. Let p be an ordinal and p = A U B , where A, B are disjoint subsets of p, and where A has the type a, and B the type /3. Then p is said to be a mixed sum of the ordinals a and P.
Different ordinals can be a mixed sum of the same two ordinals a and p. E.g. if a and P are both = w, then w w is a mixed sum of a and p, but also w, for w can be represented as the union of the set of odd and the set of even numbers, which both have type w.
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8.10 Lemma. Let a and b be positive integers, a an ordinal. Then every mixed sum S of w" . a and w" b is 5 w" (a b).
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Proof. Let S be a mixed sum A U B, where A and B are disjoint sets with types w" . a resp. w" . b. The initial segment W of type w" of
S is = (W n A) U (W n B). By 8.8 one of these summands has type w". W.r.0.g. we assume that W n A has type wQ. Then we construct a new order 5 in S. We shift the elements of W n B behind those of W n A, so that S with 5 has an initial segment consisting of the elements of W n A. And behind this segment follows a segment consisting of the elements of W n B. The rest is unchanged. Then the type of S with its original order is the type of S with 5 . Now the type of S\(W n A) with 5 is a mixed sum of wQ . ( a  1) and w" . b. If we apply induction on the sum a b we can conclude that the last mixed sum is 5 w" ( a  1 + b). And then S has a type <w"+wa.(al+b)=wa.(a+b).
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Now we obtain the main theorem on mixed sums of ordinals. A proof of this was given by Neumer [125],where he remarked that this theorem was also observed by' carruth [12]. 8.11 Theorem. Every mixed sum of ordinals a, ,8 is
< a @ P.
Proof. If a or ,B is 0 , the assertion is trivial. So we assume that both are > 0. Using the normal form of ordinals we can represent a and p in the form wan  b,, a = wQO . a0 .  . wan a, and p = wQO bo  where a0 >    > a, are ordinals and where ao, . . . ,a,, bo, . . . ,bn are nonnegative integers. The above representations show that a (resp. P ) is isomorphic to an ordered sum A (resp. B ) of disjoint linearly ordered sets Ao,. . . ,A, (resp. Bo,. . . ,Bn) over the argument ( 0 , . . . ,n)<, where A, (resp. B,) has the order type w"" a , (resp. wa".b,) for u = 0 , . . . ,n. We apply induction on the number n. For n = 0 our assertion is proved by 8.10. Suppose now that n is > 1 and that the theorem is proved for n  1. Let then p be a mixed sum of a and P. Then the ordinal p is isomorphic to a linearly ordered set ( M , 5 ) ,where M is the union of two disjoint sets A and B, where A has the order type a and B the order type P, both with respect to the restriction of 5 onto A resp. B. Let now T be the set of all elements of M\(Ao U Bo) that are 5 some element of A. U So. Then we define an order j * i n M by: The union A. U Bo is an initial segment of ( M , d * ) , T is an initial segment of M\(Ao U Bo),and the rest of M forms a final segment of M: Here all three segments shall have the order induced by 5 . Then it is clear that the order type of ( M , 5 ) (and then also that of p ) is the type of
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4.8. CANTOR'S NORMAL FORM
141
(M, 3 ) ,for the elements of T were "absorbed" before by A. U Bo. By the induction hypothesis it now follows that the set M\(Ao U Bo) with the restriction of 5* has an order type T := ( ~ " l . a ~ + ~ . . + w " n  a , ) $ ( w " ~ bl+..+wanb,). And so finally M has by 8.10 a type 5 w"O (ao bo) T . So the theorem is proved for n, and by induction then for all n E N. Later we shall need the following simple theorems:
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8.12 Theorem. Let W be a wellordered set of type w" . n, where n E N , and x an element of W. Then (W > x) has an order type 2 wa.
.+
Proof. W can be represented as an ordered sum W = Wl+. W,, where each Wv, Y = 1,.. . , n , is a segment of type w". If x is in a Wi with i < n, the assertion is trivial. If x 6 Wn it follows from 8.7 since (Wn > x) has type w". 8.13 Theorem. Let p,p and a be ordinals and p p$p
+
< w". Then
Proof. We have p = p' p", where p' (resp p") is the sum of those summands wv of the representation of p in Cantor's normal form for which v 2 a (resp. v < a ) holds. Then p $ p is = (p' $ p") $ p = p' $ 6, where /G = p" @ p is < w", as the sum of finitely many ordinals of a form wp with ,G' < a. Since p' $ K < p w" holds, we are done.
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Chapter 5 Universally ordered sets In this chapter we treat the question which concerns the existence of N,  universally ordered sets. These are posets P which have the property: Every poset T of cardinality N, is isomorphic to a subset of P. In Section 4.3 we considered this question for the linearly ordered case. Here embeddability was reduced to the requirement that the embedding function cp : T + S fulfills: x < y (in T)+ cp(x) < p(y) (in S).Now, in the general (partially) ordered case also x 11 y cp(x) 11 cp(y) must hold to make cp[T], with the order induced by P, isomorphic to T. In order to construct N,  universally ordered sets of cardinality t, we apply a method of "successively adjoining IFpairs". These are studied in the next section.
5.1
Adjoining IFpairs to posets
The following two definitions are essentially the same as certain definitions of Cuesta Dutari [16], 1171:
1.1 Definition. Let X be a poset, A an initial and B a final segment of X, such that x < y holds for all x E A and y E B. Then we call ( A ,B) an IFpair of X. The set of all IFpairs of X is denoted by P(X). Here A or B or even both can be empty or the whole set X. In a linearly ordered set X every cut (A, B) is also an IFpair, but of cou.rse, not conversely. In the construction process for universally ordered sets we use an ordering in the set X U P ( X ) , whose restriction on X is the given order of X. 1.2 Definition. Let (X, <) be a poset, where X is disjoint to the set P(X).Then we extend to an order, which also is denoted by 5, of X U P ( X ) as follows: Let x, y be elements of X, (A, B) and (A', B') IFpairs of X. Then we define the strict order < of by
<
<
(l)x
(2) x
< (A,B)  X
E A,

(3) ( A , B )< X  X EB, B n A' # 0. (4) ( A ,B ) < (A',B') Evidently (4)is equivalent to (4') ( A ,B ) < (A',B') 3x E X such that ( A ,B ) < x From (4)we obtain:
< (A',B').
1.3 Theorem. Suppose ( A ,B ) < (A',B') for two IFpairs of a poset A' and B' B hold.
X,then A
c
c
Proof. By (4) there exists an element x E B n A'. This is > all elements of A, and A A' follows from x E A'. On the other hand B contains x, which is in A'and thus < every element of B1.Then the final segment B also contains B'.
c
We now have to verify that Definition 1.2 yields indeed an order in
X U P ( X ): 1.4 Theorem. The relation < of 1.2, augmented by the identity relation o n X U P ( X ) ,i s a n order relation o n this set.
Proof. It follows immediately that 5 is reflexive. It is also antisymmetric: For two IFpairs ( A ,B ) and (A',B') we cannot have ( A ,B ) < (A',B') and (A',B') < ( A ,B ) . For otherwise we would have an element x E B f l A' and an element y E B' r l A. Then X ( E A') < y ( B') ~ and y ( A) ~ < X ( E B ) would follow with contradiction. To verify the transitivity let x, y, z be elements of X and Ci = (Ai,Bi) for i = 1,2,3 IFpairs of X. We have ( 5 ) x 5 y 5 z (in X U P ( X ) )===+ x 5 z by ( I ) , (6) x ~ ~ ~ C ~ & X < ~ E A ~ ~ X E A ~ + X < C ~ (7) x 5 C 1 5 y + x E A 1 a n d y ~ B ~ + x < y , (8) C 1 5 x 5 y + x E B 1 * y E B 1 + C 1 < y , (9) x 5 C ~ L C 2 + x E A 1 ~ A 2 = +  x E A 2 + x < C 2 , (10) CI 5 x 5 C2 x E Bl n A 2 + Cl < C2, (11) Cl 5 C2 5 x + x E B2 G BI x E B1 + Cl < x, (12) Cl < C2 < C3 + By (4') there exist elements a,b in X with C1 < a < C2 and C2 < b < C3. Then by (7) we obtain a < b, and then by ( 6 ) a < C3. This and Cl < a yield C1 < C3 by (10).
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5.2. AN Ha UNIVERSALLY ORDERED SET
5.2
Construction of an set
145
N, universally ordered
2.1 Definition. A poset P is said to be N ,  universal or N , universally ordered, if every poset of cardinality N , is isomorphic to a subset T of P, where T, of course, is equipped with the restriction of the order of P.
B. Jonsson [92] proved : For every ordinal a there exists an N , universally ordered set of cardinality t,. His proof is nonconstructive. A constructive proof was published by Crawley and Dean [15]. They proved that certain free lattices F,(3) are N ,  universally ordered sets of cardinality t,. In this section we also shall present a constructive proof of the existence of such sets, which was published in [79]. In this connection Kurepa [I071 had introduced the following notion of q,,  set: 2.2 Definition. A poset E is said to be an q,,  set, if every subset X C E with 1x1 < N , has in E an extension in every direction. This x1 < N , there holds: If xo is means precisely: For every X C_ E with 1 an element which is not in X, and if X o := X U { x o ) is a poset, whose restriction on X coincides with the restriction of the order from E onto X, then there exists an element xT, E E, such that the mapping f, which puts f ( x o )= xT, and f ( x ) = x for x E X, is an isomorphism of X o onto a subset of E.
From the last definition some elementary properties of q,,  sets can immediately be seen: An qaa  set is dense, and so there are no neighboring elements. It has no first and no last element. For these qor,  sets Kurepa formulated the following two theorems: 2.3 Theorem [107]. An q,,  set is N ,  universally ordered.
Proof. Let E be an q,,  set and S a poset of cardinality N,. We take a wellordering of S such that S is representable as S = {s,/v < w,). Suppose that for an ordinal p < w, we have already constructed an isomorphic mapping f p of {s,[v < p) (where this set is equipped with the induced order of S) into E. Then there exists an extension fp+l : {s,lv p} + E because E is an q,,  set. If for a limit ordinal X for all v < X mappings f , are already constructed such that for p < Y < X the function f , extends f p , we define f x to be their limit
<
mapping. By transfinite induction we so construct a tower of mappings fi, : {s,lv 5 p ) + E, which extend with p. The mapping f,, yields an isomorphism of S onto a subset of E. 2.4 Theorem [107]. Every maximal chain of a n qaa  set E is a n q,  set.
Proof. Let M be a maximal chain of an qaa  set E, and A , B subsets of M which are neighboring : Every element of A is less than every element of B, and there is no element x of M with a < x < b for all a E A, b E B. Then A or B must have cardinality 2 N, because otherwise there would exist an element between A and B in the q,, set E. This would not be in M since A and B are neighboring, and then M would not be maximal. Thus M fulfills the condition for an q,  set. In the following we present two further properties of q,,  sets. First we define (resp. recall): 2.5 Definition. If S and T are arbitrary sets and f : S + T a mapping, x E S, we say: f leaves x fixed if f (x) = x. If A S and f (x) = x holds for all x E A we say: f leaves A firced. If P is a poset, a subset D C P is said to be strictly den.se in P if there holds: For every two elements a, b of P with a < b there exists an element d E D with a < d < b. If T is a subset of a poset P, and if a and b are elements of P\T, we say: a and b are in the same position to T iff (T < a ) = (T < b) and ( T > a ) = ( T > b). Then, of course, also (T 11 a) = ( T 11 b) holds.
Now there follows: 2.6 Lemma. Let T be a subset of a poset P, and let a,b be elements of P \ T with a < b, which are in the same position to T. T h e n every element z E P which satisfies a < z < b i s also in the same position to T as a and b.
Proof. Let z E P satisfy a < z < b, then z is not in T, for otherwise we would have the contradiction x E ( T < b)\(T < a) = 0. Now (T < x) C ( T < b) = (T < a) C ( T < z) holds. Then ( T < z) = (T < a) follows, and analogously (T > z) = ( T > a).
2.7 Theorem [79]. Let (E,I)be a n qaa  set and D a subset of E , which i s dense in E. T h e n D (with the restriction of 5 onto D) i s also a n q,,  set.
5.2. AN N, UNIVERSALLY ORDERED SET
147
Proof. First we see that the set D is also strictly dense in E since E is dense. Let X D and 1x1 < N,, xo 4 X, and let an order 50be given in XU {xo) whose restriction onto X coincides with the restriction of I onto X. Since E is an qa,  set there now exists an eo E E and an isomorphic mapping fo : X U {xo) + X U {eo) with respect to the orders and I r (Xu {eo)), which leaves X fixed and maps xo onto eo. Let now el be an element which is not in E. We introduce an order I l i n X U {eo) U {el) as follows: Il ( X U {eo)) is the same order as 5 1 ( X U {e")). Further we put eo <1 el, and we define the rest of
so
r
r
x.
We had eo < I el and, since f l is isomorphic, then also eo = fl(eo) < fl(el) = ei. Since D is dense in E there now follows the existence of an element d E D with eo < d < e;. Now eo (and el) and ei are in the same position to X. By 2.6 then also d is in the same position to X as eo. Then the mapping fi : X U {eo) + X U {d), which leaves X fixed and maps eo onto d, is an isomorphism with respect to the order I . The mapping fi o fo : X U {so) + X U {d) is now an isomorphism with respect to l o a n d I , which leaves X fixed and maps xo onto d. Thus D is an vaa:  set. 2.8 Theorem 1791. Let (E,I)be an qa,  set, and a,b elements of E with a < b. Then the open interval (a, b) := {x E Ela < x < b) is again an qcva set.
Proof. Let X 5 (a, b) and IXI < N,, xo 4 X, and let X U {xo) be a poset with order
Now IX U {a, b)l is < N,, and since E is an qaa  set, there exists an element e E E and an isomorphism f : ( X U {a, b)) U {xo) + ( X U {a, b)) U {e) with respect to and 5 which leaves X U {a, b) fixed and maps xo onto e. Since f is < preserving, a <' xo <' b implies a = f (a) < f (xo) = e < f (b) = b so that e E (a, b). The restriction f X U {xo) is then an isomorphic mapping with respect to <'and 5 into the interval (a, b), which leaves X fixed and maps xo onto e E (a, b). Now we consider the case where xo is one of the elements a, b. Suppose xo = a; the case xo = b is analogous. We choose an element x i where , the which does not belong to E and order X U {x:) with I* restrictions of and <*onto X coincide, and where x: is with xo(= a ) in the same position to X. Now x: is different from a and b, and so the first part of our proof yields that there exists an element e E (a, b) and an isomorphic mapping f * : X U {x:) + X U {e) with respect to I*and which leaves X fixed and maps x: onto e. Now we replace here x: by xo. Since these elements are in the same position to X we so obtain an isomorphic mapping of X U {xo) with
<'
<
<,
Kurepa [I071 had posed the question, whether there exists an q,, set of cardinality t,, if N, is regular. Subsequently such a set is constructed. First we introduce some new concepts: 2.9 Definition. Let A be an initial segment of a poset X, and A' C A a subset that generates A, i. e. A = {x E X13a E A'with x 5 a). Then A' is called a basis of the initial segment A. If IA'I k for a cardinal k, we call A' a k  basis of A. Analogously we define: If B is a final segment of X,which is generated by a subset B', i. e.: B = {x E X13b E B' with x 2 b), B' is said k. to be a basis of B, and a k  basis, if in addition to this IB'I Finally let (A, B ) be an IFpair of X. Then a pair (A', B') of subsets A' C A, B' B is called a basis of the IFpair (A, B), if A' (resp. B') is a basis of A (resp. B).It is said to be a k  basis of (A, B), if A' is a kbasis of A and B' a kbasis of B. Initial segments, final segments, and IFpairs, for which there exists a basis of cardinality 5 k, are called kgenerated.
<
<
Now we construct the N,  universal set U,.
5.2. AN
N, UNIVERSALLY ORDERED SET
149
2.10 Definition. If T is an ordinal and (T,, 5,) a poset, we define P,(T,) to be the set of all 171  generated IFpairs of T, and then further S(TT):= ((7,P )I P E P, (T,)). The reason why we list the pairs P together with an index r is that this ensures that the sets S(T,) are pairwise disjoint. Let To be the empty set 0. It is ordered by the empty relation. Let now 7 be an ordinal such that we have already defined a poset (T,, 5,) for all a 5 7. Then we put ( 1 ) TT+1:= TT U S(TT).So e.g. we have TI = ((0, ( 8 , 0 ) ) ) , T2 = TI U ( ( 1 ,P)1 P E Pl(Tl)).We order T,+l by I,+lin the sense of 1.2: (2) For x, y E T, we put 5T+1 y 57 Y. (3) For x E T,, y = (7,P ) with P = ( A ,B ) E PT(TT) we put x <,+I y x E A. (4) For x = (7,P) with P = ( A ,B ) E PT(TT) and y E TT we put x 1,+1 y +=+ y E B. (5) For x = (A,B ) and y = (C,D ) E P,(T,) we put ( 7 , ~17+1 ) (7,~) B n c # 0. If X is a limit ordinal for which posets (T,, I,) have been defined for all a < X,then we put Tx := U,


( 6 ) TT = u { S ( T , ) ~<~T ) . Since the sets S(T,) are pairwise disjoint by construction it follows from (1) and ( 6 ) that for every ordinal T the sets T, and S(T,) are disjoint. By transfinite induction we have so obtained a tower of posets (T,, 5,). And finally we define for every ordinal a :
(7) U, := Tu, and equip it with the order L,, . Our definition yields immediately that the U, form a tower of posets: 2.11 Theorem. If a and ,8 are ordinals with a < ,8, then U, is a subset of Up, and its order is the restriction of the order of Up to U,. And for limit numbers X one has Ux = Uu<xUu. 2.12 Theorem. Let a be an ordinal and wy = cf (w,). Then U, is an qyr  set.
Proof. Let X 5 U, with 1x1 < Ny and xo $ X. Let further ( X U {xo}, l o ) be a poset, where the restriction l o r X coincides with the restriction of U, onto X. We have to show: There exists an isomorphic mapping (with respect to lo and 5 ) f from X U {xo) oat0 a subset of U, with f (x) = x for x E X. Let ~ ( 1 x 1 )be the initial ordinal for the cardinal 1x1. Because of 1x1 < Ny there exists an ordinal p < w, such that X C T,. And since the Tv increase with v, there exists also a p which in addition to this satisfies ~ ( 1 x 1 5 ) p. The element xo determines in X the IFpair ( X
2.13 Theorem [79]. U, is an q,,
 set if N, is regular
Now there follows:
2.14 Theorem. For every ordinal a there holds: U, is an N, universally ordered set.
Proof. For regular N, our assertion is already proved by 2.3 because then U, is an q,,  set and thus N,  universal. Let now (X, 5,) be a poset of singular cardinality N,. Then there exists a bijective mapping of X onto w, and then also a representation X = {x,lv < w,). We put X, := {x,lv < w,} for p < a . Then x = U{X,lp < a}. There exists an isomorphism fo of Xo (with 5,) into Uo since Uo is an 700  set. Suppose that r is an ordinal < w, and that we have already defined a tower of isomorphic mappings f, : X,, + f EXp] Up for a11 p < r. Then we construct an isomorphic mapping f, of X, (with 5,) into U, as follows: If 7 is a successor ordinal a 1 let Xu+l\Xu = { U ~ I L < K } . Then, since is an q,+l,,+l  set we can successively construct isomorphic mappings cpo of Xu U {ao) into Uu+l, cplof Xu U {ao,a l ) into UU+l,. . .
+
5.2. AN N,
UNIVERSALLY ORDERED S E T
151
and so on, where every cp, extends the mappings cpp with p < v.The limit of the mappings cp,, L < 6,is then an isomorphic mapping fT+1 of X,+l into U,+l. If T is a limit ordinal and f , defined for all v < T , then we define f , to be the limit mapping of the f,, v < T . By transfinite induction we so obtain a tower of isomorphic mappings f , : X , + U,, T < a. Their limit mapping f , : X ( = X,) + U, is then an isomorphism of X onto a subset of U,. Before we determine the cardinality of U, we mention two lemmas.
<
2.15 Lemma. Let X be a poset with 1x1 2" where k is a n infinite cardinal. T h e n the set P of kgenerated IFpairs of X has cardinality [PI 2k. If Y i s a poset with lYl = N o , then the set of all finitely generated IFpairs of Y i s countable.
<
Proof. P has at most so many elements as there are pairs (A,B ) of subsets A, B of X with cardinalities k . The set of all subsets A X with cardinality [ A / k has a cardinality the cardinality of the set of all mappings of w(lk1) into X , which means its cardinality is (2k)k = 2k. Analogously we have for B also only at most 2k possibilities. Then /PI 2k . 2' = 2k follows. The rest is an easy consequence of the fact that the set of all finite subsets of a countable set is countable.
<
<
<
<
<
2.16 Lemma. For the sets T, of Definition 2 1 0 there holds: For T
2 wo we have ( 1 ) ITTI 2 21'1.
Proof. ( 1 ) is satisfied for T = wo because of ITuoI = No < 2N0.This follows immediately from the fact that all Tv with v E N are finite. Assume now that ( 1 ) holds for some T 2 wo. From Definition 2.10 and 2.15 we conclude IT,+1 I ITTI IS(TT)I 211' 2IT1 = 2IT+l1, and so ( 1 ) also holds for T 1 instead of 7. Let now X be a limit ordinal and let ( 1 ) be proved for all T < A. Then lThl C T < h IT,I ET<* 211' 5 IXI .21hl = 2'1. And so the proof follows by induction.
+
<
<
<
+
<
We determine the cardinality of
2.17 Theorem. IU,I
+
= t, (=
U, :
xu<,2 N u ) .
Proof. We consider the representation U, = u{T,Iv < w,). For a = 0 then lUol = No immediately follows. For a > 0 ( 1 ) yields IU,I 5 C { I T , I I U< w,) 5 C { ~ ~ ~ I I V < w,) = t,. On the other hand, if w, is regular, then U, is an q,,  set, and then it contains an q,  set which by 4.3.7 has a cardinality 2 t, . If w, is singular, then a is a limit numbe; and we have by 2.11 U, = U{U,+lIu < a ) . Now IU,I 2 IU,+ll = 2Nv holds for all v < a, and this entails IU,I 2 t,. 2.18 Remark. Subsequently to the construction of the sets U, we remark that for ordinals of the form w,+l one could have performed a somewhat simpler construction, which also would have led to an N , universally ordered set V,+1. Namely one can alter Definition 2.10 so that the sets PT(TT) are replaced by the sets of the N,  generated IFpairs of T,. If all other things are unchanged we would arrive at the sets in quite the same way as before to the sets U,+l. Our construction had the advantage that one could immediately see Theorem 2.11.
5.3
Construction of an injective <preserving mapping of U, into H,
We supplement Definition 4.1.9 by the following: 3.1 Definition and Theorem. Let a be an ordinal. W e define 0, (resp. E,) to be the set of all transfinite sequences s = (sVIu < w,) E 2((w,)) for which there exists an index p such that all s, with p 5 v < w, are = 0 (resp. = 1). W e put F, := 0, U E,. Then F, is embeddable i n Ha.
Proof. 0, = u { M p I p < w,), where Mp is the set of those sequences (s,) E Fa, for which s, = 0 for v 2 p. Of course, every M p is embeddable in H a , and then also 0, is that by 4.4.6. Analogously also the set E, is embeddable in H a , and then also F,. 3.2 Lemma. Let a be an ordinal, T a subset of H, with cardinality IT1 < Nr = cf (N,). Then T has a supremum s := supT E 0, and an infimum i := inf T E E,.
Proof. The existence of s and i follows from 4.2.10. Let s = (svl u < w,). If s would not be in 0, there would exist a strictly increasing
5.3. INJECTIVE <PRESERVING MAPPING OF U, I N Ha
153
sequence v,, K < w,, which is cofinal in w, and for which s,, = 1 for K < w,. Because of s = s u p T there exists for every K < w, an element v,, Then the set t, = (t,,lv < w,) in T with t,, = s, for all v C := { t , l ~< w,) is cofinal in (H, < s), and ICI IT/ < N, holds. On the other hand (H, < s) has a cofinal subset V of order type w,, namely the set of sequences b, = (b,,lv < w,),where b,, = s, for v < v, and b,, = 0 for v, < v < w,. By 3.1.10 the regular w, is the least ordinal which is cofinal in (H, < s), and this contradicts ICI < N,. Analogously to this it follows that, if i = (i,lv < w,) would not be in E,, the set of indices v with i, = 0 would be cofinal in w,. Then we could construct a subset of T which is coinitial in (Ha > i) and of cardinality IT1 < N, which is impossible.
<
<
<
The next two lemmas contain most of the proof idea of the main theorem of this section:
<,
3.3 Lemma. In the following TT,S(TT), are used in the meaning of 2.10. Let r be an ordinal < w,+l, f : T, + HV+lan injective < preserving mapping (with respect to <,and the standard order by first differences of HU+l). Let k be the cardinal Nu, We define k(HU+l):= HU+lU Pk(HU+l), where Pk(HUtl) is the set of all kgenerated IFpairs of Hv+17 and order it according to 1.2. (H,+l is disjoint to Pk(H,+1).) In the following we denote this order of k(HU+l) by
<
Proof. We put f * (x) := f (x) for x E T,. Let now (7, P ) be an element of S(T,) where P = (A, B ) is a 171  generated IFpair of T,. Then 171 2 IB'I, such that A' there are subsets A' A and B' B with IA'I generates the initial segment A and that B' generates the final segment B. Then we define f * (7, P) to be the IFpair ( I S (f [A],F S (f [B]). We have
c
c
<
(1) I S ( f [A']) = W f [A]) and F S ( f [B'I) = F S ( f PI) Indeed, IS(f [A']) I S ( f [A]) is trivial, and if x E I S ( f [A]) holds,
then we have x I f (a) for an element a E A.This is 5, a' for an element a' E A', and then f (a) 5 f (a') and x E IS(f[A1]) results. The rest follows analogously, and we have obtained f * ( r , P ) = (IS(f[AI),FS(f[BI)) E 471(Hv+l) C Pk(Hv+l), for 171 I Nu = k holds. So f is extended from T, to the function f * on the set T,+l. We verify that f * is <  preserving with respect to and Ik. (It is not necessarily injective.) For x E T, and P = (A, B) E P,(T,) we have (compare with 2.10, (3)): x <,+I (7,P) x E A f ( x ) E PEA1 f ( 4 E IS(f[AI) f*(x)(= f ( 4 ) < k ( W f [A]),F S ( f [BIN = f *hP ) . If (7, P ) < x holds, it follows that x E B + f (x) E f [B] + f (3) E FS(f P I ) ( W f P I ) , F S ( f [BIN < ( f ( 4 =) f * ( d hence f *(T,P ) < f*(x). Let now P = (A, B ) and Pl = (A1, Bl) be 171  generated IFpairs of T, with (7, P ) <,+I (7, PI). Then we have B f l A1 # 8, and so there exists an x E B n A1. Let A; A1 be a subset of cardinality 5 171, that generates the initial segment A1, and let B' C_ B be a subset of cardinality 5 171, that generates the final segment B. Then there exist elements b E B' and a E A;, with b x a. This implies f (b) 5 f (x) 5 f (a). Then f (x) E I S (f [A;]) n F S ( f [B']). Hence this set is nonempty and because of IS(f [A;]) = IS(f [A1]) and F S ( f [B']) = FS(f[B]) (by (1)) we also obtain I S ( f [Al]) n F S ( f [B]) # 8, so that f *(., PI = ( W f [A]),FS(f[BI)) < ( W f [All),FS(f [Bll)) = f *(T, Pl) holds. So we have obtained that f * : TT+1+ IC(H,+~)is a <  preserving and S k ) mapping which extends f. (with respect to Different elements of T,+l can have the same f * image. But this can only happen if these elements are in S(TT)because f *is injective on T, and by construction. And there holds: If (7, P ) and (7, PI) are elements of S(TT)with the same f * image, then P and Plmust be incomparable since f * is <  preserving.
*
*
*
*
*
3.4 Lemma. Let k := Nv, /C(H,+~)= HV+lU Pk(Hv+l), equipped such with the order
sk,
Proof. We define a linear order
in k(HV+l) by
5.3. INJECTIVE <PRESERVING MAPPING OF U, I N H,
155
a) For x, y E H u + 1 C k(HU+l) we put x 51y e x 5 y in Hu+1. b) For x E HU+l,(A, B ) E Pk(HU+l)we put (A, B ) <1 x, if x $ A, resp. x <1 (A, B), if x E A. c) For (A, B) and (C, D ) E Pk(Hu+l)we put (A, B ) <1 (C, D) e A C C or (A = C and D C B). Then 51 is the union of <1 and the identity relation on k(Hu+l). We verify that ilis an order relation. If this is proved this order is automatically also linear. Only transitivity is nontrivial. We verify it for all possible cases: Let x, y, z E HU+land (A, B), (C, D), (E, F ) E Pk(Hv+l). Then we have 1) x
Analogously, again by 3.1, the set of all k  generated final segments B of HU+l,equipped with the order 2,is embeddable in H,+l. Then, if A is a fixed k  generated initial segment of H,+l, the set A*of all k  generated IFkpairs (A, B ) E Pk(HvS1)has an order type 5 h,+l. Let Jkbe the set of all these A*, ordered by 2 . Now Pk(H,+1) is the union of the sets A* of type h,+l over the linearly ordered argument Jk, which also has order type 5 h,+l. By 4.4.6 then Pk(HV+l)has an order type 5 hU+l,and finally this also holds for k(HU+l)= HU+lu Pk(HvS1). The existence of the function g is now clear.
<
Since f * and g are both results of 3.3 and 3.4:
< preserving we obtain, combining the
3.5 Lemma. The composition h := g o f * : TT+1 + HU+lis preserving with respect to the orders and 5 .
<
h is not yet injective and therefore we shall alter it to obtain a function which is also injective. 3.6 Lemma. With the concepts of 3.3, 3.4, 3.5 it follows: There exists an injective <  preserving (with respect to <,+l and 5 ) mapping fT+1 : TT+1+H,+l, which satisfies (1) For x,y E we have h(x) < h(y) f ~ + l ( x< ) ~T+I(Y). Proof. For y E h[T,+l] we define C(y) := hl(y). Then {C(y)l y E h[TT+l]) is a partition of T,+l into subsets with the same h  image. Since h is <  preserving, comparable elements of T,+l have different h  images, and so every class C(y) is an antichain of TT+1. By the way : The classes C(h(x)) with x E T, contain only one element, namely x. Every class C(y) has at most the cardinality t,+l because of T < w,+l and 2.15 which entails 1 5 2Nv= IH,+l 1, and so there exists an injective mapping cpy : C(y) + HU+l,which, of course, is <  preserving since C(y) is an antichain. For y E h[T,+l] we now define My := {(y,x)Ix E HU+l)and with these sets the ordered sum C = C{Myly E h[TT+l]). Let 5, be its order. Roughly speaking: We construct the linearly ordered sum of copies of H,+l over the linearly ordered set h[TT+l] as index set. Then C has an order type 5 hV+lby 4.4.6. We now define a mapping f:+l : TT+l + HU+las follows: Every x E TT+1 is in the class C(h(x)), and then we put fJ+l(x) = (h(x),cph(,)(x)) E C. Then fJ+l is an injective <  preserving (with respect to <,+land 5,) mapping of T,+l into the linearly ordered set
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(C, I,), which has an order type 5 hV+l.There further exists an injec.Then finally f,+l := e o f:+l : tive <  preserving mapping e : C + HV+l T,+l + HV+lis injective and <  preserving with respect to and
I
If elements x,y E TT+1satisfy h(x) < h(y), then fJ+l(x) follows, and then further f,+l(x) < f,+l(y) (in HU+l).
<, f:+l(y)
We can resume, renaming now the f of 3.3 by f, : 3.7 Lemma. Let T be an ordinal < w,+l, f, : TT + HV+lan injective and <  preserving (with respect to 5, and I)mapping. Then there exists an injective <  preserving (with respect to and I ) mapping f,+l : TT+1+ HV+l,for which (1) of 3.6 holds.
In 3.7, f,+l is not necessarily an extension of f,. But we have the following: 3.8 Lemma. We introduce a strict linear order <<, in T, by x <<, y w fT(x) < fT(y) in HVtl and analogously in TT+1: a <<,+I b f,+l(a) < fT+1(b) in HV+l.Then <<,is a linear extension of I , and , <<,+la linear extension of 5,+lsince f, resp. f,+l are <  preserving. Then <<,is the restriction of <<,+lonto T,.
Proof. For x, y E T, we have x <<, y w f,(x) < f,(y) in Hv+1. This yields, since f *extends f,, f*(x) < k f*(y) inHV+1UPk(H,+1), and since g is <  preserving this entails h(x) = g(f * (x)) < g(f * (y)) = h(y). By 3.6 this finally leads to f,+l (x) < f,+l (y), which means to x <<,+I y. Now we can prove the main result of this section: 3.9 Theorem [79]. For every ordinal a there exists an injective  preserving mapping cp, of U, into H,.
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Proof. First we prove the theorem for successor ordinals a = v 1. In Definition 2.10 we introduced the sets TTfor all ordinals 7.For To = 0 we trivially have a linear order coin To whose order type is 5 hV+l.If for a fixed ordinal T < w,+l we have a linear order <<,in T, of type hV+l which extends <, we obtain by 3.7 and 3.8 that there also exists a linear order <<,+lin TT+1 of type 5 h,+lwhich extends <,+l and <<, . If X is a limit ordinal w,+l,and if <,is a linear order in T, for all p < X which extends the order <,of T,, so that (T,, 5,) has an order type 5 h,+l, where the relations <<, increase with p, then we define <
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on Tx as the union of the relations K P ,p < A. It follows directly from the construction that this yields a linear order <
B, for vl < y < p, and where n,<,B, is a oneelement set or empty. In the first case we call the blockdegression pointcontaining, in the second pointless.
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It follows easily that the above blockdegression can only be pointless if p is a limit ordinal. If in particular U is a dyadic splitting of a linearly ordered set, then we assign to every maximal blockdegression (Bv)v<x, where X is a limit ordinal, that dyadic sequence of length A, of which the sequences cp(B,), v < A, are initial segments. At the beginning we had only considered pointcontaining blockdegressions, but also the others can give useful insights into the structure of linearly ordered sets. 3.6 Theorem [67]. Let S be a linearly ordered set of cardinality Nv+l, which is free from w,+l and from w:+~. Let m be the least cardinal > Nu holds. (So, i n particular, m is 5 Nu.) for which Let U be a dyadic splitting of S into segments, D the set of those sequences which are ascribed to the maximal blockdegressions of U. Let D' be the set of those sequences of D, which have a changenumber < w(m). Then D'has cardinality 5 Nu. Of course D has at least elements. And thus the set of sequences with changenumber 2 w(m) has also cardinality I Dl. (Roughly speaking: ((Nearly all" dyadic sequences of D have change number 2 w(m).)
Nr
Proof. If d = (x,),<~ is a sequence of D and a an ordinal < c(d) we define the atrunc dla of d to be the initial segment of d, whose range of definition is the union of the first a constancysegments of d, hence dla = (x,lv E UT
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Analogously the supremum yl of the lengths of the sequences of Nu. So the D l ( l ) is < w,+l. And then IDl 1 = ID1(0)l ID1(l)l is equation
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(I) lDT1 5 N V is satisfied for T = 0 and T = 1. We assume now that (1) holds for all T < p, where p is an ordinal < w(m). If p is a successor number, then Nu. For if a sequence of Dp is given, there holds IDp/ 5 lDp1 1 Nu it arises from a sequence of Dp1 by attaching to it only digits 0 resp. only digits 1. And then the set of all sequences of Dp which arise in this way from one of Dp1has (similarly to Dl before) a cardinality Nu, so that indeed IDPI 5 IDp11  Nu Nu follows. If p is a limit ordinal, then the sequence dlp is uniquely determined by the complex (dlo),

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If instead of Nu+1 we have a regular limit cardinal, we obtain the following variant of 3.6:
3.7 Theorem [67]. Let S be a linearly ordered set with a regular cardinality Nu, where v is a limit ordinal, and where S is free from w, and w:. Let m be the least cardinal, such that a product of m cardinals, which are all < Nv, is 2 Nv. Let U be a dyadic splitting of S and a < w(m) a given ordinal. Then the set D' of all dyadic sequences of D(U), which have a changenumber < a, has a cardinality ID'/ < Nu. If the GCH is assumed we have m = Nu.
Proof. The proof is quite analogous to that of 3.6: Like there, we define sets Dr. Then by transfinite induction we obtain IDTI < N, for all v < w(m), and then ID'[ I UT<, D,I < Nu follows.
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As an application of Theorems 3.6 and 3.7 we obtain the following statement, which is a partial result of a theorem of Erdos/Rado [36]: 3.8 Theorem. Assuming the GCH there holds: Every infinite linearly ordered set S, which has no well and no inversely wellordered subsets of cardinality 181, and for which IS[ is regular, contains for every cardinal k < IS1 a well and an inversely wellordered subset of cardinality k.
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Proof. If IS1 = Nu+1 holds, where Nu is regular, then the m of 3.6 is = Hv because of the GCH, and there exists a dyadic sequence in the D(U) of a dyadic splitting U of S, which has a changenumber 2 w,, and this yields our assertion by 3.4, c). If IS1 is a regular limit cardinal Nv, then the m of 3.7 is = Nv, and thus 3.7 proves our assertion.
Theorems 3.6 and 3.7 can formally be generalized a bit more after the following theorem, which exhibits a connection between arbitrary representations of a linearly ordered set by dyadic sequences and representations which arise from a dyadic splitting. 3.9 Theorem. Let S be a set of dyadic sequences, of which no one
is a proper initial segment of another one, so that we have a linear order in S according to the principle of first digerences. Then there exists a dyadic splitting U of S into segments, so that the sequence s' E D(U), which corresponds (in the sense of 3.3) to an element s E S, is a subsequence of s, and so that S is isomorphic to the set {s'ls E S ) by S +s'. Proof. Let T be the least ordinal for which there exist sequences s E S, which have different T  components s(T). Then we define blocks B[O], B[1] (of a splitting of S, which we also define): B[O] := { s E Sls(r) = O), and B[1] := {s E S ~ S ( T=) 1). In the same way as we have split S into B[O] and B[1] we now split B[O] into sets B[OO] and B[01], and B[1] into sets B[10] and B[11],always provided that the sets to be split have at least two elements. In general we proceed as follows: Let ( v ~ ) ~ be < , a dyadic sequence for which a block B[vplp < K] and an ordinal 7[vplp < K] are already defined. Let then T := T [ V ~Ip 5 K] be the least ordinal such that there exist sequences s E B[vplp < rc] with different T components ~ ( 7 ) . Then we put B[vpIp < K,O] := { s E B[vplp < K] and S(T)= 0) and B[vplp < ~ , 1:= ] {s E B[vplp < K] and S(T)= 1). By the way, in this formalism the ordinal T from the beginning is T[ 1, the T of the empty sequence. If X is a limit ordinal, for which blocks B [ ~ p l p< K] are defined for all K < A, which form a descending sequence with increasing K, we put B[vplp < XI := n,<xB[.pIp < ~ 1 , if the latter intersection is nonempty. By transfinite induction we so obtain a dyadic splitting U of S (into blocks B[. . .I).
Our construction shows that the sequence s' := which we ascribe to s is a subsequence of s, and that s + s' defines an isomorphic mapping of S onto S' := {s'ls E S ) . The change number of a subsequence of a dyadic sequence s is of course always 5 that of s. Then we can establish the following generalization of Theorems 3.6 and 3.7, which has no relation to dyadic splittings:
3.10 Theorem [67]. Let S be a set of dyadic sequences, of which no one is a proper initial segment of another one, so that S is equipped with the linear order of first diferences. 1 ) If IS1 > Nu+1 and i f S has no subsets of type w,+l, w ; + ~ ,and i f m is the least cardinal for which NT > Nu holds, then there exist at most Nu sequences i n S with a changenumber < w ( m ) . 2) If IS1 N u , where Nu is a regular limit cardinal, and if S has no subsets of type w,, w;, and i f m is defined as i n 3.7, then for every a < w ( m ) the set of sequences of S, that have a changenumber < a, has a cardinality < Nu.
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Supplementing Theorems 3.6 and 3.7 we still have a look at the situation where we have a singular cardinal N,. Before we mention a theorem of Hausdorff [82],see also [36],Lemma 1.
3.11 Theorem. Let N , be a singular cardinal, w, := cf(w,), so that w, = C{wa,, lu < w,),where the w,, , u < w,, are initial ordinals < w,. For u < W , we define S,, to be a linearly ordered set of type wz,. Then we define a linearly ordered set S as the ordered sum S := C{S,,Iu < w,). ( W e can assume that the S,, are pairwise disjoint.) Then every wellordered subset of S has an order type 5 w,, and every inversely wellordered subset has an order type T* with some T < w,. Proof. Let W E S be wellordered. Then W intersects every summand S,, in only finitely many elements because S,, has no infinite ascending chain. This implies that the type of W is 5 w,. Let I S be an inversely wellordered set. Then I has no infinite ascending chain, and thus it intersects only finitely many of the summands S,,. So for its type T* there holds T < w,. Based on the last theorem we can now prove:
3.12 Theorem. Let N , be a singular cardinal. Then there exists a dyadic splitting of the set S of 3.11, i n which all dyadic sequences have
6.3. THE CHANGE NUMBER OF DYADIC SEQUENCES
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a changenumber 1 3. Proof. For v < w, we define a set T,, as a set of dyadic sequences as follows: T, contains exactly all sequences which have digits 1 at the first v positions, and behind this block of 1's follows a block of at least one digit 0, but fewer than w, digits 0, and behind the zeros is just one digit 1. Illustration of T,, : 1111111....01 1111111....001 1111111....0001 Then the set T,, has order type w;,, and the ordered sum T := C{T,, Iv < w,) has the order type C{w;, Iv < w,), the same as that of S. By 3.9 there exists a dyadic splitting of T (and then also of S ) into segments, in which all dyadic sequences (are subsequences of sequences of the T,, and thus) have a changenumber 3. At the end of this section we have a look at the wellordered sets in connection with possible changenumbers. There holds:
3.13 Theorem. Every wellordered set (and by symmetry also every inversely wellordered set) has a dyadic splitting into segments, i n which every corresponding dyadic sequence has the changenumber 1 2. Proof. Let W be a wellordered set. W.r.0.g. we can assume that W is an ordinal p. Let then S be the set of all dyadic sequences s of the following type: s begins with v < p digits 1, and behind them follows exactly one digit 0, (so that S looks like this: 0, 10, 110, 1110,...... . ). Then S has the order type p. By 3.9 then there exists a dyadic splitting of S, which consists of subsequences of the sequences of S, which then all have a changenumber < 2, and this also holds for W. In 4.3.4 we had seen that every linearly ordered set is isomorphic to a set of dyadic sequences which is ordered according to the principle of first differences. Of course then one is interested to find such representations where the dyadic sequences have length as short as possible. In this connection we introduce the notion of dyadic depth :
3.14 Definition. Let (S,5 ) be a linearly ordered set. If D is a set of dyadic sequences which is isomorphic to S, we denote the supremum
of the lengths of the sequences of D by 1(D). Then we define the dyadic depth 6(S) of (S, 5 ) to be the minimum of all l(D), where D is a set of dyadic sequences which is isomorphic to S. From the definition it is immediately clear that for every ordinal X the lexicographically ordered set 2((X)) of dyadic sequences of length X has a dyadic depth 5 A. But it turns out that it is also 2 A. Before we prove this we mention a simple fact: 3.15 Remark. Let S and T be linearly ordered sets, f : T + S an isomorphism, S = SoU Sl and T = To U TI with To n TI = 0 , where So resp. To is an initial segment of S resp. T. Then f 1 To is an embedding of To in So,or f 1 TI an embedding of TI in S1. For if all values of f [To]are in So,then the first assertion holds, in the other case the second.
3.16 Theorem. Let X be an ordinal, T := 2((X)) ordered according to the principle of first differences. Let S be a set of dyadic sequences, ordered by the principle of first differences, which is isomorphic to T. Then S contains a dyadic sequence of length 2 A.
Proof. By 3.9 there exists a dyadic splitting U of S, for which the set D(U) of corresponding dyadic sequences is isomorphic to S, and where every sequence of D(U) is a subsequence of a sequence of S. So we are done if we have proved that D(U) has a sequence of length 2 A. So for sake of simplicity we can assume S = D(U). If p 5 X holds and if (avlv < p) is a dyadic sequence, we denote the set of all sequences of S (resp.T) which have a, as v component for v < p, by S(a,lv < p) (resp. T(aUlv< p). Then we have S = S(0) U S ( l ) and T = T(0) U T(1). Here S(0) and T(0) are initial segments of S resp. T , and S ( l ) and T ( l ) the corresponding final segments of S resp. T . By 3.15 T(0) is embeddable in S(0) or T ( l ) in S(1). In the first case we put a 0 := 0, in the second a 0 := 1. Suppose now that we have constructed a dyadic sequence a o , . . . ,a,, . . . lv < p with the property: For every r < p the set T(a,lv 5 r ) is embeddable in the block B, := S(a,lv 5 r ) of height r 1 of U. For p = 1 this induction hypothesis is fulfilled due to the choice of a o . If p is a successor ordinal K 1 then T(a,lv 5 K) is embeddable in the block B, := S(a,lv 5 K)of height ~ + of1U. Then by 3.15 one of the
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two sets T(a,lu K, a,+l = 0), T(aUlv5 K, a,+l 1) is embeddable in a block B,+lof height ~ + of2 U, where this embedding is a restriction of the embedding of T (a, 1 v 5 K). And so one can prolong the sequence a,, v < p, by an element ap E (0, I), such that T(a,lv p) is embeddable in a block of height p 1 of U. If p is a limit ordinal, then T(a,lv < p) = n{T(a,lv 5 K)IK< p) is embeddable in ~ { B , [ K< p), and the last intersection is a block of height p of U. So by transfinite induction we construct a descending sequence of blocks B,, v < A, of height v 1 of U. The set of dyadic sequences which correspond to U has then also a sequence of length A, and this finishes our proof.
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In connection with the dyadic depth the question arises, whether a linearly ordered set that has dyadic depth X also has a subset of type X or A*. In the literature several false statements occur in the surrounding of this question, among others some false "proofs" in connection with the Suslin problem. We present a counterexample to the above question:
3.17 Example. Let I be a subset of R of cardinality N1, say I = {i,lv < wl) and equip I with the restriction of the usual order of R. For every v < wl we put M(i,) := 2((v)) and equip this with the order by first differences. Let S be the ordered sum C{M(i,)li, E I). Then, of course, the dyadic depth 6(S) is 2 sup 6(M(i,)) = sup{vlv < wl) = wl by 3.16. But S has no subset of type wl, and symmetrically no subset of type wy . Indeed, if T would be a subset of type wlof S = {(i,, x) lv < wl and x E M(i,)) and hence wellordered, then the set of first components of the elements of T is wellordered, but countable, since I, like R, has no uncountable wellordered subset. Also every 2((v)) with v < wl has no well or inversely wellordered subset of cardinality N1, so that M(i,) n T is also countable for every v < wl. Then we would obtain that T = ~ { M ( i , ) l v< wl) is countable, a contradiction.
6.4
An application in combinatorial set theory
In this section we introduce some fundamental concepts of combinatorial set theory, in particular the arrowrelation of Erdos/Rado [35]: 4.1 Definition. Let a, b, c be cardinals. Then a + (b, c ) means ~ the following: If A is a set of cardinality a, and if the set [AI2 of two
element subsets of A is divided into two disjoint sets K1, K2, then there exists a subset B 2 A with IBI = b and [BI2 K1 or a subset C A with ICI = c and [CI2 K2. Iffor x # y we have {x,y) E Ki, we say: x and y are irelated. This abstract definition can be described also in the following intuitive manner: If for x # y of A the set {x, y) belongs to Ki, we say: The edge {x, y) has color i. The above statement can then be reformulated as: There exists a monochromatic simplex in color 1 with b vertices, or one in color 2 with c vertices. )~ of course the negation of a + The notation a $, ( b , ~ means ( 4 cI2.
c
c
c
In this context the following fundanlental theorem of Erdos [40] holds: 4.2 Theorem. Let a and c be infinite cardinals, where c is the least cardinal for which aC> a holds. Then we have a+ 4(a+, c ) ~ .
Proof. Let A be a set of a+ elements, [AI2 = K1 U K2, where Kl nK2 = 0. We assume that A has no subset B with cardinality a+ for which [BI22 Klholds. Then we are done if we have proved the existence of a set C with cardinality c and [CI2 C K2. First we define a subdividing function for A. If T is a subset of A with at least two elements, we choose a maximal subset M(T) E T such that all twoelement subsets of M(T) belong to Kl. The existence of such a set M ( T ) follows easily by applying one of the maximad principles. Now we choose a fixed wellordering for every M(T) : M (T) = {xulv < T ) , where r is an ordinal depending on T. We call elements x, y of A irelated, if {x, y) E Ki. Then every element of T\M(T) is Zrelated to at least one element of M(T), since this set is maximal. For v < T let Tv be the set of all elements of q M ( T ) which are 2related to xu, but lrelated to all x,, with p < v. Then Z(T) := {{xv)lv < r) U {Tvlv < r and Tv # 0) is a partition of T. So we have defined a subdividing function for A, and this induces now a splitting of the set A. Next we prove: (*) For every a < U ~ Cthere ) holds: (1) The set Buof blocks of height a has a cardinality a. For a = 0 the set A is the only block of height 0, and for a = 1 there exist 2 . IM(A)I a blocks of height 1 due to our construction (and assumption).
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Now we assume that p is an ordinal < w(c) for which (1) is already proved for all a < p. Case 1. p is a successor number T 1. Then for every block B of height T the partition 2 ( B ) has at most a blocks, so that on the whole there exist a  a = a blocks of height p. Case 2. p is a limit ordinal. B, Then every block of height p is an intersection ~ V < p B where V, is a block of height v. For these intersections we have at most alp1 = a possibilities, for (pI is < c. Thus (1) follows by transfinite induction for all a < w(c). We can now conclude that there exist blocks of height w(c). For otherwise (*) would yield IAl C{IB,I la < w(c)) a . c = a, for c is 5 a. And so we have obtained a contradiction. Let now B be a block of height w(c). Then there exists a strictly decreasing (with growing index) sequence of blocks Bo > B1 > . .  > Bu > .  Iv < w(c), where Buhas height v. In every Bu there exists an element e,, which is not contained in BU+l,and then also e, is contained in no Bp with Y < p < w(c). Now all elements of C := {e,lv < w(c)) are pairwise Zrelated, which proves our statement.
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Later we need the following partition theorem of Erdos/Rado [35] which is very similar to 4.2: 4.2' Theorem. Let a,b be infinite cardinals. Then (ab)++ (a+, b+)2.
Proof. The proof is a variant of Ghat of 4.2. Let A be a set with /A1 = (ab)+and[AI2 = Kl UK2.We assume that A has no subset B with cardinality a+ which satisfies [BI2C Kl .Then we prove the existence of a set C E A with ICI = b+ and [CI2 C K2. For every T C A which has at least two elements we choose a maximal subset M(T) 2 T such that all twoelement subsets of M (T) belong to Kl . Then 1M (T)I a. We choose a wellordering for every M(T). Consider a fixed M(T) = {xuIv < p). We call elements x # y of A irelated if {x, y) E Ki. Then every element of T\M(T) is 2related to at least one element of M(T). For v < T let T, be the set of all elements of T\M(T) which are 2related to xu, but 1related to all x, with p < v. Then 2 ( T ) := {{x,)lv < p) U {T,lv < p and T, # 0) is a partition of T. And 2 defines a subdividing function and and a splitting U of S. Let T be the height of the splitting U. In complete analogy to 4.2 T 2 w(b+) follows. Indeed, in the case T < w(b+) we would
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have 171 b and IAl 1531 ~ { a l ~ l l < l v7) ab . 171 ab b = ab, a contradiction. So there exists a strictly decreasing sequence Bo > B1 > . . > B, > . . . ,where B, is a block of 53 of height v for v < w(b+). We choose an element b, E B,\B,+l for every v < w(b+).Then the set C := {bUlv< w(b+)) has cardinality 2 b+, and its elements are pairwise 2related. Another important partition result, whose substance is due to Sierpinski [162], is: 4.3 Theorem. 2 ' ~ + (Nu+1, Nu+l).
Proof. We consider the set S := 2((w,)) of dyadic sequences of length w,, taken with the order Llby first differences. This linearly ordered set has, due to 4.2.6, no well and no inversely wellordered subset of cardinality Nu+1. Now we take also a wellordering l 2 o f S. For x and x # y of S we put {x,y) E Kl if x Il y and x 1 2 y, or y x. And we define K2 := [ q 2 \ K l . y Then there is no subset T of S of cardinality Nu+1, such that all its twoelement subsets belong to Kl (resp. K2). For otherwise in T the orders $and I 2 h a d to be equal, which is impossible since T has like S no wellordered and no inversely wellordered subset of cardinality N,+l in the order Il . A similar idea can be applied in the proof of the following theorem of Erdos/Rado [36]: 4.4 Theorem. Let w, be a singular initial ordinal, wy := cf(w,). Then N, + ($+I, N, ). Proof. We consider the set S of 3.11 with the order which it had in 3.11, and which we denote here by . In addition to this we consider a wellordering 1 2 i n S. The definition of K1 and K2 is the same as in 4.3. Since by 3.11, S with l l h a s no wellordered subsets of cardinality N,+l and no inversely wellordered subsets of cardinality N,, our assertion follows. In the following we present another important theorem of the partition calculus of set theory which also has applications in the theory of posets. First recall the graphtheoretical notions of 2.5.4. We supplement this by: 4.5 Definition. If (V, C) is a, graph and a E V we put C(a) := {x E Vl {a, x) E (2) and C*(a) := C(a) U {a). The cardinality IC(a)1 is called
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the degree of the element a. It is the cardinality of the set of edges (= elements of (2) which contain a. For M V we put C ( M ) := U{C(a)laE M ) and C*(M):=MU
C(M)A subset of V, of which all elements are unlinked, is called independent. A subset T of a set S which has the same cardinality as S is also called a full part of S. 4.6 Lemma. Let (V,(2) be a graph, where IVI = k is a n infinite cardinal, such that every full part F of V has a n element of degree k in the graph (F,E r l [ F ] ~ ) . T h e n V has a n infinite subset whose elements are pairwise connected.
Proof. There exists an element a1 E V =: Co with degree k. Then C1 := C(al) has cardinality k. Analogously there exists an element a2 E Cl, so that the set C2 of elements of C(al) that are connected with a2 has cardinality k. If generally we have for an n E N already defined sets Cl  > Cn with ICnI = k and elements a,+l E C, for v = 1,.. . ,n  1, we have an element a,+l E Cn which is connected with all elements of a full part Cn+l E Cn. By induction we so construct a set {al,an, . . .) of N o elements of V, which are pairwise connected.
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4.7 Lemma. Let (V,E ) be a graph, where k := IVI i s a regular infinite cardinal, and where every element of V has a degree < k . T h e n V has a n independent full part.
Proof. We choose a0 E V, a1 E V\C*(ao), a2 E V\(C*(ao)U C*(al)), . . . . If in general for an ordinal p < w(k) already elements a, are defined for v < p, we choose a, E V\ U {C*(a,)lv < p}. This is possible since every summand C*(a,) has a cardinality < k and because k is regular. By transfinite induction we so obtain a set {avlv < ~ ( k ) } , which is an independent full part of V. The same statement as that of 4.7 appears in another formulation as:
4.7' Lemma. Let (V,C) be a graph where k := IVI i s a regular infinite cardinal, and suppose that every full part of V has two connected elements. T h e n V has a n element of degree k. Now we can prove the regular case of the theorem of Dushnik/Miller/Erdijs [30]:
4.8 Lemma. Let k be a regular infinite cardinal. Then k + (k, follows. For the special case k = No this is Ramsey's theorem (for the infinite case). Proof. Let (V, (2) be a graph with IVI = k. If there exists a full part of V of independent elements, we are done. In the other case every full part F of V contains two connected elements. By 4.7/, applied to F, then F has an element x, which is connected with k elements of F. By 4.6 now V has an infinite subset of pairwise connected elements.
A rather immediate consequence of 4.8 is: 4.8'Theorem. Let M be an infinite set and [MI2= K1 U . . . U Kn, where n E N . Then there is an infinite subset T M and an i E ( 1 , . . . ,n) with [ T ]E~Ki.
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Proof. It suffices to prove this for [MI = No. For n = 1 (resp. n = 2) the statement holds by 4.8. Suppose that it is proved for a fixed n E N and that [MI2 = K1 U  .U Kn U Kn+1. Then 4.8 entails that there is an infinite set T C M with [ T ]G~(K1 U . . . U Kn) or [TI2E Kn+l. And by induction hypothesis then there exists an i E (1, . . . ,n 1) and an infinite T * E T with [ T * ]C_~Ki.
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We shall extend 4.8 also for nonregular cardinals. The next theorem contains most of the proof of that. 4.9 Theorem [30]. Let (V, (2) be a graph where IVI = k is infinite, and where every element of V has a degree < k. Then there exist No elements, which are pairwise connected, or k independent elements. Proof. Suppose that there are no No elements which are pairwise connected. Then we are done, if we have proved: (1) There exists an independent full part I of V. According to 4.7 we only have to consider the case where k is a singular cardinal. Then there exists a representation (2) k = C{kvlv < w,),where w, = cf(k), and where the summands kv are regular cardinals > N,, which then also must be < k. First we prove the following, which contains the main idea of the proof:
6.4. APPLICATION IN COMBINATORIAL SET THEORY
187
(3) Let m be a regular cardinal with N, < m < k. Then there exists an independent subset M C V with 111.11 = m and IC(M)I < k. Let T be a subset of V of cardinality m. Due to 4.8 then there exists an independent subset A C T with IAl = m, for T has, like V, no infinite subset of connected elements. For v < w, we then define: A, := {x E A1 the degree of x is k,). Since every element of V has a degree < k,we have by (2) A = u{AV(v< w,), and thus (/A[=) m C{IAVIIv < w,). Since m is regular and > N, this entails that one of the summands /Av/must have cardinality m. Let M be such an A,. Then IC(M)I 1 M 1 . k, < k holds, and (3) is proved. Applying (3) we obtain: There exists an independent set I. C V of cardinality ko with IC(Io)l < k. Further there exists an independent subset Il V\C*(Io) (the last set has cardinality k) of cardinality kl with JC(Il)I < k. If in general p is an ordinal < w,, and if independent sets I, of cardinality k, are defined for all v < p,such that V\ U {C*(I,)lv < p) still has cardinality k, then by (3) there exists an independent subset I, C V\ U {C*(I,)lv < p) of cardinality k, for which IC(I,)I < k holds. By transfinite induction we so construct a sequence of pairwise disj~int independent sets I, for v < w,, where also for v < p all elements of I, are not connected with all elements of I,. Finally we then have: I := ~ { I , l v< w,) has cardinality C{kVlv < w,) = k, and I is independent, so that (1) is proved.
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There follows: 4.10 Theorem of Dushnik/Miller/Erdos [30]: For every infinite cardinal k we have k + (k, No)2.
Proof. Let (V, C) be a graph with IVJ = k. We have to show: There exists an independent set of k elements of V, or there exists an infinite subset of V, whose elements are pairwise connected. If every full part F of V has an element which is connected with k elements of F, our assertion follows from 4.6, according to which we obtain infinitely many elements of V which are pairwise connected. So we assume that there is a full part F of V for which every element has degree < k. From 4.9 (with F instead of V) we obtain the existence of No pairwise connected or of k independent elements of F V.
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The before mentioned partition theorems have interesting consequences in the theory of posets. We list same of them: 4.11 Theorem. Let (P, 5 ) be an infinite poset. W e call here two elements a # b of P connected, i f they are comparable, resp. independent, i f they are incomparable. Then there hold: a) If P has a cardinality a+, and i f c is the least cardinal for which aC > a holds, then P has a chain of cardinality a+ or an antichain of cardinality c. Also: P has an antichain of cardinality a+ or a chain, of cardinality C.
b) If P has an infinite cardinality k, then P has a chain of cardinality k or an infinite antichain. Also: P has an antichain of cardinality k or an infinite chain. Proof. This is a direct consequence of 4.2 and 4.10.
Another application yields a theorem of Lindenbaum 11121: 4.12 Theorem. Let S be an infinite set with two wellorderings L1 and L2 . Then there exists a full part of S, over which these two orders coincide. A direct consequence of this is that if i n S there are finitely many wellorderings, then there exists a full part, i n which they all coincide. Proof. If a and b are two different elements of S, we put a R b +=+ a <1 b and a <2 b hold both, or none of them holds. Then there is no infinite set of elements which are unrelated in R. For this would entail the existence of elements a1 ,a2, . . . of S with a1 2 a2 >2 a3 >2  , which is impossible since a wellordered set has no infinite strictly decreasing sequence.
We mention without proof a generalization of 4.12 (see [74]): 4.13 Remark Let S be a set of cardinality Nu+1 and c the least cardinal for which NC, > N u holds. And let W be a set of wellorderings of S where IWI < c holds. Then there exists a full part of S, over which all wellorderings, which belong to W , coincide. This statement is tight.
6.5. COFINAL SUBSETS
6.5
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Cofinal subsets
After we have proved 2.5, which considered linearly ordered sets, we now address the (only partially) ordered sets and study how their cardinality depends on the cardinality of their well (resp. inversely well) ordered chains and antichains. In this context the cofinal (resp. coinitial) subsets of a poset are of importance. First we introduce several concepts (similar to those of [55]): 5.1 Definition. Let (P, 5 ) be a nonempty poset. We define S+(P) (resp. S(P)) to be the supremum of all ordinals T such that S has a subset of type T (resp. T*, the reverse of T), and put d+(P) := jSS(P)I and d(P) := 16(P)I. A subset C P is said to be cofinal in P, if for every x E P there exists a c E C with x c. Dually C is called coinitial, if for every x E P there exists a c E C with c 5 x. (So cofinal, resp. coinitial, sets of nonempty sets are never empty.) Then C C P is cofinal in P iff P = u{(x]~xE C). A set B P is called a n  base of P, if B is cofinal and coinitial in P. Let n(P) denote the smallest cardinality of a T  base of P. We define no(P) (resp. nl(P) ) to be the least cardinal k such that P has a coinitial (resp. cofinal) subset of cardinality k. Then it is trivial that n ( P ) is 2 ;ro(P) and 2 nl(P), and further .,(P) 5 no(P) + nl(P). So, if one of the cardinals n(P), no(P), n1 (P) is infinite, then we obtain n(P) = max{.rro(P),nl(P)). Finally we put ao(P) := sup{no(T)IT C P), a l ( P ) := sup{nl(T)IT C P), and a ( P ) := sup{~(T)lTC P). If P has a first (resp. last) element a, then {a) is a coinitial (resp. cofinal) subset of P, and then no(P) = 1 (resp. xl(P) = 1). These cases are not very interesting, and so the importance of the concepts no(P), nl(P) results mainly in connection with posets P which have no first (resp. last) element.
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The substance of the following theorem is due to John Ginsburg [55]: 5.2 Theorem. Let (P, 5 ) be a poset, which is not linearly ordered. ( ~ )dually /PI 5 a o ( ~ ) ~ + ( ~ ) . Then IPJ5 C ~ ( P ) ~ and
Proof. In every subset T E P with IT1 2 2 we choose a cofinal subset C(T) of minimal cardinality. Then we have
(1) T = C(T) U u{(T < c)lc E C(T)). Here some summands ( T < c) could be empty, and the sets (T < c) need not be pairwise disjoint. But, starting from (I), one can find a partition 2 ( T ) of T into the singletons of C(T) and into nonempty sets T: G (T < c), c E C' G C(T) which are pairwise disjoint. If we have done so for all T 2 P with at least two elements, we have constructed a subdividing function 2 to P (see 1.14). By 1.16 2 induces a splitting U of P. P is the block of height 0 of the splitting, the elements of 2 ( P ) are the blocks of height 1 and so on. Let T be the height (see 1.11) of the splitting U. We prove T 5 6(P). For this purpose suppose that Bo > B1 > . . . > Bu >    , v < p, is a strictly decreasing sequence, where every Bu, v < p, is a block of U of height v. For v 1 < p all elements of Bu+lare < some xu of C(Bu) 2 Bu,and we choose for every v 1 < p such an element xu. And if p is a successor ordinal p 1 we further choose an arbitrary element xp E Bp. For every v 1 < p all elements of B, with v 1 K < p are < xu. Then also all x, with v < a < p are by construction less than the chosen element xu. And so the xu, v < p, form an inversely wellordered set of type p*. So we have p 5 6 ( P ) , and therefore we obtain T 5 6(P). We abbreviate a l ( P ) =: k. First we consider the case where k is infinite. Now P is the only block of height 0, and the set of blocks of height1 has cardinality 5 2 k = k, since k is infinite. For it contains k many sets, which the singletons of C ( P ) and the at most J C ( P )I each contain strict lower bounds to an element of C(P). So for v = 1 the following induction hypothesis is satisfied: (2) The set of blocks of height v has cardinality kluI. If v is < T, then every block of height v is split into k blocks of height v 1, and so the total number of blocks of height v + 1 is  kl"l . k = kI"+ll. If (2) holds for all v with 0 < v < A, where X < is a limit ordinal < p, then every block of height X is an intersection n{B,lv < A), where Bu is a block of height v. For every v < X we have at most klVl5 klXlpossibilities, and so there exist at most (klXI)IXI= klxl such intersections, and thus (2) also holds for A. By transfinite induction then (2) is proved for all v with 0 < v < T. Finally there follows by 1.12:
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(3) lPl lUl C{klU'lv < 7). For v < r we have Ivl 171 5 d(P). Then the last sum in (3) is <  kd (PI . I 1 < kd(PI . d (P) = kd (PI. 
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6.5. COFINAL SUBSETS
191
Suppose now that Ic is finite. Every antichain A of P is its unique cofinal subset, so that rl(A) k. So every antichain of P has at most k elements, and then the width w of P is k. Also P has no subset of type wo. Indeed, such a set would have a cofinal subset of cardinality No k, a contradiction. By Dilworth's theorem now P is a union of w disjoint chains, that are inversely wellordered. Let A* be the type of the longest of them. Then we have X 6 (P),and it follows I PI w. IAI kd (P). This is kd(P) because of k 2 2; here we use that P is no chain. So the theorem is proved.
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The question arises why we excluded the case where P is linearly ordered. Here one has simple counterexamples. Let e.g. P be a finite chain with at least two elements. Then I PI > 1 = al (P)~(') since here q ( P )= 1. The last theorem has several interesting corollaries. First we recall the concept wellfounded poset of 1.9.8: 5.3 Definition. A poset satisfies the descending (resp. ascending) chain condition if it has no subset of type w: (resp. wo). So a poset is wellfounded iff it satisfies the descending chain condition.
5.4 Theorem [55]. Let P be a wellfounded poset, which is no chain, and k := al(P)infinite. Then IPI = al(P).And duzlly: If P is a poset which is no chain and satisfies the ascending chain condition, and i f ao(P)is infinite, then lPl = ao(P). Proof. In the proof of 5.2 we obtain from (3) with the additional information that P has no subset of type w:, /PI IUI < ~ { k l " l l v< wo) = k since T 5 6(P) < wo. That IPI k holds, is trivial.
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5.5 Theorem [55]. Let P be an infinite poset, but no chain. Then there follows I PI < 2u(P).
Proof. First we mention that a(P) is infinite. For the infinite set P has an infinite antichain or an infinite chain, and in the latter case a subset of type wo or wg. In all of these cases there exist subsets T C P where r ( T ) , and then also a(P),is infinite. Further it is immediately clear from the definition that ao(P)and q ( P )are < a ( P ) . Next we prove: (1) d+(P) al(P)=: k. Assume the contrary, k < d+(P). Then P contains a subset T of type w(k+). This T has a cofinal subset of type w(k+), and we obtain
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the following contradiction: k+ 5 .rrl(T) 5 a l ( P ) = k . The dual of (1)
5.6 Theorem [55]. Let P be a nonempty poset. Then there exists a cofinal subset C C P which is wellfounded. And dually: There exists a coinitial subset D P, which satisfies the ascending chain condition.
Proof. If M is a maximal antichain of P, we denote the set of those elements of P which are greater than some element of M by U(M). Now we choose a maximal antichain A. of S. If U(Ao) is empty we are done, since then A. satisfies the conditions for C. In the other case we choose a maximal antichain A1 of U(Ao), and so on. In general we proceed as follows: Suppose that for an ordinal p we have already defined nonempty and pairwise disjoint antichains A, for v < p such that the following holds: If v 1 < p holds, every A,+1 is a maximal antichain of U(A,), and for limit ordinals X 5 p, for that n { u ( A , ) l ~ < A) is nonempty, Ax is a maximal antichain of this set. By transfinite induction we so construct a set of pairwise disjoint antichains A,, v < 7, of P. This construction procedure must indeed stop at some ordinal 7 because in every step of the construction the rest of P is diminished. There cannot exist elements in P which are not surpassed by elements of u{A,Iv < r), for then n{U(A,) Iv < 7) would be nonempty, and the construction had to go on. So C := u{A,Iv < r) is cofinal in P. Also C is wellfounded: If xo > xl > . . . > x, > . . . would be an infinite decreasing sequence in C, every x, is in exactly one antichain A,(,) and there follows: X, > 2, p < U (< Wo) K, > K,, and then the K,, v < WO, would form an infinite decreasing subset of r, which is impossible.
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Now we can prove the following lemma of Ginsburg [55]:
5.7 Theorem. For every poset P there holds: a) .rro(P) W ( P ) ~  ( ~ ) ,and dually b) .rr~(P)I W ( P ) ~ + ( ~ ) .
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Proof. It suffices to prove b). By 5.6 P contains a cofinal subset C which is wellfounded. Then every chain in C is wellordered, and
6.6. SCATTERED SETS
193
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all chains in C have cardinality d+(P) := b. Further all antichains of C have cardinality w(P) := a. We partition the set [pI2of all twoelement subsets of P into the set A (resp.B) of those sets {x, y) where x is comparable (resp. incomparable) with y. From 4.2' we obtain (ab)++ (a+, b+)2. Then C must by 4.11 have a cardinality 5 ab, and this entails ( P ) ICI ab.
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Using the last theorem Ginsburg [55] gave a proof for the following theorem which contains a theorem of Kurepa [106]:
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5.8 Theorem. Let P be a poset with width w(P) 2, d(P) := sup{klP contains a well or inversely wellordered subset of cardinality k). Then
\PI 5 W ( P ) ~ ( ~ ) .
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Proof. For subsets S C P one has of course w(S) 5 w(P), d(S) de(P), d+(S) 5 d+(P). By 5.7 a) there follows ./ro(S) w(s)~(') w(P)~('). Then we have a o ( P )= sup{./ro(S)IS 5 P) w(P)~('). From 5.2 we obtain /PI q ( ~ ) ~< +(w(P)~('))~+('). (') If one of the cardinals d (P), d+ ( P ) is infinite their product is 5 d(P), and we are done. If both are finite, P has only finite chains, so that all of them have cardinality 5 d := d(P). Let MI be the set of minimal elements of P, M2 the set of minimal elements of P\Ml, and so on. Then P is a union of antichains MI,. . . ,Md, and then its cardinality is w(P).d w(P)~(').
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5.9 Corollary. /PI
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5.10 Corollary [55]. Let P be a poset which is not linearly ordered. If GCH is assumed and d(P) < cf(w(P)), then [PI= w(P) holds. If here in addition IPJ= k+ holds, then P contains an antichain of cardinality Ic+.
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Proof. /PI w(P)~(') = w(P) follows by assumption. Of course, also w(P) /PI holds. If \PI = Ic+ and if P would have no antichain of cardinality Ic+,5.8 would entail Ic+ = I PI = w(P) 5 Ic, a contradiction.
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6.6
Scattered sets
A dense linearly ordered set, which has at least two elements, contains a subset of type ho. This simple fact follows e.g. by 4.3.13. As a
counterpart to this concept we now consider posets which have no subsets of type ho. In this context Hausdorff ([81], p. 95) had established the following: 6.1 Definition. A poset is said to be scattered, if it has no subset of type ho. An order type is called scattered if it has a scattered realization. An example for scattered sets are the well (resp. inversely well)ordered sets because they already don't contain subsets of type w; (resp. wo).
Hausdorff had investigated only linearly ordered scattered sets. A more general study was performed in the paper [I] of AbrahamIBonnet. We put the last definition into a more general context, using the concept of forbidden subtype: 6.2 Remark. For an order type T we defined in 1.9.6 a poset to be has no subset of type T. The property T  free is of course hereditary. That means: If a poset is T  free, then every subset of it is also T  free (with respect to its induced order).
T
 free if it
6.3 Theorem. Let T be a linear order type, such that for a realixation T of T the following holds: Every open interval of T has a subset of type T . Then the class of T free linearly ordered sets is closed under ordered sums. More precisely: Let I and sets Mil i E I, be T  free linearly ordered sets, where the M i are pairwise disjoint, then the ordered sum S := CiCI Mi is also T  free.
Proof. We assume indirectly that S has a subset T of type T . Then there holds: (I) T has at most one point in common with every summand Mi. For if T has at least two elements in common with an Mi, say a , b with a < b, then the whole open interval ( a ,b) of T must be a subset of Mi, and since (a, b) contains a subset of type T, also the set Mi would contain a subset of type T with contradiction. Now we map every element x E T onto the uniquely defined i E I for which x E Mi holds. Due to (I) this yields an injective and < preserving mapping of T onto a subset of I. This contradicts the fact that I has no subset of type T.
6.6. SCATTERED SETS
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The sets Ha of Definition 4.3.3 have by 4.6.3' the property, that every open interval has once again the type ha. And so 6.3 yields: 6.4 Corollary. For every ordinal a the class of ha  free linearly ordered sets is closed under ordered sums.
In the special case a = 0 this is a theorem of Hausdorff (1811, p.95), which states: 6.5 Theorem. A n ordered sum of scattered linearly ordered sets over a scattered linearly ordered argument is again scattered. In connection with the last theorems we introduce the ring concept: 6.6 Definition. A class R of posets is called a ring if the following holds: If I and Mi, i E I, are elements of R , where the Mi, i E I, are pairwise disjoint, then the ordered sum Mi also belongs to R . A class R of order types forms a ring, if there holds: If I and Mi, i E I, are posets for which the types t p I and tpMi, i E I , belong to R , and where the Mi are pairwise disjoint, then also the type of the ordered Mi belongs to R. sum
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So we have obtained that the class of ha free linearly ordered sets, and in particular the class of scattered linearly ordered sets, forms a ring. Another interesting theorem of Hausdorff ([81], p. 95) is the following:
6.7 Theorem. Every linearly ordered set S is an ordered sum of scattered sets over a dense argument I. We prove 6.7 together with the next theorem, of which 6.7 is a special case: 6.8 Theorem. For every ordinal a there holds: Every linearly ordered set S is an ordered sum of ha  free sets over an argument I, which has the property that for every two elements a, b of I with a < b the open interval (a,b) contains a subset of type ha.

Proof. For elements a, b of a linearly ordered set S we put a b if a = b or if the closed interval with ends a and b is ha  free. Then w is an equivalence relation: If a < b < c are elements of S with a b and b C, then also a c holds, for if [a,c] would contain a subset M of type ha, then already [a,b] or [b, c] would contain at least two elements N

of M and then also a subset of type h, with contradiction. The other cases are trivial. Further we have: (I) Every equivalence class is h,  free.
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Let C be an equivalence class, x one of its elements. Then ( C x) is h,  free. This is trivial, if C has a greatest element. If it has none we choose a wellordered cofinal subset of ( C x), which has x as first element, say {x,lv < T), where xu < z p for v < p < T. Then ( C x) is by 6.4 h,  free because it is an ordered sum of h,  free intervals [xu,xu+l) over the wellordered (and thus h,  free) set of all v < T. Analogously it follows that ( C < x) is h,  free, and then also C is h,  free. For if a set M E C would have type h,,this also would by 4.6.3' hold for M n ( C < x) or M n ( C > x). We have obtained that S is partitioned into a set I of h,  free classes Si,i E I,which are segments of S. Then I has the natural order given a < b for all a E St,b E S". Now by: For S', S" E I we put S' < S" I has the property that everyone of its closed intervals [a, b] contains a subset of type h,. For if an interval [a, b] of I would not do that, the ordered sum C{SiJi E [a, b ] ) would be h,  free due to 6.4. Then the elements of the classes S, would be equivalent to those of Sb,which means they had to be in the same class, but S, and Sbare different. This yields a contradiction.
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Hausdorff gave another characterization of scattered linear order types. They form the least class of order types, which contains the ordinals and is closed under ordersummation and inversions. First we mention:
6.9 Definition. If S is an arbitrary class of order types, we define , the Ti, i E I,are C ( S ) as the class of all order types '&I ~ i where order types of S, and where I is a poset, whose type is in S. We define the rzngclosure R(S) of S as the least class of order types which contains S and is a ring. It always exists because the class of all order types is a ring, and then R(S) is the intersection of all rings that contain S. Of course the ringclosure of a class of linear order types contains only linear order types. One can obtain all linearly ordered scattered sets by starting from the ordinals and their inverses by successively adjoining certain ordered
6.6. SCATTERED SETS
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sets which are ordered sums of sets which were already defined in earlier steps. Precisely there holds the following structure theorem: 6.10 Theorem. Let w, be a regular initial ordinal. Let Co be the class of all linearly ordered sets which have a type T or T* for some regular ordinal r < w,. If for an ordinal ,u < w, we have already defined a class Cp of linearly ordered sets, we define Cp+l to be that class of all linearly ordered sets that are isomorphic to an ordered sum C{Mili E I},where I and the Mi, i E I, are i n Cp, and where the Mi, i E I, are pairwise disjoint. If A is a limit ordinal 5 w,, for which classes C,, v < A, are already defined, we put CA:= u{C,IV < A). B y transfinite induction we so have defined classes C, for all v 5 w,,which form an ascending (with v) tower of classes. Now there follows: C,, is the class of all scattered linearly ordered sets of cardinality < N,.
Proof. We consider the statement: ( I ) All sets of Cvare linearly ordered, scattered, and have cardinality < N,. For v = 0 then ( I ) is satisfied. Suppose now that ( I ) holds for a Then M is isomorphic to an fixed v < w, and that M is a set of ordered sum of scattered sets of C, over an index set, which is in C, and thus M is also scattered by 6.5. Since N, is regular, M has a cardinal < N,. So ( I ) also holds for v 1. For limit ordinals X ( I ) holds, if it is valid for all v < A, since then every element of Cx is also in a C, with a v < A. And the construction shows that the classes C,, v < w,, form an ascending tower. Finally we have obtained by induction: (11) C, has only scattered linearly ordered sets of cardinality < N,. We still have to show: (111) C,, contains all scattered lingarly ordered sets of cardinality < N,. Suppose therefore that T is a scattered linearly ordered set of carb iff a = b, or if dinality < N,. For elements a, b € T we put a the interval with ends a, b is embeddable in a set of C,,. Now ,., is an equivalence relation: Let a , b, c be elements of T with a < b < c, the other cases are analogous or trivial. Then [a, b] is embeddable in a set of C, , and then also in a set C,,. with 61 < w,, and similarly [b,c] is
+

embeddable in a set of a C,, with ~2 < w,. Now [a, c] = [a, b] U [b, c] is embeddable in a set of C, E Cumwhere K = rnax{Kl, K ~ ) 1. Before proving (111) we need the following: (IV) Every equivalence class K of is a segment of T and isomorphic to a set of some C, with Y < w,. The first is clear. Let now x be an element of K. If K has a last element 1, then x 1 holds, and ( K 2 x) = [x, I] is embeddable in a set of some Cp with p < w,. Suppose now that K has no last element, and that {x = xo < XI <   < x, < . IT < w6) is a cofinal subset of K, where w6 is a regular initial ordinal. Here S < a holds since ( T and) K has cardinality < N,. Every interval [x,, $,+I), T < w6, is embeddable in a set of a C,(,) with a V(T) < w,. Then all intervals [X,,X,+~)are embeddable in sets of C, where a := SUP{V(T)~T < wg) which is still < w, because this is regular. Also w6 is in (Co C,,and then the ordered sum C{[x,, x ~ + ~ <) wS) ~ T = ( K 2 x) is in C,+I. Analogously (K < x) is in a class C, with a p < w,. Now K = ( K < x) U ( K 2 x) is in C, where v = max{a 1,p ) 1, and (IV) holds. Finally there cannot be two equivalence classes of which are neighboring segments of T for their union would be isomorphic to a set of some Cc with < w,, and then both segments had to be subsets of the same equivalence class with contradiction. So we have obtained that the set of equivalence classes, which is a partition of T into segments, is dense (with respect to the natural order inherited by T). But then this set can have only one equivaJence class, since otherwise it would contain a subset of type ho. This would also hold for T in contradiction to the fact that T is scattered. So the unique equivalence class is T. With (IV) the proof is complete.
+

N
c)
+
+

<
Another characterization of scattered linearly ordered sets via dyadic sequences was given in [78]. First we mention the following: 6.11 Lemma. Let U be a dyadic splitting of the set Q of rational numbers (with their natural order) into segments. Let B be a block of U which has more than one element. Then there exists a block C C B such that both blocks of Z(C) (according to 1.8) have more than one element.
Proof. 1) If B has no first and no last element, then already both blocks of Z(B)have more than one element. 2) Suppose that B has a first, but no last element. If the left subsegment (see 3.3) of B has a last element, then the right subsegment
6.6. SCATTERED SETS
199
R of B has no first and no last element. By 1) then both blocks of Z ( R ) have more than one element, and C := R fulfills our assertion. 3) The case where B has no first, but a last element, is symmetric to 2). 4) Suppose now that B has a first and a last element. Then one of the blocks of Z ( B ) has no first or no last element, and so we are done by 2) and 3). Now there follows: 6.12 Theorem. Let U be a dyadic splitting of the set Q of rational ~ degressions (see 3.5) numbers into segments. Then there exist 2 N block > Bv > . . . , v < wo, to U , whose corresponding dyadic Bo > B1 > sequences ao,a1,. . . have a change number wo.
>
Proof. By 6.11 there exists a block C E U with Z(C) = {C(O), C(l)), where C(0) and C(1) both have more than one element, and where C(0) (resp. C(1)) is the left (resp. right) subsegment of C. Suppose now, that for a fixed n E N we have for every finite dyadic sequence ao, . . . ,am with m 5 n, already defined blocks C(ao,. . . ,am)6 %,which all have more than one element, and which satisfy C(ao) >  . . > C(ao,. . . ,a,). If now a o , . . . ,a, is a fixed dyadic sequence, then by 6.11 we can choose a block C* & C(ao,. . . ,a,) for which Z(C*) = {Cl, C,), where Cl (resp. C,) is the left (resp. right) subsegment of C*, and where Cl and C, both := Cl (resp. have more than one element. Then we put C(ao,. . . , C,) if a,+l = 0 (resp. = 1).By induction we so have defined for every dyadic sequence ( a v1 v < u0) blocks 3 C(a0,. . . ,a,) > . . . C(ao) > C(ao,a l ) >   > These form a part of a block degression to U, whose corresponding dyadic sequence s has (aVIv< w0) as subsequence. If the latter has change ~ number > wo, then this also holds for s. Since there exist 2 N dyadic sequences of length wo which have a change number 2 wo, the proof is complete. An easy consequence of 6.12 is:
6.13 Theorem. Let (S,5 ) be a linearly ordered set which is not scattered. Then for every dyadic splitting of S into segments there holds: W e have at least 2N0block degressions whose corresponding change number is infinite.
Proof. S has a subset T which is isomorphic to the set Q of rational numbers. If U is a dyadic splitting of S into segments, {T n BIB E U)
defines a dyadic splitting UT of T. Every block degression of UT has a corresponding dyadic sequence which is a subsequence of the dyadic sequence which corresponds to some block degression of U. And so 6.12 entails 6.13. Now we can establish another characterization of scattered linearly ordered sets: 6.14 Theorem [78]. L is a scattered linearly ordered set ifl there exists a dyadic splitting of L such that all dyadic sequences of the splitting have finite change number.
Proof. We define the classes C, of linearly ordered sets inductively as in the proof of 6.10, but now for all ordinals v, not only for v < w,. To this purpose we only have to supplement the former definition by : For every limit ordinal A, for which C, is already defined for the v < A, we put CA:= U{C,Iv < A). We consider the following statement:
(*) For every linearly ordered set of C, there exists a dyadic splitting such that all its dyadic sequences have finite change number. Then by 3.13 (*) holds for v = 0. Assume now that (*) holds for a fixed ordinal v. If then T is a linearly ordered set of C,+l, T has a Ti as an ordered sum, where I and the T,, i E I, representation T = are in C,. By induction hypothesis, I and the Ti have dyadic splittings in which all dyadic sequences have finite change number. Then we can attach to each sequence s ( i ) , which corresponds to an i E I in the splitting of I, the sequences of the splitting of Ti.This furnishes a dyadic splitting of T in which all dyadic sequences have finite change number, and thus (*) holds for v 1. Finally (*) also holds for limit ordinals A, if it is proved for all v < A. And so (*) is proved for all ordinals v by transfinite induction.
xiGI +
6.15 Remark. One can pose the question whether 6.14 could be sharpened so that every dyadic splitting of a scattered set into segments has only dyadic sequences of finite changenumber. But this is false. Already very simple scattered sets furnish counterexamples: We consider the set S := {:In E N ) U (0) U {f In E N). Its order type is wo 1 w:, and so it is scattered. We can define a dyadic splitting U of S into segments, of which the following sets Si, i E w, are blocks of height i. We put So := S,S1:= (So< I), sz:= (S1 > $), s3= (S2< s4= ( 5 3 > L)3 '
+ +
i),
6.6. SCATTERED SETS
201
i),
S5= (S4< S6= (S5 > $), S7= (S6< $) and so on. The idea of this construction is that we have: If Y is even (resp. odd), then SV+lis the left (resp. right) block of 2(SV).
The sequence So,S1, $2, S3,.. .then determines a blockdegression of U (with n{Svlu < wo) = {0)), whose corresponding dyadic sequence is O,l,O,l,O,l,. . .which has change number wo. In the class of scattered linearly ordered sets we have a remarkable structure theorem of Laver. Before we formulate it we.define in the class of linearly ordered sets a quasiorder 5, supplementing Definitions 1.9.2 and 1.9.6:
6.16 Definition. For two linearly ordered sets (Ll, 51) and (L2,5 2 ) we put (L1, 51) 5 (L2,5 2 ) iff there is a <  preserving function from L1 in La. It follows immediately that 5 is a quasiorder. In analogy to 1.9.8 we call a quasiordered class wellquasiordered, abbreviated wqo, if it has no subset of type w*, (i. e. no infinite strictly descending chain) and no infinite antichain. Now Laver's theorem states:
6.17 Theorem [111]. The class of scattered linearly ordered sets is wqo (with respect to the quasiorder of 6.16). The proof of this theorem is long and complicated, and so we refer the interested reader to the original paper.
Chapter 7 The dimension of posets Due to the fact that the linearly ordered sets are much more intuitive than the (only partially) ordered sets one is interested to establish relations between the general ordered sets and linearly ordered sets. One possibility was already treated by Dilworth's theorem, which considered coverings of posets by chains. Another possibility is given by studying linear extensions of posets. This leads to the dimension theory of posets which was founded by Dushnik/Miller in the paper [30] in 1941. We begin with some remarks on the topology of posets.
7.1
The topology of linearly ordered sets and their products
Linearly ordered sets are generalizations of the real line R, which besides its ordertheoretical aspects also has topological features. So it is not surprising that also general linearly ordered sets give rise to the introduction of corresponding topological concepts. First we compile several notions of general topology and consider some connections between this theory and the theory of posets. We recall the concept of topological space: 1.1 Definition. A topological space is a pair (X,Q), where X is a set and 0 a set of subsets of X, which contains 0 and X, and has the following properties: 1) For any two sets 01, O2 of Q the intersection O1no2also belongs to 0. 2) If Oi, i E I, are elements of 0, then their union UiEIOi also belongs to Q. E.g. there holds: If (P, 5 ) is a poset, 3 (resp. 5 ) the set of its initial (resp. final) segments, then (P, 3) (resp. (P,5 ) ) are topological spaces.
In the linearly ordered sets one has a very natural topology which is completely analogous to the usual topology of the real line: 1.2 Definition. Let (S, 5 )be a linearly ordered set. Sets which are open intervals (a,b) of S, or of type (S < x) or (S > x) for elements
x E S, or which are = 0 or = S are called open. Further a subset of S is said to be open, if it is a union of segments of the before mentioned kind. (So open intervals are also open sets.) Let Q be the set of open subsets of S. It is easily verified that (S,13) is a topological space: If 01,O2 E Q , j J) where we have representations O1 = u{Sili E I), O2 = ~ { T j l E the Si and Tj are open segments of S. Then O1 n O2 = u{(Si n Tj) 1 i E I,j E J). Here each summand Si r l Tj is open as can easily be verified. So condition 1) is satisfied, and 2) is trivial. Q is called the order topology of (S,5). 1.3 Definition. A subset T of a topological space (X,O) is called UiEIOi, where the Oi are in Q, then compact, if there holds: If T there exists a finite subset F E I, such that T UiEFOi. We formulate this also as follows: Every open covering of T has a finite subcovering.
c
c
1.4 Theorem. Let (S,<) be a dense linearly ordered set which has no gaps, but a first element a and a last element z, Q its ordertopology. Then (S, Q) is a compact space.
Proof. The proof runs quite analogously to the proof for the compactness of closed intervals of the real line: If S is covered by the union of open sets UiEIOi then there exists the supremum s of the values x E S, for which the initial segment (S 5 x) can be covered with finitely many sets of the Oi. Also s is contained in one of the Oi, and then it must be = z. And so we have obtained a finite subcovering. We now define a kind of linearly ordered sets which can be considered as generalizations of the real line R: 1.5 Definition. A dense linearly ordered set which has at least two, but no first and no last element and no gaps, is called a linearly ordered continuum. We call a cartesian product C = C1 x . x C, of n E N linearly ordered continua Cl, . . . ,Cn in short an ncontinuum. If S, is an open segment of C, for v = 1,. . . ,n, the cartesian product P := S1x .   x S, is called an (n dimensional) open segment of C. Every subset of C which is a union of open segments of C, is called an open set of C.We denote their set by 0. Again it follows easily that (C, 0) is a topological space. If I, = [a,, b,], v = 1,. . . ,n, is a closed interval of C, for v = 1,. . . ,n, the product Il x . . . x Inis said to be an ndimensional closed interval of C.
7.1. TOPOLOGY OF LINEARLY ORDERED SETS
205
If M is a subset of C, a point h E C is an accumulation point of M, if every open set which contains h also intersects M. A subset M of C is closed if all of its accumulation points belong to M. It follows immediately that M is closed iff its complement C \ M is open. We call M bounded, if there exists an n dimensional closed interval X{[a,, b,]lv = 1,. . . ,n) which contains M. If T is a subset of an ncontinuum Cl ,and C2 an mcontinuum, where n and m are natural numbers, a mapping f : T 4C2 is said to be continuous, if for every point x E T and every open set 0 of C2 which contains f (x) there exists an open set U of Cl, which contains x, with f [ U r l T ] 2 0. An open set 0 of an ncontinuum C is connected if it has no representation 0 = O1 U 0 2 , where O1 and 0 2 are nonempty disjoint open sets of C. And a set K is a connectivity component or region of an open set 0 if it is a maximal connected open subset of 0. Now a lot of classical invariance theorems of the topology of the euclidean nspace Rncan be generalized to ncontinua. We list some of them without proofs (and don't need them in the rest of the book): First we mention the following generalization of the separation theorem of Jordan/Brouwer/ Alexander:
1.6 Theorem [80]. Let K be a bounded closed subset of an ncontinuum C and f : K + C an injective and continuous mapping. T h e n the sets C \ K and C \ f [K] have the same cardinal number of connectivity components. A special case of 1.6 is the following generalization of the JordanBrouwer separation theorem, which for n = 2 was previously proved by LottgenIWagner [113]:
1.7 Theorem. Let B be the boundary of a closed ndimensional interval XF,l[a,, b,] of an ncontinuum C = Cl x .  . x Cn, i.e. the subset of those points (xl,. . . ,x,) of it, for which at least one xu is = a, or = b,. And let f : B 4C be a n injective and continuous mapping. T h e n C\f [B] has exactly two connectivity components. In another formulation: f [B] separates C into two regions. Also Brouwer's theorem of the invariance of the open set can be generalized to ncontinua:
1.8 Theorem 1761. Let 0 be an open subset of an ncontinuum C,
f :0
+ C injective
and continuous. Then f [0]is an open subset of
C. The last theorem entails a generalization of Brouwer's theorem on the invariance of dimension: 1.9 Theorem [175]. If Cm is an mcontinuum and Cn an ncontinuum, where m,n are natural numbers with m > n, then there doesn't exist a bijective and continuous mapping of Cm into Cn.
Brouwer's fixed point theorem was generalized to ncontinua by Dyer [31]. Another proof was given in [76]: 1.10 Theorem. Let P be an ndimensional closed interval of an ncontinuum C and f : P + P a continuous mapping. Then there exists a point x E P with f (x) = x.
7.2
The dimension of posets
Now we introduce the concept of dimension for posets, which was established by DushnikIMiller [30].We mention first: 2.1 Theorem. Let (P,5) be a poset. Then there exists a set of linear orderings (P, i E I, such that 5 is the intersection of the
si),
orderrelations li,i E I. Proof. For every pair (a, b) of incomparable elements of P we choose two linear extensions 5 , b and Sbafor which there holds a 5 , b b and b 5 b a a. Such linear extensions exist due to the theorem of Szpilrajn and Lba,where (2.3.2). Then the intersection of all linear orderings (a, b) runs through the set of all pairs of incomparable elements of (P, 5 ) is the relation 5 .
sab
Generally the set of linear extensions, which was used in the last proof, has many more elements than necessary. And since every nonempty set of cardinal numbers has a least one, we can now define: 2.2 Definition. Let (P,5)be a poset. Then the least cardinal d for which there exists a set of d linear extensions of 5, whose intersection is 5, is called the dimension of (P,L), in symbols dim (P, 5 ) or in short dim P.
7.2. THE DIMENSION OF POSETS
207
It can be expected that there are some common features between the above ordertheoretic concept of dimension and the topological one. In this connection there holds:

2.3 Theorem. Let Rn be the euclidean nspace, equipped with the product order 5 : For x = ( x l . . . ,xn) and y = (yl,. . . ,y,) of Rn we have x y xu yv for v = 1,. . . ,n. Then dim(Rn) = n.
<
<
This theorem is contained below in Theorem 2.11. The following is immediately clear: 2.4 The0rem.a) A nonempty poset has dimension 1 i f it is linearly ordered. b) If P is a totally unordered set (each two elements of P are incomparable) and [PI 2, then dim P = 2.
>
<
<*
Proof. b) Let be an arbitrary linear order on P and its reverse order. Then their intersection is the identity relation = on P. One could guess that the dimension of a poset P is a kind of measure of how much the order of P deviates from a linear one. But 2.4 b) shows that the order of a set can differ very much from a linear one, whereas the dimension nethertheless can be very small. Concerning the dimension of products we have: 2.5 Theorem. Let (P,<) be the orderproduct of linearly ordered i E I . Then there holds dim (P,<) 111. sets (Si,li),
<

Proof. For i E I we define an order Oi on P as follows: For x = and y = (yi)iEI of P we put: (1) xOi y xi yj holds,
<
>
<
<
<
<
this would by (1) entail yOjx and then also yLjx. Together with xLj y we would obtain x = y, contradicting x j > yj. So we must have xi 5 yi for all i E I, and then also niErLi is a subset of I,and both orders are equal. The last theorem cannot be sharpened so that dim (P, 5 ) = 1 1 1 results. If e.g. all Si are oneelement sets, P also has only one element and therefore the dimension. 1 whatever I is. But if all Si have at least two elements we obtain equality, as we shall see a bit later. A simple statement is expressed in: 2.6 Theorem. Let (P, I ) be a poset and T a subset of P. Then dim(T, 5 T ) 5 dim(P, 5 ) holds.
r
One can pose the question whether every cardinal number can be the dimension of a poset. This is true indeed and will result from the following reflection of DushnikIMiller [30]:
2.7 Lemma. Let S be an arbitrary set, A the set of its atoms (= oneelement subsets) and B the set of its antiatoms c, = S\{x), where x E S. Then for the set A U B, equipped with the order by inclusion, there holds dim(A U B ) ISI.
>
Proof. Let IS[be 2 2, the other case is trivial. The order C of AUB is the intersection of linear orderings Si,i E I, where III is the dimension of A U B. For every x E S the sets {x} and c, are disjoint and thus incomparable. Therefore there must exist an i E I with c, < i {x). Let y E S be an element # x. Then there is an index j E I with cv < j {y). Now i and j are different. Otherwise, because of x E cy and y E c,, we would have C,
< C,.
So we have obtained that every
A bit later the statement of 2.7 will be sharpened. First we determine the dimension of power sets. These are isomorphic to certain ordered products: 2.8 Theorem. Let S be a nonempty set. For every i E S we put Si := {0,1) and equip this set with its natural order (0 < 1). Let then
7.2. THE DIMENSION OF POSETS
209
X := XiEISi be the cartesian product of the Si, i E I , with the product order. Then the power set q(S)with the inclusion order is isomorphic to X . A n orderisomorphism between these sets is given by the mapping f T , which ascribes to every subset T 5 S its characteristic function fT on S, which is given by fT(x) = 0 if X $ T, fT(x) = 1 i f X E T.
c
Proof. f~ is of course bijective. And if we have sets TI 2 Tz 5 S, then fT, (x) = 1 implies f ~ (x) , = 1, and then fT, fT, follows.
<
An easy consequence of the previous theorems is now the theorem of Komm [loo]: 2.9 Theorem. For every nonempty set S there holds dim(q(S),C) = IS1 (= dim({O,l)S). Proof. By 2.7, 2.6, 2.8 and 2.5 we have IS[ WS), = dim((0, 1IS) ISI.
c)
<
< dim(AU B, C) < dim
The last proof now also yields, sharpening 2.7, the following theorem of DushnikIMiller [30]: 2.10 Theorem. Let S be an arbitrary set, A the set of its atoms, B the set of its antiutoms. Then the set AUB, equipped with the inclusion order, has the dimension IS]. Also 2.5 can now be sharpened to: 2.11 Theorem. Let (P,<) be the ordered product of the linearly ordered sets (Si,Si),i E I, where every Si has at least two elements. 11. Then dim(P, <) = 1 Proof. Let Ti be a twoelement subset of Si for i E I. Applying 2.5, 2.6 and 2.9 then there follows I I\ dim XiErSi dim XiEIz = dim ({O,1l1) = 14.
>
>
With 2.11 we obtain a new possibility of geometrical flavour, mentioned by Ore [132], to characterize the dimension of a poset: 2.12 Theorem. Let (P,<) be a poset. Then its dimension is the least cardinal k, such that P is embeddable i n an ordered product of k linearly ordered sets. Proof. Let
<
si
is embeddable in the ordered product XiEI(P, li) by mapping x E P onto (x) := (xi)iEI, where xi = x for every i E I. This is of course injective. If x and y are elements of P with x 5 y, then also x
2.13 Definition. Let (P,5) be a poset. In a former definition we introduced for x E P the principal ideal (XI as the set of all elements of P which are 5 x, and dually [x). Now we define the Pouzetset D ( P ) of P by D ( P ) := {(xllx E P}U{P\[x)lx E P), and equip it with the order C . So D ( P ) contains all principal ideals of P and the complements of the principal filters. Further we denote the set of initial segments of P by J ( P ) and consider it as ordered by the inclusion C . It can easily be seen that every set P\[x), x E S, is an initial segment of P and hence in J(P).Then D(P) C J(P) follows. Now the theorem of Pouzet [I431 states:
2.14 Theorem. Let (P,5) be a poset. J(P)= CCN(P).
Then dim D ( P ) = dim
The proof results from the following four lemmas:
Lemma 1. Let 5i be a linear extension of D(P). Then Ci := {x E PI (P\[x), (XI) E li} is a chain of D(P). Proof. Let x and y be elements of Ci7i.e. (P\[x), (x]) and (P\[y), (y]) are in 5i . We have to show that x is comparable with y. We assume
7.2. THE DIMENSION OF POSETS
211
the contrary x 11 y. Then y E ( P 11 x) 5 (P\[x)) and further, since (XI.Then also (y] Si (XI P\[x) is an initial segment of P, (y] E P\[x) holds because Si extends . Analogously (XI Si (y] follows and thus (XI = (y]. This implies x = y and yields a contradiction to our indirect assumption x 11 y.
si
Lemma 2. Let
si
Proof. We have to show: For every x E P there holds x E Ci for at least one i E I. The same: For at least one i E I there holds (P\[x), (XI) E
<
Proof. Let now the order of 9 ( P ) be the intersection of linear orders li, i E I, where 1 1 1 is the dimension of 9 ( P ) . Then by Lemma 2, S is covered by the union of 1 1 1chains Ci, i E I, so that CCN(P) (111 =) dim 9 ( P ) holds.
<
Lemma 4. Let P be covered by chains Ci,i E I, where /I/= CCN(P). Then the set J ( P ) of all initial segments of P is isomorphic to a subset of the ordered product XiEIJ(Ci), where J(Ci) is the set of all initial segments of Ci, which is a chain because of Lemma 1. The latter product has by 2 5 a dimension III, and so we further obtain dim J ( P ) 5 II(= CCN(P).
<
Proof. We consider an initial segment B E J ( P ) and map it onto the family ( B f l Ci)iEI. Then each component B n Ci of this is an initial segment of Ci, and so we have a mapping f : J ( P ) + XiEzJ(Ci). f is injective. For if B and B' are different elements of J ( P ) , then for at least one index i E I there holds B n Ci # B' n Ci, otherwise we would have B = UiEI(B n Ci) = uiEI(Bfn Ci) = B' with contradiction. f is also isotone: If .B C B' holds, then for all i E I we have B nCi C B' n Ci.
Finally f also preserves incomparability: Let B and B' be incomparable. Then there exists an element b E B\Bt,and a component Ci with b E B n Ci, but b 4 B' n Ci, so that B n Ci B' fl Ci. Th.us f (B) is not f (B'). Analogously f (B') is not 5 f (B).
<
Now the rest of the proof of Theorem 2.14 follows: By Lemma 3 we have CCN(P) 5 dim D(P); this is 5 dim J(P)because of D(.P) J(P), and dim J(P)is 5 CCN(P) by Lemma 4.
2.15 Remark. Pouzet's Theorem 2.14 contains the theorem 2.10 of Dushnik/Miller as a special case: Let P be a set with = as order relation, so that P is totally unordered in the sense of Definition 1.9.8. Then CCN(P) = IPI. Further we have, with the definitions of 2.13 3(P) = !J3(P). And now 2.14 yields dim D ( P ) = dim 3(P) = dim !J3(P)= /PI by 2.9. Here D ( P ) is the set which contains all oneelement subsets of P and their complements in P. From 2.14 we now can easily derive the following theorem of Hiraguchi [86]:
<
2.16 Theorem. Let (P,<) be a poset. Then dim P CCN(P) holds. If i n particular P is finite, dim P is 5 the width w(P) of P.
<
Proof. P is isomorphic to a subset of J(P)by 1.9.9. Then dim P dim J(P) = CCN(P) follows by 2.14. The rest follows with Dilworth's Theorem 2.5.6.
Using 2.14 Pouzet gave an answer to a question which was posed by Wolk [179]. First we define a special kind of posets, which will be treated later in more detail:
2.17 Definition. A poset is called partially wellordered, in short a pwoset, if all of its antichains and all of its descending chains a0 > a1 > . . . are finite. Wolk had asked: Is it true that all pwosets have a dimension The following theorem of Pouzet [I431 gives a negative answer:
< No?
2.18 Theorem. W e consider the ordered product P = w, x w, of the set w,, equipped with its natural order, with itself. Then the Pouxetset D(w, x w,) is a pwoset and has dimension N,. Proof. First we show that P has no infinite antichain. Otherwise P would have a denumerable antichain {(al, bl), (a2,b2), . . .) where its
7.3. Rl3LATIONS BETWEEN THE DIMENSION
213
elements are enumerated in such a way that the first components increase (in the wide sense), and so we can assume that a1 a2 5 a3 . . . holds. Since the pairs (ai, bi) are incomparable we then must have bl > b2 > b3 > . . , which is impossible, since there is no infinite decreasing chain in the ordinals. Also P has for the same reason evidently no infinite decreasing chain, and thus P is a pwoset. Let now A be an antichain of 2) (P).Then A has only finitely many sets (XI with x E P, because also their elements x form an antichain. Also A contains only finitely many sets P\[x) with x E P. For again their corresponding elements x form an antichain. (xl 5 x2 implies P\[xl) P\[x2)). Thus A is finite. In the set of the (XI there is no infinite descending chain in 2)(P), for this would imply that the corresponding x would form an infinite descending chain in P. Also in the set of the P\[x) there does not exist an infinite descending chain P\[xl) > P\[x2) >  . , for this would imply [xl) c [x2) c . . . and XI > x2 > . . . , which is impossible. So we obtained that 2)(P)is a pwoset. By 2.14 and 2.5.8 there follows dim D(w, x w,) = CCN(w, x w,) = N, 
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7.3
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Relations between the dimension of a poset and certain subsets
In 2.12 we characterized the dimension of a poset P as the least cardinal d such that P is embeddable in an ordered product of d linearly ordered sets. Now every linearly ordered set of cardinality N, is embeddable in H, of 4.3.3, and then we can reformulate the content of 2.12 in the form: 3.1 Theorem. If (P,5 ) is a poset of dimension d with cardinality [PI = N,, then P is embeddable in the ordered product ~ , (of d d factors Ha).
In the following we need a set which contains H, and has a compact ordertopology. The easiest way to obtain such a set is to fill the gaps of H, and introduce a first and a last element. 3.2 Definition. For an ordinal a we define K, as the set C,,which is augmented by the first and the last element of 2((w,)). It can easily
be verified that K, is dense and without gaps (compare with 4.7.32), and H, is dense in K,. With its ordertopology now K, is a compact space by 1.4. Of course the posets of finite dimension are of special interest. It turns out that the property to have a dimension 5 n, where n E N, is of finite character. We present two proofs of this. The first one of [68] consists of two embedding theorems and gives some additional insight. 3.3 Theorem [68]. Let (P,I) be a poset, such that every finite subset of P has a dimension I n, where n E N. Then also P has a dimension 5 n.
Proof. Let P be infinite, otherwise the assertion is trivial. Let N, be a regular cardinal > 2IPI. We consider the sets HE and KE, the ordered products of n factors H, resp. K,.Within this proof we understand under an embedding a function f : T + HE which is an isomorphic mapping of a subset T E P onto a subset of Hg, where for each v = 1,.. . ,n and different elements s, t E T the vcomponents (f (s)), and (f (t)), of f ( s ) resp. f (t) are different. We call here an embedding f : T + HE finitely extendable, if there holds: For every finite subset E C P\T there exists an embedding fE : T U E + Hg , which extends f. 3.3' Theorem. Let f : T + HE be a finitely extendable embedding, e an arbitrary element of P\T. Then there exists a finitely extendable embedding f, : T U {e) + HE, which extends f. Proof. Let C be the set of those finite subsets of P\T which contain e. For every E E C there exists an embedding f E of E which extends f. Then we define: An element a E Kg is called an accumulation point of {f E (e)1 E E C) if for every open subset U of Kg, which contains a and for every E E C there exists a set E* _> E in C with fE*(e) E U. The set of the E*, E E C, is then cofinal in (C, C). And now the last definition can, roughly speaking, also be formulated as: For every open U with a E U there are "cofinally many7' sets E E C, for which fE(e) E U.) First we prove: (I) The set {fE (e)IE E C) has an accumulation point a = (al, . . . ,a,) in Kg. If (I) would be false, there would exist for every x E KE an open set U,, which contains x, and a set Ex E C such that f ~(e)* 4 Ux for all
7.3. RELATIONS BETWEEN THE DIMENSION
215
>
E* E, of C . Since Kg is compact, there exist finitely many points XI,.. . ,xk in KE such that K: is covered by U,, , . . . , U,, . For the union E* := EzlU . U E,, of the corresponding sets of C we then would have fE*(e) $ KE, which of course is impossible. Thus (I) holds. Case 1. a E HE. Then we have: (11) For each E E C there is an E' E in C with fE1(e) = a. Otherwise we would have: (111) There is an E E (2 such that f E l (e) # a for all E' E of (2, and then there would exist an open subset U(a), which contains a, but none of the (due to 1fZ1 5 2IPI < N, less than N, many) fEt(e), E' E of (2. This follows from the fact that each elementcharacter of the components a, of a is (w,, wt). And this is a contradiction to the property of a to be an accumulation point of the fEl (e). So (11) entails that the function f e y given by fe T = f and fe(e) = a, satisfies 3.3'. Case 2. At least one component a, of a is in K,\H,, and then a # fE(e) for all E E (2. We put X := f [TI U u{~E[E]IE E (2). Then for each x = (xl,. . . ,x,) E X we have x, # a,. By construction a, is neither least nor greatest element of K,, and then a, has an elementcharacter (see 4.3.8 and 4.3.9) (w,, w;) or (up, wt ) with a regular initial ordinal wp. W.r.0.g. we assume the first case. Then there exists an a: < a, in H, such that the interval [a:, a,] of K, contains none of the x,, x E X. Otherwise a set of fewer than N, many x, would be cofinal in (H, < a,) with contradiction. Now the point a', which results from a by exchanging the component a, by a: has the same order position to the elements of X as a. In finitely many steps we so can alter all components of a which are not in Ha to elements of H,, but' in such a way that the point a" E HE, which is so obtained, is still in the same position to the elements of X as a. Now the function fe, which is defined by fe 1 T = f and fe(e) := a", satisfies 3.3'. 3.3" Theorem. Let X be a limit ordinal, T, E P for v < X and T, C T, for p < v < A. Let further f, : T, + P be a tower of embeddings which each are finitely extendable. Let f : TA:= U{T,Iv < A) + P be the limit mapping of the f,, v < A. Then also f is finitely extendable. Proof. Let E = {el,. . . ,ek) C P \ T be a finite set. For each v < X there is an embedding f,* : T, U E + HE which extends f,. First we
>
>
>
r
prove:
(I) There is a cofinal subset C of X such that for pl
< pa < A,
and
all v = 1,.. . ,n the following holds: The mapping, given by (fp, ( e d v $ (fpz ( 4 ) V for IF = 1,' ' ' ,k is <preserving and hence an isomorphism. Roughly speaking: The linear order in the set S := {(fp(e,)),l~ = 1,.. . ,k) is the same for all p E C. There are at most k! linear orders in the kelement set S, and so one of them must belong to cofinally many p < A, and (I) follows. We call here (other than before) a point (pl,. . . ,pk) of (K;)~ an accumulation point if for each system of open subsets O1,. . . ,Ok of K, with p, E 0, for IF = 1,. . . , k there is a cofinal subset A 5 X such that f;(e,) E 0, for IF = 1,.. . ,k and all p E A. Since K; and (K;)~ is compact (with the product topology) there is an accumulation point (a1, .. . , a d . Let a,, be the vcomponent of a,. Since a, is an accumulation point of f;(e,) its vcomponent a,, is an accumulation point of the set (of vcomponents) {fp(e,)),lp E C). Then we obtain: (11) If for one (and then also all by (I)) p E C we have (f;(e,,)), < (f;(en2)), then also a,,, a,,, holds. This follows immediately from the accumulation point property by indirect assumption. If all a,,, K = 1,.. . ,k, v = 1,.. . ,n, would be in Ha we would be done. For an a,, E H, is = (f,(e,)), for cofinally many p E C. If an a,, is in K,\H, we shift if to a position a',, E H, such that the a,,I K = 1,.. . ,k, preserve the position (coordinatewise) to the elements a,, < a,, of f [TI, which the a, had before and so that e,, < e,, holds. This can be done so that in (11) a&, < a',,, is valid. That this is possible follows in an analogous manner as in the proof of 3.3', because or each a,, has, as an element of K,, an element character (w,, (wp,w;). Let f *be that mapping which coincides with f over T and puts f (e,) := a, for IF = 1,.. . ,k. It satisfies the statement of our theorem 3.3N. In some more detail we show: For t E T and IF E (1, . . . ,k ) there holds: (111) t < e, + f (t) < a,. For then for p E C we have (fp(t) =) f (t) 5 inf{fp(e,)lp < A) 5 a,. For a, < inf{f,(e,)lp E C) is not possible by construction of a,. So we
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WE)
7.3. RELATIONS BETWEEN T H E DIMENSION
217
<
already have f (t) a,. If a, 4 Hg also f (t) < a, follows. If a, E HE holds, a, has to be = f,(e,) for cofinally many v, otherwise a, could not have the above mentioned accumulation point property. And then f (t) = f,(t) < f,(e,) = a, follows for cofinally many v, so that (1) holds. (IV) ei < ej + ai aj. For ei < ej implies f,(ei) < fp(ej) for all p E C,and for each v = 1 , . . . ,n the v component of ai cannot be greater than that of aj.
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If finally we choose a representation P = {p,l~< wp), where P is a.The Theorems 3.3' and 3.3" together yield by transfinite induction a tower of embeddings f, : {p,l~ p) + Hg,p < wp. Their limit mapping fUB is now an embedding of P into Hg.
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We present now a proof of 3.3 which was suggested by H.A.Jung: Second Proof of Theorem 3.3. If S is a subset of P we call an ntuple (21,. . . ,5,) of linear orders <,on P an norder on S, iff their intersection is 1 S. Let 3 be the set of all finite subsets of P. For the finite subsets C of 3 we define f2 as follows: We consider the union UC (of finitely many finite sets L E C) and equip it with n linear orders 5 1 , . . . ,<, whose intersection is f UC. For I E C let then f2(I) be that norder on I which results from 51,.. . , by restricting these linear orders on I. According to Rado's Selection Lemma (see 8.1.6) there is a mapping f * : J + A = ~ J { A E~ J), ~ I where Az is the set of all norders on I), such that there holds: For each finite subset C of J there is a finite set 9Jl with C 5 9Jl C 3 for which
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<
<,
(*) f *(I) = fm(I) for I E C. For every finite I C P (i.e. I E 3) an norder on I is uniquely defined by f *. Now also f * defines an norder for the whole set P, since the definitions for the finite subsets of P are compatible: Let Il and I2be finite subsets of P. Then there is a finite set %I satisfying 3 %I {Il, 1 2 ) =: C for which (*) holds. In particular we have f*(Il) = fm(Il) and f*(I2) = fm(Iz). Herewith fm(Il) and fm(12) are those norders on Il resp. I2which result as restrictions of the norder of U r n , which is defined by fm and also by f*. So for a, b E Il n I2we have a b in Il +==+a b in I2since both are
> >
<,
<,
<,
the restrictions of the 5, of Uf137 2 Il U 1 2 . And so f * defines an norder on P. Another proof of 3.3 was given in the book [45];this makes use of the ultrafilter theorem. The Theorem 3.3 yields the following corollary: 3.4 Theorem. For every natural number n there exists a countable set B, of order types of dimension n, such that every poset of dimension n contains a subset of a type T E Bn.
Proof. It is obvious that the set B, of all order types that belong to finite posets of dimension n E N is countable, for even the set of all order types that belong to finite posets is countable. Then by 3.3 every poset of dimension n contains a finite poset of dimension n, whose type is in B,. Now the question arises whether the last theorem can be sharpened so that Bn can be replaced by a finite set of order types. For n = 2 this is trivial. For every poset of dimension 2 contains two incomparable elements, and so the order type of a twoelement antichain forms such a finite set B2 of order types. But already for n = 3 the answer to the above question is negative, as we shall prove a bit later. First we consider the "descending staircase" set:
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3.5 Theorem. Let P be the set of all points (x, x) and (x, x 1) of R~ where x is an integer. W e order P with the restriction of the product order of Z x Z : (a, b) 5 (c,d) e a 5 c and b 5 d. Then (P,<) has the dimension 2. See the illustrations in Figure 8 and 9.
Proof. Z x Z has the dimension 2, and then its subset P has a dimension 5 2. It is also 2 2, because P contains incomparable elements.
Figure 8
7.3. RELATIONS BETWEEN THE DIMENSION
219
The diagram of (P,5 ) (in the sense of 1.7.6) is
Now the following theorem (Baker/Fishburn/Roberts [7],see also [68],in this paper n should be replaced by 2n on page 188 line 12) gives an answer to the question which was raised before: 3.6 Theorem. Let n be a natural number 2 3 and let C be the set { a l , . . . ,a,, bl, . . . ,b,}, which is equipped with the following orderwhich contains all (a,, b,), u = 1,. . . ,n, and all (a,, bVl) relation with v = 2 , . . . ,n, and ( a l , bn) and of course all ( p , p ) with p E C. SO in the sequence a l , bl, a2, b2,. . . ,a,, b,, a,+l := a1 every element is comparable exactly with its predecessor and successor. (The comparability graph, which is defined later, is then a circle with 2n vertices and 2n edges.) Then (C,<) has the dimension 3.
<,
The following Figure 10 (for n = 5) illustrates the structure of the order of C:
Figure 10 Proof. Since C has incomparable elements, its Cimension is at least 2. We now assume that it is 2 and construct a contradiction. Then we can further assume that C is a subset of the euclidean plain R2 with the order induced from the product order of R2,where all 2n points p E C have pairwise different first and pairwise different second components.
Let wl, w2,. . . ,w, be an enumeration of the set {al, ..,a,} of the a's such that the first components of the elements wl , . . . ,w, strictly increase. This implies that the set of corresponding second components strictly decreases.Then there holds: (1) For 1 5 i upper neighbor.
< j  1 5 n the elements wi
and wj have no common
For if z would be an upper neighbor of wi and wj, it would also be an upper neighbor of wi+l. (In the following Figure 11, z would be situated in the hatched area.) And no element of C has three lower neighbors.
Figure 11
Let vl be one of the two upper neighbors of wl. According to (1) the lower neighbor of vl which is different from wl must be w2. Now w2 has exactly one upper neighbor which is different from vl, say v2. This has besides w2 exactly one lower neighbor, and this must be ws. For it cannot be, wl because this would entail C = {wl, vl, w2, v2}. Continuing in this way we finally obtain that w, has besides v,1 exactly one upper neighbor v,. This has exactly one lower neighbor different from w, in the set {wl , . . . ,w,I}, and this can only be wl since to the elements ~ 2 , . .. ,w,1 we have already ascribed two upper neighbors. But now wl and w, would have v, as a common upper neighbor, and this contradicts (1). We have proved that the dimension of C is 2 3. It remains to prove that it is also 5 3. This (can easily be verified but it) is a consequence of the following theorem of Hiraguchi, according to which the dimension of a poset P increases by at most 1 if a new element is added to P. Indeed,
7.3. RELATIONS BETWEEN THE DIMENSION
221
if one of the points of C is omitted, the remaining poset has dimension 2 according to 3.5. 3.7 Theorem [86] (Hiraguchi 1956) Let (P*,<*) be a poset, a E P*, which is the restriction 5* P. Then P := P*\{a) ordered by dim (P*,5 *) 5 dim (P,5) 1.
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Proof. Let d be the dimension of (P,5 ) and P # 0. Then 5 is the intersection of linear orderings Li, i E A, on P, where A is an initial segment of the class of ordinals and /A/ = d is the dimension of P. In particular 0 E A. We consider the initial (resp. final) open segment of a in P* :
I := {x E P I x
<* a) and F := {x E P I a <* x),
and then further the initial (resp. final) segment of Co := (P,Lo) which is generated by I (resp. F ) : I*:= {x E P I x Lo y for some y E I) and F* := {x E P I y Lo x for some y E F). Let further B := P \ (I*U F * ) be the set of all elements which in Co are between I*and F*.The following Figure 12 illustrates this: Co with order Lo : Figure 12
We now define two linear orders L*, L** on P* : L* and L** each consist of five consecutive segments, each of which is linearly ordered by (the restriction of) Lo : (P*,L*) has the consecutive segments I, {a), I*\I,B, F*,and
(P*,L**) has the consecutive segments I*,B, F*\F,{a), F. For 0 # i E A we further choose orders Lf on P*,which are linear extensions of <*,and for which Lf P is the order Li. (There are usually more than one possibility to find a suitable position for a in La.) For 0 # i E A now (P*,Lf) has the three consecutive sfegments
r
{x E P I x Lf a), {a), {x E P I a Lf x),
where in the first and third of these segments Lz coincides with Li. The linear extensions L*, L**, and the Lz, 0 # i E A, contain I*, and so L*is a subrelation of I n : = L* n L**n n{LtlO # i E A). : But also Inis a subrelation of (and then =) I* Let x, y be different elements of P with x 5, y. Then xL*y and xL**y hold and this entails x Lo y, as can easily be verified in all cases (x E I*resp. x E B resp. x E F*, y E I*resp. y E B resp. y E F*). y also entails xLzy and then further x Li y for all i with x y hold. 0 # i E A. Then x 5 y and x If x Ina holds for an x E P, we have xL*a and thus x E I, so that x <* a holds. If a Sn x holds for an x E P we have aL**x. This yields x E F and a <* x. and this order is So we have obtained In(C and then also) = now the intersection of the d 1 linear orders L*, L**, Lz, i E A\{O).
<*
+
s*,
In the dimension theory of posets one mainly investigates relations between the cardinality of the ground set, the dimension of the poset itself, and the dimension resp. cardinality of certain subsets. We present some of the most important ones. First we introduce a concept: 3.8 Definition. Let (P,5) be a poset, a # b two elements of P. We say: a and b are positionequivalent, iff there holds: For P' := P\{a, b) we have (P' < a) = (PI < b) and (PI > a) = ( P I > b) (and then of course also (P'lla) = (P'llb)). In this context we have the following reduction lemma of Trotter [170]:
3.9 Lemma. Let (P, 5) be a finite poset, a # b elements of P, which are positionequivalent. a) If P\{a) is not linearly ordered, then dim(P\{a)) = dim(P). b) If P\{a) is linearly ordered, and a and b are comparable, then P is also linearly ordered, and then both sets have dimension 1. c) If P\{a) is linearly ordered, and a and b are incomparable, then 1 = dim(P\{a)) < dim(P) = 2.
Proof. a) Suppose dim(P\{a)) = cl, and the order P\{a) is the intersection of the d linear orders L1,. . . ,Ld on P\{a). We have d 2 2, and then we define linear orders LF, . . . ,L: on P as follows: For v = 1,.. . ,d  1 we take the order of L, and insert a directly above b to obtain LE. In L: we take the order of Ld and insert n directly below b. Then the order of P is = ~$,,Lc, so that d i 4 P is d. It is 2 d by 2.6.
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b) is trivial, and c) follows with 3.7 The last lemma enables us to delete certain elements in posets without lowering the dimension. By such reductions often situations become more transparent. Another lemma of that kind is: 3.10 Lemma. Let (P,<) be a poset with /PI 2 2 and x an isolated element of P, which means one that is incomparable with all elements of P\{x) := T. Suppose that T with its induced order is not linearly ordered. Then d := dimT = dim P. Proof. Let {Li(i E I) be a set of linear orders on T with [I/= d, for which 5 1 T is = n{Lili E I). We have /I/2 2.We choose a fixed j E I and define a linear order L; := L j U {(t, x) It E T) on P. (Roughly speaking: One puts x as last element behind (T, Li).) For i # j we put L% := Li U {(x, t) It E T). Then 5 is the intersection n{Lf li E I), and so dim P is d. Of course dim P is also (dimT =) d.
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These lemmas are now applied in the next theorem of Trotter: 3.11 Theorem [170]. Let X be a finite poset which is no chain and no antichain, A an antichain of X with IX\A/ = 2. Then dim X = 2. Proof. dimX 2 2 is trivial. Let X\A = {a, b). If one of these two elements, say a, is incomparable with all elements of A , but b not, let X I , . . . ,X i be all elements of the antichain A ~ { a which ) are comparable with b and xi+l, . . . ,x, the rest of the elements of X. Our notation assumes that this rest is not empty, but in this case things become easier.Then X I , . . . ,xi are all < b or all > b. For if there would exist elements y, z in A U {a) with y < b < z then we would have y < z, and this contradicts the fact that A U {a) is an antichain. W.r.0.g we assume that b is greater than 21,. . . ,xi. Then we take the following two linear orders L1, L2 of X: We define L1 by the sequence X I , . . . ,xi, b, xi+l,. . . ,x,. And L2 is given by the sequence x,, . . . ,X I , b. Then Ll n L2 is the order of X, which now has dimension 2. So we still have to discuss the situation where each of a and b is comparable with at least one element of A. Now a (and analogously b) cannot be greater than an element of A , and at the same time smaller than an element of A , and so we have to treat, due to symmetry, only two cases:
1) a is greater than an element of A, and b is greater than an element of A . 2) a is greater than an element of A, and b is less than an element of A. If A has elements which are isolated in X, we can omit them as long as at least two elements of A remain, without changing the dimension. Further we apply the reduction process of 3.9: If a is greater than three elements X I , x2,x3 of A, then, due to the pigeonhole principle, there are two of them which are < b, or two which are > b, or two which are incomparable with b, for b cannot be less than one of the three and greater than another of them. In each of these three subcases there are two elements of {x1,x2,x3) which are posisitionequivalent, and then one of the three can be omitted by 3.9. Iterating this reduction process we can assume that a is greater than exactly one or exactly two elements of A, and by symmetry (between a and b) we can reduce the Cases 1) and 2) as follows: Case 1. a is greater than exactly one or exactly two elements of A, and the same for b. Case 2. a is greater than exactly one or exactly two elements of A, and b is smaller than eactly one or exactly two elements of A. Now we see that after these reductions A has at most four elements. For each of its elements is comparable with a or b, and these two elements are each comparable with at most two elements of A. If A would have exactly four elements we could make another reduction, for then two of its elements had to be comparable with a, the other two with b, and then the two elements, which are comparable with a, are positionequivalent, so that one of them could be omitted. So finally we can assume w.r.0.g. that A has two or three elements. If IA1 = 2 holds we have 1x1 = 4,and then dim X 5 2 easily follows. Let now A := {xl, x2, xs). If a is greater than two elements, say X I , x2 of A, and if b is greater than these both, or smaller than both, or incomparable with both, then again XI and $2 are positionequivalent, and we can omit one of them without increasing the dimension. After all reductions we then have obtained that in case1 (resp. case 2) X has the same dimension as a set which is isomorphic to a subset with the following diagram of Figure 13 (resp. Figure 14):
7.3. RELATIONS BETWEEN THE DIMENSION
x1
572
23
31
Figure 13 b
These sets have dimension 2, the first one by 3.5. The order of the second set is the intersection of two linear orders, namely X I , b, 2 2 , a, x3 and b7x3,x2,x1,a. We mention an easy property of posets: IAl
3.12 Lemma. Let (X, 5) be a yoset, A an antichain of X with 2 2 for which IX\AI = 1 holds. Then dimX = 2.
Proof. Let x be the element of X\A. If x is incomparable with all elements of A, then X is an antichain and we have dimX = 2. So we assume that x is comparable with an element of A, w.r.0.g. we can assume that x is greater than an element of A. Then all elements of A are smaller or incomparable with x. We put A1 := (resp. A2) the set of those elements of A, which are < x (resp. 11 x). Then we define a linear order L1 on X as follows: L1 has Al as initial segment, then follows x, and A2 is a final segment. We define a linear order L2 on X by: L2 has A2 as initial segment, then follows A1 as a segment, and x is the last element of (X, L2). But L2 f A1 resp. L2 t A2 are the reverse orders of L1 f A1 resp. L1 f A2. See the Figure 15:
Figure 15
Now Ll n L2 is the order 5
3.13 Theorem (Trotter 11701). Let X be a finite poset, A an antichain of X with IX\AI 2 2. Then dim X 5 IX\AI. Proof. If X is a chain or antichain the assertion is trivial. In the other case we put X\A = {xl,. . . ,x,). Then by 3.11 the poset {xl, x2) U A has dimension 2. Now we add successively to this set the remaining elements 2 3 , . . . ,x, if n 2 3. Then the set {xl, x2,x3) U A dim({xl, x2) U A) 1 by 3.7. After n  2 steps, has a dimension always applying 3.7, we obtain: X = {xl,. . . ,x,) U A has a dimension < d i m ( { x l , x z ) ~ A ) + n  2 = 2 + n  2 = = n = IX\AI. Now a theorem of Hiraguchi 1861 follows in an elegant way (see 11701):
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3.14 Theorem (Hiraguchi 1955). Let X be a finite poset with n elements. Then its dimension is 5 LgJ.
>4
Proof. Let A be an antichain of X with w elements, where w is the width of X. For X = A the statement is trivial. If IX\AI = 1 we have /A[2 3, and then by 3.12 dimX = 2 151. If finally IX\AI 2 2 holds we have by 2.16 dimX 5 w = IAl and by 3.13 dim X IX\AI. Summing up the left and the right sides of these inequalities we obtain 2 . dimX 5 IAl IX\AI = 1x1= n.
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Before stating the next theorem we need a new concept, which was introduced in a paper of Bogart [lo]:
3.15 Definition. Let (X, 5) be a finite poset, C a chain of X. Then a linear extension L of 5 is said to be a lower (resp. upper) linear extension of with respect to C, if there holds: If x E C, and if y is incomparable with x, then y L x (resp. x L y) holds.
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Concerning the existence of lower linear extensions we have:
3.16 Theorem. Let (X, 5 ) be a finite poset, C a chain of X. Then there exists a lower (and analogously an upper) linear extension of 5 with respect to C. Proof. Let X I , . . . ,x, be the elements of X\C. Then the set I := (C < xl) is an initial segment, F := ( C > xl) a final segment, and B := ( C 11 xl) a segment of C, all elements of which are between those of I and those of F. We adjoin $1 to C and put it between I and B. Then Cl := C U {xl) is linearly ordered and a lower linear extension of C with respect to C. If for an i < n already a lower linear extension Ci := Cu{xl,. . . ,xi) of C is constructed, we put xi+l between
7.3. RELATIONS BETWEEN THE DIMENSION
and the rest of Ci. Then Ci+1 := C U {xl, . . . ,xi+l) is a (Ci < (on Ci) with respect to C. After n steps lower linear extension of X = C U {xl, . . . ,x,) is a lower linear extension of 5 with respect to C.
<
Trotter [I701 proved:
3.17 Theorem. Let X be a finite poset, A an antichain of X, and X\A # 0. Then dimX 5 2 width(X\A) + 1. Proof. Let w := width(X\A). Because of Dilworth's theorem there exists a representation of X\A = Cl U . . . U Cw as union of w disjoint chains. For every i = 1,.. . ,w there exists a lower linear extension L2iW1 and an upper linear extension Lai of with respect to Ci. Further we define a linear extension LzW+lof 5 to X, in which A is a segment (see f A is the reverse order of L1 f A, the rest is 2.3.3) and where arbitrary. Then is the intersection n{Lili = 1 , . . . ,2w 1) := I. Indeed, if a 5 b, then also a I b holds trivially. Let now a and b be incomparable in 5 . We distinguish three cases: 1) If a and b are both in X\A we have a E Ci for some i E (1, . . . ,w).Then b L2il a and a Lzi b hold, so that a, b are also incomparable in I. 2) If a and b are both in A, then in L1 and in L2w+1they appear in reverse order and are thus incomparable in I. 3) If one of a, b is in X\A, the other in A , say w.r.0.g. a E A, b E X\A, then b is in some Ci. It follows a L2il b and b Lpi a, and again a and b are incomparable in I. So I is the order 5 of X, whose dimension is now 5 2w 1.
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3.18 Theorem (Trotter [170]). Let (X, 5) be a finite poset, M the set of its maximal elements, and M C X. Then dim X < width(X\M) + 1. Proof. Let t := width(X\M). Then by 2.16 the dimension of X \ M is 5 t , and so its order is the intersection of t linear orders L1,. . . ,Lt on X\M. We extend every Li to a linear order Lf of X. Further we define a linear order Lt+1 on X as follows: We take some linear extension of 5 1 X \ M and place all elements of M on its top in the order which is reverse to that one in which they appear in Lt. It follows immediately that 5 is the intersection n{Lili = 1,.. . ,t 1). Finally we mention a theorem of V. Novgk [127], see also [75]: 3.19 Theorem. Let (P,5 ) be a poset, D M ( P ) its DedelcindMacNeille completion. Then D M (P)has the same dimension as (P,5).
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Proof. Let d be the dimension of (P, 5 ) and {li li E I) a set 1 1 = d and = niEl
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3.20 Remark. In [I271 NovAk proves a more general theorem, which we mention here without proof, namely: If P is a poset and D a subset of P which is adense and &dense in P (see 1.10.8), then dim P = dim D.
7.4
Interval orders
Starting from a linearly ordered set one can investigate the set of its closed intervals with the inclusion order E . Using this order Dushnik/Miller [30] obtained a new characterization for posets of dimension 2 2. 4.1 Definition. Let (P, <) be a poset, a < b and c < d elements of P. Then we define [a, b] 5 [c,dl ++ [a*,b] E [c,dl. (The same: e c a < b 5 d). It is clear that this defines an order in the set of closed intervals of P.
<
4.2 Theorem ([30]). Let (P, <) be a poset. Then the following two conditions are equivalent: 1) P has a dimension 5 2. 2) P is isomorphic to a set of closed intervals of a linearly ordered set, which is ordered by inclusion. Proof. Suppose 1) holds. Then the order 5 is = Ll n L2, where L1 and L2 are linear orders on P. Let f : P + P' be a bijective mapping of P onto a set P' which is disjoint to P. Instead of f (x) we also write XI.
We order P I by L' where for x, y E P we put x'L'yl ++ yLax. Then we define a linearly ordered set C as an ordered sum C := (PI, L') (P, L1). Let 5 denote the linear order of C.
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7.4. INTERVAL ORDERS
229
Now we map every x E P onto the interval I(x) := [x', x] of C. This yields an isomorphism of (P,5) onto a set of intervals of C. We verify this: Suppose x y (in P). Then xLly and $Lay hold, and the last relation implies y1L'x' (in P I ) . Now there follows y' 3 x' 3 x 3 y and then [x', x] E [y', y]. So the mapping x + I(x) is 5  preserving. Let now x and y be incomparable elements of P. W.r.0.g. we can assume that xLly and yL2x hold, where the last relation implies xlL'y'. Then we have x' 5 y'. Here x' is strictly less than y' with respect to 5 because of x' # y', and so [x', x] is no subset of [y', y]. Also x is strictly less than y in C, and thus [y', y] is no subset of [x', XI. Therefore the imageintervals of x and y are incomparable. And I) + 2) is verified. Now we prove 2) + 1) : Let 3 be a set of closed intervals of a linearly ordered set (T, ST). Let then A (resp.B) be the set of first (resp. last) elements of the intervals of 3. We equip A with the order
<
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r
Among the sets of dimension 5 2 is an important class of posets, namely the generalized trees, which have been defined in 6.1.6. 4.3 Theorem. If (T, 5 ) is a generalized tree, it is either a chain or it has dimension 2. In particular every tree has dimension 5 2.
Proof. First we prove our statement for finite trees and apply induction on the cardinality of T. For trees (T, 5 ) with IT I 5 2 the assertion is trivial. Let now n be a natural number 3, such that the theorem is already proved for all trees with 5 n elements, and let T be a tree with n + 1 elements. Let a be the least element of T and T := T\{a), and let al, . . . ,ak be the immedate successors of a. For each K = 1,.. . ,k
>
the set TK of all elements of T, that are
> a,
is again a tree, and
<
since T, has n elements, its dimension is 5 2, so that its order 5, is Then we consider the the intersection of two linear orders OK1and 0K2. following two linear orders L1, L2 on T: (Our notation assumes k 2 2, but the case k = 1 follows similarly.) (T,L1) is the linearly ordered sum {a) (TI,011)  . (T,, Okl), (T, La) is the linearly ordered sum {a) (TKl0k2) . . . (TI, 012). In L2 the segments TI,. . . ,Tk appear in the reverse order to that, in which they appear in L1, so that the elements of Ti are in the order L1 n L2 incomparable with all elements of Tj if i, j are different elements of (1, . . . ,k). Further for every K = 1,. . . ,k the intersection (L1 n L2) T, is equal to . So the intersection Ll n L2 is exactly the order of T. Let now (T, <) be an infinite tree. Then all finite subtrees of it have a dimension 2, and then also T has dimension 2 by 3.3. If finally (T, 5 ) is a generalized tree, it suffices for the same reason to prove the assertion for the case where T is finite. Then each principal ideal (T x) with x E T is (not only linearly ordered, but also) wellordered. We add a new element e to T, which is less than all t E T, so that e becomes the minimum of the enlarged set. Now T U {e) is a finite tree; so it has a dimension 5 2. And this holds a fortiori for T.
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<, <
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Chapter 8 Wellfounded posets, pwosets and trees In this chapter we consider a class of posets which fulfill certain finiteness conditions. For this reason they have a lot of features in common with finite posets. A forerunner of this class is the algebraic concept of Hilbert and others of the descending chain condition in algebra, which is satisfied e.g. in the set of ideals of a ring, if every descending sequence of ideals is finite.
8.1
Wellfounded posets
1.1 Definition. A quasiordered set (W, <) is called wellfounded, if W has no infinite descending chain {ao,al, ..,a,, ..In < w ) , i. e. with ao > a 1 > a 2 > *   . If W is wellfounded, and if in addition every antichain of W is finite, W is called wellquasiordered, abbreviated wqo, or a wqoset. If here W is not only quasiordered but also ordered, W is said to be a partially wellordered set, abbreviated pwoset.
Another possibility to define the notion "wellfounded" for posets is given by the following characterizations, which are immediately clear: 1.2 Remark. A poset is wellfounded if each of its chains is wellordered. A poset is wellfounded if each of its nonempty subsets has a minimal element.
Every finite quasiordered set is wellfounded, also every wellordered set. The property to be wellfounded is of course hereditary. And it turns out that some important notions for finite posets, which have been defined in 1.7.1 (see also 1.7.3), can be generalized to wellfounded sets:
1.3 Definition. Let W be a nonempty wellfounded poset, Lo the set of its minimal elements. Since W is nonempty also Lo is nonempty. If for an ordinal v sets L,, p < v, have already been constructed, we define L, to be the set of all minimal elements of W \ u{LPIp < v) and
call it the v  level of W. This is also denoted by W(v). There exists a least ordinal a for which La is empty (and then also Lp = 0 for all p > a). This a is denoted by h(W) or Ht( W) and called the height of W. And so we have h(W) = {vlL, # 0). The elements x of L, are said to have height h(x) = v. So we have h(W) = {h(x)lx E W). We also write h(x, W) for h(x), if there are more sets in consideration, of which x is an element, and in which x has possibly different heights.. A maximal chain of W is called a branch of W, and an initial segment of a branch is said to be a path of W. Then every branch and every path of a wellfounded poset is wellordered. The height of a branch or path is also called its length. If ,f? is an ordinal 5 h(W) we define the ,6  stump of W to be the union u{L,Iv < p). We compile several simple facts about wellfounded sets, which are partially in analogy to 1.7.2: 1.4 Theorem. Every level L, of a wellfounded poset W is an antichain of W. The levels L,, v < h(W), are pairwise disjoint, and so they form a partition of W into antichains. Let p and v be ordinals with p < v < h(W), and x E L,. Then there holds: a) There exists an xp E Lp with xp < x. b) For every y E LP we have y < x or y 11 x. c) For elements x,y of W there follows: x < y h(x) < h(y). d) For every a E W there holds h(a) = U{(h(x) 1)lx < a ) =: R. e) For a E W we have h(a) = h(W < a ) . f) If two diferent elements a,b of W have the same set of predecessors (W < a ) = (W < b ) , then h(a) = h(b), and a and b are incomparable.
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Proof. a) Lp is the set of minimal elements of W \ U{L,lr; < p ) =: S, and we have x E S, which implies a). b) An element y E Lp is minimal in S, and so x E S cannot be less than y. c) By definition y is a minimal element of W\ U {L, lv < h(y)). Then an element x < y cannot be in this set, and so x E L, must hold for some v < h(y). This v is = h(x), and c) is proved. d) For x < a in W we have h(x) < h(a) and thus h(x) 1 5 h(a), so that h(a) R holds. On the other hand, let ,6 be an ordinal < h(a).
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8.1. WELLFOUNDED POSETS
233
Then by a) we have an element x < a in W with height h(x) = P. Then h(x) + 1 > p, and thus R is > /3 for all ordinals ,!?< h(a), which means R 2 h(a). e) a E W holds by definition: h(W < a) is the least ordinal which is > h(x) for all x < a, and so h(W < a) is the least ordinal which is 2 h(x) + 1 for all x < a. This implies h(W < a) = ~ { ( h ( x ) l)lx < a} = h(a) by d). f) This follo~vsimmediately by e).
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Here we recall 1.7.5, which shows that it can happen that elements a,b of a wellfounded poset can satisfy a a b but h(b) > h(a) + 1. Every time a new class of ordered sets is introduced the question arises of what can be said about the structure of the chains and antichains of the sets of the class. In 1.2 we already had mentioned a fundamental property of the chains of wellfounded posets. Now we investigate them in greater detail to learn more about the order types of the wellordered chains. A first question which suggests itself, is: If W is a wellfounded poset, does there exist a chain passing through all levels of W ? It can easily be seen that this is not true. If W is a union of disjoint chains Cn, n E N, where each C, has n elements, and where for m # n all elements of C, are incomparable with all elements of C,, then W is wellfounded, and has height w, but no chain of type w. A weakened form of the above question has a positive answer, namely there holds: 1.5 Theorem. Let W be a wellfounded poset, x E W an element of height J. Let hl < . . . < h, = J be finitely many ordinals < h(W). Then there exists a chain {xl, . . . ,x, = x) of W , such that x, has height h, for v = 1 , . . . , n . In particular there holds: Let x,+, E W be an element with height T n, where r is an ordinal and n E N . Then there exists a chain C W, which contains x,+, as last element and for every nonnegative r m. integer m < n an element x,+, of height h(x,+,) So C passes through all finitely many levels L, of W with T a 5 r n.
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Proof. The assertion is an immediate consequence of 1.4, a).
A class of wellfounded sets W , in which a chain exists which passes through all levels of W , is that of those wellfounded sets in which each
level is finite. For the proof of this we use Rado's Selection Lemma (see [I471 and [148]), which has several useful applications in set theory. In the following proof of it we use the notation X << Y for: X is a finite subset of Y. 1.6 Theorem. Let A and I be nonempty sets and Ai << A for i E I. Suppose that for each L (< I, we are given a choice function f L : L + A such that fL(i) E Ai for i E L. Then there exists a choice function f * : I  + A such that, given any L << I, there is an M with L C M << I and f*(i) = fM(i) for i E L.
Proof. If I is finite then f * = fI satisfies our assertion. Now let I be infinite. Let R be the set of all families (Xi1 i E I ) such that Xi 5 Ai for i E I and, given any L << I, there is an M with L 5 M << I and fM(i) E Xi for i E L. (This entails that the Xi are nonempty.) By hypothesis, (Ail i E I ) E C2 (put M = L). We consider the following (natural) order in R : (yil i E I) 5 (Zil i E I ) C Zi for i E I.
x
We prove that (R, 5 ) has a minimal element by using Zorn's lemma. Let A ={Ckl k E K ) be a chain of R, i.e. K a linearly ordered index set where i < j (in K) + Ci < Cj (in 0 ) . Let Ck = (Xkil i E I ) for k E K. Then there holds for every i E I : The family (Xkilk E K ) is monotonically decreasing with k, and since the sets Xki are finite this entails that for every i E I there exists an index k*(i) such that all Xki with k 5 k*(i) are equal, say = X;, and this is nonempty. The family (X:l i E I)belongs to R : Let E (< I . For k 5 min{k* (i)1 i E E ) we have Xki = X: and so there exists a set Fkwith E 5 Fk<< I and f f i (i) E (Xki =) X: for i E E. Of course, (XtI i E I ) is a lower bound of A, and thus by Zorn's lemma, R has a minimal element P = (Pi I i E I). Next we prove that all sets Pi, i E I, contain exactly one element. Let j be a fixed element of I. By construction Pj is nonempty. Let now a and b be elements of Pj. We wish to show a = b. We consider the following statement (I), which will turn out to be false: (1) For every L (< I there exists an M with L M (< I, such that fM(i) E Pi for all i E I and fM(j) # a. If (1) would be true, then the family (PJ i E I ) , for which Pi' = Pi for all i # j of I and Pj = Pj \ {a) holds, would also be in R, and it
8.1. WELLFOUNDED POSETS
235
would be less than a minimal element of fl with contradiction. So (1) is false, and thus we have: (2) There exists an (L =) Ia<< I such that for every M with Ia C M << I and fM(i) E Pi for i E I there holds fM(j) = a. Symmetrically there holds: (2') There exists an Ib<< I such that for every M with Ib2 M << I and fM(i) E Pi for i E I there holds fM(j) = b. But now IaU Ibis a finite set M with IaU IbE M << I, and so by (2) and (2') we have fM(j) = a = b. So all Pi,i E I, have exactly one element pi. Finally we put f*(i) := pi for i E I. Due to (Pi1 i E I ) E S2 now f * satisfies our assertion. If we make use of topological concepts, in particular of Tychonoff's product theorem, we can present another elegant proof of 1.6, which was found by Gottschalk [59]:
Second proof of 1.6: Let C be the set of all finite subsets of I. For L E C let EL be the set of all points p = (pili E I ) E X :=X{Ai li E I) for which there exists a B E C with B L where for each i E L we have pi = fB(i). (For i E I \ L the element pi is arbitrary.) By assumption every EL is nonempty. Provide each Ai, i E I, with the discrete topology. (Every subset of Ai is open and also closed.) With this Ai is a compact space, since Ai is finite. The set X :=X{Ai(i E I), equipped with the product topology, is then also compact by Tychonoff's theorem. ELis a closed subset of X. For every cartesian product P :=X{Y,li E I),where for i 6 L we have Y, = Xi, and where Y,is a oneelement subset {pi) of Ai for i E L, is closed; and EL is a union of finitely many such sets P.
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(I) The set (ELIL E C) has the finite intersection property, which means the intersection of finitely many of its elements is nonempty. Indeed, let L1,. . . ,L, be finitely many finite subsets of I. Then E' := ELIn.. . n EL, = {p = (pi)iEI E XI For every v = 1,.. . ,n and each i E L, we have pi = fBv (i) for a B, L, of L). Now L := L1 U    U L, is finite and a superset of the L1, ..,L,. We put pi := fL(i) for i E L, and Pi arbitrary E Ai for i E I \ L. Then (pi)iEI belongs to E'which is hence nonempty, and (I) is proved.
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A compact space has the property that if an intersection of a set C of closed sets is empty, then already a finite subset of C has an empty intersection. Therefore (I) entails n{ELIL E C) # 0. Let q = (qi)iEr be a point of this intersection. We define f * ( i ):= qi for i E I. Now f * satisfies our assertion: If L << I is given and i E L, then, since q E EL, qi = fM(i)for some M E with M L.
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We can now prove the theorem of Wolk which was mentioned before:
1.7 Theorem [179]. Let W be a wellfounded poset of height h with leveldecomposition {Lala < h), in which every level L,, a < h, is finite. Then there is a chain containing precisely one element from each level L, of P.
Proof. Let H = {hl, . . . ,h,)< be a nonempty finite subset of the ordinal h. Then by 1.5 there exists a chain {xl ,. . . ,x,) of W with h(x,) = h, for v = 1,.. . ,n. In this manner we construct a mapping f H : H + W, where for every a E H we have f H (a) (= x,) E L,, and where the fH(a), a E H, form a chain. Due to 1.6 there exists a function f * : h + P, such that for every H << h there is an M with for a E H. (*) H M << h and f*(a) = fM(a) Now f *(a)E L, for a E h, and these elements form a chain: If a, b E h we can take H = {a, b ) and an M satisfying (*). Then f *(x) = f M (x) for x 6 H, and then f *(a) and f *(b) are comparable since f M [{a, b)] is a chain. The last theorem can be considered as a generalization of Konig's Infinity Lemma, which corresponds to the case h = w. In the last theorem the assumption that all levels of W are finite cannot be omitted, as the example before 1.5 shows. Pouzet [I411 exhibited the following counterexample in which additionally all maximal chains have the same height as W:
1.8 Example. Let W be the set of pairs (n,m) of nonnegative integers n, m with m < n. This set is ordered by (n, m) 5 (n', m') a n = n' and m 5 m', or n < n' and m 2 5 m'. Then (W, 5 ) is wellfounded and of height w. But if C is a chain of W, it cannot intersect each level. Indeed, let (n, m) be the least element of C, and (n, k) the greatest element of C which has n as first component. Then t is < n, and every immediate successor (n', m') of (n, k) in C must have m' k 2. Then
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8.1. WELLFOUNDED POSETS
C has no element with second component k height k 1.
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+ 1 and thus no element of
In the following we investigate in more detail the types of chains and antichains of wellfounded posets. In this context we introduce a useful notation which was defined by Milner and Sauer. In the pioneering paper [I201 they give a very detailed discussion on this subject. 1.9 Definition [120]. Let a,@, y be ordinal numbers. Then a + [p,y] means that the following statement is true: Whenever (W, <) is a wellfounded poset of height a, then either there is a chain C C W of type P, or there is an antichain A 2 W such that the order type of the set of ordinals {v E alA n W(v) # 0) is y, where W(v) is the set of those elements of W which have height v. The negation of a + [P,y] is, of course, denoted by a n [P,y]. We remark already here that the main theorem of the rest of this section will be a theorem of Milner/Sauer and Pouzet, stating that a + [a,w] holds for all ordinals a. For the case of denumerable ordinals a this was first proved by Diana Schmidt [155]. The symbol a + [P,y] is suggested by the symbol a + (P,y)2 of the partition calculus [38], (with which it should not be confused). This is defined as follows, supplementing Definition 6.4.1 which concerned cardinals: 1.10 Definition. For ordinals a,P, y the symbol a + (P,y)2 means that the following is true: Whenever (S,5 ) is a wellordered set of type a, and the set [SI2of twoelement subsets of S is partitioned into two sets KO,K1, then either there is a set B C S of order type P such that [ B ]C ~ KO,or there is a set C 2 S of order type y such that [CI2C K1. The symbol a + (P, y)2 is stronger than a + [P,y] : 1.11 Theorem. Let a,P, y be ordinals for which a + (P, y)2 holds. Then also a + [P,y] is valid.
Proof. Let W be a wellfounded poset of height a. Then for every v < a we choose an element x, of height v. For p < v < a we put {p, v) E KOiff x,, x, are comparable, and this means xp < x, since x, is on a lower level than x,. And we put {p, v) E K1 iff x, and x, are incomparable. Due to our assumption and 1.10 with a for S we obtain: a of order type P such that all x, with v E B There is a set B are comparable, and then they form a chain of type P, or there exists a
set C a of type y such that all xu with v E C are incomparable. And then the indices of the levels of W which contain elements of {x,lv E C ) form a set of order type y. By the way, the partition relation Ic + (Ic, No)2 of 6.4.10 entails a + ( a , w ) and ~ thus by 1.11 a + [a,w] for the case where a is an initial ordinal. This will later be generalized in 1.16' on all ordinals a . Now we mention several elementary facts on the height function, which are rather obvious, but should nevertheless be checked carefully: 1.12 Lemma. Let W be a wellfounded poset, and V 5 W a subset which is equipped with the restriction of the order of W . Let x E V. Then for the heights of x in W resp. V we have
(1) h(x, W) 2 h(x, V). Proof. If h(x, V) = 0 holds, (1) is trivial. Let now (1) be proved for all values h(x, V) < p, and let y E V have height h(y, V) = p. Then by 1.4 a) for all v < p there exists a y, in the v  level L,(V) of V with y, < y. By induction hypothesis then y, E Lp(,)(W) for some p(v) 2 v. Since y > y, holds for all v < p we obtain by 1.4 c) h(y, W) > h(y,, W) = p(v) 2 v, and thus h(y, W) p, so that (1) also holds for h(x, V) = p. And our statement is proved.
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1.13 Lemma. Under the assumptions of 1.12 we have h(V) 5 h(W). Proof. h(V) is the least ordinal which is > h(x, V) for all x E V, and h(W) is the least ordinal which is > h(x, W) for all x E W. So 1.12 implies 1.13. If we consider a wellfounded set W and a subset V of it with the induced order, so that V is also wellfounded, usually the heights of an element x E V in W and in V are different. In the following lemma we have a situation where they are equal: 1.14 Lemma. Let W be a wellfounded poset, x E W an element with height h. W e denote the v  level of W by L,, that of the subset (W < x) by l,, for v < h. Then there holds 1, = (L, < x) for all v < h.
Proof. For v = 0 the statement is trivial. Let now p be an ordinal < h such that the assertion holds for all v < p. Let m E I,. Then m < x holds, further m $! U{L,Iv < p), because otherwise by induction hypothesis m would be in an L, with v < p, and then also in I , with
8.1. WELLFOUNDED POSETS
239
contradiction. Now m is a minimal element of T := W\ U {L,lv < p) and thus in L,. For if an element y of T would be < m E I,, then also y < x would hold, and then by 1.4 c) y would already be in some I, with v < p and then, due to the induction hypothesis, also in the corresponding L,, contradicting y E T. Hence m E (L, < x). On the other hand, let m E (L, < x). Then m $ 1, for v < p, since otherwise by the induction hypothesis also m E L, would hold for a v < p, which is impossible. Therefore m E (W < x)\ U {l,lv < p) =: U, and m is minimal in U and thus in 1,. For if we would have u < m for an element u E U,then by 1.4 c) u E L, would follow for a v < p. This would by induction entail u E 1, for some v < p, contradicting u E U. 1.15 Lemma. Let 6 be an indecomposable ordinal wX, where X is a limit ordinal. Let (vili < cf (A)) be a strictly ascending sequence of successor ordinals which has X as supremum, so that also sup{wv~i< cf (A)) = wX. Then S := C{wVi li < cf (A)) = wX.
>
Proof. wX u{wVi\ wvilli < cf(X)). Here each wUi\ wViI has order type wV< F'urther the summands in the set union are pairwise disjoint, and so its order type is 2 S. Of course, it is also 5 S and thus = S.
The main theorem of this section, which we address now, was established independently by Milner/Sauer [I201 and Pouzet [I411 with entirely different proofs. Here we present the proof of Pouzet. 1.16 Theorem. Let W be a wellfounded poset of height 6. Then there exists an infinite antichain of W all of whose elements have different heights, or there exists a chain of W which has the same height 6 as W.
Proof. If 6 = 0 holds, W must be empty, and then the statement is trivial. Also the case 6 = 1 is trivial. Suppose now 6 > 1, and that the assertion is proved for all 6' < 6. We make the assumption that there is no infinite antichain of W such that all elements of it have different heights. We distinguish two cases: Case 1. 6 is not indecomposable. Then we have a representation 6 = 6' + 6", where 6' and 6" are both < 6. Let W" be the set of all x" E W for which the height h(xl', W) of xI1in W is 2 6'. We have h(W") = 6", and then by induction hypothesis W" has a chain C" of type 6". Let m be its least element. Then we have m x for an element x E W(6'). According to 1.4 e) the wellfounded
>
set (W < x) has height 6', and so by induction hypothesis (W < x) contains a chain C' of type 6'. Now C' U C" is a chain of W of type 6' 6" = 6. Case 2. 6 is indecomposable and > 1. Then by 1.15 we can represent 6, if it is > w, as a sum 6 = C{Gili < cf (6)), where (&),i < cf (S), is a strictly ascending sequence of ordinals . sup{Si li < cf (6)) = 6. (If 6 = wV+' we can take = w"  i 6.  w *~ with for i < w. If 6 = w we put Ji := i for i < w.) Let dj). Wi := {X E WI C j < i 6) h(x, W) < CjZi Then Wi is wellfounded and has height Si < 6. By Induction hypothesis there is a chain Ci E Wi of type 6i. Now A := u{Cili < cf (6)) has height 6, for it is 2 Si for all i < cf (6). All elements of A have different height in W. A has no infinite antichain since otherwise A would intersect infinitely many chains Ci, and then also infinitely many levels of W, which is excluded by our assumption. A fortiori this entails that all levels of A, these are antichains, are finite, and so from 1.7 there follows the existence of a chain C of (A and) W which has exactly one element of each level A(v) with v < 6. And so also C has height 6.
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In the notation of 1.9 the last theorem states:
1.16' Theorem. 6 + [6, w] for all ordinals 6. In connection with 1.16 Pouzet established a theorem, which also implies a proof of 1.16, but gives more insight into the structure of wellfounded posets, so that it deserves some interest in itself. The main part of its proof is treated in the following technical Lemma 1.18. First we introduce a new concept.
1.17 Definition. Let 6 be an ordinal > 0. Then by 4.8.5, 6 has a unique representation as 6 = wffO . . w f f k as a sum of finitely many indecomposable ordinals, where a0 2 . 2 ak are ordinals. Then we define l(6) to be the last summand wffkin this normal form representation of 6. If 6 is a successor number, then ak = 0 and w f f k = 1. If W is a wellfounded poset and 6 an ordinal < h(W), then W(6) denotes the 6  level of W.
+ +
1.18 Lemma [141]. Let W be a wellfounded poset with height function h, 6 < h(W), X E W(6) and 1x1 11(6)1. Then W has a subset A 2 u{W(C)(C < 61, which satisfies 1) [ A n W(c)I 1 for a l l c < 6,
<
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241
8.1. WELLFOUNDED POSETS
2 ) h ( A < x ) = 6 for all x E X . Proof. We apply induction on 6. For 6 = 0 the set A = 0 satisfies 1 ) and 2 ) , here we have h ( 0 ) := 0. Also the case 6 = 1 is trivial. Let now 6 be > 1 and the lemma be proved for all 6' < 6. We distinguish three cases: Case 1. 6 is not indecomposable. Then we can write 6 = 6' 6", where 6" = l ( 6 ) . Let W" be the set of all x E W which have a height h ( x ) 2 6'. Then we have ( 1 ) X G W1'(6"). Indeed, for x E X we have h ( x , W ) = 6 = 6' 6" and on the other hand h ( x , W ) = 6' h ( x , W"), so that 6" = h ( x , W " ) , and thus ( 1 ) follows. Further we have ( X I 5 I1(Sr')I (= 11 ( 6 )I because of l(6") = l ( 6 ) ) . Now our induction hypothesis yields: ( I ) There is a set A" C W" which satisfies 1 ) and 2 ) (with W,6 replaced by W",6"). So by 1 ) we have IA"I 5 16"l. Now we choose for each xu E A'' an element x' E W ( S 1 )with x' 5 XI'. Let X' be the set of all x'so chosen. Then X' 5 W ( 6 ' ) and IX'I 5 jA1'I. Further we have I6"I 11(6')1 , since in the normal form representation of 6 = 6' 6" the term l(6') is a summand which precedes l ( 6 ) = S", so that 6" 5 l(6') holds. So we have IX1I(5 lA"I 16"1) 5 11(6')1. Now, by induction hypothesis, applied on W,6' and X I , we obtain: (11) There exists a set A' E u{W(C)IC < 6') which satisfies 1 ) and 2 ) , with 6, X replaced by S', X I . In particular we have ( 2 ) h(A1< x') = 6' for all x' E XI. Finally A := A' U A" satisfies our assertion: 1 ) is trivially fullfilled, x in A" and x' xu due to ( I ) and (11). Let now x E X . For every x" in XI C W (6') we have by 1.4 e) and ( 2 ) h ( A < x ) 2 h(A' < X I ) h ( x U A") , = 6' h(xl', A"). Then h(A < x ) is also the supremum of the values 6' h ( x U ,A"), which is 6' 6" = 6, and so 2 ) holds. Case 2. 6 = w. Here we have 1x1 5 Il(6)I = Iwl = No. For every x E X we choose an infinite subset Y, E w such that these sets are pairwise disjoint. Since w x w is equipotent to w this is easily done. Next we choose a subset Ax (W < x ) n {yl h ( y , W ) E Y,) which contains an element of each level W ( n )for n E Y, and has height w. This can be achieved as follows: We represent each Y, as an ordered sum Y, = C{I,lrn E N ) , where
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<
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<
<
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+ >
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<
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Imis a segment of Yx with m elements. Then by 1.5 there exists an melement chain Cm of W which has exactly one element in common with every level W(n),n E I,. Now the set Ax := ~{C,lm E N ) has height 2 m for all m E N, and so its height is w. Finally A := u{AX(xE X ) satisfies 1) and 2). Case 3. 6 is an indecomposable ordinal > w, which means 6 = wV for an ordinal v > 1. Here l(6) = 6 holds. Then we choose a strictly increasing sequence (6ili < cf (6)) of ordinals < 6 for which there holds sup{Sili < cf (6)) = 6 : If v is a successor ordinal p 1, and hence 6 = wP+l ,we take Si := WP . (1; 1) for i < w (= cf (6)). Here = wP for all i < w. If v is a limit ordinal, we take a strictly increasing sequence of successor ordinals vi, i < cf (v)(= cf (6)), whose supremum is v, and 6i := wVi. Here 1(&) = In any case we have by 1.15, C{6ili < c f (6)) = 6. Next we choose a system of subsets Xi, i < cf (6), of X, which satisfies Xi X j for i < j < c f (6), and
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+
a) u{Xili
< cf(6)) = X,
and b) lXil 5 11(6i)l.
<
Now we define for each i < cf (6) the set Wi := {x E Wl Cj,6j h(z, W) < Cjsi6j), and put := Wi UX. Then we have Xi G Wi(6,). 6, and Xi. So we Due to b) we can apply our induction hypothesis on obtain a subset Ai 2 % which satisfies 1) and 2) for di and Xi instead of 6, X. Finally we put A := u{Aili < cf (6)), and this now fulfills our assertion: 1) is clear, and since an element x E X is in some Xi with i < c f (S), it is also in all Xj with i < j < cf (S), and so h(A < x) is 2 Si for all i < cf (6), which means = 6.
K
R,
In 1.18 the assumption 1x15 11(6)I cannot be omitted. The following simple example confirms this: 1.19 Example. Let W be the union of N1 disjoint chains of type
+ 1, where the elements of different chains are incomparable. Then W is wellfounded and has height w + 1. Let X := W(w). Suppose now
w
indirectly that A is a set C LI{W([)~<< w), which satisfies 1) and 2) of 1.18. Then, due to 2), for x E X holds: (A < x) must have height w, and therefore it contains an element x' < x. The set X' := {x'lx E X ) has cardinality N1(= 1x1) because x + x' is injective. Since the Nlelements of X' A are distributed over the countably many levels of W there
8.1. WELLFOUNDED POSETS
243
exist two elements (also N1 many) in X' which have the same height, and this contradicts 1). The technical lemma 1.18 now implies as a special case: 1.20 Theorem [141]. Let W # 0 be a wellfounded poset of height h(W) =: 6. Then W has a subset A which (with the induced order) has the same height 6 as W , and which for every v < 6 has at most one element y of height (in W ) h(y, W) = v.
Proof. We adjoin to W an element x which is > all x E W and put W' := W U {x). We apply 1.18 on W'. We have 6 < h(Wf) = h(W) 1 = 6 1 and h(x, W1) = 6, so that X := {x) C W1(6). Of 11(6)1 holds. Thus by 1.18 there exists a set course, also 1x1 = 1 A 5 W' with h(A < x) = 6, which has at most one element of height v (in W' and in W) for each v < 6. In the last theorem we have simpler assumptions as in Lemma 1.18, but the induction process in the proof of the lemma needed more detailed assumptions.
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At the end of this section we list some theorems without proof, which are proved or mentioned in the paper [I201 of MilnerISauer. Here K denotes infinite initial ordinals and K+ the initial ordinal of the successor cardinal of IKI. Other greek letters denote ordinal numbers, n E N. fin+ [ ~ n , @for ] n < w , P < wl, c f > ~ w. a * [w, K 1] for a < K+. (For a < wl this was proved by Diana Schmidt [155].) a * [ K , c ~K + 11 for Q < K+. a + [ K , w ~ ]for K 2 ' 0 , ~ < K+. a + [ w ~ + ~ , Kfor ] K 5 2 ' 7 , ~ < K+. a * [KW, w 11 for a < K+. (Galvin [50]) a < w1 5 cf K implies K + [a,K].
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1.21 Remark. A simple fact, which is contained in the second of the above theorems, is that there are wellfounded posets having arbitrary height but which do not contain any infinite chains. This follows by induction: For the ordinals n E N the sets (0,. . . ,n1) (with their natural order) have height n and no infinite chain. Let now a be an ordinal such that for all P < a there is a wellfounded poset (Pp,sp)of height P, which has no infinite chain. W.r.0.g. we can assume that the PR,P< a, are pairwise disjoint. Then the union U :=
< a), equipped with the order U{Sa IP < a), is wellfounded and has height 2 P for all P < a. If a is a limit ordinal, then U has height a. If a is a successor ordinal P + 1 we add a new element to U U{PplP
which is greater than all elements of U. 'The enlarged set has height a, but no infinite chain.
8.2
The notions wellquasiordered and partially wellordered set
Two specializations of the concept "wellfounded poset", which was defined in 1.1, are of great importance, namely partially wellordered set and tree. The last notion was already defined in 1.9.8 (see also 6.1.6), but trees will be investigated in detail later. By definition the partially wellordered sets arise from the wellfounded posets by implying a second finiteness condition. The wellfounded quasiordered sets and the wqosets can be characterized by using forbidden subsets: A quasiordered set is wellfounded, if it has no subset of type w*, and it is wqo, if it has no subset of type w* and no subset of the type of an antichain with No elements. A property which is defined by forbidden subsets is hereditary, and so we can state: 2.1 Remark. If W is a wellfounded quasiordered set, then every subset of W with the induced quasiorder is also wellfounded. And if W is wqo, each of its subsets is also wqo. The concepts of 2.1 were introduced in the early 1950 s at nearly the same time and independently by several authors (Higman [85], [84], ErdosIRado 1371, Kruskal [102], [loll, Michael [119], Rado 11491). We mention here also the paper [I241 of NashWilliams and the two very informative survey articles of Milner [I211 and Pouzet 11421. The preoccupation with trees, which are also wellfounded, was e.g. promoted in a lot of papers by D. Kurepa, beginning with [104], and in connection with Suslin's problem, which we treat later. The pwosets have many applications in different parts of mathematics. We mention two important ones in graph theory. We refer to the definition of graph which was introduced in 2.5.4. For general informations on graphs see e.g. the book 1231 of Diestel.
8.2. THE NOTIONS WQO AND PWO
245
2.2 Definition. A subgraph of a graph (V,E ) is a graph (V',E ' ) , where V' V and E C E' holds. We say: (V', El) arises from (V, E) by contraction of an edge { a , b) of E, if V' = V\{a, b) U { c ) , where c is a vertex which is different from all vertices of V, and where E' contains all edges which link elements of V\{a, b), and further all twoelement subsets { c , x ) , for which { a , x ) or {b, x ) are in E. If a graph G'arises from a finite graph G by successively contracting edges, it is called a minor of G. For this we use the sign G' 4 G. Also G itself is considered as a minor of G. Then the relation 4 is a quasiorder in every set of finite graphs. For a finite graph H we further put H 4 G, if H is isomorphic to a minor of G.
In this context Robertson and Seymour proved between 1986 and 1996 in a series of more than twenty papers the famous minor theorem: The class of finite graphs is wellquasiordered by the minorrelation 4 . Of course, we cannot go into the details here. It comprises a theorem of Kruskal [I021 for graphtheoretical trees: In graph theory a tree 5 is defined as a connected graph which has no circle. Here a graph G is said to be connected, if for each two different vertices a, b of G there is a finite sequence a = a l , . . . ,a, = b, such that {ai, ai+i) is an edge of G for i = 1,. . . ,n  1. And a circle is a sequence of vertices a l , . . . ,a, with n 2 3, which are pairwise different by exception of a1 = a,, such that {ai, ai+i) is an edge of G for i = 1, . . . ,n, where we have put a,+l := a l . Then Kruskal's theorem states: The class of finite trees is wellquasiordered by the minorrelation. This theorem was generalized by NashWilliams 11221 on infinite trees. Recently Daniela Kiihn [I031 published a proof of his theorem which is considerably shorter than the original proof. There are some more possibilities to characterize wqosets, and we present the most important ones here: 2.3 Theorem and Definition. Let Q be a quasiordered set. Then the following conditions are equivalent: 1) Q is wqo. 2) For e v e y Qsequence (u,),~, , which means for every sequence (a,),<, of elements of Q , there exists an infinite subsequence such that ani 5 anj for i < j < w. Such a sequence is called a perfect subsequence. 3 ) For every Qsequence (a,),<, there exist two indices i < j such that ai 5 aj holds.
A Qsequence (an)nE,, for which there exist two indices i < j with ai 5 aj, is called good. If not it is called bad. So 3) can be stated as : Every Qsequence (an)nEwis good. 4) There is no infinite strictly increasing sequence of final segments of Q (with respect to the order E by inclusion). 5) There is no infinite strictly descending sequence of initial segments of Q. 6) Any subset A of Q has a finite coinitial subset.
Proof. We make a ringproof. 1) + 2): Suppose Q is wqo, and (an),<, is a Qsequence. Then there exists an index no such that a,, is minimal in the set {a,ln E w) because Q has no infinite descending chain. For the same reason the set {a,ln > no) has a minimal element anland so on. Then in the sequence a,,, an, ,. . . we have ani (5 or I I ) a j for i < j< W. Thus by putting bi := ani for i < w we have a subsequence (bi)i<, of (a,),<, where bi(5 or 11 ) bj holds for i < j < w. For i, j E w we put i R j iff i < j + bi 5 bj. By Ramsey's theorem (see 6.4.8) there exists an infinite subset N' E N such that i R j for all i # j of N'or i(nonR) j for all i # j of N'. In the last case we would also have bil 1 bj for all i 3 of N' and thus an infinite antichain, which is impossible. Therefore the first case must hold and we are done. 3) is trivial. 2) 3) + 4). Suppose indirectly that Fo c Fl c . .  c Fn c .  . is a strictly increasing sequence of final segments of Q. For i E w we choose an element xi E Fi+l\Fi. Due to 3) there exist i < j such that xi 5 xj. This implies x j E Fi, and this contradicts x j cj Fj+l\Fi. 4) 5). If (I,),<, would be a strictly descending sequence of initial segments, (Q \ I,),<, would be a strictly increasing sequence of final segments in contradiction to 4). 5) ==. 6). Let A be a nonempty subset of Q. Then the set M of those elements which are minimal in A (with its induced order) is coinitial in A and satisfies M E A E F ( M ) , where F ( M ) is the final segment generated by M. Also M is finite. For if M has an infinite subset {m,lv < w),then the initial segments IS(.[m,ln 5 v < w)), n E w, form an infinite sequence of initial segments of Q which is strictly descending because of mn $ IS(m,ln 1 5 v < w) contradicting 5). 6) ==. 1). Let A be an antichain of Q. Then A has a finite coinitial subset B E A. So for every a E A there exists an element b E B with
+
*
*
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8.2. THE NOTIONS WQO AND PWO
247
b 5 a. This b must be = a because b < a cannot hold in an antichain, and therefore A is C B and finite. If Q would have an infinite descending chain D of elements a0 > a1 >  .. , then D had to contain a finite coinitial subset of D, which of course is impossible. The property, which was mentioned in 2.3.2, can be generalized to more general transfinite sequences: 2.4 Theorem. Let (Q, 5) be a wellquasiordered set, and let (u,Iv < w,) be a transfinite sequence of elements of Q, where w, is an initial ordinal. Then there exists a subsequence (x,lv < w,) of (u,Iv < w,) such that p < v < W, xp 5 xu.
Proof. Let U be the set of all 2  element subsets of w,, and 23 (resp. C) that subset of U, which contains all those {p, u ) E U for which p
* < <
The importance of the concept pwo stems among others from the fact that every linear extension of a pwoset P yields a wellordering of P. We prove this with the following lemma: 2.5 Lemma. Let (P, 5 ) be a poset which has an infinite antichain
A. Then there exists a linear extension of 5 which is not a wellordering. Proof. We choose a linear order LAof A, which is no wellordering, e.g. one in which A has a subset of type ho. Then by 2.3.3 there exists a linear order on P which extends < and whose restriction on A is LA . And so it cannot be a wellordering. Now we prove the characterization theorem of Wolk [179] for pwosets, which justifies the choice of the naming: 2.6 Theorem. A poset (P, 5 ) is pwo i f every linear extension of
< is a wellordering of P.
Proof. Let (P,5 ) be pwo and < ~ linear a extension of < . Suppose indirectly that xl > L 2 2 > L . > L xn > L    , n E N, is a strictly
decreasing sequence in < L . Then we consider the set X := {x,lu < w) with the relation R (of strict comparability with respect to 5 ) : For x # y of X we put xRyex
2.7 Theorem. A poset (P, 5 ) is wellfounded if there exists a linear which is a wellordering of P. extension of
<
Proof. If we have a linear extension of 5 which is a wellordering, then a fortiori (P, 5 ) cannot have an infinite strictly decreasing sequence, and is thus wellfounded. On the other hand let (P,<) be wellfounded. And let L,, u < h(P), be the levels of P. Each of them is an antichain of P, and we choose a wellordering <,in L, for each v < h(P). Then we form the ordered sum C{(L,, I,)lv < h(P)}. This furnishes a wellordering on P which extends the order of P. This is an immediate consequence of 1.4 c).
<
Now we consider the question whether the properties wellfounded (resp. pwo) are also productive, which means: If all factors of a cartesian product X have such a property, does X have this property as well. Here we only have that the properties wellfounded, wqo and pwo are "finitely productive" . 2.8 Theorem. Let Wl, . . . ,Wn be wellfounded quasiordered sets. x Wn is also wellfounded. Then the ordered product X := Wl x
Proof. Let (xil, . . . ,xin), i E N, be a weakly descending sequence of X,which means for i < j in N we have (xil,. . . ,xi,) 1 ( ~ ~ 1. .,,xjn). .
8.2. THE NOTIONS WQO AND PWO
249
Then for every v E {I,. . . ,n) the sequence is weakly descending, and thus there exists an index i(v) € N such that all xi, with i i(v) are equal, say = x:. Then for all i max{i(v)lv E {I,. . . ,n)) a11 (xil,. . . ,xin) are = (XI,.. . ,x:), and thus X is wellfounded.
>
>
2.9 Remark. If in 2.8 we have more than finitely many wellfounded sets, their ordered product is not necessarily also wellfounded. If e.g. Wn is the linearly ordered set {0,1) (with 0 < 1) for n E N , then the ordered product Wl x W2x .  .x Wn x  .  is not wellfounded. The infinite set of sequences (111.. .), (01111.. .), (001111.. .), (0001111.. .),. . . is strictly descending. 2.10 Theorem ([119], [123]). Let Wl, . . . ,Wn be finitely many wqosets. Then their ordered product X := Wl x x Wn is wqo. Then also pwo is finitely productive. Proof. For i E N let (ail,. . . ,ann) be an element of .X. By 2.3, 3) we are done if we can prove that there exist two indices i < j such that (ail,. . . ,ain) 5 (ajl,. . . ,ajn) holds. The set {ailli E N ) of first components is wqo, and so there exists by 2.3, 2) an infinite subset Nl 5 N, such that the sequence {ail li E Nl) is increasing with i. Analogously there exists an infinite subset N2 Nl such that {ai21i E N2) is increasing with i, and so on. After finitely many steps we obtain a subset Nn 5  5 Nl such that for every v = 1,.. . ,n the sequence {ai,li E Nn) is increasing with i. This entails that the elements (ail,. . . ,oin), i E Nn, form a sequence in X which increases. By 2.3, 2) then X is wqo. In connection with 2.8 and 2.10 the question arises whether the property to have only finite antichains is productive. An easy counterexample shows that this is not the case, even if we have only two factors:
2.11 Example. Let 5 be the usual order in the set N of the natural numbers and <*its reverse order. Then the orderproduct (N, <) x (N, <*) has an infinite antichain, e.g. the set of all pairs (x, x) with x E N. Indeed, if x < y in N, then (x, x) 11 (y, y) holds because of y <* x. The following theorem was observed by Wolk [179]: 2.12 Theorem. Let (P,<) be a poset of finite dimension d Then there holds:
> 0.
P is pwo
P is embeddable in an ordered product of finitely many
wellordered sets.
Proof. Let P be pwo. The order L: is = f&
and this also holds for all of its subsets. Considering wqosets instead of pwo sets does not yield a real gain in generality, as the following trivial theorem shows: 2.13 Theorem. Let (X, 4) be a quasiordered set and (S,<) its corresponding ordered set (of equivalence classes according to 1.4.9). Then there holds: X is wqo e=+ S is pwo.
Proof. Let X be wqo and let Cl, C2,. . . be a sequence of S. We choose an element ci E Ci for every i E N. Then there exist indices i < j with ci 4 cj and this entails Ci Cj. And thus S is pwo by 2.3, 3). Suppose S is pwo, and cl, cz, . . . a sequence in X,Ci the ci containing class of S. Then there exist indices i < j with Ci < Cj and this implies
<
c+ 4 cj. In wellfounded posets (W, 5 ) we have the trivial relation Ih(W)I In the much more special situation of pwosets we even have:
I W I.
<
2.14 Theorem. For the height h of an infinite pwoset P there holds
lhl = lPl.
Proof. P is the union of its I hl levels, which are antichains of P and thus finite.Therefore Ihl must be infinite. Each nonempty level contains at least one element of P; so there is an injective mapping of the set of IPi follows. On the other hand we have [PI all levels into P, and lhl < Ihl No = Ihl, and so we obtain Ihl = IPI. 
<
8.3 Partial ordinals The class of order types is quasiordered by the embedding relation, see 1.9.6. This is reflexive and transitive, but not antisymmetric. However the class of the types of wellordered sets, which is nothing else than
8.3. P A R T I A L ORDINALS
25 1
the class of ordinals, is wellordered. This fact can be embedded into a more general context. Following Wolk [I791 we define: 3.1 Definition. An order type of a pwoset is called a partial ordinal. Of course, every ordinal is also a partial ordinal. If a and T are partial ordinals, we put a 5, T if the following holds: If S (resp. T) is a poset of type a (resp. T ) , then there exists an isomorphism of S onto an initial segment of T.
The relation 5, is a subrelation of the relation 5 of Definition 1.9.6, which was not antisymmetric. Now it follows immediately that <,is reflexive and transitive. We shall prove that it is also antisymmetric. First we show: 3.2 Theorem (Wolk [179]). Let P be a pwoset, f an isomorphism of P onto an inital segment f(P) of P. Then f [PI = P.
Proof. Let f0 be the identity function on P, and if for an n E w we have already defined f we put f n+' := f ( f n ) . We suppose indirectly that f [PI c P holds. Then ( 1 ) fn[plC fn'[PI holds for n = 1. Suppose that ( 1 ) holds for a fixed n E w. Since f is injective this entails f [ f n[P]]C f [ fnl [PI], so that ( 1 ) also is valid for n 1 and then by induction for all natural numbers n 2 1. Further there holds for every n E w :
+
( 2 ) f * [PI,n E w, is an initial segment of P. For n = 1 this is true by hypothesis. Suppose now that (2) holds for a fixed n 2 1 of w, and let x E fn+'[P],y E P and y < x. We have to ] thus x E f [ P ] . Ftom show that y E fn+'[p].We have x E f [ f n [ P ]and y < x we obtain y E f [PI and then f  l ( y ) < f  ' ( x ) E fn[P].Since f [PI is an initial segment of P we also have f ( y ) E f [.PI. Finally Y = f ( f  l ( y ) ) E f [fn[pll= fn+l[p1* Now ( 1 ) and ( 2 ) yield that ( f n [ P ] l nE w) is a strictly decreasing sequence of initial segments of P, and this contradicts 2.3, 5).
'
The next two theorems are listed as corollaries in [179]: 3.3 Theorem. If P and Q are pwosets, and each is isomorphic to an initial segment of the other, then P and Q are isomorphic. And therefore the relation 5 of Definition 3.1 is also antisymmetric.
Proof. We have an isomorphism f : P + Iq,where Iqis an initial segment of Q, and an isomorphism g : Q + Ip,where Ipis an initial segment of P. Then g[f [PI] is a subset of Ipand moreover an initial segment of this set, and then also of P. Since it is isomorphic to P it is = P by 3.2. So we obtain g [ f [ P ] ]= P. This entails f [ P ] = Q, for otherwise g[f [PI]would be a proper subset of P. 3.4 Theorem. If A and B are distinct initial segments o j a pwoset P and A is isomorphic to B, then A and B are incomparable sets. Proof. If A would be a subset of B, A would also be an initial segment of B. Then B would be isomorphic to its initial segment A and hence = A by 3.2, contradicting A # B.
Instead of the relation 5, which was introduced in 3.1 one could also have a look at the following relation 5 : If a and T are partial ordinals If S and T are posets with types a resp. T, then S is we put a 5 T isomorphic to a subset of T. Then 5 is reflexive and transitive and thus a quasiorder. But 5 is not antisymmetric, as the following example shows: 3.5 Example. We consider the following two pwosets A and B, whose diagrams are sketched in Figure 16, where an element x of A resp. B is an element y of the same set, if x is situated strictly left of y:
<
Figure 16
Each of these two sets is embeddable in the other, but they are not isomorphic. A counterexample with finite sets is, of course, not possible because an isomorphism of a finite set into another finite set and an isomorphism of this into the first set is only possible if both sets have the same cardinality. The counterexample exhibits a difference between the ordinals and the partial ordinals because for ordinals there holds: If a and T are
8.4. THE THEOREM OF DE JONGH AND PARIKH ordinals, and if a is isomorphic to a subset of to a subset of a,then a = T .
T,
and if
T
253 is isomorphic
The ordered sum of wellordered sets over a wellordered index set is again a wellordered set. This fact can be generalized to: 3.6 Theorem. Let (I,5 ) be a pwoset and (Pi,Li), i E I, pairwise disjoint pwosets. Then the ordered sum (see 4.1.1) (C, <) := CiEI(Pi, li)is PWO
Proof. Let (a,ln E w ) be an w  sequence of elements of C. If the set A := {a, In E w ) intersects infinitely many of the sets Pi, then there exists an infinite subset B := {&In E w ) of A, such that every two of the b's are in different sets Pi. If then bm E Pi, bn E Pj, then bm 5 bn holds iff i 5 j, and so the order of C, restricted to B, is pwo like that of I, and so there exist two indices k < 1 with bk bl. If A intersects only finitely many of the sets Pi: then one of them, say Pj, contains a, for infinitely many n E w . And since Pj is pwo, there exist again indices k < 1 with ak 5j (and then also 5) al.
<
8.4
The theorem of De Jongh and Parikh
In 2.6 we saw that every linear extension of a pwoset is wellordered, so that its order type is an ordinal. The question arises whether the supremum of the set of the ordinals which occur as such types is also an order type of a linear extension of the pwoset. The theorem of de Jongh and Parikh [21] gives an affirmative answer to this question. We mention first a theorem which also has interest in itselE
so)
4.1 Theorem [21]. Let (Y, be a poset, Z Y equipped with the be an order on Z which satisfies restriction
r
so
Proof. It is trivial that T is reflexive and transitive. We prove its antisymmetry: Suppose indirectly that a, b are different elements of Y for which aTb and bTa holds. From aTb follows the existence of a sequence a = alTaaT. . .Tan = b, where n is an integer 2 2 and where every sign T is to be replaced by one of the signs
(*) a = alTa2T. . .Tan = bTa,+lT. . Tam = a, where every T is to be replaced by one of the signs
r
Proof. By 2.3.2 the transitive hull of extended to a linear order of Y by 2.3.2.
5 ULz is an order and can be
We now define several concepts which are essential for the proof of the theorem of de Jongh and Parikh: 4.3 Definition. Let (X, 5) be a pwoset. We define o(X, S ) , in short o(X), as the supremum of all ordinals T, such that (X, 5 ) has a linear extension (which is wellordered by 2.6) of type T. For x E X we put L(x) := {y E Xly < x or y 11 x} (=: (X x)) and l(x) := o(L(x)) = sup{~lL(x)(with the order induced from 5 ) has a linear extension of type T}. An element x E X is said to be superfEuous, if o(L(x)) = o(X). Later it turns out that superfluous elements don't exist in pwo's, but for the moment we have to take into account the possibility that they do exist.
8
In this context we have the following: 4.4 Lemma 1211. Let (X, 5 ) be a pwoset. Then there exists a subset Y C X with o(Y) = o(X), which has no superfluous elements.
8.4. T H E THEOREM OF DE JONGH AND PARIKH
255
Proof. We define inductively sets Xo,XI, . . . and elements xo, X I , . . . E X as follows: Xo := X. If X has a superfluous element xo, so that xo) and have o(Xo) = o(Xo xo) holds, then we put X1 := (Xo o(X1) = 4x0). If in general we have already defined Xn 2 X for an n E w such that o(X,) = o(X), and if Xn still has superfluous elements, we choose one of them, say x,, and put Xn+l := (Xn x,). Then o(Xn+l) = o(Xn x,) = o(X,) = o(X). This construction stops after finitely many steps. For we have for the sequence xo, X I , . . . the relation: i < j xi $ xj, because all elements of Xi+l, and a fortiori all of Xj are xi. According to 2.3, 3) then the construction has to break off, say at Xm =: Y, which now has no superfluous elements, but satisfies o(Y) = 0(Xm)= o ( X ~  = ~ ). . . = o(X0) = o(X).
2
2
2
2
2
4.5 Lemma. Let (P, <) be a poset, T an ordinal, and I,, v < T, initial segments of P with ~ { I , l v< T) = P, where for p < v < T we have IpE I,. For v < T we put By := I,\ U {IpIp < v ) and call it the v  block. Then we define an extension <*of as follows: y For x,y E P we put x x y, or x 6 and for the blocks Bt and B, which contain J resp. q we have J < q. Then is an order relation on P.
<*
<
<
<*
Proof. By construction we have (1) If x E P is in BE,then J is the least ordinal with x E It.
(2) If x E Bg, y E Bq and x
< y, then J < q.
Indeed, the initial segment I, contains y and then also x , and so (2) follows from (1). Also there holds: (3) I f x ~ B ~ , y ~ B , a n d x < t*hye ,n J < q .
< <
For if also x y holds we are done due to (2). And if x 6 y, then by definition J q follows. In the rest of the proof we always suppose x E Bg and y E B,. By definition <*is reflexive. y and y x. By (3) we have J <*is antisymmetric: Let x q 5 J and thus = q. Then x and y are in the same block, and we have x
<*
<*
<*
< <
<
<
<*
<*
<
< <
<*
<
Now we can prove the main theorem of this section: 4.6 Theorem of de Jongh and Parikh [21]. Let (X, 5 ) be a pwoset. Then there exists a wellordering <'of X extending such that tp(X, 5') = o(X, 5). Proof. We use the definitions of 4.3. First we can assume that X has no superfluous elements. For otherwise we can find by 4.4 a subset Y C X with o(Y) = o(X), which has no superfluous elements. If then our theorem is proved for Y, we have a wellordering Lyof Y extending 51 Y with type o(Y) = o(X), and this can by 4.2 be extended to a linear order l x o f X,which by 2.6 is automatically a wellordering. This has a type 2 o(X). This type is trivially also o(X) and then = o(X). And this shows that it suffices to prove: There exists a subset S 2 X with a wellordering lswhich extends 5 S and has type o(X). If o(X) is a successor ordinal the assertion is trivial, since here supremum is the same as maximum. So let X := o(X) be a limit ordinal and W7 := cf(X). From the representation of X in Cantor's normal form we obtain X = p wa, where p is a sum of finitely many powers wV which all are 2 w", and where a is > 0 since X is a limit ordinal. First we show that we can restrict ourselves to the case p = 0 : To this purpose we choose an element z of X with l(x) > p. This is possible since evidently o(X) = sup{l(x)lx E X}. Since z is not superfluous we have l(x) < X = p w", and so we have 1(x) < p for a < w". For z we obtain: (1) o(X 2) w". Suppose the contrary. Then o(X 2 x) < q holds for an q < w". W is a Every subset W of X with a wellordering which extends union of a subset of L(x) and a subset of ( X x), and so its order type is a mixed sum of an ordinal 5 l(x) < p+[ and an ordinal o(X x) < q. And so its order type is by 4.8.11 and 4.8.13 ( p + < ) @ q5 p @ < @ q= p @ (< @ q) < p w", since @ q < W" holds, and we have a contradiction to o(X) = p w". So it remains to prove that ( X 2 x) has a subset T which can be wellordered in extension of 5 f T such that it has the type wa, since a wellordered subset of L(x) of type p,which extends 5, can then be prolonged by T to furnish a wellordered subset of X , extending 5, of type p wa = o(X).
<
<
r
+
+<
+ > >
+
+
+
<
> <
<
<
>
8.4. THE THEOREM OF DE JONGH AND PARlKH
257
Due to ( 1 ) we can from now on assume X = wa. Let (r,lv < w,) be a strictly ascending sequence of ordinals whose supremum is o ( X ) = X = w". For v < w, let u, be an element of X with l(u,) 7,. Then by 2.4 the sequence (u,Iv < w,) has a subsequence (t,lv < w,) such that p < v < w, =+t p 5 t, holds. In particular then {t,lv < w,) is a chain of X, and sup{l(t,)Iu < w,) = X = w". Since the t, are not superfluous all l(t,), v < w,, are < X = w", and their set is cofinal in w". We finally choose a subsequence ( x v l v < w,) of (t,lv < w7), for which we have:
>
( 2 ) p < v < w, w,) = w" = A .
*xP <
XU
and l ( x p ) < 1(x,), and sup{l(x,)lv
<
To this purpose let E be the set of all those t, (with u < w, and) for which I ( & ) > l ( t p ) holds for all p < v. Their set E , ordered by magnitude, yields a sequence (x,lu < w,) which satisfies ( 2 ) . We distinguish two cases:
+
Case a) a is a successor number p 1, so that w" = wp w. Then cf X = w , because the set {wp nln E w ) has X = w" as supremum. Using ( 2 ) we choose for every i E w an element yi E {xvlv < w,) := F and a number ni E w such that for all i E w there holds: First we choose no = 0, then a suitable yo in J , then nl so that ( 3 ) is fulfilled for i = 0, which is possible since X has no superfluous elements.Then yl E F , n2, y2 E F, n s , . . . are determined successively. From ( 3 ) we now obtain for every i E w the existence of a subset Wi 2 L ( y i ) with a wellordering extending 5 Wi of type wp . ni. We put := L ( Y ~ + ~ ) \ L (for Y ~i) E w. Then contains all elements of X, which are yi and yi+l. And for i < j < w we have yi < yj since the y's are in the chain F and due to ( 3 ) . We have Wi+1 n # 8. Otherwise Wi+l c L ( y i ) would hold because of Wi+1 2 L ( Y ~ + ~ But ) . Wi+i has type . ni+l > l ( y i ) by ( 3 ) , a contradiction. So for i E w we can choose an element pi+l E Wi+1n for which now also pi+l 1: yi holds. Finally for i E w we and put Fi+1 := (Wi+1> pi+l) This final segment of Wi+1 is C has an order type 2 wB by 4.8.12. Now we consider the ordered sum S := C{Fi+1 li E w ) . The summands are pairwise disjoint, the L ( y i ) are initial segments of X, and so S is by 4.5 a wellordered set of type
x+i
>
2
r x+l
x+l
> wp . w
= wa, whose order is an extension of
S. Thus case a) is
proved. Case b) a is a limit ordinal. Then we choose successively for every i E wy an element yi E X and an ordinal wVi such that for all i E wythere holds: (4) wvi < I(yi) < wVi+l 2 Ti. Similarly to case a) we can determine successively by transfinite induction elements yi € {xVIv < wy) and ordinals vi, beginning with vo = 0, yo, y,yl, ... , which satisfy (4). Due to (4), there exists for every i < wy a subset Wi C L(yi) with a wellordering on Wi of type wVi which extends 5 Wi. And we have: (5) For i < wy there holds Fi+1:= Wi+1\L(yi) = Wi+l\(Wi+l n L(yi)) has type wvi+l. And
r
(6) &+I C L(yi+l)\L(yi). Indeed, we have tpWi+1 = wVi+l and tp(Wi+1n L(yi)) 5 I (yi) < wvi+l , and then (5) follows from 4.8.7. Now, using 4.5 we see that the on this ordered sum C{Fi+lli< wy) is a wellordered extension of set. By (6) its summands are pairwise disjoint and thus the sum has the type ~ { w ~ + + < l l vw7) = wa = o(X).
<
4.7 Remark. We mentioned before that superfluous elentents don't exist. Now, where we can use the last theorem, this easily follows: If x would be a superfluous element of X, then there exists a set W C L(x) with a wellordering <,which extends 5 r W of type o(X). Then we can in the set W U {x) which extends 5 on this set introduce an order and <,,, and in which x is the last element. This set would have the type o(X) 1, which is impossible.
<,
+
8.5
On the structure of J(P),where P is wellfounded or pwo
In many cases it gives useful insight into the structure of an ordered set if we consider the set of its initial segments with its natural order by inclusion. For this we use the following notation: 5.1 Definition. Let (P,5 ) be a poset. Then the set of all initial segments of P is denoted by J(P).And this set is always considered as equipped with the order by inclusion.
8.5. ON THE STRUCTURE OF J(P)
259
Of course there are intensive connections between the order of P and the order of J ( P ) , and in general every time, in introducing a new concept, the question arises whether the properties of P are shared by J ( P ) . We investigate this now for the notions wellfounded and pwo. First we mention a version of Konig's Infinity Lemma: 5.2 Theorem. Let Si, i E w, be pairwise disjoint nonempty finite sets and R a relation on U := u{Sili E w), for which the following holds: For every i E w every element of Si+1is i n relation R to at least one element of Si. Then there exist elements ai E Si for i E w such that ai+1R ai holds for all i E W.
Proof. For every n E w and x, E Sn there exists a set {xo,. . . ,x,) holds x ~ for i < n. We with xi E Si for i = 0,. . . , n such that x ~ + ~ R call this set a path from s o to x,. Since So is finite it follows from the pigeonhole principle that the set of all paths from elements of So to elements which are not in So has an infinite subset of paths which all share the same element, say ao,of So.Then we consider all paths from a0 to elements which are not in So.Infinitely many of them share the same element, say a1,of S1.Continuing in this way we can successively find elements ai E Si for i E w such that ai+1R ai holds for all i E w. An initial segment I of a poset P is called finitely generated, if there exists a finite subset F I which is cofinal in I. Now there follows a theorem of Birkhoff [9]: 5.3 Theorem. Let (W, 5 ) be a wellfounded poset. Then the set 5 of its finitely generated initial segments, equipped with the order 2 , is wellfounded.
Proof. Suppose indirectly that (Inln E w) is a strictly descending sequence of elements of 5. For n E w let Mn be a subset of Inof minimal size which generates In.Then Mn is finite, and every element of Mn+l is 5 an element of Mn. Next we intend to reduce the system of the Mn to a sequence of pairwise disjoint sets. We call a sequence So,S1,. . . of subsets of W special, if the following holds: (*) All Sn,n E w , are finite and nonempty, and every x,+l E Sn+1 is 5 an element xn E Sn. So (Mnln E w) is special. If an element e E Mo is contained in infinitely many of the Mn, we consider the subsequence (Ao = Mo),A1, A 2 , .. . of all those sets Mn which contain e. Then we
delete e from all sets A, and obtain a sequence go(= Mo\{e}), B1, B2,. . . , which is still special. Indeed, for every b,+l E B,+l there exists a bn E Bn with bn+1 5 bn. Indeed, bn+1 is an element x E A,, and x is # e, because otherwise b,+l < e E A,+l holds, and then An+1 would not be minimal, for b,+l could be omitted since all elements which are < b,+l are also < e. Continuing in this manner we can eliminate in finitely many steps all those elements of Mo which occur in infinitely many of the Mn with n > 0. And in every step the new sequence remains special. So we obtain a special sequence Co,Cl, . . . , which begins with a subset Co C MO,such that every element of Co occurs only in finitely many of the C,. Then (Cnln E w) has a subsequence (D,ln E w) with Do = Co, which is special, and where Do is disjoint to all D, with n > 0. Iterating this process we finally arrive at a sequence (E,ln E w) which is special, and where all sets En are pairwise disjoint. But now by Konig's Infinity Lemma 5.2 there exist elements a, E En with a,+l < a, for n E w. They form an infinite decreasing chain in W, a contradiction.
<
5.4 Remark. In in 5.3 we replace the set 5 by the set J(W) of all initial segments, the statement no longer remains true. Let e.g. Q be the set of rational numbers with the identity relation {(x, x)lx E Q) as order, then (Q, =) is wellfounded, but the set of its initial segments contains all subsets of Q, and in this we have infinite decreasing sequences. With a stronger assumption one can prove that J(W) is wellfounded. In this context Higman [84] proved:
5.5 Theorem. A poset P is pwo if3(P) is wellfounded.
Proof. Suppose that P is pwo. We assume indirectly that 3(P) is not wellfounded. Then there exists a sequence of initial segments (Inln E w) of P which is strictly descending. Now we choose for n E w an element x, E In\In+1. Then x, is also not in I, for m > n, and this entails x, 6 x, because otherwise x, had to be in I,. But now P could not be pwo because of 2.3, 3). Suppose that P is not pwo. Then there exists a strictly descending infinite sequence of elements of P, or P has an infinite antichain. If (%,In E w) is a strictly descending sequence of P, then the sets (P < x,), n E w, form a strictly descending sequence of elements of J(P),which is thus not wellfounded.
8.5. ON THE STRUCTURE OF J ( P )
261
And if {x,ln E w) is an infinite antichain, we define Into be the set of all elements of P, that are 5 some element of {x,, %,+I, xn+2, . . .). Then the initial segments In,n E w, form a strictly descending sequence of J(P).For m < n < w entails, that x, is not 5 an element of {xn,xn+l,. . .) so that I, > In holds. And again J(P) cannot be wellfounded. The property of an ordered set P to be pwo does not entail that also J ( P ) is pwo. The standard counterexample was given by R.Rado [I491 and independently by Kruskal (unpublished): 5.6 Example. Let P be the set w x w. We introduce in P the following relation S p : For every x E w and y 5 z of w we put (x, y) Sp (x,z), and for every (x,y) E P we put (u,v)
+
p
<
+
<
+
+ <
component sum 5 x. Finally every antichain of P is finite. If the different points (al, bl), (a2,b2),. . . would form an infinite antichain, where we can assume w.r.0.g. that a1 < a:! <  . . holds, then we would have a1 bl > a, for every integer n 2 2. But then the infinitely many a, could not all be different.
+
The next theorem presents a situation where pwo is carried over from a set X to J ( X ) :
5.7 Theorem (NashWilliams [123]). Let k be a positive integer and W,, 6 = 1,.. . ,IC, wellordered sets, X := Wl x . x Wk equipped with the product order. Then its set 3(X) (of .initial segments) is pwo.
Proof. Every initial segment of X has evidently the form Il x .  x Ik, . . . ,Ink), where I, is an initial segment of W, for 6 = 1,. . . , k. Let (In1, n E w, be a sequence of elements of 3(X). Since J(Wl) is pwo there exists by 2.3, 2) an infinite subset Nl 2 w such that the sets Inl,n E Nl, increase with n. Analogously there exists an infinite subset N2 Nl n E N2, increase with n. In k steps we obtain such that the sets In2, . . ,Ink), n E Nk, an infinite subset Nk 2 w such that the ktuples (In1,. increase with n. By 2.3, 2) our statement follows. 5.8 Remark. In [I791 Wolk remarked that the last theorem shows that the set P of the counterexample of Kruskal/Rado of 5.6 has infinite dimension. Otherwise P would by 2.12 be embeddable in a product of finitely many wellordered sets, and then J(P) would be pwo, with contradiction.
8.6
Sequences in wqosets
If we have two w  sequences consisting of elements of a poset P, there exist several possibilities to introduce a quasiorder between these sequences which stand in relation to the order of P. In section 4.1 we already studied the ordered product and the lexicographic product of orders. Here we investigate another relation, which, contrary to the before mentioned notions, is only a quasiorder. 6.1 Definition. Let (Q, 5 ) be a quasiordered set. If a is an ordinal and a,, u < a, are elements of Q (not necessarily pairwise different), we call A := (avlu < a) an a  sequence over Q. Its length is a. If a
8.6. SEQUENCES IN WQOSETS
263
is a finite ordinal, this sequence, of course, is called finite. Here a = 0 is admitted and describes the empty sequence. If P < a holds, the sequence (a,lv < p) is called the P  segment of (avlv < a ) and a, the v  component of it. A subsequence of A is a family (apl,u E B), where B 2 a holds. If now A = (a, lv < a ) and B = (b,lv < p) are sequences over P (with possibly different lengths a and P), we put A 5 B e there exists a <  preserving mapping g of a into @ such that a, 5 bg(,) holds for all v < a. Intuitively speaking: A 5 B holds, iff B has a subsequence S of the same length as A, such that A is 5 S "comp~nentwise'~.To a certain extent 5 generalizes the notion of product order. It is immediately clear that 5 is reflexive and transitive. But, contrary to the product order, it is not antisymmetric. If e.g. A and B are w  sequences of 2((wo))which both have infinitely many 0's and l's, then we have A 5 B and B 5 A. If W is a finite sequence over a quasiordered set Q, it is in short called a word over Q. 6.2 Theorem. Let P be a wellfounded quasiordered set. Then the set m ( P ) of all words over P is wellfounded.
Proof. Let S be a nonempty set of nonempty words of m ( P ) , S' the subset of those words of S which have minimal length 1. The set of the 0  components of the words of S'is 2 P and thus wellfounded, and so it has a minimal element mo. Let Mo be the set of all words of S' which have mo as 0  component. Then the set of 1  components (if 1 2 2) of the words of Mo has a minimal element ml. Let MI be the subset of those elements of Mo which have ml as 1  component, that means which begin with mo,ml. After 1 steps we have obtained a word (mot.. . ,mll), which is minimal in S, and so m ( P ) is wellfounded. by 1.2. An important theorem of Higman [84] is now: 6.3 Theorem. Let Q be a wqcset. Then the set m(Q) of all words over Q is wqo.
Proof (by NashWilliams [123]) Suppose that m(Q) is not wqo, so that the set B of bad sequences in !D(Q) is nonempty. Since m(Q) is wellfounded by 6.2 the set of 0  components of the sequences of B has a minimal word Wo. Let now B(Wo) be the set of all sequences (Wo,. . .)
of B which begin with Wo. Then the set of all 1  components of the elements of B(Wo) has a minimal element Wl. Let then B(Wo,Wl) be the set of all sequences of B (Wo) that have Wl in the 1 component. By induction we so obtain an w  sequence Wo, Wl,.. . of words, for which by construction there follows: (1) Wo, Wl, . . . is a bad sequence. Indeed, for i < j < w there exists a sequence S in B whose icomponent is Wi and whose jcomponent is Wj, so that Wi 5 Wj cannot hold since S is bad. All words Wn, n E w, are nonempty, otherwise their sequence would be good. So for all n E w the 0  component w, of the word W, exists. And then the w  sequence wo, wl, . . . (of elements of Q) has an increasing subsequence by 1.3, 2), and so we have: (2) There exists a strictly increasing mapping f : w + w such that In E w) increases with n.
(W ),(
We define for n E w word Xf),( which arises from Wf),( by omitting from Wf),( its 0  component wf(,). Finally there follows: (3) The sequence w01WI,.   1 Wf(0)1, Xf(0), X f ( l ) l Xf(2), . . . is good. For Xf(o) is strictly shorter, and then also strictly less than Wf(0). But this was a minimal word among all words which prolong Wo,..  ,W~(O)I to a bad sequence. Now (3) leads to a contradiction. For (3) implies the existence of two words A, B in the sequence of (3), where A is left from B, but A 5 B. Then A and B cannot be both in the segment Wo, . . . , Wf (o)l because of (1). Also they cannot be both in the segment Xf (o), Xf (11,. . . . For then also Wf (01, Wf (11,. . . would be good, contradicting (1). If finally A would be in the segment Wo, . . . ,Wf (o)l and B in the segment Xf(o),Xf(l),... , say B = Xf(m) for an m E w, then a fortiori A 5 Wf (,) would hold, contradicting (1). By the way, in [84] a problem of Erdijs [39] is solved.
6.4 Definition. A word ( X I , . . be of sum s, if x1 . . x, = s.
+ +
. ,x,)
over N is (of course) said to
An application of Higman's theorem yields immediately with 2.3, 2) the following theorem, which will be used later:
8.6. SEQUENCES IN WQOSETS
265
6.5 Theorem. W e consider the set N of natural numbers with its natural order. For n E N let w, = (xnl,. . . ,x,,~(,)) be a word over N with sum n, this implies l(n) 5 n. W e equip the set {wnln E N) with the order for words according to 6.1. Then there exists an infinite subset N* of N such that the set {wnln E N*) is strictly increasing with n. The question arises whether there is a version of 6.5 for the finite case, where instead of N proper initial segments of N are considered. Indeed, one can obtain a finitistic version of 6.5 by applying a compactness argument. There holds: 6.6 Theorem. For every m E N there exists a natural number k(m) such that the following holds: If for every natural number n 5 k(m) a word w, = (xnl,. . . ,x,,~(,)) over N of sum n is given, then there exists a strictly increasing set of size m of words W i l 5 . 5 wi,.
Proof. Suppose indirectly that there exists a natural number m, for which no k(m) exists which satisfies the assertion.Then for every n E N there exists a mapping fn which ascribes to every natural number x 5 n a word w, over N of sum x, such that there is no melement subset of (1,. . . ,n), over which f , is strictly increasing. Of course f,(l) is the word wl = 1 for all n E Nl := N. From the pigeonhole principle it follows that there is an infinite subset N2 E Nl, such that all words fn(2), n E N2, are equal, say = 202. Analogously there exists an infinite subset N3 C N2 such that for n E N3 all words fn(3) are equal = w3 and so on. Then we define f by f (n) := w, for n E N. Now wl, w2,. . . has no increasing subset of length m, since this would also be an increasing subset with respect to a function fn, which by construction is impossible. The last proof is a pure existence proof. It does not contain any information about the growth of the function k(m). In the paper [62] a constructive proof for the existence of k(m) is given. Finally we mention a theorem of Laver [Ill]. Its proof is long and complicated, and so we omit it here:
6.7 Theorem. There is no infinite set of pairwise incomparable order types of scattered linearly ordered sets. And there is no infinite strictly descending sequence TO > TI > ... > Tn > . . . , n E N, of order types of scattered linearly ordered sets. In short: Every set of scattered linear order types is wellquasiordered.
8.7
Trees
Now we wish to study trees in more detail. In particular we investigate the relationship between set splittings and trees. Since every principal ideal of a tree is wellordered, a tree is wellfounded. But, contrary to the situation of a pwoset, a tree can have infinite antichains. E.g. a poset P that has a first element x, which has infinitely many upper neighbors, which are pairwise incomparable, and which has no other elements, is a tree. On the other hand, a pwoset need not be a tree; every finite poset is pwo. In 1.7.1 we defined the concept "height" for elements of finite posets P, and in 1.3 we extended this notion to elements of wellfounded posets. So also in a tree every element has a welldefined height, and the tree itself has a height. Also the concepts branch (resp. path, resp. Pstump) of a tree are now special cases of the general definition in 1.3. A branch of a tree is a maximal chain and thus it contains with an element x also all predecessors of x. In this context we have a simple fact:
7.1 Theorem and Definition. Let B = {bUlv< T) be a branch of a tree T , where for p < v < T we have b, < b,. Then b, has the height v and belongs to the vth level L, of T.
< <
For x E T the set of elements y E T with y x is a path, which we call the path ending i n x. It has the form {x,lv h(x)) where x, E L,. W e call xu the v  predecessor of x. (So x is its h(x)  predecessor.)
Proof. For v = 0 the assertion holds since bo must be the least element of T. Let p be an ordinal < T, such that the assertion holds for all v < p. Then bp is minimal in the set of those elements of T which are > bu for all v < p because B is a maximal chain of T . Therefore b, E L, holds, and our theorem is proved by induction. If a set B = {b,lv < T)< is a branch of a set W which is only wellfounded, we have no such conclusion: If W is the pentagon (see Figure 18) then {bo, bl, b2) is a branch of W , but ba has height 3.
8.7. TREES
Figure 18 bo
This also shows that in a wellfounded set an element x can have more than one path ending in x, which in trees is not possible. The concept "tree" is closely related with the concept "setsplitting", which was defined in 6.1.7, and this is one of the main reasons for the importance of trees. Now we consider this relationship in detail. To this purpose we first define:
7.2 Definition. A branching point of a tree is an element which has at least two immediate successors. We call a tree T a splitting tree if the following two conditions are satisfied: 1) Every element of T which is not maximal is a branching point. 2) If X is a limit ordinal and {b, lv < A)< a chain of T, where b, has height h(bu) = v for v < A, and if the set A of elements of T that are > b, for all v < X is nonempty, then A has a least element. A tree T which satisfies 2) is called semichaincomplete. So, a tree is semichaincomplete, iff every nonempty bounded chain of T has a supremum. (It is not required that every nonempty chain of T has a supremum.) If T is a tree, which is not necessarily a splitting tree, then the set A which is defined in 2) can have many elements which are minimal in
A. Now the equivalence between setsplittings and splitting trees can be described:
7.3 Remark. Let U be a splitting (see 6.1.7) of a nonempty set S. In U we have the order by inclusion. But in order to establish the analogy between splitting trees and setsplittings we now put for blocks B1, B2 of U: B1 5 B2 ++ B1 1Bz.
Then (U, <) is a tree, and S is its least element (in the order 2 ) . Indeed, let Bv E U be a block of height v. (For the concept "height" of a block see Definition 6.1.10.). Then for every p < v there exists exactly one block Bp of height p with Bp > Bv. And so the set {BpIp < v) of all predecessors (with respect to 5)of Bv is wellordered. Thus (U, <) is a tree. Since U satisfies 2) of 6.1.7 also 1) of 7.2 is fulfilled. And since U satisfies 3) of 6.1.7 also 2) of 7.2 holds. Thus (U, 5) is a splitting tree. On the other hand, let T be a splitting tree. Then we define U to be the set of all principal final segments [a), a E T, of T. If B E U, say B = [a) for an a E T, we define the set Z ( B ) (corresponding to 6.1.7, 2)) as the set of all segments [b), where the elements b are immediate successors of a. And then the properties 1) and 2) of Definition 6.1.7 are fulfilled. By 2) of 7.2 U contains also all nonempty intersections of final segments [a), so that 3) of 6.1.7 is satisfied. In 6.1.10 we defined the concept "height" for the blocks of a setsplitting U. If this is considered as a tree with order , then the elements of this tree also have a height as defined in 1.3. Now these heights coincide, so that the corresponding definitions are compatible.
<
Next we wish to study linear extensions of trees. We begin with a general concept, which is a weakening of the notion positionequivalent of Definition 7.3.8:
7.4 Definition. Let P be a poset. For elements a,b of P we put a b +=+ (P < a) = (P < b), (in words: a and b have the same set of predecessors). N
is an equivalence relation. And the It follows immediately that set of minimal elements of P is an equivalence class  for all x of it we have (P < x) = 0. Equivalent elements of P are incomparable, and so every equivalence class is an antichain of P. By 1.4. f) equivalent elements of a wellfounded poset, in particular of a tree, have the sam.e height. If we choose a linear order in every equivalence class of a tree we have a natural method to extend the order of the tree to a linear one: N
7.5 Definition and Theorem. Let (T, 5 ) be a tree, and suppose that in every equivalence class K of T a linear order S K is given. Then we define a standard linear extension <*of 5 in the following way: Let x and y be in T. If already x 5 y holds we put also x 5*y.
8.7. TREES
269
<
If
and y are not comparable in we consider the paths xu,. . . ,xc = x)< and {yo,. . . ,y,, . . . ,Y7) = y)< which end in x (50, . resp. so that xu is in the level L, for v 5, and analogously for y,. Due to the fact that x and y are not comparable in <, there exists a least index v for which xu # y, holds. Indeed, if such a v would not exist, one of the paths, ending in x (resp. y) would be an initial segment of the other, and this would entail x y or y x. So, in the case where x and y are not comparable in <, we consider the first index v where the above paths differ. Then xu and y, are in the same equivalence class K, and we put x <* y xu
<
<
<
<*
<
<
<
7.6 Remark. It is by no means so that every linear extension of the order of a tree is a standard linear extension of the above kind. To see this we can e.g. use 2.3.3, in which the following was proved: If A is an antichain of a poset P, and if we introduce an arbitrary linear order
The linear order of 7.5 preserves some features of the tree T . First we mention:
7.7 Lemma. Let (T, <) be a tree, <*a standard linear extension of 7.5. For s E T let [ s ) be the final segment of (T,5 ) containing all t E T with s t. Let a and b be elements of T which are in the same equivalence class K of T and which satisfy a < K b. Then for every x E [a) cnd every y E [b) there holds x <* y .
< in the sense of Definition
<
Proof. We consider the paths of x and y. The first ordinal v for which x, # y, holds, is the height of a and b, and so x, = a and y, = b. By definition we have x <* y a < K b, and this implies our assertion. Now there follows an important relation between the order of a tree and the linear order of a standard linear extension of it:
7.8 Theorem. Let a be an element of a tree (T,5 ) and <*a standard linear extension of 5. Then the final segment [a) of a i n (T, 5 ) is a segment i n (T,<*) which has a as first element.
Proof. All elements b of T which are > a, are also >* a by definition, and so [a) 2 ( T >* a) holds. Let now x and y be elements of [a) with x <* y.We have to show that every element x which satisfies x <* z <* y, also fulfills z a. Suppose indirectly that x 2 a is false. Then there exists a least index K such that for the K  predecessors of a and x we have a, # x,. The elements x and y are (2 a) 2 a, and therefore in [a,). And z 2 x, is of course in [z,). If a, < K z, holds where K is the class containing a, and z,, then by 7.7 every element of [a,) is <*every element of [z,),and we would obtain that x and y are both <* z, a contradiction. If x, < K a, holds we conclude analogously that z is <* x and <* y, which again is a contradiction.
>
Finally we consider an obvious method to embed a tree into a semichaincomplete tree:
7.9 Theorem. Let T be a tree with levels L,, v < h(T). If X is a limit ordinal and B := {bVlv < A)< a chain of T with bv E L, for v < A, which has upper bounds, but no minimal upper bound in T , we introduce a new element e and put e >* bv for v < A, and e < *x for all upper bounds x of B, and e is incomparable with all other elements of T. If a and b are new ekments we put a <* b, i f there exists a x E T
8.7. TREES
271
<*
<
with a <* z <* b. For a,b E T we of course put a b a b. Then the set T* := T U E, where E is the set of new elements, is a semichaincomplete tree with respect to the order I*. The height of a new element i n the tree T*is a limit ordinal.
Proof. First we have to show that the relation <*is an order relation. If a,b,c are in T* and a <* b <* c, we have to prove a <* c. For this purpose one has to consider eight cases, depending on the situation, whether a,b,c are in T or E. Each of these is trivially verified. If a,b are in T*, we cannot have a <* b and b <* a. If from a and b at least one is in T , this is trivial. So let a and b be both in E, and suppose that a <* b and b <* a hold. Then, by definition, we must have elements y and z in T satisfying a <* y <* b and b <* z <* a. This entails y < z and z < y with contradiction. So <*is antisymmetric. We still have to prove that T* is a tree. If a new element e is introduced in th.e tree T , we first see that Te := T U {e) is also a tree, in which (Te < e ) = (T < e ) holds by construction, so that it is wellordered. And e has in Te as height a limit ordinal, namely the height of (T < e). The successors of e in T*, which have a height h(T < e ) n with n E w in T , obtain in Te the height h(T < e) n 1. All other heights remain unchanged. Let {A,, Y < K ) , be the set of limit ordinals < h(T) with Ap < A, for p < Y < h(T). Then also the set T, augmented by the new elements b, for which ( T < b) has type w = Ao, is a tree TIAo]. Successively we add the new elements b, for which TIAo] < b has type XI, and so on. With transfinite induction we so obtain an ascending sequence of trees, whose union is again a tree, and = T*. By construction it is clear that T* is semichaincomplete, and that the new elements have as height in T* limit ordinals.
+ +
+
Comparing the size of chains and antichains of T resp. T* we have the following statement:
7.10 Theorem. Let (T, <) be a tree and (T*,I*) as defined i n 7.9. Then there holds: a) Every chain of (T*,<*) has a length which is equal to the length of a chain of (T, I ) , or such a length +l. b) Every antichain of (T*,I*) has a cardinality which is also the cardinality of an antichain of (T, 5). Proof. a) is an easy consequence of the fact that, in a maximal
chain {a,lv < T ) < of T*, only elements a,, where v is a limit ordinal, can belong to T*\T. b) Let B be an antichain of T*\T, so B contains only new elements. For each b E B we choose an immediate successor (with respect to <*) b' E T of b. Then we have: (I) B' := {b'lb E B ) is an antichain of cardinality IBI. For if bl, b2 E B we cannot have bi < bi because this would entail bl <* bi < b; and thus bl <* bi. Now b2 is an immediate predecessor of b;, and T* < bb is linearly ordered. So bl 5 b2 would follow with contradiction to the fact that B is an antichain. Now every antichain of T* has the form A U B, where A E T and B T*\T. Each element a of A is incomparable in 5 with every element b, and a > b' b'of B'. Indeed, a < b' would entail (a <* b' and) a would imply a > b. Each of these is impossible in the antichain A U B. So A U B' is an antichain of T with the same cardinality as A U B.
<*
8.8
Aronszajn trees and Specker chains
A description of properties of a tree suggests considering the cardinality of the chains, the antichains and the levels of the tree. A typical result in this area is Konig's Infinity Lemma, which now can be reformulated in the form: 8.1 Theorem. Let T be an infinite tree, all of whose levels are finite. Then there exists an infinite branch. Subsequently to this the question arises if there exist generalizations to higher cardinalities. The first interesting case is then: Given a tree with N1 elements, all of whose levels are countable, does there exist a branch of type wl? Aronszajn found a counterexample, which we present a bit later. (It was communicated by him in a letter to Kurepa.) We mention before some remarks on the set of wellordered subsets of R. 8.2 Example. Let 2D be the set of all wellordered subsets of R , each of them is ordered by the restriction of the natural order of R. The ~ c. set 2D has cardinality 2 N =: 1271 2 c is trivial since 27 contains all oneelement subsets of R. On the other hand, if W E 2D, then W has the form {xvlv < a ) < , where a
8.8. ARONSZAJN TREES AND SPECKER CHAINS
273
is a countable ordinal. And then El is the union ~ { E l ~
<
8.3 Example. A fortiori it follows from 8.2 that the set of all wellordered subsets of the set Q of rational numbers has cardinality c. Its cardinality is also 2 c, for it contains the set of all wellordered subsets of N, which is identical with the set of all subsets of N, and so has cardinality c. Contrary to the situation in N, every ordinal < wl is the type of a wellordered subset of Q. (Later this fact is important in the proof for the existence of an Aronszajn tree.) In this example we define T as the set of all wellordered subsets of Q. If A and B are elements of T we put A 5 B if A is an initial segment of B. It can easily be verified that T with this relation 5 is a tree, since the set of initial segments of a wellordered set is also wellordered. It follows easily that all levels L, with Y 2 w have cardinality 2 c. But in spite of the fact that all levels L,, v < wl, of T are nonempty, there cannot exist a branch of length wl. For if 93 = {BVlv< wl)< would be such a branch, we could choose an element b,+l E Bv+1\Bu for every v < wl, and the set {b,+l lv < wl} would be a wellordered subset of Q of type wl, which is impossible. The Aronszajn tree will be a strict subset of the tree T just defined.
<
8.4 Definition. If K is an infinite initial ordinal, an Aronszajn Ktree is a tree of height K which has no branch of length K, and is such
that each of its levels has cardinality
< 1 ~ 1 . Instead
of Aronszajn wl
tree one also says Aronszajn N1 tree or in short Aronszajn tree. In order to construct an Aronszajn tree we have to modify the construction of the tree of 8.3 so that it satisfies the additional requirement, that each level is countable, but the height remains wl. We recall: 8.5 Definition. If K is an ordinal and X, = {xp(p < K)< a wellordered set, and if v K we call the wellordered set X, = {xplp < v)< the v  segment of X,.
<
8.6 Theorem. There is an Aronszajn wl tree T.
Proof. We construct sets L,, v < wl, successively by induction. These will become the levels of the tree T which we construct. We put Lo := (01, and L1 := {{r)lr E Q). Generally for Y < wl the set L, will
be a set of bounded subsets of the set B3, of all wellordered subsets of Q which have type v. For ordinals v < wl we consider the following property (P,), which will be proved for all v < w l by induction: (P,) The sets L, are already defined for all p 5 v as countable bounded (in Q) subsets of B3, (of 8.2). And further there holds: If p < v and X, E L,, X, E L, is called a prolongation of X,, if for all K with p 5 K 5 v the K segment X, of X, belongs to L,. If now p < v, X, E L,, and if r is a rational > supX,, then there exists a prolongation X, of X, with sup X, = r. It is clear that (P,) holds for v = 0 and v = 1 (recall tp(0) = 0). Let now a be an ordinal < wl such that (P,) holds for all v < a. Then we define L, and prove (P,). We distinguish two cases: (I) If a is a successor number p 1, we put L, := {X U {r)lX E Lp, and r is a rational > sup X). Intuitively speaking: All wellordered sets E Lp, which all have type P, are prolonged by a rational, and this leads to wellordered subsets of Q of type a. Then also (Pa ) holds: For let 0 < p 5 P, X, E L,, r > supXP a rational. If p = p, then X,(= Xp ) E Lp has by construction a prolongation Xp U {r) E L,. Now suppose p < P. Then, due to (Pp), there exists a prolongation Xp E Lp of XcLwith supXp = r' for some r' with supX, < r' < r of Q. Then X, := Xp U {r) E L, is a prolongation of X, with sup X , = r. And thus (Pa)holds. (11) Let now a < w l be a limit number, p < a, X, E L,, r > SUPX, a rational. Then we choose a strictly increasing sequence (avlv < w)< of ordinals whose upper limit is a, and with p < a 0 . Further we choose a strictly increasing sequence (r, lv < W) < of rationals whose upper limit is r, and where ro > sup X, holds. By induction hypothesis we conclude that there exist sets X,, E L,, where X,, is a prolongation of X, with is a prolongation sup X,, = ro, and where for every v E w the set X,,,, = r,+l. of X,, with sup X,,,, Let X, := u{X,, Iv < w}, and let L, be the set of all these X, so obtained. Then L, is countable since u{L,Iv < a) and Q are countable sets. And by construction (Pa)holds. By induction we have obtained that each set L, with a < wl is nonempty and conntable. The set T := U{L,Iv < wl), equipped with the order "is an initial segment of ", is a tree. For it has 0 as least element,
+
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275
and each of its elements is a wellordered set, whose set of predecessors, which are initial segments of it, is wellordered. For this tree T we have: (111) For each v < wl every element of L, has height v. So L, is indeed the v  level of T. And thus T has height wl. For v = 0 this is trivial. If this statement holds for a fixed v < wl, it holds by construction also for v 1. For every element of L,+l arises from an element of L,, which by induction hypothesis has height v, by attaching exactly one element to it. And if A is a limit number, such that the statement holds for all v < A, then the construction of LA ran in such a way that only such elements were included in LA,which had height A. Thus (111) is proved. Finally it follows (as in 8.3) that T has no branch of type wl. If 23 = {B,Iv < wl}< would be such a branch, we could choose an element b,+l E B,+l\B, for every u < wl, and the set {b,+llv < wl} would be a wellordered subset of Q of type w l , which is impossible.
+
Next we shall construct a Specker chain. With this set Specker was able to solve a problem of Erdos/Rado [36]. This chain has interesting applications in the partition calculus of set theory (see e.g. a result of Galvin/S helah [48]). We define: 8.7 Definition. A linearly ordered set (S,5 ) is called a Specker chain, if IS1 2 N1, and if S has no subset of type w l , no subset of type w;, and no subset of cardinality N1 which is embeddable in R. Concerning the possible cardinalities of a Specker chain we have: 8.8 Remark. A Specker chain has cardinality 5 2 N (= ~ IRI). This is already a consequence of the fact that a Specker chain has no subsets of type wl, w; and of the embedding Theorem 3.3.7 of Hausdorff according to which a Specker chain is embeddable in HI, whose cardinality is 2 N ~ . Also it is clear that every subset of a Specker chain that has N1 elements is again a Specker chain. 8.9 Lemma. Let (S,5 ) be a linearly ordered set which has a countable subset D which is dense i n S. Then S is embeddable i n R.
Proof. The set X of elements x E S, which have an immediate successor y in S, is countable. For since D is dense in S, x or y belongs to D. The set X+of these x E X which are in D is countable, and also the set Xof those x E X which are not in D is countable, since their corresponding elements y are in D, and form a set equipotent to
X.Analogously the set Y of those elements y E S which have an immediate predecessor in S is countable. Let now D* be that subset of S which contains D U X U Y and the least and greatest element of S, if such an element exists. Then D* is again countable and dense in S, and there exists a <  preserving mapping f : D* + Q . If now x E S\D*, then ( S < x) has no greatest, and ( S > x) no least element, and the corresponding statement holds for their f  images f [S < x] and f [S> XI. Since R has no gaps there is an element z between these sets. We put f (x) := x , and obtain in this way an embedding of S in R. An essential property of Specker chains is expressed in the following statement, which is an immediate consequence of 8.9:
8.10 Theorem. is dense in S.
A Specker chain S has no countable subset which
Starting from an Aronszajn wl tree T we can now construct a Specker chain by applying a standard linear extension of 7.5:
8.11 Theorem. There exists a Specker chain. Proof. Let T be an Aronszajn wl tree. (It is not necessarily the same as in 8.6.) We choose a linear order < K in every equivalence class K (see 7.4) of T and define a corresponding standard linear extension <*of the order I of T according to 7.5. We prove that (T, I * ) is a Specker chain. First we show: (I) (T, I * ) has no subset of type wl, and symmetrically no subset of type w!. Suppose indirectly that a set W 5 T has type wl with respect to the linear order I*. Let v be an ordinal < wl. Then the set {x E W I h(x) < v) is countable (as a subset of the union of countably many levels Lp, p < v, of T) and thus not cofinal in (W, I*). Therefore W has a nonempty final segment Fvof cardinality Nlsuch that all elements of Fvhave height > v. There exists an element a, of the level L, such that N1 elements of F, are in [a,) (= {x E Tlx 1: a,)). This follows from the pigeonhole principle since each of the N1 elements of F, is 2 one of the countably many elements of L,. By 7.8 [a,) is also a segment of (W, I * ) , and moreover a final segment, for otherwise it could not contain N1 elements of F, since this has type wl. So we have obtained: (11) For every v < wl there exists an a, E L, such that [a,) contains a nonempty final segment S, of (W, I * ) .
8.8. ARONSZAJN TREES AND SPECKER CHAINS
277
This entails now that for p < v < wl we have a, < a,. For S, n S, is = S, or = S,, and hence # 0, and for an element z E S, n S, we have z 2 a, and z 2 a,, so that a, and a, are comparable and then a, < a,, so that {a,lp < w l ) would be a chain of type wl of (T, a contradiction. Next we show: U) is (111) Let U be an uncountable subset of T. Then U (with not embeddable in R. Suppose the contrary. Then there also exists an uncountable set U' 5 U which is dense and not embeddable in R. To see this we consider (in analogy to the construction in 3.1.15) the relation p which is defined by: For x, y E U with x < y we put {z E Ulx < z < y) is countable. spy Then p is an equivalence relation in U, and every equivalence class is countable. This is an easy consequence of the fact that cf(U) < wl and coin(U) = y* for a y < wl. Therefore there must be N1 equivalence classes. We choose an element u E U in each of these classes. The set U'of these u is uncountable and dense. So for the rest of the proof we can assume that U is (= U') dense and uncountable. U such that D is strictly dense (IV) There exists a countable D in U. By our indirect assumption U is isomorphic to a subset U* of R, say by a mapping cp. In R we have countably many open intervals (rl,r2) where rl < 72 are rationals. If such an interval contains elements of U* we choose one of them and collect them in a set D*, which then is countable. D* is strictly dense in U*. For let ul < u2 be elements of U*. Since U* is dense, like U, there exists an element u E U* with ul < u < u2 and further two rationals rl and r 2 with ul < rl < u < 72 < u2. By construction of D* there also exists an element d E D* with (ul <) rl < d < r 2 (< u2), and SO D* is strictly dense in U*. Then also U has a countable subset D := c p  l ( ~ * ) which is strictly dense in U, and (IV) is proved. The set of heights {h(x)lz E D) is a countable set of ordinals < wl, and so its supremum is still < wl. This entails that there exists an a < w l such that all z E D are in the a  stump T l a of the tree T. The set U := U \ T l a has cardinality N1 since T l a is countable as the union of countably many levels of T.
s),
<*r
For u E U let u, be the a  predecessor of u in T . Then {u,~u E U) L, is countable, and so, by the pigeonhole principle, there exist two elements u,t in U with equal a  predecessors u, = t,, so that u and t E [u,). Now there is no x E D with u <* x <* t. For this would by 7.8 entail that also x E [u,) and thus h(x) 2 a holds. But all x E D have a height < a. Summarizing we have obtained: There are two elements u, t E U, for which there is no element of D which is strictly between them relative to . This is a contradiction to (IV), which arose from the indirect assumption that (111) would be false. So (111) and the theorem is proved.
<*
8.9
Suslin chains and Suslin trees
In the first volume of Fundamenta Mathematicae M. Suslin posed in 1920 the following problem: Let C be a linearly ordered set without gaps and steps such that every set of nonoverlapping intervals which each have more than one element is countable. Is C embeddable in R ? [In the original text: . . . estil nkcessairement un continue linkaire (ordinaire) ?] This problem had great impact on set theory, and many papers dealt with it and with related questions. This is not astonishing since here properties of the real line are in discussion. The set R has a countable subset which is dense in R, e.g. the set Q. And as a consequence of this R also has the property that every set of disjoint segments, which each have at least two elements, is countable. So it makes sense to compare these two properties also in the general case and to determine if one involves the other. One direction is nearly trivial:
9.1 Theorem. Let (P,<) be a linearly ordered set which has a countable subset D which is dense i n P. Then every set of disjoint segments, each of which contains at least two elements, is countable.
Proof. Let 6 be a set of disjoint segments of P that all have at least two elements. In every S E 6 we choose an element d(S) of D.Then by mapping each S E 6 onto d(S) we obtain an injective mapping of 6 into the countable set D, and then also 6 is countable. In connection with the question whether 9.1 admits an inversion we define:
8.9. SUSLIN CHAINS AND SUSLIN TREES
279
9.2 Definition. A Suslin chain (or Suslin line) S is a linearly ordered set for which there holds: Every set of disjoint segments of S, that each have at least two elements, is countable, but S has no countable subset which is dense in S.
So one version of the Suslin problem can be formulated as: Does there exist a Suslin chain ? It is known that the existence of a Suslin chain is consistent with ZFC, and even with GCH. And this was also proved for the negation of this statement. In this context we refer to papers of Jech [89], Tennenbaum [169], Solovay and Tennenbaum [165], and Ronald Jensen [91], [go]. Jech and Tennenbaum discovered models of set theory in which a Suslin line exists, and Solovay and Tennenbaum proved that existence of a Suslin line is not provable in ZFC. Jensen proved that a Suslin line exists in the constructible universe. We also refer to the book [22] of Devlin and Johnsbraten. Concerning the cardinalities of Suslin chains we have: 9.3 Theorem. A Suslin chain has no subset of type wl and no subset of type w;. The cardinality of a Suslin chain is 2 N1 and 5 2 N ~ .
Proof. Suppose indirectly that {aL,lv< wl)< is a subset of a Suslin chain S, where for p < v we have ap < a,. Then the intervals [a,, a,+l], v < wl, where v has an odd number as last summand in the Cantor normal form of v, are disjoint, and their set is uncountable. This yields a contradiction. The case of wT is symmetric. Now IS1 5 2N0follows by the Theorem 6.2.9' of Hausdorff/Urysohn. If S would be a countable linearly ordered set, then S itself is a subset of S which is dense in S, and then S could not be a Suslin chain. So IS1 2 N1. In the preoccupation with the Suslin problem a related concept turned out to be of importance, namely: 9.4 Definition. A tree T is called a Suslin tree if T is uncountable but all of its chains and all of its antichains are countable.
A Suslin tree with N1 elements is also an Aronszajn N1 tree. On the other hand we cannot prove that an Aronszajn tree is also a Suslin tree, for it is not excluded that an Aronszajn N1 tree has uncountable ant ichains. Another version of the Suslin problem is: Does there exist a Suslin tree. For in the following we shall see that the existence of a Suslin chain
implies the existence of a Suslin tree and conversely. The relationship between Suslin chains and Suslin trees was established by Kurepa [104] in 1935. An elementary property of Suslin trees is: 9.5 Theorem. All levels L,, v
< wl, of a Suslin tree are nonempty.
Proof. Each level of a Suslin tree T is an antichain and thus countable. If there would be only countably many levels of T, then T would contain only countably many elements. 9.6 Theorem. Let S be a Suslin chain. Then there exists a subset 3 of the power set P ( S ) of S, where the elements of 3 are closed intervals of of S or = S, so that 3 is a Suslin tree with respect to the order reverse inclusion.
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Proof. The idea of the proof is to construct a blocksystem 3 (see 6.1.3) of N1 closed intervals of S. Let T < wl and suppose that we have already defined a set of different intervals I, := [a,, b,] of S for v < T such that each two of these intervals are disjoint or comparable. Then the set C := {a,lv < T) U {bVlv< 7) is a countable subset of S. It is not dense in S. And so there exists an interval I, = [a,, b,] which contains no element of C. So it satisfies: (1) For every v < T holds: ITis disjoint to I, or a proper subset of it. By transfinite induction we so obtain a blocksystem 3 of N1 closed intervals I,, v < wl, of S. Next we prove: (2) Every chain of (3,>) is wellordered.
A chain of (3, >) has the form {I,la E A), where A is a subset of wl,where the I,, a E A, are pairwise comparable. If a1 < a 2 are in A , then I,, > I,, must hold by (1) since I,, r l I,, = 0 is impossible, for it is = I,, or = I,,. This implies that {I,la E A), equipped with the order
>, is isomorphic to A , and thus wellordered.
Now we can prove: (3) Every chain of ( 3 , 3 is countable. Suppose the contrary. Then by (2) there exists a wellordered chain of 3, which can be described in the form {I,lv < wl), where for p < v < wl there holds Ip> I,. For every v < wl we choose an element xu
8.9. SUSLIN CHAINS AND SUSLIN TREES
281
in I,\I,+l,and then x, is also in none of the sets I,, with p > v. Now for every v < wl one of the following two possibilities holds: a) x, is < all elements x,, with p > v, or b) x, is > all elements x,, with p > v. Then there are N1 indices v, for which a) holds, or there are N1 indices v, for which b) holds. In the first case S would have a chain of type wl, in the second a chain of type w:. By 9.3 each of these is impossible, and (3) is proved.
(4) Every antichain of 3 is countable. For every antichain A of J consists of disjoint segments of S, which each have at least two elements, and so A is countable. If J has a first element, then J is already a Suslin tree because J is uncountable. If it has not yet a first element, we add to J the set S which is (and hence 5 ) all I E 3.Then the extended set is a Suslin tree.
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Before we prove the opposite direction of 9.6 we mention first some lemmas.
9.7 Lemma. Let T be a Suslin tree. Then there exists o subset Tl of T , which is still a Suslin tree and satisfies: (1) Every element of T has uncountably many successors.
Proof. We consider the set S of those elements of T that have only countably many successors. Let M be the set of minimal elements of S. Then M is an antichain and thus countable. Further S is = u{(S 2 m)lm E M ) and then also countable. So finally Tl := T \ S is still uncountable and, of course, a Suslin tree. In 7.2 we defined the concept semichaincomplete tree. We can restrict our consideration to these trees, for there holds: 9.8 Lemma. Let T be a Suslin tree satisfying (1). Then there exists a Suslin tree T* > T which is semichaincomplete and still satisfies (I).
Proof. We consider the tree T* which is defined according to 7.9 by adding to T the set E of new elements. Then T* is semichaincomplete, and (1) is satisfied for T*, since every element of E = T*\T has an immediate successor in T, which by assumption has uncountably many successors. After these modifications we can prove the counterpart of 9.6:
9.9 Theorem. Let T be a Suslin tree. Then there exists a Suslin chain.
Proof. Due to 9.7 and 9.8 we can assume that T satisfies (1) and is semichaincomplete. In the following we construct a subtree L of T, which has other levels than T , and so we define: For v < wl let L, denote the v  level of L. First we show:
(I) Every x E T has an infinite set of successors which form an antichain. Let x E T. If the set S(x) of successors of x would have only finite antichains, then S(x) would have a chain of cardinality N1, which is impossible. This follows from the partition relation N1 + (N1, of 6.4.10, applied on the comparability relation in S(x). So we can choose a denumerable antichain in every successor set S(x),and can extend it to a maximal antichain C S(x) by applying Zorn's lemma, so that we obtain: (11) For every x E T there exists a maximal infinite antichain A(x) E S(x), which, of course, is denumerable. We put Lo := {m), where m is the least element of the tree T , then L1 := A(m), L2 := u{A(x) lx E L1),and so on. In general we proceed as follows: Suppose that X is an ordinal < wl, such that the sets L,, v < A, have already been defined. If X is a successor number K, 1, we put LA:= U{A(x)lx E L,). Let now X be a limit ordinal. We consider all sequences (x,lv < X)<,where x, E L, and where for p < v we have xp < xu. (These are paths in the tree L which is coming into being.) Since T is semichaincomplete and since (1) holds, there exists sup{x,lv < A) (in T). And now we define LA as the set which contains all these suprema. With transfinite induction all sets L,, v < wl, are thus defined, and we put L := u{L,(v < wl). Next we verify:
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(111) For every v < wl there holds: (*) L, is an antichain. And u{L,Ip
< v) is a subtree of T
For v = 0 and v = 1 (*) is trivial. Let now X such that (*) holds for all v < A. Case 1. X is a successor ordinal v 1.
+
< wl
be an ordinal
8.9. SUSLIN CHAINS AND SUSLIN T m E S
283
Let a and b be different elements of LA.They have immediate predecessors a' (resp. b') in L,. If these are equal, a 11 b follows immediately. If a' # b' we have by induction hypothesis that L, is an antichain and thus a' 11 b'. And then also their successors a (resp.b) are incomparable. Thus LA is an antichain. It is easily seen that now also (*) holds for v = A. Case 2. X is a limit ordinal. For two different elements a,b of LAlet a = sup A, b = sup B, where A = {a,lv < A)<, B = {bUlv< A)< with a,,b, E L, for v < A. These sequences represent the wellordered sets of predecessors of a (resp. b) in L. (A,resp. B,is the path of L, which ends in a, resp. b.) From the sequences A and B no one can be an initial segment of the other because they have the same length. And since they are different, there exists a first index T < X with a, # b,. These elements are in the antichain L, and so we have a, 11 b,. This, together with a a, and b 2 bT implies a 11 b, and thus LA is an antichain. By construction now (*) holds for v = A, and by induction (111) is proved.
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(IV) Each L,, v
< wl, is nonempty.
Let this be proved for all L, with v < p, where p is an ordinal < wl. Since the L, are antichains and thus countable, their union U := U{L,Iv < p) is also countable, and for every x E U the set of its predecessors (T < x) (in T) is countable. This finally also holds for I := u{(T < x)lx E U),which is an initial segment of T. This entails that the heights of the elements of I form a countable subset of wl, which thus is bounded by an ordinal p < wl. Therefore there exists an elements t in T with height (in T) > p. Now t is > m =: t(0). Also t is > an element of L1 = A(m) because A(m) is a maximal antichain, and then t is comparable with an element t(1) E L1 and then also > t(1) because its height exceeds the heights of the elements of L1. This t(1) is uniquely defined since all predecessors of an element of a tree are comparable. Suppose that we have already ) t for L < v holds. defined elements t ( ~ E ) L, for L < v such that t ( ~ < , then t is If v is a successor ordinal K 1, then we have t > t ( ~ )and comparable with an element t ( ~ 1) = t(v) of the maximal antichain . the height of t in T is > p, it exceeds A ( ~ ( K )of) elements > t ( ~ )Since the heights of the elements of L,, so that t(v) is also > an element of A ( ~ ( K ) )And . t(v) is uniquely defined since in a tree an element cannot have two incomparable predecessors.
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+
If v is a limit ordinal, then we have t 2 sup{t(~) IL < V) =: t(u) E L,, which exists since T is semichaincomplete. In any case L, is nonempty, and (IV) is proved by induction. In the tree L every element x has by construction denumerably many immediate successors. We order their set N, by a linear order I,,so that , has ) the order type 7 of Q. Now we can define a Suslin chain (Nz, I C. Its elements are the maximal chains (= branches) of (L, I ) . These are wellordered sets, and so we can order their set by the principle of first differences, since no one of these maximal chains can be a proper initial segment of another one. We denote the corresponding order by 5: If c and c' are different elements of C, there exists a least ordinal u;, for which the K  predecessors (in L) c, and c; of c resp. c' differ. Since T is semichaincomplete, u; is a successor ordinal a 1. For otherwise c, and c; had both to be the supremum of the set of their (common) c, 5, c; for that x of the level La predecessors. Now we put c 5 c' of which c, and c; are immediate successors. It can easily be verified that 5 is indeed a linear order on C, this corresponds largely to the ordering by first differences. (V) (C, 5) is dense. Let a # b be elements of C with a 5 b, (a, b) their open interval in C. Then there exists an element t E L for which the final segment [t) (a, b). For let K be the least ordinal for which the u; of t in L is predecessors a,, b, of a (resp.b) in L satisfy a, <, b,, where x is their common immediate predecessor in L. Then there exists a t, of the same level L, with a, < t, < b,, for tp(N,) = 7.Every t E C that begins with the path (in L) ending in t, then satisfies [t) E (a, b). Thus (C, 5 ) is dense. If now (ai, bi), i E I, is a set of pairwise disjoint intervals of C, we choose elements ti, i E I, with [ti) C (ai, bi). These final segments (in L) [ti), i E I, are pairwise disjoint, and this implies that the ti, i E I, form an antichain of L, and thus I is countable. (VI) C has no countable subset which is dense in C. Let B be a countable subset of C. Every branch b E B has as height an ordinal < w l , and then there exists an ordinal p 5 w l such that all b E B have a height < p in T. Now for every element x E L with height p the final segment [x) of x in L contains by (11) infinitely many elements of L, but no element of B, so that B cannot be dense in C. This proves (VI). And so (C, 5) is a Suslin chain.
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Chapter 9 On the order structure of power sets In the class of posets the power sets deserve great interest. They occur in many mathematical considerations, and every poset is isomorphic to a subset of a power set (see 1.9.9). With regard to this, order theory is nothing else than the theory of power sets and their subsets. When a poset is investigated, a first overview can be obtained by studying the structure of its chains and antichains, in particular the maximal ones. Their order type, resp. cardinality, determines a great deal of the properties of the set under consideration.
9.1
Antichains in power sets
If S is a finite set with n E N elements, we can immediately determine the maximal chains of the power set Q(S) of S: 1.1 Theorem. Let S be a set with n E N elements. Then every maximal chain of p ( S ) has the form (8, {al), {al, az), . . . ,{a1, . . . ,a,)), so that every linear order {al,. . . ,a,)< of S yields a maximal chain of NS). Since an nelement set admits exactly n! permutations, the number of maximal chains of p ( S ) is n!. The question concerning the maximal antichains of := '$({I,. . . ,n)) is more complicated. In particular one is interested to know those antichains which are not only maximal, but also of maximal size. In other words: Which antichains have as cardinality the width of p, ? A maximal antichain of p, is of course every subset which contains exactly all k  element subsets of (1,. . . ,n) for a fixed k E (0,. . . ,n). We call this set the klevel of p,. In a famous paper [166] Sperner determined in 1928 the width of p, and its maximalsized antichains. His paper is one of the pioneering publications in the field "Combinatorics of finite sets". This discipline has many correlations with order theory. We refer here to the textbooks [5] of Anderson and [32] of Engel/Gronau.We introduce a notation which is analogous to that of (E) for numbers n, k .
p,
1.2 Definition. If S is a set and k a cardinal, then (j?) denotes the
set of all kelement subsets of S. Before we proceed we recall some elementary facts about the binomial coefficients (b). 1.3 Remark. Let n and k 5 n be natural numbers. Then (b) is 1.2.....k We further put (8) = 1.The number (b) is defined as n'(nl)""'(n(kl)). the cardinality of the set of all k  element subsets of an n  element set. Then (b) = (Ek). This follows e.g. by considering the function f, which maps every k  element subset K of an n  element set S onto its complement S \ K. This function is bijective, and so (f) and (Ek) have the same number of elements. It follows For k > 0 there holds the formula (b) = ( H I )  *. immediately from the definition.' The factor is > 1 exactly for k 5 And so we obtain from the last formula, that the values (Z), (y), . . . , (b) strictly increase until k := If n is even, then the numbers of the sequence . . . ,(E) strictly decrease. If n is odd = 2k+l, then (F) = (;+J, (b), and from k 1 on the binomial expressions strictly decrease.
q.
[$I. +
Before we deal with Sperner's theorem we prove the very useful LYMinequality. First we remark: 1.4 Lemma. Let S be a set with n E N elements, X C S a subset with 1x1 = s 5 n. Then the number of maximal chains of q ( S ) that contain X is = s! (n  s ) ! .
Proof. A maximal chain of F ( S ) which contains X is obtained by a permutation al, . . . ,a, of the elements of S, for which {al,. . . ,a,) = X. There are s! possibilities to find such an arrangement of the elements of X, and for each of them there are (n  s ) ! possibilities to prolong it with sequences whose elements are in S \ X. On the whole we so obtain s! (n  s)! maximal chains which contain X. Now we can prove the LYMinequality. This acronym refers to three mathematicians who had come across related things, Lube11 [115], Yamamoto [I811 and Meschalkin [ll7]. 1.5 Theorem. Let S be a set with n E N elements and O an antichain of q ( S ) . Then there holds (1) C{(L,)'~X E O) 5 1 (LYMinequality).
9.1. ANTICHAINS IN POWER SETS
Proof. We have (2) C{lXl! . (n  IXI)!IX E U)
< n!.
For the summand IX I! (n  [XI)!is the number of maximal chains of ';P(S), which contain X. In the sum (2) no chain is counted twice, since every maximal chain of '$(S) contains at most one X of the antichain U. The left side of (2) counts a set of maximal chains of '$(S), and is X )! hence 5 n! by 1.1. Division by n! yields C{IXI!.(nI IX E 24) 1, ! and this is the same as (1).
<
One can refine the statement of 1.5 in the following way: 1.5' Theorem. Let the assumptions of 1.5 be satisfied, and let pi be the number of elements of U which have i elements. Then C{pi/(9)li = 0,. . . ,n ) 5 1. Mind: pi/(?) is the proportion of elements of U i n the ilevel of ';P(S).
Proof. From (2) there follows C{i! (n  i)!.pili = 0,. . . , n ) 5 n!, and division by n! yields the assertion. We can now prove Sperner's theorem [166]: 1.6 Theorem. Let S be a set with n E N elements, U an antichain of ';P(S), k := LgJ. Then I%/ (E) holds. In particular: a) If n is even, and then = 2k for some k E N , then U = ( i ) . S b) If n is odd, say = 2k 1, then U = ( i ) or U = (k+l).
< +
(r),
Proof. For every i E (0,. . . ,n ) we have (9) 5 and then the left side of (1) is 2 IUI  (:)I, and this is 1 by (1). Thus IUI 5 (F). Let now U be an antichain of '$(S) with (F) elements. Then (1) yields for this U : 1 2 C{(Ll)'jX E I(} 1111(:)I = 1. So the sum is = 1. It is the sum of = (F) summands ( L l )  l , each of which is 2 (;)I. But this So every X E U has size is only possible if every summand is = (:)I. k, if n is even, and then U = ( i ) holds, so that a) is proved. Let now n be odd = 2k 1. Then every X E U must satisfy ( L l ) = (:),and thus X has size k or k f l . And we have U (f) U which means:
<
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(3) 'U = Uk U Uk+1 with Uk C (f) and Uk+1 2 If %k+l is empty, we are done. So suppose that it is # 0. For the set B := (:)\Uk we obtain:
(4) All k  element subsets of a set A E Uk+1 belong to %. For such a subset cannot be in Uk since U is an antichain. For B E 23 let r B be the number of elements A E Uk+1 with B c A. Then there holds r B k + 1 for all B E 23. For such a set A can only have the form B U {x), where x E S \ B , and we have IS\B I = k + 1. Now we count the number of pairs (B, A) with B E % , A A Ek+1 and B 2 A in two ways and obtain by (4):
<
(k + 1). By (3) we have 1231 = (F)  lUkl = IUk+ll, and SO the left side of (5) is a sum of 1231 = IUk+ll numbers rg k 1. But then all r ~B ,E %, must be = k 1, so that we have: (6) Each B E 23 has Ic 1 supersets in Uk+1 (5)
C{~BIB E B) = l%+il.
+
< +
+
Now it turns out that Uk+i contains with some set (and it was supposed to be nonempty) also all sets of For let X E Uktl Then every set which arises from X by exchanging an element of X by an element of S \ X is again in Uk+l. Indeed, if X' = (X\{x)) U {y) with x E X, y E S\X, then X\{x) E 23 by (4), and then X' E by (6). By finitely many exchanges one can transform X into every other set of (:+,), so that this set is Uk+l. Now contains already (;+,) sets, and so it is = Uk+1 = U. Sperner's theorem has a lot of generalizations, and there also exist several other proofs of this theorem or parts of it. We mention here a proof of F'reese which under the assumptions of 1.6 proves that IUI (E) and a) of 1.6 holds.
<
1.7 Proof. (by Freese [46],1974). We consider the set A of all maximalsized antichains of '@(S). Their size is the width w of !$?(S), and so, by Dilworth's theorem, '@(n)can be covered with w disjoint chains C1, . . . ,C,. Let 23 and 9 be elements of A. Then 2' 3 (resp. 9 ) has exactly one element Bv (resp. Dv) in common with each C,, v = 1,. . . ,w. And then {max{Bv, Dv) = Bv U DVIu= 1,. . . ,W) is a TU  element antichain of '@(S) (see 2.6.6 and 2.6.13) which is the supremum in the lattice A of maximalsized antichains of !$?(S) of 93 and 9. This lattice is complete (like every finite lattice), and so it has a greatest element !TI. Let .rr be a permutation of S, and let X := {XI,. . . ,X,) E A. Then also X, := {.rr[X1],. . . ,.rr[X,]) is a w  element antichain and thus E A. The mapping I3 : A + A, which puts II(X) := X, is an isomorphism of A, and so it maps the greatest element !TI of A onto itself.
9.1. ANTICHAINS IN POWER SETS
289
If now M E Dl, then TIM] must again be in Dl, and it has the same cardinality as M. Among the sets .rr[M],where T runs through the set of all permutations of S, are all subsets of S, that have the same cardinality as M, and therefore Dl contains with an element M also all elements of that level of q ( S ) , to which M belongs, which means: Dl is a union of levels of q ( S ) . But it cannot be the union of at least two levels, since every set of the lower level of these would be a strict subset of a set of the higher level. So Dl is a level of q ( S ) , and then of the maximal cardinality (2) = w. Since Dl is maximal each welement antichain contains in case a) no set with more than k elements and symmetrically no with fewer than k elements, and a) of 1.6 follows. Another proof of 1.6 was given by Lovasz [114]. In Sperner's original proof a concept was used to which the next definition refers: 1.8 Definition. Let R be a set of k  element subsets of an n element set S, where k n are nonnegative integers. Then the upper shadow VR of R is the set {X C SllXl = k 1 and X > K for a set K E R). The lower shadow ARof Ris the set {X SIIXI = k1 and X c K for a K E R. (Instead of upper (resp. lower) shadow also the name shade (resp. shadow) is in use.) So the upper (resp. lower) shadow of R consists of all those subsets of S which have exactly one element more (resp. fewer) than a set of R. For the upper and lower shadow of a set system we have two elementary inequalities:
<
+ c
1.9 Theorem. [I663 Under the assumptions of 1.8 there holds: a) IVRl 2  IRl for k < n. for k > 0. b) Pal 2 "R1
2
Proof. a) We count the number of pairs ( K , X ) with K E R, K C X S and IX\KI = 1. For each K E R there exist n  k elements in S\K, such that each of them can be added to K in order to obtain a (k 1)  element superset E S of K. The number of the above pairs is thus = (n  k) . 1R1. On the other hand we have for each X of such a pair k 1 many k  element subsets, which are not necessarily in R. So the number of the above pairs (K, X ) is 5 IVRI  (k 1). And this yields (n  k) . 1R1 5 IVRI . (k I), which proves a).
c
+
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b) In this case we count the number of pairs (X, K ) with K E R, X c K and 1x1= k  1. Each k E R has k subsets of size k  1, and so the number of pairs is = 1R1 k . On the other hand we have for each X of such a pair (X, K )  hence with X E AR  , n  ( k  1) elements in S\X, and thus n  k t 1 many k  element supersets of X within S which are not necessarily in R. This entails lARl  (n  k 1) 2 lRl k and b).
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In the following we prove some generalizations of Sperner's theorem. One of these is a lemma of Littlewood and Offord. First we mention a notion of Kleitman 1981 and one of its theorems: 1.10 Definition. Let P be a finite poset with a raak function r. Then a set C of maximal chains of P is said to form a regular covering of P by chains, if there holds: For each rank number k every element of rank k occurs in the same number of chains of C.
Then there holds the following theorem, which was proved by Kleitman 1981 in a more general context: 1.11 Theorem. Let P be a finite poset with a rank function r, and let C be a regular covering of P by chains. For x E P let NT(,) be the number of elements of P which have rank r(x). Then there holds for every subset T C_ P : IT1 I m a x c ~ C{NT(x) e 1% E C nT I .
Proof. S := CcEe C{NT(,) Ix E C n T) = CsET NT(,)  I@[, where C1is the set of those C E C, which contain x. Since C is a regular covering of P by chains, the number lC1l is the same for all elements of rank r(x), which means it is = Now the sum S is = CZET lCl = IT1 . lCl. N.c.1 This implies that at least one of the lCl summands of S is 2 IT[,and this proves the assertion.
m.
With the last theorem one can now prove a theorem of Erdos [33]:
1.12 Theorem. Let 2' 3 be a subset of !)3(S), where S is a set with n E N elements, such that 23 has no chain with k+l elements. Then 1231 is 5 the sum of the k greatest binomial coeficients (?), (i E (0, . . . ,n)). Proof. In order to apply 1.11 we first construct a regular covering of V(S) by chains of maximal length. Let a l , . . . ,an be an arrangement of the elements of S. To this there corresponds the maximal chain C := (0,{al), {al, a%),. . . ,{al,. . . ,an)) of the initial segments. We so obtain
9.1. ANTICHAINS IN POWER SETS
291
n! chains, which form the set C. Every k  element subset T C S is in exactly k! . (n  k)! chains of C (by 1.4), and therefore C is a regular covering of P by chains. According to 1.11 we have 1231 5 maxcEe C{N,(,) lx E C f l 23). Now, by assumption, every C E (5 has at most k elements in common with 23, so that in the last sum at most k elements x contribute to C N,(,).Therefore 1231 is 5 the sum of the k greatest N,(,). 1.13 Remark. If we take k = 1 in 1.12, then this means that 23 is an antichain. And the conclusion yields 1231 5 (F) with k := [$I, which is again a part of the statement of 1.6.
In the paper 1981 Kleitman established some conditions which are equivalent to the property that a finite poset has a regular covering by chains. F'rom this we later need the following: 1.14 Theorem. Let P be a nonempty finite poset with a rank function, and for i E w let ni be the number of elements of P which have rank i. Suppose that there is a regular covering C of P by chains. Then P has the socalled LYMproperty. That means: If A is an antichain of P with pi members of rank i, then 5 1.
xi
Proof. Let A be an antichain of P and Ai the set of its members of rank i, so that [Ail = pi. Since C is a regular covering of P every element of A, is in the same number of chains of C, which means in chains. Let Ci be the set of those chains of C which contain an element of Ai. = pi . since no chain of C has more than one element of Then the antichain Ai. For different indices i the sets Ci axe disjoint since A = is an antichain. Therefore we obtain 5 1C1, which entails our assertion.
5,
xi
5)
In the following we prove a theorem which has some resemblance with Sperner's theorem, but which is no direct consequence of it. 1.15 Theorem. Let S be a set with n E N elements, j an integer with j 5 [$I. Then the set P of all subsets of S with j elements has elements is (f). width (y), and the only antichain of P which has
(7)
<
Proof. We consider the set C of all chains (0, {al), {al, az), . . . ,{al,. . . ,aj)), where the system (al,. . . ,aj) runs through all combinations of j different elements of S. Then C is a regular covering of P. Indeed, every i(5 j)  element subset of S is contained : I : ) of them. The in the same number of chains of C, namely in i! . (
<
number of elements of P with rank i is ni := (1) w , where w is the width of P. Now we can apply 1.14: Let A be an antichain of P with w elements. And let pi denote the number of elements of A which have i elements, and thus rank i. By 1.14 now there follows:
( 1 ) x { E l i = 0 , . . . ,j ) 5 1 and a fortiori C { p i / ( y ) l i = 0 , . . . ,j ) 1, and w = C { p i l i = O,...,j} Of course also w (and then also =) holds. + For each i with 0 i < j the inequality ( 1 ) entails 1, and so 1  f$ 1 and Then pi = 0 must hold since otherwise we can divide by pi and obtain the contradiction (7) (1). So all pi, i = 0 , . . . ,j  1 are = 0 , and thus A must be a subset of (f'), and of course it is then equal to this set.
< (y).
&+
> < <
< (7)
&
ft < &.
<
<
From 1.15 we extract the following specialization: 1.16 Corollary. Let S be a set with n E N elements, i an integer < Then the set ( f ) U equipped with C , has width w = (;+,), and is the only antichain with w elements.
LIJ.
With 1.16 we can prove a relation between "neighboring" ranks of a finite power set and a partition theorem:
1.17 Theorem. Let S be a set with size n E N , k := LIJ , i an integer < k. Then there exists an injective mapping gi of ($) into Further there exists a partition of u { ( $ ) [ o i [ ; I ) into )(; saturated chains.
< <
(5,).
Proof. According to 1.16 the set (f) U (f+,) has width (l+l). By Dilworth's theorem then this set can be decomposed into (n ) disjoint chains. Each of these must contain exactly one element of (i+l) and no or exactly one element of ($). If an element x E ($) is in one of the chains, we map it to the greatest element of that chain. So we obtain an injective mapping gi satisfying the assertion. With the mappings go, . . . ,gk1 we now construct a partition of 5 := { T C SllTl k } into saturated chains. This contains (as longest chain) {@,90(@),91(90(@)),.. . ,gkl!gk2(.  .,go(@)..))
<
If in general Ai+1 is an element of ($+I), which is not the gi  image of an element of ($), we take its gi+l image Ai+2,from this the gi+2 
9.1. ANTICHAINS IN POWER SETS
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image, from this the gi+s  image and so on until we have reached an element of (f). In this manner we obtain a decomposition 9 of 5 in disjoint saturated chains. The statement of 1.17 can be strengthened to a partition of the whole set P ( S ) into saturated chains, which also satisfy a symmetry condition: 1.18 Definition. Let P be a finite ranked poset with rank function r. Then the elements X I , . . . ,xh (where h is an integer 2 1) of P form a symmetric chain C of P, if the following two conditions are satisfied: 1) For i = 1,. . . ,h  1 the element X i + l is an immediate successor of xi. The same: C is a saturated chain. 2) r(xl) r(xh) = r ( P ) (= the highest rank).
+
The second property states, roughly speaking, that C is symmetric to the middle rank. For it includes that C contains with an element of rank i also an element of rank r ( P )  i. In connection with a prize problem which was posed by the Dutch Wiskundig Genootschap in 1949, de Bruijn, Tengbergen and Kruyswijk [19] proved the following: 1.19 Theorem. Let S be a set with n E N elements. Then the power set P(S),equipped with its natural rank function r, can be partitioned in symmetric chains.
Proof. If S has only one element x, then the set which contains only the chain (0,{x)) is a partition of q(S) in symmetric chains. Suppose now that the theorem is proved for a fixed number n E N, and that S is a set with n 1 elements. W.r.0.g. we can assume S = {I,. . . ,n 1). Let C be a partition of P ( { l , . . . ,n)) in symmetric chains. Let C2 (resp.Cl) 2 be the set of those chains C = {Cl,. . . ,Ct)< E C: for which t (resp. t = 1) holds. For C E C2 we define two symmetric chains of Y({l,. . . , n 1) : C' := {Cl, . . . ,Ct, Ct u {n 1)) and C" := {Cl u {n I), . . . ,Ct1 U {n + 1)). Indeed r (C1) r (Ct U {n 1)) = r (Cl) r (Ct) 1 = n 1, and r((C1 U {n 1)) r(Ct1 U {n 1)) = r(C1) 1 r(Ct1) 1 = n 1. The set Cl can only be nonempty, if n is even, because of the symmetry of its chains. For C E Cl we put C' := {C, C U {n 1)). Now it can easily be seen that u{{C' U C1')IC E C2) U {C'IC E El) is a partition of V((1,. . . ,n 1)) in symmetric chains.
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1.20 Remark. In [19] the authors proved a more general theorem, namely: Let D be the set of positive divisors of a natural number n. Then the relation alb ( a divides b) defines an order in D, and D is ranked by: For a divisor d of n we put r ( d ) := the number of prime factors of d, counted with their multiplicity. So e.g. r(36) = r ( 2  2 . 3 3) = 4. Then D has a partition in symmetric chains. With 1.19 we obtain again a proof for a part of Sperner's theorem. Namely the following holds:
1.21 Theorem. Let S := (1,. . . ,n) and let 9 be a partition of v ( S ) i n symmetric chains, w the width of v ( S ) . Then 1 9 1 = (E) = w follows for k := 151. Every antichain of P ( S ) of size w has exactly one element i n common with every chain of 9. Proof. Every chain of 9 contains exactly one element of (f), and every element of (f) is in exactly one chain of 9.Thus we have 1 9I = (E). Let U be an antichain of Q(S) of size w. Each X E U is in exactly one of the chains of 9 , so that (E) 5 w = IUI 5 191 = (Z), and thus IUJ = 1 91holds. This proves the assertion. One of the earlier generalizations of Sperner's theorem was given by Erdos/Ko/Rado [41] in 1961, where the cardinality of antichains is investigated which satisfy additional conditions. In 1972 Katona [94] gave another proof of their theorem which used a new method, namely the application of cyclic permutations:
1.22 Definition and Lemma. Let S be a finite set with at least two elements. Then we define a cyclic permutation of S as a relation p in S, for which there exists an indexation a l , . . . , a n of the elements of S, such that aipai+l holds for i = 1,.. . ,n  1 and anpal. Intuitively speaking: It is possible to arrange the elements of S in a circle so that, starting from a1 and moving forwards in the positive sense, one reaches a2,. . . ,an in this succession and comes back to al. So a1 is not distinguished in comparison with the other elements of S. The number of cyclic permutations of an n  element set S is (n I)! For there are n! permutations of S, and each cyclic permutation is determined by n permutations: ( a l ,  . . , a n ) ,(a2,...,an,al), (a3,.=,an,al,a2),...,(an,al,..,anl) determine the same cyclic permutation.
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1.23 Theorem of Erdos/Ko/Rado [41]. Let U := {A1,. . . ,A,) be an antichain of the power set of a set S which has n E N elements, where every two sets of the antichain have a nonempty intersection. Further it is assumed that there is a natural number k 5 $, so that [Ail 5 k holds for all i = 1,... ,m. Then we have m 5 (:I:), and this is tight.
Proof (by Katona [94]). First we show that m can be = (;I:). To this purpose let Al, . . . ,A, be all k  element subsets of S that contain a fixed element x of S. There are as many as there are possibilities to choose k  1 elements in S\{x), nameiy (;I:) many. Before we start with the main part of the proof we first mention that it suffices to assume that [Ail = k holds for all i = 1,. . . ,m. Indeed, if some [Ail are < k, we can replace them by subsets of S which have size k. To this purpose we take a partition of P ( S ) in symmetric chains and replace each Ai with size < k by the uniquely defined subset of size k which contains Ai and which is an element of that chain of the partition which contains Ai. (By the way: Here also 1.17 could be applied.) The chosen supersets of these Ai are distinct since the chains of the partition are disjoint. So it is sufficient to prove 1.23 under the specialized form: Let A1, . . . ,A, be m pairwise distinct k  element subsets of (1,. . . ,n), where k 5 $, and where the Ai have pairwise a nonempty intersection. Then m 5 (:I:). The proof applies the method of counting in two ways. We determine the size p of the set P of all pairs ( a , A), where a is a cyclic permutation of (1,. . . ,n), and where A is an Ai of U which is encompassed by a. That means: A is a segment of a in the circle representation. of a with positive orientation. First we determine the size p* of the set P*of all pairs ( a , K), where a is a cyclic permutation of (1,. . . ,n) and K a k  element subset of (1,. . . ,n), which is encompassed by a . We obtain p* = (n  I)!  n = n!. For by 1.22 there are (n  I)! possibilities for a, and for each a there are n sets K, which are encompassed by a. By reasons of symmetry here every k  element subset K 5 (1,. . . ,n) appears in the same number of pairs of P*,which means in p*/(F) of them. And so for every i E {I,. . . , m ) the above number p of pairs (a, Ai) is = n!/(F). The number p is the m  fold of this, since for i we have m possibilities. So we have obtained: (1) p = m n!/(F). Now we apply a second form of counting. First we prove:
(2) Every cyclic permutation a of (1, . . . ,n) encompasses at most k of the sets Ai. Indeed, let a be a cyclic permutation of {I,. . . ,n) and Ah a set of {A1,. . . ,A,), which is encompassed by a. On the circle representation Ah determines a certain sector. The rest of the sets of {A1,. . . ,A,), which are encompassed by a, must have sectors, which intersect the sector of Ah. Among these let Al (resp. AT) be that set, which in the positive orientation is farmost to the left (resp. right) of Ah. Then the sectors of Al and AT must still have a nonempty intersection because of Al n A, # 0. Now the following is clear: All Ai which are encompassed by a , have sectors whose beginning b is between (in the wide sense) the beginning of Al and the beginning of A,. Therefore b can only be one of the k numbers of Al, and this proves (2). Now (2) yields the following estimation for IPI :
7r'
(3) IPI 5 (n  I)!  k. From (1) and (3) we conclude m . n!/(E) k n  n1 5 ; (k)  (k1).
< (n  I)! . k , and further
At the end of this section we also consider antichains in infinite power sets. Before we introduce an obvious concept: 1.24 Definition. Let p > 0 be an ordinal. It has by 4.8.5 a representation in Cantor's normal form as p = wffO a0 . . . wan . an, where n is a nonnegative integer, a 0 > . > a, is a strictly decreasing sequence of ordinals, and ao, . . . ,a, are natural numbers. If a, = 0 holds we have wan = 1 by definition, and then the last summand in the normal form is a,. Then p is called even (resp. odd) if a, is even (resp. odd). If a, > 0 or p = 0 we also call p even. It is clear that even (resp. odd) nonnegative integers are also even (resp. odd) ordinals.
+ +
1.25 Theorem. Let S be an infinite set of cardinality N,. Then the power set V(S) has antichains of cardinality 2Na := c. Moreover V(S) has 2' antichains of cardinality c.
Proof. Evidently it suffices to consider the case where S = w,. We partition the set w, in the twoelement segments S, := {v, v + 1), where the v are even ordinals < w,. Let IM be the set of all subsets T 5 w,, which have the property: T contains exactly one element from each S,. Evidently we have IT I = N,.
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If now A and B are different sets E M, there is at least one v, such that A contains an element e, E S,, which is not in B. Then B contains that element d, of S, which is # e,, and d, is not in A. Thus we have that A\B and B\A are nonempty. So A and B are incomparable, and ?JXis an antichain of !j3(S). The cardinality of M is 2Na,for we have N, segments S,, and for each of them we have two possible choices to obtain an element of M . So M is equipotent to the set of all mappings of {S,lv < w,) in (0, 1). There also exist 2Cantichains of ')3(w,) which have cardinality c. For we can represent t)32 as M = {M,lv < w(c)). Now every subset of M, which contains (among others) the sets Mv with even v, has cardinality c. For there are c sets Mv with odd v, and each of their 2' subsets can be added to {M, lv < w(c) and v even) to obtain an antichain of !J3(w,) with c elements.
9.2
Contractive mappings in power sets
A choice function f in a set S ascribes to every nonempty subset S an element f (T) E T. In 6.1.14 we mentioned a generalization of this concept, which was introduced by Kinna/Wagner [96], namely generalized choice functions on a set S, which ascribe to every T E S, that has at least two elements, a proper subset of T. In this section we consider some combinatorial properties of power sets, in which we have a generalized choice function. First we define for general ordered sets:
T
2.1 Definition. Let P be a poset. A mapping f : P + P is said to be contractive, if for each x E P we have f (x) L: x. We call f strictly contractive or regressing, if f (x) < x holds for all x E P, which are not minimal elements, and for these we have f (x) = x. We are chiefly interested in the case where P is a set of sets which contains with a set X also all subsets of X, and where the order of P is the set inclusion S . There are many examples in mathematics, where from given sets certain special kinds of subsets are of interest. E.g. in topology one has the notion open kernel of a subset T of a topological space  this is the greatest open subset of T. Another example is the linear kernel, which was introduced in 2.4.1.
In many cases a contractive mapping f : P + P of a poset has the addional property that for every T C P also f (f (T)) = f (T) (Idempotency) holds and also A B + f (A) f (B) (Isotonicity). If these conditions are satisfied f is called a kernel function, and this is the counterpart of a closure operator. In the following we require neither idempotency nor monotonicity from our contractive mappings!
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2.2 Definition. Let X be a set of sets, which contains with a set T also all subsets of T. In X we have the order . Let f be a contractive mapping on X. We define a regressive chain of length r E N to be a set > f r ( X ) ) , where X is {f(X), f 2 ( X ) ,...,f r ( X ) ) > (so that f ( X ) > an element of X, and where the iterates of f are defined in the usual manner: f 1 := f and f n + l ( x ) := f ( f n ( X ) ) . A fixed chain (of f ) is a chain ?2J 2 X such that f (X) = X holds for all X E 9, and a constant chain (of f ) is a chain C X for which all sets f (C), C E C, are equal.
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The main theorem of this section states that for every n E N there exists an h(n) E N, such that for every contractive function f in the power set !Q({l,. . . ,h(n)) there is a chain of size n which is regressive, fixed or constant. For its proof we need several preparations. 2.3 Definition. If A, B are finite subsets of N satisfying A 2 B we define the position p(A, B ) of A in B as follows: If A = 0 we put p(A, B ) := 0. If A # 0, say A = {al,. . . ,ak)<, then there is a uniquely defined bijection p of B onto the initial segment (1,. . . , IBI) of N which is <  preserving. We put xi := p(ai) for i = 1,.. . ,k , and define p(A, B ) := {xl,.. . ,xlc). The following example illustrates this: Let B := {3,4,7,9,11,14) and A := {4,9,11), then p(A, B) = {2,4,5). And this p(A, B ) indicates that A contains the second, the fourth and the fifth number of B. If A is a oneelement subset of B, then its unique element is the p(A, R ) th element of B. The class of the contractive mappings has a subclass which is easier to handle: 2.4 Definition. Let (2 be a nonempty set consisting of finite subsets of N such that B E (2 and A 2 B implies A E (2. So the set {IBIIB E (2) is an initial segment D of w. Now a contractive mapping f : (2 + (2 is called special if, given any t E D, for all T E (2 with IT1 = t the position
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of f (T) in T is the same, hence only depending on t, so that here we can define the position function .rr off on D by: For t E D we put n(t) := p(f (T),T), where T is an arbitrary set of C with IT1 = t. Of course, I.rr(t)1 = If (T)I for every T E C with [TI = t. We call f nspecial (for an n E N), if for the set U of all n  element sets E C there holds: For each A E U the position p(f (A),A) is the same set. So, a special contractive mapping f is n  special for all n E N, and conversely. If e.g. f maps every nonempty set E E C onto its least element and .rr puts
0 onto 0, f is special, and its corresponding position function ~ ( 0= ) p(0,0) = 0 and ~ ( 1 = ) (1).
In the following we perform a reduction process, by which we can reduce in some sense general cont~activemappings to special ones. For this we need a form of Ramsey's theorem, which we now present. 2.5 Theorem. Let n,p,q be natural numbers, where p and q are Then there exists a natural number r(p,q,n) (which only depends on p,q,n) with the following property: If S is a set with IS1 2 r(p, q, n), and if [SIn= A U B, then there S with IPI = p and [PIn A, or there exists a exists a subset P subset Q C S with IQI = q and [Qln C B.
2 n.
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Proof. For n = 1 and all p,q 2 1 the statement is true. Here r(p, q, 1) := p q  1 fulfills the assertion: If IS[ 2 p q  1 holds and S = A U B , then A must contain at least p elements or B at least q elements. Further we have:
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(1) If p = n or q = n holds the assertion is fulfilled.
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In the first case we put r(n, q, n) := q. If now IS[ q and [Sjn= AUB with A n B = 0 holds, then there are n elements in S, whose set is in A, or otherwise all nelement subsets of S belong to B, and [SIn B holds. Symmetrically we conclude the same in the second case. Suppose now that n 2 2 holds, and that the theorem is proved for n  1 instead of n (and all p and q which are 2 n  1). We prove the validity of the assertion for n by induction on the sum of p and q. Since p and q are 2 n, the least such sum is 2n, and here the assertion holds by (1)So we have to show, under the assumption that the theorem holds for n  1 : There exists a number r (p, q, n) which satifies the statement,
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if it is already proved that a number r(pl,q', n) exists for the case where p' q' < p q. Let now natural numbers p,q with p > n < q be given, so that by induction hypothesis numbers pl := r (p  1,q, n) and ql := r (p, q  1,n) exist, which satisfy the assertion. We now verify it for the triple (p, q, n), and then our theorem is proved by induction. We show: There exists a number r(p, q, n) which fulfills the statement, and for which holds:
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(2) r(p,q,n) I r(pl,ql,n  1) + 1 := 8 To this purpose let S be a set with at least s elements and [SIn= A U B and A f l B = 8. Let e be a fixed element of S, and S' := S\{e). We consider the decomposition [S'Inl = A' U B' , where A' is the set of those (n  1)  element subsets T of S', for which T U {e} E A, and B' is the set of those (n  1)  element subsets T of St,for which T U {e) E B. Because of (2) we have IS'[ 2 r(pl, ql, n  I), and so, due to the induction hypothesis, there exist 1) a set PI of pl elements in St, such that [plInlC A', or 2) a set Q1 of ql elements in S', such that [QlInl C  B'. Suppose that 1) holds. Due to pl = r(p  1,q, n) then there follows U [B n [P1ln) : There are q elements by considering [P1ln= (A n [P1ln) of PI,such that all the n  element subsets of their set belong to B (and then we are done) or there exists a subset X C PI with 1x1= p  1 and [XIn & A. In this case from X C Pl and 1) we obtain (*) [XI"' [PI]"' & A'. Let X , := X U {e). Then /X, I = p. It follows that [XJn 2 A. Indeed, let N & X, and IN/ = n. For e $ N we have N C X and N E [XIn C A and thus N E A. For e E N we have N = {e) U T, where T is an (n  1)  element subset of X and St. So T E A' by (*), and by definition of A' now T U {e} = N E A. The case 2) is symmetric to case I), and so (2) is proved, and with it the theorem. The last theorem is easily extendable to the situation, where we have not only two, but finitely many classes: 2.6 Theorem. Let k E N and n,pl,. . . ,pk be natural numbers with p, 2 n for 6 = 1,.. . , k . Then there exists a natural number
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r(pl,. . . ,pk, n) =: r , for which there holds: If S is a set with at least r elements and i f [Sin = A1U  . .UAk, then for at least one IE E ( 1 , . . . ,k ) there exists a set P, C S with IP,I = p, and [P,In E A,.
Proof. For k = 2 the assertion holds by 2.5. Let now k be an integer 2 2, for which the statement is already proved, so that a number r(pz, . . . ,pk+l, n) already exists.We show: There exists a number 7'031,. . . , ~ k + 1 , 45 r(p1,r(p2,. . . , ~ k + l , n ) , n=: ) 8. Let S be a set with IS/ = s and [SIn= Al U . . . U Ak+1, where the summands are pairwise disjoint.We consider the decomposition [Sln = A1 U (A2U  .  U Ak+1)in only two classes. Here 2.5 can be applied, and there exists a set PI E S with IPII = pl and [P1ln 5. A1, or there exists a set Q1 E S with lQll = r(p2,. . . ,pk+l,n) and [QlIn E A 2 U .  ' U A k + l . In the last situation there exists an i E ( 2 , . . . ,k 1 ) and a Pi E Q1 with lPil = pi and [Pi],E Ai, and this proves our assertion by induction hypothesis.
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From the last theorem we obtain immediately by putting all p, equal and = p : 2.7 Theorem. Let p,lc,n be natural numbers with p 2 n. Then there exists a number R := R ( p , k , n ) E N such that the following holds: If S is a set with IS/ = R, and if [Sin = Al U .  U Ak, then there exists a set P 5 S with /PI = p and a K E ( 1 , . . . ,k ) for which [PIn A,. Now we apply the idea which was used in the proof of [63],Satz 3.1 to obtain the following reduction lemma: 2.8 Lemma. Given n E N , there exists a natural number m := m(n),such that the following holds: If M := ( 1 , . . . , m } and if f : v ( M ) + v ( M ) is a contractive mapping, then there exists a subset S C_ M with IS1 n such that the restriction o f f to '$(S) is a special contractive mapping (in the sense of
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2.4) 
Proof. In this proof let R ( p , k , n ) always be the least number satisfying 2.7. We put mo := n,and define recursively numbers mo, . . . ,m, as follows: For i = 1,. . . ,n we put: ( I i ) mi := mi1, 2i, i ) , SO that ml := ~ ( m ~ , 2 I ) l, .,. . ,m, := R(rn,l, 2,,n). Then m := m, satisfies our assertion. To see this we put M =: Mn := (1,. . . ,m,).
If T is an n  element subset of M, the set f ( T )has one of 2, possible positions p l , . . . , P p in T, namely so many as there are subsets of T. Then we consider the representation [MIn= U1 U . . . U U2n,where for i E ( 1 , . . . ,2") '& is the set of all those n  element subsets T of S, for which f ( T ) has the position pi in T. Due to (1,) then there exists a subset Mnl Mn of size m,1 such that [Mn1lnis a subset of one of the '&, and then Mn1 is n  special. Suppose that we have already defined sets M, Mnl  . Mi for an i with 0 < i < n, so that lMjl is = mj and ( j 1 )  special for i 5 j < n. Then, similar to the case i = n, and due to ( l i )there exists a set Mi1 C Mi of cardinality mi1, which is ispecial. Recursively we so obtain a sequence Mn Mn1 . Mo, where for each i < n the set Mi is (i 1)  special, and also j  special for all j with i < j 5 n. Finally Mo is i  special for i = 1,. . . ,n. That means: f is a special contractive mapping over Mo =: S. Now S has size mo = n and so satisfies our assertion.
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For the case of special mappings we now prove: 2.9 Theorem [66]. Let C be the set of all finite subsets of N and f : C + C a special contractive mapping. Then there is an infinite fixed chain, or for every r E N there is a regressive chain of length r or a constant chain of length r.
Proof. According to 2.4 the special mapping f is determined by its position function T . We distinguish two cases: l t for Case 1: There is an infinite subset W C N such that I ~ ( t )= every t E W. For the sets T E C with IT1 E W we then have f ( T )= T. Thus the sets Mw := (1, . . . ,w), w E W, form an infinite fixed chain. Case 2. There exists a number u E N such that I ~ ( t )
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By the pigeonhole principle, then there also is an infinite set X N such that for all x E X we have In(x)l = q, where q is a fixed nonnegative integer < b. The case q = 0 is trivial: In this case for all x E X we have n(x) = 0 and f (M) = 0 for the M E (E with (MI = x, and this yields an infinite constant chain. So we may assume q > 0. To every x E X we ascribe a (q 1)  vector v(x) = (21,. . . ,xq+1) as follows. Let n(x) = {sl,. . . ,sq)<. Then v(x) is defined to be the vector of the lengths of the pieces into which n(x) partitions the interval [0,x] : v(x) := (XI,.. . ,xq+l) := (s1, S2  51,S3  S2,. . . ,sq Sq1, x  sq). Hence we have n(x) = (xl,xl +x2,. . . , X I . . xq) for x E X. We introduce a partial order in the set {v(x)lx E X ) as follows: If x', x" E X we put v(xl) v(x") iff xi x: for all i  components, i = 1,.. . ,q + 1, of x', XI'. Now from Higman's theorem 6.3 there follows, since the set of all words over N is wqo:
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(1) There is an infinite subset Y with x' xr' we have v(xl) v(xU).
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E X such that for all x', x"
E Y
We can prove (1) also without referring to Higman's theorem, making use of an idea of Bernd Voigt: For i E (0, . . . ,q 1) we define a set Ai [ x ]as~follows: Let x', XI' be elements of X with x' < x". We put {x',x") E A. iff x: x; for all v = 1,.. . ,q 1. If {XI,x") E A. does not hold, there is a first index i E (1,. . . ,q 1) with xi > x: and then we put {x', x") E Ai. The set [x12is= u{Aali = 0,. . . ,q+ 1). In a graphtheoretical formulation: The sets Ai, i E (0,. . . ,q+ 1), define an edge coloring of the complete graph, whose vertices axe the numbers x E X, with q 2 colors. Now 6.4.8' entails that there exists an infinite subset Y N, such that [yI2is a subset of an Ai with i E (0,. . . ,q 1). And this i can only be i = 0, since in N we have no infinite descending chains. So (1) is again proved. Now, given r E N, we can construct a constant chain for f of length r. Let yl, . . . ,y, be the first r elements of Y (in its natural order). We shall define recursively sets M(yr),M(Y,~),. . . ,M(yl) ,so that they have the same f  image.We abbreviate y := y, and x := yr1 and consider the position functions n(x) = {J1,. . . ,Eq) and n(y) = ( ~ 1 ,... ,qq). Let M(y) be an arbitrary set of y natural numbers. Then the set f (M(y)) the qih,. . . , and the element of M (y). Now we contains the define M (x) as follows: M (x) contains the J1 last elements (in the order
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vP,
qkh
of N ) which are 5 the rlih element of M(y), and for every i = 2,. . . ,q the ti  ti1 last elements of M(y), which are 5 the rl" element of M(y), and finally the x  Jq last elements of M(y). Then M(x) has t1 C{ti li = 2, . . . ,q) x  tq= z elements, and its tfhelement is the same as the rljh element of M(y). Since J1 5 71 and Ji 5 Q  holds for i = 2 . . . ,q and x  Jq 5 y  qq this construction is possible. Now f (M(x)) is = f (M(y)). So we constructed to M (y,) a set M(y,l). In the same manner we construct to the latter set a set M(yT2) with f (M(Y,~))= f (M(y,l)), and we continue this construction until we have reached a set M(yl) with the same f  value. Then the sets M (yl) , . . . ,M (y,) form a fixed chain of length r with respect to f.
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2.10 Remark. Subsequently to the last theorem the question arises whether the statement can be sharpened so that there also exists an infinite chain of E, which is fixed, regressive or constant. A simple counterexample is given by the mapping f : E + E, which maps every nonempty T E E onto the greatest element of T. For special contractive mappings now there follows: 2.11 Theorem [66]. Given n E N , there is a number h*(n) E N such that the following holds: If g is a special contractive mapping on v(h*(n)), then there exists a chain of length n which is fixed, constant, or regressive.
Proof. Assume that the statement is false. Then there exists a number n* E N such that for every k E N there is a special contractive mapping f k on p(k) such that there is no chain of length n* which is fixed, constant, or regressive for fk. We enumerate the set E of all finite subsets of N in a sequence of type w ; E = {Fili E N). The set {fk(Fl)lk E N ) is finite, and so there is an infinite set of indices Ic, say Nl, such that all sets fk(Fl), k E Nl, are equal. Further there exists an infinite subset N2 C Nl such that a11 sets f k (F2),k E N2, are equal, and so on. Now we define a mapping f : E + E as follows: For i E N let f (Fi) be the common value fk(Fi) for k E Ni.Then f is a special contractive mapping on (5 since the mappings f k have the same property. Further we see: (1) There is no chain of length n* which is fixed, constant, or regressive for f.
9.2. CONTRACTIVE MAPPINGS IN POWER SETS
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Otherwise such a chain would have the corresponding property also for one of the mappings f k which gives a contradiction. So ( 1 ) would be valid. But it contradicts 2.9, and therefore the indirect assumption is false and the theorem proved. Together with the reduction Lemma 2.8 the last theorem now immediately yields one of the main theorems of this section: 2.12 Theorem [66]. Given n E N , there is a number h ( n ) E N such that the following holds: If S is a set of h ( n ) elements, and i f f : p ( S ) + p ( S ) is a contractive mapping, then there exists an n element chain of p ( S ) which is fixed, constant, or regressive. 2.13 Corollary. If S is a set of h(n) elements and if f : p ( S ) + p ( S ) is regressing, then there is an n  element chain of p ( S ) which is constant or regressive.
The Theorem 2.11 sharpens the following two theorems of Rado [I461: 2.14 Theorem [146]. Given n E N , there is a positive integer h s ( n ) such that the following statement holds: If S is a set of h 3 ( n ) elements, and i f f ( X ) , for every subset X of S, is a subset of X, then there always are subsets X o , . . . ,Xn of S such that X o c . . . c X n and ~ ( X OE )f(x1)E ... E f(xn).
This theorem is a consequence of 2.12. For if there is an n+l element chain of p ( S ) , which is fixed or constant, this is clear. And if there is a regressive chain { f n ( X ) ,. . . ,f 1 ( X ) ,f ' ( x )= X ) , then X , := f n  v ( X ) for v = 0 , . . . ,n satisfies the assertion. 2.15 Theorem [146]. Given positive integers n and j, there is a positive integer h 2 ( n ,j ) such that the following holds: If S is a set of h2(n,j ) elements, and i f f ( X ) ,for every subset X of S, is a subset of X having at most j elements, then there always are subsets X o , . . . ,X n of S such that X o C .  .C X n and f ( X o ) = = f(Xn).
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Proof. We take h 2 ( n ,j ) := h ( k ) (of 2.12), where k = max{j 2, n ) . Then by 2.12 there is a chain X o c . . C Xj+1, over which f is fixed, regressive or constant. The first two possibilities cannot occur, because in these cases f (Xj+1) had to contain at least j 1 elements.
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2.16 Remark. The proof of (2.11 and) 2.12 was nonconstructive. A contradiction was deduced under the indirect assumption that no
bound h * ( n ) (resp. h ( n ) ) would exist. Also the proofs of 2.14 and 2.15 were nonconstructive. Rado [I461 had posed the question whether there exists a constructive proof. In [62] such a proof of 2.12, and then also of 2.14 and 2.15, was given. The Theorem 2.12 has also a geometric interpretation. Its content can be reformulated in graphtheoretical concepts. To this purpose we first define:
2.17 Definition. For n E N the directed cubic graph Cn has as vertices the 0 , l  sequences ( e l , . . . ,en) (ei E { 0 , 1 ) for i = 1,. . . ,n ) , and its directed (or oriented) edges are the pairs (b,a ) , where for a = ( a l , . . . ,a,) and b = (bl, . . . ,b,) we have ai 5 bi for i = 1,. . . ,n, so that a 5 b holds in the product order. The naming is explained by the fact that the vertices of Cnare the vertices of the unit n  cube [0,lIn of the euclidean n  space Rn. There is a natural bijective mapping between Cn and the power set !& := p ( ( 1 , . . . ,n ) ) , namely that one which ascribes to the n  tuple ( E ~ .,. . ,en) the subset T of those i E ( 1 , . . . ,n ) for which ei = 1 holds, in other words: T is that subset of ( 1 , . . . ,n ) , which has ( e l , . . . ,en) as characteristic function. This in mind one can reformulate the substance of 2.13 as follows: 2.18 Theorem. For each n E N there exists an m = m(n) E N such that the following holds: If from each vertex ( e l , . . . ,E,) of Cm that is diferent from (0,. . . , 0 ) exactly one directed edge leads to a vertex which (in the product order) is strictly less than ( e l , . . . ,e m ) , then there exists a directed path of length n, i. e. a subset of vertices v l , . . . ,v,, where {vi, vi+l) is a directed edge of the graph Cm for i = 1,. . . ,n 1, or there is a vertex e, which is the common endpoint of n edges {ai,e),where the ai, i E ( 1 , . . . , n ) form a chain of Pn.
Until here we had considered contractive mappings in power sets. Of course one can widen the reflections on more general posets. Klaus Leeb had posed the problem to find bounds such that any contractive mapping on a poset must have a monotone (k 1)  chain if the size of the poset exceeds those bounds. For posets of bounded width this problem was solved by G.W.Peck, P.Shor, W.T.Trotter, and D.B.West [136]. Before presenting their theorem we introduce some concepts which are used in their proof:
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2.19 Definition. Let (P,2) be a finite poset, and g : P + P a contractive mapping, also named contraction (so that g(x) I x holds for all x E P ) . A monotone kchain is defined to be a chain {xl, . . . ,xk)< of k E N elements for which g(xl) I . . . I g(x,+). If for w, k E N a finite poset has width at most w and has a contraction with no monotone (k + 1)  chain, it is called a (w, k)  poset. Let f (w, k) be the smallest integer such that there is no (w, k)  poset with that many elements. In other words: Every contraction on every poset that has at least that many elements, but width at most w, has a monotone (k 1)  chain.
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Then there holds:
+ l)IC. Proof. First we prove f (w, k) I (w + 1)" Assume that P is a (w, k) 2.20 Theorem [136]. f(w,k) = (w
 poset and that g is a contraction on it with no monotone (k+ 1)  chain. We proceed by induction on k. If k = 1, and then P a (w, 1)  poset, every element of P must be a minimal element. For, if an x E P would not be minimal we could choose a minimal element m I g(x). Then {m, x) would form a monotone 2  chain since m = g(m) 5 g(x) holds, a contradiction. Since the minimal elements of a poset are pairwise incomparable, now /PI I w follows and thus f (w, 1) 5 w + 1. Now assume k > 1, and that the assertion holds for all natural numbers < k. For each x E P let G(x) be a longest monotone chain in P which has x as greatest element. Then x is said to have eflective height IG(x)I  1. SOeach element of P has an effective height < Ic. Let M be the set of elements with effective height k  1. If M is deleted from P, then the restriction g 1 P \ M is a contraction on P \ M having no monotone k  chain. Indeed, it is impossible that there exists an element x E P \ M with g ( x ) E M. For such an element a longest monotone chain G(g(x)) could be prolonged by adjoining x as greatest element, thus producing a monotone (k 1)  chain in P. All elements of P \ M have effective height < k  1 in P, so that there are no more monotone k chains, and therefore P \ M is a (w, k  1)  poset. So P \ M has by definition a size < f (w, k  l ) , and this is = (w 1)"' by induction hypothesis. Thus IP \ MI I (w l)kl  1 holds. Next we find an upper bound for /MI.Let F be the set of fixed points of g in M.
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(1) Two fixed points x, y of F cannot be comparable, and therefore
M contains at most w fixed points of g. Indeed, if x, y would be comparable fixedpoints of g in M, say y < x, then a monotone k  chain G(y), having y as greatest element, could be prolonged to a monotone (k+ 1)  chain by attaching x as a new greatest element. For the same reason there follows: (2) There are no two comparable elements x, y E M which have the same g  image. Finally no element x E M is mapped by g onto an element y # x of M. For this would entail x > y, and again as before this would yield a monotone (k 1 )  chain. Then by (2) at most w elements of M \ F have the same g  image, which has to be in P \ M. This entails 1M \ FI w IP \ MI 5 w ((w 1)"'  I ) , so that finally IPI = IP \ MI + / M I = IP \ MI + IF1 + 1M \ FI (w l)k'  1 w w . ((w l)k'  1) = (W l ) k 1, SO that we have f (w, k) 2 (w l)IC. In order to show that also 5 holds we construct by induction on k a poset Pk having width w and size (w l ) k  1, and a contraction gk on it with no monotone (k 1 )  chain; then Pk is a (w, k )  poset and f (w, k) 2 (w l ) k .Here PI C . . C Pk will hold. Let PI be a single antichain of size w. This is a ranked poset where each element has rank 0. Let go map each of these elements onto itself. Then go has no monotone 2  chain, and we are done for the case k = 1. Suppose now that k is an integer > 1 and that for all K < k sets P, and ma,ppings g, have already been defined which satisfy the required conditions. Then we add to Pk' w disjoint chains to the top of Pkl, each having (w l)lC'elements, so that every new element of Pk \ Pk' lies above every element of PkV1.The new elements form (w + l)kl ranks of w elements each. Now Pk has (w 1)  1 (w + 1 )kl .w = (w l ) k 1 elements and width w. We put gk 1 Pkl := gk1. And let the minimal elements of Pk\ Pk' be fixed points of g k . Let R be one of the (w 1)"'  1 remaining ranks of Pk\ Pk'. Then all of its elements are mapped by g k onto a single element of Pk1, and this so that elements of distinct R's have distinct g  images. This is possible since Pk' has the same number (w l)kl 1 of elements as the number of those ranks of Pk\ Pk', which don't contain minimal elements of this set. Hereby the uppermost w ranks of Pk map to the w minimal elements of Pk'. The next highest w ranks of Pk map to the w elements of Pk' at rank 1, and so on.
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Now gk has no monotone (k 1)  chain. For by induction gk1 on no monotone k  chain, and in addition it follows that no two elements of Pk\ Pk1 can appear in a single monotone chain. For let x > y hold for two elements x, y of Pk\Pk1.Then they are in the same maximal chain C of this set. If y is the minimal element of C, it is fixed, and x has a smaller gk  image. If x an.d y have their gk  images in the same rank of Pkl,these are incomparable, and if they are in different ranks of Pk1, then by construction gk(x) < gk(y) holds. Pk1 has
9.3
Combinatorial properties of choice functions
In this section we consider the special case where such contractive mappings on sets of sets are investigated, which ascribe to each nonempty set a oneelement subset of it, or what is essentially the same, a single element of it. These mappings are nothing else than choice functions. Before we consider a special case of 2.11 we introduce some concepts:
3.1 Definition. Let D be an initial segment of N and f : D + N a function with f (x) 5 x for all x E D. The graph of f : D + N is the set of all pairs (x, f (x)) with x E D.
Then we define the concept doubly monotonic fchain of length 1 in the following sense: It is a sequence pl, . . . ,pl of 1 points of the graph of f ,where pi = (xi, f (xi)) holds for i = 1,.. . ,I with xl <  < xl, and where the sequences f (xi) and xi  f (xi) are monotonically increasing (the same: 5  preserving) with i. Of course, pl is called the endpoint of this chain. The octant oct(p) of a point p = (p1,p2) E R~ is the set which , which pl < ql , p2 5 q2 and contains all points q = (ql ,q2) of R ~for < 91PI < 0 1 holds. See Figure 19.
Figure 19
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In this section we put p q for two points of R2, iff q E oct(p). This defines an order in R2,and relative to this the concepts chain and antichain. 3.2 Theorem [72]. Let n E N . Then for each mapping f : {1,2,. . . ,2n) + N with f ( x ) 5 x for x E (1,. . . ,2n) there exists a doubly monotonic f  chain of length n+ 1. And this statement is tight. Proof. For n = 0 the assertion holds. Suppose now that the theorem is valid for a fixed n E N. Under this assumpton we first prove: (I) If v is an integer with 0 v 5 2n  1 and f , : D, := {1,2, . . . ,2n v ) + N a function with f,(x) 5 x for x E D,, for which no doubly monotonic f ,  chain of length n + 2 exists. Then there are v +1 points pl, . . . ,p,+l which are endpoints of a doubly monotonic f ,  chain of length n+ 1. For v = 0 , ( I ) holds by the induction hypothesis concerning n. Let now ( I ) be proved for a fixed v and f,+l : DU+l := {1,2,. . . ,2n v 1 ) +N a x for x E DU+l,for which no doubly monotonic function with f,+l(x) f,+l  chain of length n 2 exists. Let f , := f,+l {1,2,. . . ,2n v ) . By induction hypothesis (on v ) there are at least v 1 endpoints pi of doubly monotonic f ,  chains of length n 1 (they are also f,+lchains). We take those points pi of them which have the v 1 lowest x  components and choose the arrangement of the pi := ( x i ,yi) so that q <  . < x,+1 holds. Since f,+l is supposed to have no doubly monotonic chain of length n 2 it follows that each octant oct(pi) does not contain a p j , j E ( 1 , . . . ,v 1). We define sets P and R (see Figure 20) as follows: P contains those points of the graph of f,+l, that have an x  component > x,+l and a y  component > y,+l. (These points are in the parallelotope P of Figure 20.) And R contains the points of the graph of f,+l that have an x  component > x,+1 and a y  component < y,+l. (These points are in the rectangle R of Figure 20.)
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Figure 20
We define a function f * by altering f,+l as follows: We eliminate p,+l, and then we replace the points (a, b) of P by g(a, b) := (a1, b1), and the points (a, b) of R by g(a, b) := (a  1,b). The other points remain unchanged. So there arises a graph of a new function f * : D, := (1,. . . ,2n v) + N which satisfies f * (x) 5 x for x E D,. If there would exist a doubly monotonic f * chain of length n 2,then also there would exist a doubly monotonic f,+l chain of length n+2, which is excluded by our assumption. Now by induction hypothesis there exist v 1 points which are endpoints of doubly monotonic f * chains of length n 1. If pv+2 is that point of them, which has the greatest x  component, then g1(p,+2) is the endpoint of a doubly monotonic f,+l chain of length n 1. So, together with p,+l we have v 2 endpoints of doubly monotonic f,+l chains of length n 1. And (I) is proved by induction. Let now h : (1, . . . ,2n+1) + N be a function with h(x) 5 x for all x E (1,. . . ,2n+1}, p := 2n  1 and h,, : Dp := { 1 , 2 , . . . , 2 n + p ) + N the restriction of h to D,,. We assume that no doubly monotonic h,, chain of length n 2 exists, otherwise we are done. Then, by (I),there exist p 1 = 2n ends of doubly monotonic h,,  chains of length n 1. Let Q = {ql, . . . ,q2n) be their set, where the x  components xi of the points qi strictly increase with i. For p # q in Q we have p 4 oct(q) since otherwise p would prolong the chain ending in q. By induction hypothesis we have xl 5 2n. Then oct(ql) intersects the segment S, which links the points (2n+1,1) and (2n+1,2n+1), in a segment of length 2 2n. Since no qi is in oct(qj) for i # j of (1, . . . ,2n} each oct(qi) intersects S in a segment which has a subsegment of length 2 1, which
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is not contained in oct(qj) for j # i. So together u{od(qi)Ii = 1 , . . . ,2n) covers a subsegment of S of length 2 2n 2n  1 = 2"+l  1, which means it covers all of S. See Figure 21.
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Now the point (2"f l , h(2"+l)) is in o d ( q ) for some q E Q. So it prolongs the chain ending in q and produces a doubly monotonic h chain of length n 2. And the first part of the theorem is proved by induction. Another proof of this is given in [113],page 497. In order to see that the theorem is tight, we first prove:
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3.3 Lemma. Let k E N and f : D := ( 1,...,2k  1 ) + N a mapping satisfying f ( x ) 5 x for x E D. Let T be a kelement subset of D, over which f is strictly decreasing. Then T must be the set { k , k 1,...,2 k  I ) , andwe must have f ( k + v ) = k  v f o r v = O ,..., k  1 .
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Proof. For k = 1 this is trivial. So we assume k > 1. Let m be the least element of T. Then, due to IT1 = k , we must have m+k1 5 2k1, which means m k . Further there holds f (m)5 m k. And since the strictly decreasing set { f ( x ) l x E D ) must have k elements, f (m)2 k , and then also f (m) = k holds. Now m 2 f (m) = k entails m = k. Therefore T is the set of all integers of [k,2k  11, and the rest easily follows. See Figure 22 (here k = 5).
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Figure 22
The next theorem proves that Theorem 3.2 is tight, and even something more: 3.4 Theorem [72]. For each n E N there is exactly one function Fn : Tn := {1,2,.. . ,2n  1) + N with F,(x)5 x for x E Tn such that there is no (n 1)element subset of Tn over which Fn is monotonely increasing (the same: 5  preserving), not to mention Fn(x)and x 
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Fn (x). Proof. For n 6 N we define Fn as follows: For each integer m with 0 5 m < n and v < 2m we put Fn(2m v) := 2m  v. See Figure 23: So the graph of Fn is a union of n disjoint antichains of R ~ . It can easily be verified that every subset of Tn,over which Fn is 5 preserving, has at most n elements, namely at most one of each segment  11, 0 5 v < n. [2V,2V+1
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The theorem holds for n = 1. Suppose that it is valid for a fixed n E N , and let f : Tn+l := {I,. . . ,2n+1  1) + N be a mapping, satisfying f (x) x for x € Tn+1, such that there is no (n 2)  element subset of Tn+1 over which f is  preserving. A fortiori f is not doubly monotonic over the (n 2)  element subsets of Tn+1, and then we apply (I): We have Tn+1 = (1, . . . ,2n (2n  I)), and so by (I) there exist 2n points pi = (xi, yi), i = 1,. . . ,2n, which are endpoints of doubly monotonic f  chains of length n 1. Let the numbering be so that x1 < . . . < x p . Then we have yl > .  . > y p , for otherwise one of the chains ending in one of the pi could be prolonged to a chain of length n 2, with contradiction. Now it follows from 3.3 (with k := 2n) that f 1 [k,2k  11 coincides with Fn 1 [k, 2k  11. If f 1 [I,. . . , k  11 would be different from Fn 1 [1,. . . ,k  11 there would, due to the induction hypothesis, exist an (n 1)  element subset of [I,. . . ,k  11, over which f is monotonically increasing, and this set could be enlarged by an element of [k,k l] to an (n 2)  element subset over which f is still  preserving. This contradicts our assumption, and so the theorem is proved.
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3.4' Remark. A uniqueness theorem as in 3.4 no longer holds, if we take into consideration the monotonicity of f and x  f (x). If e.g. n = 2, and if we define f by f(1) = 1, f(2) = 1, f(3) = 3, then f is defined over {1,2, 22 1) and different from F2,but there is also no doubly monotonic f  chain of length 3. Theorem 3.3 can be brought into relation to Galton functions. We give an intuitive description of this. We consider a Galtonboard (see Figure 24), with which the Gaussian bell curve exp(x2) can be generated .
9.3. PROPERTIES OF CHOICE FUNCTIONS
We identify the left border line s with an initial segment of N. The way of a ball which runs through the board which we call a Galtonway  can then be described by a function f which is defined as follows: In ~ ~ (counting the vth line from above, the falling ball meets the f ( u ) point begins from left). Then we have 1 5 f (v) 5 v, and moreover (2) f (v 1) = f (v) or f (v 1) = f (v) 1 for all numbers v 1 of the range D of definition off. Therefore f is monotonically increasing over D, and that also holds for the function v  f (v). We call a function over an initial segment D of N and with values in N a Galtonfunction, if it satisfies (2) for all v E D. Theorem 3.3 can then be reformulated as:
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3.5 Theorem. Let f : D := (1,. . . , 2 n ) + N be a function satisfying f (x) x for x E D. Then there exists a Galtonfunction over (1,. . . ,2n), with which f coincides over at least n+l elements of D.
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Proof. There is a subset S C D of size n 1, over which v and v  f (v) increase monotonically, and then one can easily construct a Galtonfunction g over D, which satisfies g(v) = f (v) for v E S.
Subsequently to 3.3 one can pose the question whether 3.3 has a related theorem for infinite ranges of definition. In this context there holds a theorem of Alexandroff/Urysohn [3]: 3.6 Theorem. Let w, be a regular initial ordinal > wo and f : w, + w, a mapping which satisfies f (x) < x for 0 < x < w, and f (0) = 0. Then there exists a subset of cardinality N, of w,, over which f is constant.
Proof. In w, we introduce an order 5 (different from <) by x 5 y iff x = y or x = f(n)(y) for some n E N. It is easily seen that this defines an order relation. Moreover each y E w, has only finitely many predecessors, namely the elements f(n)(y),n E N , which are = 0 for all n of a nonempty final segment of N , since every strictly decreasing sequence of ordinals is finite. In particular this entails that ( w , , ) is a tree T with 0 as least element, which forms the 0  level Lo of T . And each y E w, has finite height, namely the least n E w for which f y Y ) = 0. If now the assertion would be false, every x E w, (and in particular 0) would have fewer than N, upper neighbors with respect to 5 . For n E w let L, be the n  level of the tree T . Now lLo1 = 1 and lLl 1 < N, holds. If in general we have already proved IL,I < N,, then L,+l is the set of all immediate successors relative to 5 of the elements of L,, and hence the union of fewer than N, sets which all have cardinality < N,. Thus, due to the regularity of N,, also ILn+1I is < N,. By induction we so obtain lLnl < N, for all n E N , and therefore also u{Lnln E w) would have cardinality < N,, which is impossible, since this set is nothing else than w, . If in 2.12 the function f is a choice function over !$3(h(n)),one can determine the least value h(n) for which 2.12 holds. Namely the following theorem of Perry shows this. Other proofs of it were given by Lovasz ([114], p.500) and by Kleitman and Lewin [97]. We present here (in a slightly modified form) the proof of KleitmanILewin:
3.7 Theorem of Perry [140]. Let k,n be natural numbers with k 2 2,, S = {I,. . . ,k), and f a choice function defined on the nonempty subsets of S. Then there is a chain Cl C  .C Cn+l of subsets of S, such that all f (C,), v = 1 , . . . ,n 1, are equal. This statement is tight.
+
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317
Proof [97]. For 1 5 i 5 k let si be the size of a maximalsized chain of subsets of S whose f  image is i. First we prove: Lemma. If under the above assumptions k = 2"+j with 0 5 j 5 2", then (1) x L l ( s i  n) 2 j 1. If the lemma is proved we are done, since (1) implies that at least one of the summands of the left side is > 0, so that the corresponding si is n + 1.
> +
>
Proof of the lemma. (1) can be put in the following form: (1') ELl si  k  n > 2 j + 1 . The lemma is true for n = 1, j = 0. Indeed, k is 2, and therefore at least one of the si is > 2, for there is at least one 2  element subset T of S, and then f (T) and f (f (T)) form a twoelement chain. All others are 1. Thus C{sili = 1,. . . ,k) k 1. Suppose now that the lemma is true for a fixed n and j  1, where j is > 0. We prove that then it is also true for n, j. That means we have to verify (1). Let S' be the set {I,. . . ,k  I), and s{ the size of a maximalsized chain of subsets of S' whose f  image is i. We have k  1 = 2" + ( j  I), where 0 5 j  1 5 2" and k  1 E N. Then our induction hypothesis yields: ! (2) k1 (szn) = k1 s{(k1).n 2 2 . ( j  1 ) + 1 = 2j1 > 0. Then for one summand we must have s{,  n > 0, and thus (3) sio > n. Consider now Srl:= {1,2, . . . ,k)\{io), and let s: be the size of a maximalsized chain of subsets of S" whose f  image is i. Then the induction hypothesis yields (4) C{s:li E S")  ( k  1)  n 2 j  1. For every i = 1,.. . ,k we have si I si and s: I si. Now we add io to S" and note that the set S itself increases the length of the maximalsized chain with image f (S).We obtain (5) Eel si  k n = 1 E{s:li E S1') s{,  (k  1) . n  n, by (3) and (4) the righthand side is 1 2 j  1 1 = 2 j + 1. And so (1) is proved. Summarizing we have obtained that the lemma holds for our fixed n and all j 5 2". In particular it is valid for j = 2", in which case we have k = 2j. Here (1) becomes EL1si  k . n k 1 and thus
>
> +
>
>
+
+ > +
> +
+
c:=,
+
(6) si  k  (n 1) L: 1, which is the lemma for n 1 and j = 0. So the lemma is proved by induction, and with it the main part of the theorem.
+
In order to see that the theorem is tight we consider the mapping
Fn : Tn + N of 3.4, where Tn = {1,2, . . . ,2n  1). If M is an m  element ) ) ~ ~ (in the subset of Tn with m 2 1, we ascribe to M the ( ~ ~ ( mmember usual order) of M. Then for each chain C1 c . . c Ct of the power set p(T,), over which this choice function is constant, we must have that the Fn(IC, I), i = 1,. . . ,t , form a monotonely increasing sequence. This has by 3.4 at most n elements, and so t 5 n follows. Some of the previous theorems can be interpreted as statements about subtrees of finite power sets. For there is a close connection between regressing mappings in finite power sets and subtrees of these. Referring to this we introduce some concepts: 3.8 Definition. Let S be a finite set and f a regressing mapping of (!Q(S),)I( into itself. Then we define an order relation 5 on p ( S ) by: For A, B C_ S we put A 5 B *There exists an n E w such that A = f n(B), where f is the nth iteration of f; in particular f O(B) = B. It is immediately clear that (Q(S),5 ) is a tree, the tree defined by f. The order 5 is a suborder of E, and 0 is its least element. a In analogy to a notion of graph theory we call a tree (p(S), spanning tree of the poset (p(S), E), if s t i s a subrelation of 2 for which 0 is the first element.
st)
The correspondence between regressing mappings on spanning trees of this set is now expressed by:
(p(S), E) and
3.9 Theorem. For a finite set S each regressing mapping on (p(S), E ) defines a spanning tree of this set, namely the tree defined i n 3.8. If on the other hand (!Q(S), is a spanning tree of (p(S), E), then we define f : p(S)+ p ( S ) by: f (0) = 0, and for 0 # T 2 S let f (T) be the (uniquely defined) immediate predecessor of T with respect to It. Of course f is a regressing mapping on ( p ( S ) ,E).
st)
Again in analogy to a graphtheoretical concept we define: 3.10 Definition. A tree (T, I)is a star, if all elements of it which are different from its least element o are pairwise incomparable.
9.4. THEOREMS ON INFINITE POWER SETS
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Theorems 2.13 and 3.7 can now be reformulated in the following forms: 3.11 Theorem. For each n E N there is an m := m(n) E N, for which there holds: Let S be a finite set with IS1 2 m and (p(S), s t ) a spanning tree of ( p ( S ) ,(I). Then this tree has a chain of length n, or S, which has n upper neighbors i n the tree, w,'lich there exists a T form a chain i n (p(S), C).
c
3.12 Theorem. Let S be a set with 2n elements, and (p(S), s t ) a spanning tree of ( p ( S ) ,c ) , i n which 8 has the oneelement subsets as upper neighbors, and where each ( 2 2)  element subset T S has as unique lower neighbor a oneelement subset {x) with x E T, (so that the final segments T,, which are generated by the oneelement subsets {x) of (p(S), s t ) , are stars), then there is an (n 1)  element chain of ( p ( S ) ,G), which is a subset of one of the stars T,, which partition the set of nonempty subsets of p(S).
c
+
9.4
Combinatorial theorems on infinite power sets
Some order types T have the property that each poset P with type T has the property: However P is split into two parts A, B,at least one of them still has a subset of type 7.We call types T which have (resp. don't have) this property unsplittable (resp. splittable). E.g. the linearly ordered set Q of rational numbers is unsplittable: If a subset A 2 Q has no subset of type ho = tpQ, A cannot be dense in Q, and so there exists an interval [a,b] with a < b, which is disjoint to A. But then [a, b] is G B := Q\A, and B has a subset of type ho. Other examples of such unsplittable posets are the wellordered sets whose order type is an indecomposable ordinal (see 4.8.8), e.g. the set N of natural numbers. Of course, the question arises how the set R behaves in this context. It turns out to be splittable. In the discussion of the properties of infinite power sets the sets H , of 4.3.3 and the sets C, of 4.7.31 play an essential role. These subsets of 2((w,)) are always considered as linearly ordered by the principle of first differences. Let us recall: C, is dense, and H, is strictly dense in C, (4.7.32) . Further C, has no gaps, and so it is a linearly ordered continuum in the sense of 7.1.5. The set 2((w,)) is isomorphic to a
subset of C, (4.7.33), but 2((w,)), and thus also C,, is not isomorphic to a subset of Ha (4.7.11) In particular the set Co has no gaps and no first and no last element, and the set Ho,which has the order type ho of the set of rational numbers, is dense in Co. Therefore Co is orderisomorphic to the set R of real numbers, and the sets C, are the natural generalizations of it to higher cardinalities. As a special case of 4.1.14 we have that the power set (q(w,), E) is isomorphic to (2((w,)), sp),where l p i s the product topology, and that there exists a <  preserving mapping of (v(w,), E ) in 2((w,)), if now 2((w,)) is linearly ordered by first differences. This is one of the reasons why it is of interest to study the order properties of 2((w,)) and its subset C,. We introduce a concept for isotone functions: 4.1 Definition. Let D C C, have at least two elements, and let f : D + C, be a 5  preserving function. Then j E D is said to be a jumpspot off, if at least one of the following two statements holds: a) j is not the first element of D and sup{f (x)lx E (D < j)) < f (j). Or b) j is not the last element of D and f (j)< inf{f (x)lx E (D > 3)).
In analogy to the fact that a monotonically increasing function f :
R + R has only countably many jumpspots, we here obtain for the cardinal t,, which was defined in 4.3.5:
4.2 Theorem. Let D and f be given as in 4.1. Then the set J of all jumpspots of f : D + C, has cardinality t,.
<
Proof. Each jumpspot x off defines a segment S, of C,, which has more than one element and which contains no element of f (D). Since H, is dense in C, we can choose an element ex E S, f l Ha. The mapping which ascribes to each x E J the element e,, is evidently <  preserving, in particular injective, and thus J has a cardinality 5 IH,I = t, (by 4.3.7). We mention a very simple extension theorem:
<
4.3 Theorem. Let M be a subset of C, and f M : M + C, a preserving mapping. Let S be the least segment M of C,. Then there exists a 5  preserving mapping f : S + C, with f t M = fM.
>
Proof. We put f 1 M = fM. If x E S\M, there exists a greatest x containing segment Sx of S which is disjoint to M. Then we define
9.4. THEOREMS ON INFINITE POWER SETS
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f over S, to be constant = c, where c is any element which satisfies c 2 f ( t ) for the t E M , that are < the elements of S,, and c f ( t )for the t E M , which are > the elements of S,.
<
4.4 Theorem. Let S be a subset of C, with JSI= 2Na. The set F of all  preserving functions f : S + C, has cardinality 5 2'a. Since F contains all constant functions we have IF1 2 IC,l = 2Na. Using GCH this implies that F has cardinality 2Na. If S contains a segment of C,,which has more than one element, then IF1 = 2'".
<
Proof. Let f : S + C, be a 5  preserving function, J the set of its jumpspots. If a < b are elements of H, and if the interval [a, b] of C, intersects S we choose an element of [a, b]nS. Let D be the set of all elements, which are so chosen. Then ID1 5 t, holds since the set of all pairs (a, b) E H, x H, has cardinality t,. Now f is uniquely defined by its values over the set M := J U D l whose cardinality is 5 t, by 4.2. Indeed, if t E S\M, and if t is neither the first nor the last element of S, we have s := sup{ f (x)lx E ( S < t)) 5 f ( t ) 5 inf { f (x)lx E ( S > t ) ) =: j, and since t is no jumpspot, we must have s = j = f (t). If t is the first (resp. the last) element of S, f (t) is = j (resp. = s). It follows that IF1 5 I C r l 5 IG',lea = (2Na)@a = 2ea. Let now S contain a segment of C, which has more than one element. Since Ha is dense in C, the set S has a subset of type h, = h,  h, (by 4.4.5). Then S has a subset T which can be represented as the ordered sum T = C{Tili E Ha), where each Ti has order type h,, and where the Ti, i E Ha, are pairwise disjoint. We wish to construct an injective mapping of the set of all subsets of H, into the set F. To H 2 H, we ascribe a mapping fH by: fH(x) = x if x E u{Tili E H), and over each T i , where i E H,\H, we let f be constant = an arbitrary fixed element of Ti. So every subset H H, defines a function f H E F, and different sets H define different functions fH. SOthe set of 5  preserving functions over T has at least as many elements as the power set of Ha, which means 2 2'a many. The rest follows with 4.3. 4.5 Lemma. Let T C, have cardinality IT1 = 2Na.Then the set U of all subsets of C, that have the same order type T as T has cardinality 5 2'".
Proof. For each U E U there exists a <  preserving surjective
mapping f u : T + U. The mapping, which ascribes to each U E U the corresponding fu, is injective, and since { fulU E U) has cardinality 5 2'a by 4.4 also U has a cardinality 5 2&a. We can now prove the theorem that the GCH implies that all subsets of C, of cardinality 2Na are splittable:
4.6 Theorem. Using GCH there holds: Let T C C, have cardinality 2Na. Then we can partition T i n two subsets A and B, such that T is neither embeddable i n A nor i n B. (In the terminology of 1.9.6 : tpA and tpB are < tpT.) Proof. The set U of all subsets of T that have the same order type as T is by 4.5 of cardinality 5 2ea = 2Na (due to GCH), and so we can find a wellordering of their set in the form U = {T,lv < A), where A is an ordinal 5 ~ ( 2 ~First ~ )we. choose two different elements a0 and bo in To, then two different elements al, bl E Tl\{ao, bo). If in general we have a p < A, for which elements a,, b, for v < p are already defined, we choose two different elements ap, bp E Tp\(U,
c
4.7 Theorem. Let T E Co (or T R) have cardinality 2N0.Then there exists a partition of T i n two subsets A,B such that T is neither embeddable i n A nor in B. Since from the sets 2((w,)) and C, each one is embeddable in the other (see 4.7.33), we obtain from 4.6 and 4.7:
4.7' Theorem. Every set T C 2((wo)) of cardinality 2 N can ~ be partitioned in two subsets A and B, such that T is neither embeddable i n A nor in B. And using GCH there holds: Every set T 2 2((w,)) of cardinality 2Na can be partitioned i n two subsets A,B, such that T is neither embeddable i n A nor i n B.
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From 4.6 we can derive a similar decomposition theorem for power sets of sets of cardinality N,, making use of characteristic sequences. First we establish a connection between the power set p(w,) and the set C,. 4.8 Theorem. Using GCH there follows: Let S be a set of cardinality N,. Then p ( S ) has a subset C,, such that (C,, C ) has the order type A, of C,. Since 2((w,)), with the order by first diferences, is embeddable i n C,, then also @(S) has a subset which is isomorphic to 2((w,)).
Proof. W.r.0.g. we can assume S = w,. Due to the GCH, H, has cardinality t, = N,, and therefore H, is embeddable in the N, universally ordered set (@(w,), G).So there exists a subset 4, C p(w,), so that (4,, c ) has order type h,. The set of initial segments of H, has a subset which is isomorphic to C,, and thus also 4, has a set C, C p(w,) of initial segments which is isomorphic to C,. 4.9 Theorem. Using GCH there holds: Let S be a set of cardinality &,and let T be an order type of a linearly ordered set of cardinality 2Na. Then p(w,) can be partitioned i n two sets U and 23, such that neither U nor '23 (both ordered by E ) have a subset of type T.
Proof.
It suffices to prove the theorem for the situation where S = w, holds. In this case there exist by 4.7' two disjoint subsets A and B of 2((w,)) with 2((w,)) = A U B, such that neither A nor B has a subset of type r.We consider the bijective <  preserving mapping f : 2((w,)) + p(w,), which to every sequence (xVIu < w,) E 2((wQ)) ascribes the set of those u < w,,for which xu = 1 holds. We put U := f [A] and 23 := f [B],so that p(w,) = U U '23. Now U has no subset of type T , for in this case also A had to contain a subset of type T . For '23 we conclude analogously. 4.10 Theorem. Using GCH there holds: Let S be a set of cardinality N,. Then there is a partition of the power set p ( S ) i n two subsets U and 23, such that ( p ( S ) ,CJ is neither embeddable i n ( U , 5 ) nor i n (23, C ) .
Proof. Let
be the type of the set 2((w,)), which is ordered by first differences. Then we consider the representation p(w,) = UU '23 of T
4.9. Now q(w,) has by 4.8 a subset of type tp2((w,)), but U and 23 do not. Until here we had dealt with partitions of certain posets P in two subsets, and this mainly with respect to the question of whether these sets P are splittable. Now we consider also partitions of posets, in particular of Ha, C,, U, and power sets, in infinitely many subsets and investigate how sumptuous these subsets are.
4.11 Theorem [73]. Let w, := cf(w,) and P < y, and let Ha = u{K,Iv < wp). Then one of the classes Kv contains a subset of type ha. Proof. There exists an interval [a, b] of Ha, in which a class Kv is dense. Suppose the contrary. Then there is an interval [ao,bo] of Ha, which is disjoint to KO.In general, let p be an ordinal < wp, such that for each v < p an interval [a,, b,] is defined, which is disjoint to Kv,so that the [a,, b,], v < p, form a decreasing sequence. Then it follows that {avlv < p) and {bVlv < p) are not neighboring since Ha is an q,  set. So there exists an interval of Ha which is situated between these sets, and this has a subinterval [a,, b,] which is disjoint to K,. By transfinite induction we so obtain a descending sequence [a,, b,], v < wp, of intervals of Ha. Their intersection is a nonempty segment S of H, since A := {avlv < wp) and B := {b,lv < wp) have a cardinality 5 Np < N, and are therefore not neighboring. But this yields a contradiction since S is disjoint to all K, and thus also to Ha. Let now K, be a class which is dense in an interval [a, b] of Ha. By 4.6.3' (a, b) has type ha. Due to 4.4.5 we have h, . h, = ha, and therefore [a, b] has a subset M which has the form of an ordered sum C{SiliE Ha), where the Si are disjoint segments of M, which all have the order type ha. Then Kv has at least one element in each of the Si and therefore a subset of type h,. As a corollary to 4.11 we list:
p
4.111Theorem. Let S be an 7,  set and S = u{K~Iv< wp), where < a holds. Then one of the sets Kv still contains an q,  set.
Proof. S has by 4.3.13 a subset of type ha. If now w, is regular the assertion follows from 4.11. If w, is singular, then S is also an qa+l  set by 3.3.5. Then S contains a set of type and a fortiori the assertion follows with 4.11.
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4.12 Remark. If in 4.11 we replace wp by w,, the statement no longer remains valid. For if w, is regular, then, due to 4.3.12, H, is the union of N, subsets which have no subsets of type w, (and w:), and a fortiori no subsets of type ha. If w, is singular and w, := cf(w,), say w, = sup{w,,lp < w,), where all w, are < w,,then from 4.4.3 we obtain a representaion H, = u{T,I~ < w,), where Tp has the type h,,+l, so that each Tp has no subset of type ha. Next we consider partitions of the sets C,.
4.13 Theorem. Let C, = U{KUIv < w,), where w, = cf(w,). Then one of the classes Ku contains a subset of type ha. Proof. One of the classes Ku must be dense in an interval [a, b] of C,, where a and b are in Ha. Suppose the contrary. Then in the same way as in 4.11 we construct intervals I, := [a,, b,], v < w,, which form a descending sequence, and where I, is disjoint to K,. Then the intersection D := n{I,lv < w,) is nonempty, since it contains sup{a,lv < w,). But on the other hand D is disjoint to all Ku, v < w,, and thus also to C,, a contradiction. If in 4.13 w, is regular and thus = w,, %+I. For if we assume the GCH, then IC, partitioned in its N,+l oneelement subsets.
W,
I
cannot be replaced by = 2Na = Na+l could be
Concerning the cardinalities of the classes in 4.13 we see that one of the classes must have the same cardinality as C,, even in a more general context:
4.14 Theorem. If C, = U{K,IV < w,), then for at least one u we have lKvl = 2Na(= IC,l). And further: A t least one of the classes Ku is not embeddable i n H,. Proof. We can assume w.r.0.g. that the Ku are pairwise disjoint. If now IK,J < 2Nawould hold for all v < w,, then, applying the theorem of Konig, we would have IC, ( = 2Na = C{IK, I Iv < w,) < (2Na)Na= 2Na, with contradiction. And if all Ku, v < w,, would be embeddable in Ha, then this would also hold for their union C, by 4.4.6. But this contradicts 4.7.11. For the proof of the main theorem of this section we still need some preparations.
4.15 Definition. Let a be an ordinal, w, = cf(w,), U a set with IUI = t,. For p < w, we define S,, to be the set of all sequences (u, Iv p) with U, E U for v p. So each sequence of S,, has a last element up. Then we put S : = u{S,,Ip <: w,). If we have a partial order in U, we introduce a partial order in S by a variant of the principle of first differences: If u = (uVIv5 p) and u' = (uLlv p') are different elements of S we put u < u' there is an index, and then also a first index 6 (< min{p, p')) for which us # uk holds, and us < u; (in U). It can easily be verified that the sodefined relation <, augmented by the identity relation on S, is a partial order of S. In particular it follows that if an element u E U'is a proper initial segment of an element v E U, then u and v are incomparable in S. We determine the cardinality of S:
<
<
<
4.16 Lemma. (SI= t,.
Proof. We have IUI = t, = C{2Nu Iv .< a) = sup{2NuIv < a). Case 1. w, is a regular initial ordinal. Then, due t c the definition oft,, we can find a representation U = u{A,Iv < w,) where ]A,[ = 2 N ~ , and where the A, are pairwise disjoint. We put Mv := u{A~IL v). Then IM,) = 2Nvfor v < w,, and the sets M", v < w,, form an ascending tower of subsets of U. If p < w, and if (uo,. . . ,up) is in Sp,then each u,, L 5 p, is in a set M,(,) with V(L) < w,. The set { V ( L ) / L 5 p) cannot be cofinal in the regular ordinal w,, and so there is a v < w, such that all u,, L 5 p, are in . set has cardinality M,. Then the sequence (uO,. . . ,up) is in M / + ~This (2Nv)lp+11:= k. Fkom p < w, we conclude Ip 11 = N, for an ordinal T < N,. Now k is = 2N85 t,, where 6 = max{v, 7). The set S is a subset of ~ { ~ / + ' l
<
+
<
<
<
<
<
9.4. THEOREMS ON INFINITE POWER SETS
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Now we can prove the main theorem of this section: 4.17 Theorem [73]. Let a, be an ordinal, wy := cf(w,), U a set with JUI = t, and p ( U ) = U{KuI~< w7). Then one of the sets Ku contains a subset of type T, := tp(U,).
Proof. W.r.0.g. we can assume that U is the set U, of 5.2.10. This has the advantage that we can use the additional order structure of U, in the proof. We consider the set S of 4.15, which also has cardinality t,. Since the power set p(U,) is t,  universal, there exists an isomorphic mapping g of S onto a subset g(S) of p(U,). So each g(s), s E S, is a subset of U, We define S*to be the set of all sequences (u,Iv < A), where A is an ordinal < wy and u, E U, for v < A. So S* S, and S*also contains the empty sequence a. If cp = (u,Iv < A) E S*and x E U, we (of course) put (cp, x) := (uo,.. . ,u,, . . . (V < A), x), and similar if x is before cp. For p < wy we denote the set of all sequences E S, that have (u,lv 5 p) as initial segment, by S(uo,.. . ,up) or S(u,lv 5 p).'This set is evidently a segment of S. For each S(uo,. . . ,up) of S we define: C(u,lv 5 p) is the least segment of Y(U) which contains the image set g[S(u,lv 5 p)]. It is the convex hull of g[S(u,lv 5 p)] in p(U). First we prove: (I) Let x and y be different elements of U, cp E S*,and let X E C(cp,x), Y E C(cp,y). Then x < y =+ X C Y, and x (1 y 3 X 11 Y. X is in the smallest segment of p ( U ) which contains g[S(cp,x)], and so there are sets A, B E g[S(cp,x)] with A C X E B. That means: There are two sequences cpl, cp2 E S with
>
(1) d c p , x, cpd C X L dcp,x, cp2). For the same reason there are sequences ql and
$2
E S*satisfying
(2) d c p , Y, $1) E Y E dcp,Y, @a). Let x, y E S. Suppose now x < y. Then (1) and (2) yield X g(cp,x,cp2) C g(cp,y,@d C Y.Indeed, x < Y =+ (cp,x,cp2) < (cp,~,?Irl), and g is <  preserving. If x 11 y we conclude as follows: Suppose indirectly that X and Y are comparable, w.r.0.g. we can assume X C Y. Then we would have g(y, x, cpl) C X E Y C g(cp, y, $2). But this is impossible because the sequences (cp, x, cpl) and (cp, y, $2) are incomparable and since g is an isomorphic mapping, and then g(cp, x,cpl) cannot be a subset of g(cp, y, &). So (I) is proved.
If now KOhas an element in common with each set C(uo), uo E U,, then the set of these elements forms by (I) (in this case cp is the empty sequence) a subset of KO which is isomorphic to U,, and we are done. So we can assume that there is an element eo E U,, such that the set C(eo) is disjoint to KO. Suppose that for some p < w, we have defined elements e, E U, for v < p, such that each set C(eo, . . . ,e,) is disjoint to Ku (and, by the way, then also disjoint to all K,, with p < v), then we proceed as follows: We consider the set of all sequences (eo,. . . ,e,, . . . (v < p), up) with up E U,  which prolong the sequence (e,lv < p) by an element up of U,. If Kp intersects each set C(eo,. . . ,e,, . . . (v < p), u), u E U,, it follows by (I) that then Kp has a subset of type T,, and we are done. So we finally have to discuss the situation where the above construction could be performed over w, steps. But this situation cannot happen: Suppose the contrary. Then we choose for each v < w, elements a,+l and bU+l in U,, which satisfy a,+l < e,+l < b,+l (in U,). This is possible since U, is an qrr  set. Now we define intervals I, := [A,, B,], v < w,, of p(U,) as follows: We choose an element A, E g[S(eo,. . . ,e,, a,+l)] C C(e0, . . . ,e,, a,+l), and an element B, E g[S(eo,. . . ,e,, b,+l)] C(eo, . . . ,e,, b,+l). Then A, C .€holds I, since each element of S(eo, . . . ,e,, a,+l) is < each element of S(eo,. . . ,e,, b,+l), and g is <  preserving. Now I, is disjoint to K, since K, does not intersect C(eo,. . . ,e,), due to our indirect assumption, and C(eo, . . . ,e,) contains A, and B,. By construction the I, form a decreasing sequence I, . . ,v < w,. In particular A. Al .. C lo Il A, . . . , v < w,, holds. The intersection fl{I,lv < w,) is disjoint to all Ku,v < w,, and then also to their union p(U,). But, on the other hand, the set A := u{A, IV < w,) is contained in n{I,, v < w,) and, of course, an element of p(U,), which yields a contradiction.
c
> > c
>
>
c
c
The Theorem 4.11 has an analog for the partially (and N,  universally) ordered sets U, of 5.2.10: 4.18 Theorem. Let w, = cf(w,), P < w,. And let U, = u{K,Iv wS). Then one of the sets Ku contains an rly,  set.
<
Proof. There exists an interval [a, b] of U,, in which one of the sets
Ku is dense. Suppose the contrary. Then there is an interval [ao,bo] of U,, which is disjoint to KO.Let p be an ordinal
< wp, such that
for
9.4. THEOREMS ON INFINITE POWER SETS
329
each v < p already an interval [a,, b,] of U, has been constructed, so that [a,, b,] is disjoint to Kv, and so that the intervals [a,, b,] form a decreasing sequence. Then the union of the sets A := {a,lv < p) and B := {bUlv < p) forms a chain C, and there exists a maximal chain M C of U,. Since U, is an rlrr  set by 5.2.12, the set M is an qy  set by 5.2.4. Now A and B both have a cardinality < N,, and therefore they cannot be neighboring in M. So there is an element of M between them. On the other hand the intersection n{[a,, b,]lv < wp} is by construction disjoint to all K,, and then also to their union U, .This is a contradiction. Let now Ku be dense in the interval [a,b] of U,. By 5.2.12 resp. 5.2.8 U, and the open interval (a, b) is an q7,  set, and by 5.2.7, then also Ku n (a, b) is an  set, which proves our assertion.
>
The analog to 4.11' is:
4.19 Theorem. Let S be an qaa set and S = u{K,Iv < wp}, where p < a holds. Then one of the sets Ku contains an q,, set.
Proof. If the assertion would be false, we could again construct a decreasing sequence of intervals [a,, b,], v < wp, of S, such that [a,, b,] is disjoint to Ku for v < wp. The intersection n{[a,, b,]lv < wp} would then be disjoint to U{K,~V< wp}, and thus to S. But the sets {a,lv < wp} and {b,lv < wp} both have cardinality < N,, so that they cannot be neighboring in S, and therefore there exists an element of S between these sets. Finally one can pose the question of whether it is possible to increase the number of classes in Theorems 4.13 and 4.17 while the conclusion remains valid. For a special situation this is possible, as the following theorem shows, which we mention without proof:
4.20 Theorem [69]. a) If C, = U{K,IV < w,), then there is a < w,, so that Ku has a subset of type T+T* for every ordinal T < w,+l. b) Let IS1 = t,, and P(S) = u{K,Iv < w,). The?& there is a v < w,, so that K, has a subset of type T + T * for every ordinal T < w,+l.
v
If w, is regular, then these statements are surpassed by 4.13, resp. 4.17. For singular w, this is by no means the case.
Chapter 10 Comparison of order types In 1.9.6 we introduced the 5  and the <  relation for order types, which was defined by Fraiss6 [44]. In the class of wellordered sets we have already had a complete discussion since the order type of a wellordered set is practically the same as an ordinal. (The exact definition of order type was given in 1.9.5.) In particular in the class of order types of wellordered sets we have antisymmetry, so that this class is ordered by 5 . The situation changes radically if we consider the order types of more general linearly ordered sets, and even more of partially ordered sets. Here the relation 5 is no longer antisymmetric, and so the class of all order types is only quasiordered by 5 . A modest oversight over the class of order types is given by the N,  universally ordered sets. For the linearly ordered sets the sets Ha of 4.3.3 are N,  universal, and for partial orders the sets U, of 5.2.10 are N,  universal. Here we recall 5.3.9', which states that the order of U, can be extended to a linear order in U,, with which U, has the order type h, of H,.
10.1
Some general theorems on order types
1.1 Definition. If r is an order type we define 171 to be the cardinality of a realization of T, and generally we ascribe to r all those properties which all realizations of T have, e.g. dense, countable, wellfounded and so on. A first question arises: How many order types of a fixed cardinality exist ? We consider here only the infinite case. For this we obtain the following result, which is tight: 1.2 Theorem. There are 2Na diflerent types of
linearly ordered
sets of cardinality N,.
Proof. In this proof let 2 (resp.1) denote the order type of a 2element (resp. lelement) chain and q the type ho of the set Q of rationals with their usual order. Let C be the set of all order types C{r,lv < w,), where T, = 2 for even ordinals v < w, and r, E {I,?) for odd ordinals v < w,. The chains with such an order type evidently have cardinality N,. And there holds:
(I) Two chains with such sums, which differ in at least one Tu, have different order types: Let C be a chain of type C { ~ , l v < w,), and C' a chain of type where T, arid 7; are = 2 for even v, resp. E {l,q) for odd v. Suppose that there exists an isomorphic mapping f : C + C'. Then f necessarily maps the first element of C onto the first element of C', further the second element of C onto the second of C'. If TI = 1, then also T; must be = 1, for otherwise T{ would be = q. But this is impossible since C has a third element and C' would lack such an element. In general suppose that for all v < p, where p < w, holds, we have already T, = TL and that for the segments T, of type T, of C and the segments TL of type TL of C' we have f [T,] = TL, then also T, = T; must hold. For if we have T, = 1, which implies that p is odd, then the set u { T U I ~5 v) has a first element, and then this also must hold for u{TLIp 5 v), and this entails T; = 1. If T,, = q, then for a similar reason also 7; must be = q, and the isomorphic mapping f must map the initial segment of type T, = q of u{T,Ip 5 v) onto the initial segment of type T; of u{T;Jp 5 v). By induction we so obtain that T, = T; for all v < w,, and therefore the chains C and C' have the same order type. And (I) is proved. Now the cardinality of C can easily be counted. A type of C is by (I) completely determined by the set of the Tu, where the v are odd ordinals < w,. For each such T, we have two possibilities (to be = 1 or = q). This leads to 2Na order types in C. It is easy to see that there cannot be more than 2Na order types (the types of the partially ordered sets included !) of cardinality N,. For even the set of all relations over a set A of N, elements contains only 21AxAI= elements.
~ { T L I V < w,),
We mention two simple examples: 1.3 Example. a) The type of a twoelement antichain is incomparable with each type of a chain which contains at least two elements. b) The type w: of the set of negative integers is incomparable with all types of wellordered sets of type 2 wo. For a wellordered set has no infinite. descending subset, and dually for inversely wellordered sets.
In [13] Chajoth studied how the order type of a chain can alter if we change the position of elements in a linearly ordered set, resp. if we introduce a new element in a linearly ordered set. In the following his
10.1. SOME GENERAL THEOREMS ON ORDER TYPES
333
theorems are partially transferred onto the case of general posets, using the same methods: 1.4 Theorem. One can change the order type of an arbitrary poset P by introducing a new element.
Proof. If P has no first element, we can introduce a new element e and put e < x for all x E P. Then P U { e ) has a type different from that of P. In general, there exists the greatest wellordered initial segment W of P, such that all elements of W are situated before all elements of P \ W. (If P has no first element we have W = 8.) Let T be the order type of W. If then we introduce a new element e and put w < e < y for all w E W and y E P \ W, then P ~ { e has ) another order type than P, since an isomorphic mapping of P U{e) on P had to map the initial segment W U {e) of P ~ { e )which , has type T 1, onto an initial segment of P which also must have type T 1. This contradicts the maximality of T.
+
+
Contrary to the situation of the last theorem there exist many sets, in which one can omit single elements, and also greater subsets, without changing the order type of the set. This is e.g. the case with N, Z and Q, but not with R, since the omission of a single element of R produces a gap, and R has no gaps. 1.5 Theorem. Let S be a set of infinite cardinality m . The set of all those orders 5 on S, for which ( S ,5 ) retains its order type, if one removes finitely many elements of S, has the same cardinality as the set of all orders on S.
Proof. For a poset A we define a poset A, as follows: For each x E A we choose a set A(x) of the type w of the natural numbers with x as first element, such that the sets A(x), x E A, are pairwise disjoint. We put A, := ~ { A ( x ) l xE A) and order it by: Each set A(x) is ordered as before, and for a, b E A, with a E A(x) , b E A(y), x # y in A we put a < b iff x < y. Roughly speaking: We enlarge each element x E A to a set of type w, and order the new elements in the same way as the first elements of their w  string.
(I) If now (A, 51) and (B, 5)are nonisomorphic posets, the types of A, and B, are also different. And conversely. For suppose indirectly that there is an isomorphic mapping f from A, onto B,. If we have elements x,y E A (C A,) with x < y, the
images f (x) and f (y) cannot be in the same segment B(z) of B,. For in this case there would be only finitely many elements between f (x) and f (y) whereas there are infinitely many elements of A,, in particular of A(x), between x and y. So, if f (x) E B(z), only elements of A(x) can be mapped by f into B(z), and analogously different elements of A(x) cannot have images in different sets B(z). And so f maps A(x) bijectively onto B(z). Herewith the first element x of A(x) must be mapped onto the first element z of B(z). Now f A is an isomorphism of A into B. For the same reason f' f B is an isomorphism of B into A , and so finally f f A maps A isomorphically onto B. This proves (I). If A, and B, have different order types, then A and B do also. For it is trivial that isomorphic posets A, B have isomorphic posets A, and B w. Evidently the sets A, retain their order type if one removes finitely many elements of them, since this also holds for all sets of type w. And since before we had established a bijection between the set T of order types of the posets (S,<) and the set T*of order types of the posets (S, , <), we have IT*1 = [TI, and we are done.
r
Finally we obtain a characterization of the finite linear order types: 1.6 Theorem. The finite linear order types are characterized by the
fact, that they don't alter, i f in their realizations an element changes its position.
Proof. Let T be a linear order type with an infinite realization S. If S has no first element, we can take an arbitrary element of S and move it before all other elements of S. That changes the order type. Assume now that S has a first element, and let A be the greatest wellordered initial segment of S. If this is the whole set S, then S has a type T 2 W. Then we take the first element of S and move it behind all other elements of S, so obtaining a set of type T + 1, and we are done. If S \ A is nonempty, we choose an element e E S \ A, place e behind A and before the elements of (S \ A) \{el and obtain a linear order whose type differs from that of S, since now there is an initial segment of type > tp(A). We supplement the definitions of section 4.1 by: 1.7 Definition. If a and P are order types, we call a an initial segment of p, if there exists an order type y such that ,L? = a y. And
+
10.1. SOME GENERAL THEOREMS ON ORDER TYPES symmetrically: a is a final segment of with ,8 = y a.
+
335
P, if there exists an order type y
We remark that if a , p are ordinals the equation ,8 = a+y determines y uniquely, but /3 = y a does not. E.g. w w = (w n) w for every n E N. If S is a linearly ordered set, A an initial segment and B the complementary final segment of S, we express this in the form S = A + B. An analogous definition applies to expressions S = A B C .  . , where we have finitely or denumerably many summands. Here A, B, C, . . . are consecutive (and thus also disjoint) segments of S.
+
+
+ +
+ + +
The next theorem was established by Lindenbaum. A proof of it was published by Sierpinski [160]: 1.8 Theorem (Lindenbaum) Let a and P be linear order types, such that a is an initial segment of p, and /3 a final segment of a. Then a = P. In another formulation: The equations ( 1 ) a = a + p and p = a + p imply a = P.
Proof. From (1) we obtain ,8 = a+P+p. Starting from this equation we can choose disjoint linearly ordered sets B , S, B1,R, where these sets have the order types P, a, P, p respectively such that (2) B = S B1+ R. There exists an isomorphism f : B +B1, and for the iterates f n of f, n E w , we prove: (3) B has an initial segment S f [q . . f n[S] f n+l [B]. F!rom(2)weobtainB=S+f[B]+R=S+f[S+B1+R]+R=S+ f f 2 [ ~ ]f [R] R, so that (3) holds for n = 1. Let now (3) be proved for a fixed n E N. Then B has an initial segment S f [S]  .  f [q f n + l [ ~ + ~ l= + S+ ~ ] fn+2 [Bl+ fn+l[R1, and this proves (3) for n+ 1, so that it follows by induction for all n 6 N. of denumerably Finally also the sum S f [S]  . . f [S] many summands is an initial segment of B, so that there exists a final segment T of B with ( 4 ) B = S + f[S] fn[S] T. Since S has the type a, also all fn[S],n E w, have the type a, and so we obtain with T := tp(T) from (4) for the corresponding types
+
+
[q+
+
+
+ +
+
+ + + f[s]++ fn[s]+ fn+l[s]+ + + + +
+ . . a
+
+...+
+
p
+
+
T . On the other hand we have by (1) a = a = a a $a + ( a + a + . .  )+ T = p .
10.2
+ ,8 =
Countable order types
A rough classification of the countable linear order types is given by the property of containing a subset of the type q of the rationals or not. Since 7 is the order type of an No  universally linearly ordered set, it is clear that all denumerable linearly ordered sets whose type is 2 q are embeddable into each other. Moreover they are themselves also No  universal. The linearly ordered sets that don't contain a subset of type q form the class of scattered linearly ordered sets. For this class we have Laver's theorems 8.6.7, which state that every set of scattered linear order types, that are pairwise incomparable, is finite. And there is no infinite strictly descending sequence of scattered linear order types. An ascending sequence of type wl of countable scattered order types is e.g. the set of all countable ordinals. And on the whole by 1.2 there exist 2'0 types of linearly ordered countable sets. From DushnikIMiller [29] originates the following:
2.1 Theorem. Every denumerably infinite linearly ordered set A contains a proper subset A' to which it is isomorphic.
Proof. Let A be a denumerably infinite linearly ordered set. For a, b E A we put a b iff [a,b] is finite. This yields an equivalence relation in A. The equivalence class of a is denoted by C(a). Then we consider two cases: Case 1. There exists an a E A such that C ( a ) is infinite. Then A is representable as ordered sum A = A1 C ( a ) A3 of three disjoint segments. The component C ( a ) has one of the three types w,w*,w* w. Suppose that its type is w. Then we define a mapping f : A + A by f ( x ) := x for x E A1 U A3 and f ( x ) as the immediate successor of x in C(a). Then f is an isomorphic mapping of A onto a proper subset. In the cases where C ( a ) has type w* or w* w we again let the elements of A1 U A3 be fixed, and C ( a ) can be mapped <  preserving onto a proper subset. Case 2. All congruence classes are finite. Then we must have infinitely many congruence classes. Let F be the N
+
+
+
+
10.2. COUNTABLE ORDER TYPES
337
set of their first elements. If a and b are two different elements of F, w.r.0.g. with a < b, then there exists a c E F with a < c < b, otherwise also a and b would be congruent. Thus F is denumerable and dense. Its interval (a, b) has type 7, and so A is isomorphic to a subset of (a, b), which is a proper subset of A. 2.2 Theorem [29]. Let A be a denumerable linearly ordered set, for which each <  preserving mapping f : A + A satisfies f (a) 2 a for all a E A. Then A is wellordered.
Proof. (I) A is scattered; it has no subset of type 7. Suppose the contrary, that E A is a subset of type 7. Then choose an arbitrary element b E E. The set ( E < b) has again type 7, and so there exists a <  preserving mapping f of A into (E < b), in particular f (b) < b holds, contradicting our assumption. And (I) is proved. Further we assume that A is infinite, since for finite A the statement is trivial. Now we introduce in A the following equivalence relation: For a, b E A we put a b iff the closed interval with ends a,b is wellordered. Evidently each equivalence class is a convex subset of A. (11) Every equivalence class has a first element. Suppose the contrary. Then there is an equivalence class C which has no first element. Let T be a coinitial inversely wellordered subset {cili < w) of C of type w*, so that i < j < w implies ci > cj  such a set exists since A and C is infinite. And let ri be the order type of the , which is wellordered by construction, so that the interval Ii := [ G + ~G) ri are ordinals.
c
N
First we prove: (111) There is an m E w such that a) the set {qlm i < w) has no greatest element, or b) it has a greatest element, and this occurs infinitely often in the sequence ri, i < w . Suppose the contrary. Then there exists a greatest element rn(,,) in the sequence, and this occurs only finitely often. W.r.0.g. we can assume that n(0) is the last index at which it occurs. By assumption then further there is an index n(1) > n(O), such that 7 , ~ )is the greatest element in {rnln 2 n(l)),and we can assume that n(1) is the last index n at which r n ( ~ occurs. ) This construction could be performed over w steps and would yield a sequence r n ( ~>) r n ( ~>)   > rn(q > . . , i E w, of ordinals, which is strictly decreasing, and this is impossible. Thus (111) holds.
<
In the case a) of (111) we determine recursively numbers ni E w for i E w, which strictly increase with i, as follows: no := m, and ni is the first index > i, for which Tni 2 ri and Tni > Tnj for all j < i holds. Then for each i E w there is a <  preserving mapping fi : Ii := [ciSl1ci) +In; because of ~i T,~. Let then f be the mapping which coincides with the fi over Ii and which puts f (x) = x for the x E ( C 2 co) and for all x E A, which are not in C.Now f is a <  preserving mapping of A onto a subset of A, and there are elements x with f (x) < x. This contradicts our assumption, so that in case a) the statement (11) holds. In case b) of (111) we have a sequence m = n(0) < n(1) < < n(i) < . . . , i < w, satisfying i < n(i) for i < w, for which all rn(i), i < W, are equal, and = max{rvlm I: v < w). Then we choose for each i with m i < w a <  preserving mapping fi : [ Q + ~Q) , + [c,(~+~), cn(i)) and \ C), andso that define f : A + A by f ( x ) := x for x E ( C 2 C,(~))U(A f coincides with all fi over their corresponding range of definition. Then f is a <  preserving mapping of A into A, and there are elements x with f (x) < x, contradicting our assumption. .And therefore (11) holds. This implies that every equivalence class C is wellordered, for each initial segment [c,x], where c is the first and x an arbitrary element of C, is wellordered. After we have proved (11) it follows that there is only one equivalence class. Assume indirectly that there are at least two, C1 and C2. Their first elements a1 and a2 are not equivalent, so that the classes Cl, C2 cannot be consecutive segments of A. There is at least one class between them, and its first element is between a1 and aa. Thus the set of first elements of the equivalence classes is dense and then also denumerable, and so it contains a subset of type q. But this contradicts (I). Now the unique equivalence class is nothing else but A, which is therefore wellordered.
<
<
10.3
Uncountable order types
The Theorem 2.1 cannot be transferred on uncountable linear1y ordered sets. This was proved by DushnikIMiller [29]. They established a theorem of which we here mention a generalization on higher cardinalities, which follows with the same idea of proof: 3.1 Theorem. Let C, be the linearly ordered continuum of 4.7.31.
10.3. UNCOUNTABLE ORDER TYPES
339
Assuming t, = N,, which follows from GCH, there holds: The continuum C, contains a set E of power c := 2Nm (= IC,l), which is not isomorphic to any proper subset of itself. Moreover the set E has only one automorphism, namely the identical mapping.
Proof. Let w, be the initial ordinal for the cardinality c. Let then + C,, which are different from the identity mapping id of C,, which has only fixed elements. (For this it is sufficient that f (x) # x holds for at least = c, so that it is possible to one x E C,.) By 9.4.4 5 has cardinality arrange 5 in the above manner. For each f E 5 the set Fix f of fixed elements of f is not everywhere dense in C,, that means: There is at least one open interval of C,, which is disjoint to Fix f. For otherwise f would be the identity mapping. So there is a segment S of C, with more than one element, which has no fixed element of f7 and since IS1 = c holds, we have c elements p in S with f (P) # PWe choose an element po for which fo(po)=: q0 # po holds. If p is an ordinal < w,, for which elements p, # q, have already been defined for all v < p, we choose elements p,, q, E C,\ U,<, {p,, q,) for which f,(p,) = q, # p, holds. This is possible by construction, for we have: Since f, is strictly increasing and # id the set A, := {plf,(p) # p) has cardinality c. Then also A; := A, \ ~,<,{p,, q,} and f,[A;] have cardinality c. So there are c elements p E A; for which f,(p) is different from all p, and q, with v < p. Let p, be one of them. Then f,(p,) =: q, # p, holds, and p, and q, are in C,\ U,<, {p,, 9,). By transfinite induction, p, and q, are so constructed for all v < w,. If C' = [a, b] is an interval of C,, then there is a <  preserving function, which is = id on C,\C1 and different from id on C', so that there is at least one element x E C' with f (x) # x. To see this choose two elements x,y in C, with a < x < y < b and a <  preserving f which maps [a, y] in [a, x]  this can be done using 4.7.32  and which leaves all other elements of C, fixed. Since f is one of the f, E 5 we obtain: There exists an index v for which p, E C'. And this yields: E := U{p, lv < w,} is dense in C,. Now E satisfies our assertion. Suppose the contrary, that g : E + E is a <  preserving mapping. If g would be different from the identity mapping idE of E7 then by 9.4.3 there exists a 5  preserving mapping
5 := {f,lv < w,) be the set of all <  preserving functions f : C,
g* : C, + C, which extends g. Then g* is also <  preserving since E is dense in C,, and so g* E F is a f,. This yields a contradiction since E 3 g(p,) = g*(pv) = fv(pv) = 9" !$ E. If in 3.1 we take a = 0, then Co is isomorphic to the set R of real numbers, and to is = No. Then there follows the original version of the theorem of Dushnik/Miller [29], and this now without using the GCH !
3.11Theorem. The set of real numbers has a subset E of power 2 N ~ which is not isomorphic to any proper subset of itself. And this set has only one automorphism, the identical mapping. Next we prove several fundamental theorems of Sierpinski on order types of the subsets of R. They can be generalized to higher cardinalities, practically without modifying his proofs.
3.2 Theorem [161]. Let a be an ordinal and C, the linearly ordered continuum of 4.7.31, c = 2Na the cardinality of C,, w, the initial ordinal of cardinality IC,I. Using t, = N , then there holds: a) There exists a transfinite sequence {r,lv < w,) of linear order types r, 5 t p C,, which i s strictly decreasing: For p < v < W, we have rp > rv. b) There exists a transfinite sequence {r,lv < we) of linear order types r, 5 t p C,, which i s strictly ascending with v. c ) There exists a set of 2' linear order types of cardinality c which are pairwise incomparable. For a = 0 these statements follow without the use of GCH since to= N~ holds. Proof. In the proof the method of the proof of 3.1 is again used. First we choose a wellordering of the set F of all < preserving functions f : C, , C, that are different from the identity mapping id of C,. Since their set has cardinality c, we can arrange F as F = {f,lv < w,), and, for technical reasons, we can do this in such a way that each element f of it is enumerated twice, at an even index 20 and at its successor index 2 p + 1. Next we define elements xu and y, quite analogously to the proof of 3.1: Since fo is different from id we can choose an element xo E C, for which fo(x0) = yo # xo. For v < w, the set {XI f v ( x ) = x ) is not dense in C,. For otherwise f, would be the identity mapping. So there exists an open interval (a,, b,)
10.3. UNCOUNTABLE ORDER TYPES
341
of C, such that f,(x) # x for x E (a,, b,). And therefore we have 1 {XI fv(x) # x)l = c. Suppose that p < w, and that for all v < p elements x,, y, have already been defined. Then, as in the proof of 3.1, there exist elements x, and y, with f,(xp) = y,, which are different, and different from all x,, y, with Y < p. By transfinite induction we so have defined elements x,, y, for all v < wc. If I is an open interval of C,, then there exists, as in the proof of 3.1, a <  preserving mapping f : C, + C,, which satisfies f (x) = x for x E C, \ I, and f (x) # x for at least one x E I. This f is = f2p for some P, and we have by construction f2p(xap) (= yzp) # xzp, so that E I holds. Thus we have obtained: H := {x2,1v < w,) is dense in C., Then we put Z := {xa,+llv < w,). By construction H n Z = 0. For TCZweputET:=TUH. The main idea of the proof consists in proving: (I) Let A, B be subsets of Z with A \ B # 0. Then EA (with the induced order) cannot be isomorphic to a subset of EB. Suppose the contrary, that there is a <  preserving mapping g : EA + EB. The set H, and a fortiori its superset EA,is dense in C,. Now g can be extended to a <  preserving function f : C, 4 C,.This f is different from id. For otherwise we would have g(x) = f (x) = x for x E EA and thus g[EA] = EA C EB in contradiction to 0 # A \ B = (since H is disjoint to A and B) (A U H) \ ( B U H) = EA \ EB. So f E F. And then there is an ordinal p < w, with f = f2p. We have xzp E ( H C) EAand further f2p(x2,) = yzp q! EB = BUH, for due to B C Z only elements x, are in B U H. So it is impossible that f2,[E~] C EB holds. This contradicts fip[EA]= g[EA] C EB, which follows from f2,(x) = f (x) = g(x) for x E EA,and (I) is proved. For p < w, we put Zp := { x ~ , +1P~5 v < w,), and obtain P < r < wc (x2~+1E Zp \ Z7 and) 2, c Zp. If now for ,8 < r < w, we put in (I) A = Zp and B = Zy then (I) yields: Ezs is not isomorphic to a subset of Ez,. The order types of the sets EZp,P < w,, now form a set which satisfies a).
*
V @).Then there follows: On the other hand we put Tp := { X ~ , + ~ ~ < P < r < w, xzp+l E Ty\Tp and thus Tp c Ty.By (I) this entails tp ET, $ t p E T ~ and further t p ET, (and then also) > t p ETp,so that b) holds.
*
>
c) We have 121 = 2Na. By 9.1.25 there is now an antichain U of the power set P ( Z ) of.cardinality 2(2Na).So for A, B E U we have A \ B # 0 # B \ A, which by (I) entails t p EA $ t p EB and t p EB $ t p EA. And c) is proved. The statement of 3.2,~)can partially be strengthened to the following: 3.3 Theorem [161]. Suppose t, = N,. Then there is a set of 2Na linearly ordered subsets of C , of cardinality 2Na,which pairwise difler in at most two points, so that their order types are pairwise incomparable.
Proof. For P < w, we put Qp := { x ~ ~ +Then ~ ) . the sets EQp = Qp U H, p < w,, are incomparable by (I) and differ pairwise only in two elements. 3.4 Theorem [161]. Suppose t, = N,. Then there exist subsets H and H1 of C , of cardinality 2Na,for which no order type exists which is strictly between them. tp C , is not Therefore the quasiordered set of all order types dense.
<
Proof. We put (with the concepts of the proof of 3.2) HI := H U {xl). By definition of the Tp we have To = 0 and TI = {XI). These sets are E Z, and so there follows from (I): HI = %is not isomorphic to a subset of ETo= H, in short t p Hl $ t p H. Due to H HI this implies t p H < t p HI. Now we can prove:
c
(11) There is no order type r satisfying t p H < r
< tp H1.
Suppose the contrary. Then there exists a subset M H1 with t p M = r and t p H < t p M. And there is a <  preserving function g : H += M. This can be extended to a 5  preserving function f :
ca+ ca.
If we would have f = idca this would entail g(x) = x for x E H and thus g[H] = H. But this yields H M C HI = H U {xl), which can be strengthened to H c M c H1 = H U {xl) because of tp H < t p M < t p HI. This is a contradiction because there cannot be an element strictly between H and H U {XI). Therefore f is not = idca so that f = f2p holds for an ordinal P < w(c). Due to Zap E H we have g(x2p) = f (xza) = Yap $ HI, and a fortiori g(x2p) $ M (which is HI). This contradicts g[H] C M, and so our statement is proved.
c
10.4. HOMOGENEOUS POSETS
343
In several papers S.Ginsburg [56],[57],[58] studied related questions, among others the existence of order types ( which satisfy a < ( < T, if a and 7 are given linear order types. Further he investigates decompositions of linearly ordered sets into the union of disjoint subsets which have special properties, e.g. to be pairwise incomparable, or to be pairwise isomorphic.
10.4 Homogeneous posets Usually a good oversight over the ordertheoretical properties of a poset (and also other structured systems) can be obtained by considering its automorphisms, for they reflect its symmetry features. Some of the posets of practical importance, like Z, Q, R, are very "homogeneous": All their elements have the same structureproperties. In other posets, e.g. in the wellordered ones, we have entirely different situations. We now investigate these phenomena in detail: 4.1 Definition. Let P be a poset. An automorphism (or automorphic mapping) of P is an isomorphic mapping f : P + P of P onto itself. The set of all automorphisms of P forms a group relative to the operation o of concatenation, the socalled automorphism group of P. Its unit element is the identity mapping of P. In 4.6.1 we defined a poset P to be homogeneous, if for each two elements a, b E P there is at least one automorphism of P which maps a onto b. If lPl > 1 and if for each two elements a, b E P there is one and only one automorphism which maps a onto b, then P is called uniquely homogeneous. A poset P, which has only one automorphism, the identity mapping of course, is called rigid.
4.2 Example. The set Z of integers is uniquely homogeneous. For if a, b are integers, we map every x E Z onto x (b  a). This translation of Z maps a onto b and is an automorphism. It is uniquely determined since the immediate successor (resp. predecessor) of a must be mapped onto the immediate successor (resp. predecessor) of b and SO on. For the same reason the sets Q and R are homogeneous. But here not only translations effect automorphic mappings of Q (resp. R). As
+
we have seen in 9.4.4 there are 2 N ~strictly increasing mappings of R onto R, and each of them is an automorphism of R. The fact that Q is homogeneous can be put into a more general context, namely there follows from 4.6.4: 4.3 Theorem. If X is an indecomposable ordinal the set TAof 4.5.7 is homogeneous. And as a special case of this the sets Ha of 4.3.3 are also homogeneous.
Proof. By 4.6.4 the sets TAare homogeneous if X is indecomposable. And by 4.5.10 the set T(w,) is isomorphic to Ha. For uniquely homogeneous linearly ordered sets we have a strong restriction for their possible order types. For these are subtypes of the type of R: 4.4 Theorem. Let (L, 5) be a linearly ordered set which is uniquely homogeneous. Then ( L ,<) is embeddable i n R.
Proof. For ILI < 2 the theorem is trivial. So we assume that L has at least two elements a, b. Then there is an automophism f of L with f ( a ) = b # a. W.r.o.g.we assume a < f ( a ) . Then the iterates of f generate a sequence < f W n ( a )< < f'(a) < a < f ( a ) < f 2 ( a ) <  . =< f n ( a ) <
...
Let S be the smallest segment of L which contains all elements of this sequence. Then it follows rather immediately that f maps S into itself: For each element of S satisfies f z ( a ) x fz+'(a) for some x E Z, and then also fz+'(n) f ( x ) fz+2(a). Now S turns out to be the whole set L. Otherwise we could construct an automorphism g of L by putting g ( x ) := x for x E L \ S and g ( x ) := f (x) for x E S. Then g is different from the identity mapping of L, but maps all elements of L \ S on themselves, contradicting the uniqueness.
<
<
< <
So, if we look for uniquely homogeneous linearly ordered sets, we can find them only among sets which are orderisomorphic to subsets of R. The question arises whether each such set is isomorphic to Z. But Ohkuma has proved that there do exist uniquely homogeneous subsets of R which are dense in R and of cardinality 2*0. Before proving his theorem we mention:
10.4. HOMOGENEOUS POSETS
345
4.5 Theorem. A uniquely homogeneous linearly ordered set D, which is not isomorphic to Z, must be dense.
Proof. Suppose indirectly that a and b are neighboring elements of D, w.r.0.g. a < b. Then not only a has an immediate successor (namely b ) , but also every element x E D then has an immediate successor: If f is an automorphism of D which maps a onto x, it must map b on an immediate successor of x, and so on. Analogously each element x E D has an immediate predecessor. Now D has a segment S which is isomorphic to Z, and then D \ S must be empty. For otherwise we could map each element of D \ S onto itself and each x E S onto its immediate successor, thus obtaining an isomorphic mapping f of D which is different from idD. So D = S holds , and D is isomorphic to Z with contradiction. 4.6 Theorem (Ohkuma [131]). There exists a uniquely homogeneous linearly ordered set E which is not isomorphic to Z .
Proof. The construction of the set E will be performed in such a way that E is dense in R and a subgroup of the additive group (R,+), for which every automorphism of E is a translation, i. e. a functioil f , for which there exists a constant d E R such that f (x) = x + d for all x E E. (And here, of course, d must be a difference a  b with a, b E E.) So it suffices to construct a subset E of R, which satisfies the following two conditions: (Cl) E is a subgroup of R (with +), (C2) All <  preserving continuous functions f : R + R, which are neither bounded from below nor from above (which thus form an automorphism of R since f (R) = R ) , and which are no translations of R, don't generate an automorphism of E. That means: f 1 E is no automorphism of E. Let w, be the initial ordinal ~ ( 2of ~2N0. ~ Then ) by 4.4 we can arrange the functions f of (C2) in the form {f,la < w,). The set E will be the union of a tower of subsets E, R, a < w,,which are constructed by induction, at the same time with a tower of sets S, E R. Consider the following scheme:
c
loE, is an additive subgroup of (R, +). 2O p < a +Ep E, and Sp C S',. 3O E, n S, = 0. 4O x E E, and s E S, =j % E S, for all n E Z
c
\
(0).
5O For infinite a the cardinalities IE, I and IS, I are I max{No, la[). 6' For < a there is a tp E E, such that fa (tp) E S, or fil (tp) E
a
s f f
.
For a = 0 let Eo be the set Q of rational numbers. Then we choose an arbitrary irrational number z and put So:= { ~ I xE EOand n E Z \{O)). Now Eon So= 8. For otherwise there would be a y E Eo, which is = for an x E Eo and an n E Z \ (0). But then ny  x = z would follow, and z would not be irrational. Thus 3) holds for a = 0, also 4), for if x E Eo and s = E So (with y E Eo,n # 0 in Z), then $ (with m E Z\{O)) has the form (nx y z)/nm, which is an element of So. So for a = 0 the conditions l06' are satisfied. Suppose now that a is an ordinal < w,, and that the properties lo 6' are satisfied for all a' < a (instead of a ) . Then we define E, and S, first for the case where a is a successor ordinal /3 + 1. For m E Z and a 6 E p we put Xp(m,a) := {x E RI fp(x) = mx + a), and Xp := ~{Xp(m,a)llml# 1 in Z, a E Ep), Yp := U{Xp(l,a)la E p ) , Zp := U{Xp(1, a)la E Ep), and Wp := R \ (Xp UYp U Zp U Sp), so that (0) R = Xp U Yp U Zp U S p U Wp.
+ +
<
<
By 5' we have ISB[ 1/31 resp. No, so that in any case there holds: (1) ISPI < 2No. The function gp, which is defined by gp (x) := fp(x) x for x E R, is strictly increasing, and so the set Xp(1, a) = {x E Rlfp(x) = x + a) contains exactly one element. This entails, due to lEp 1 5 max{No, 1/31} < 2No : (2) IZp I I lEpl < Po. Next we prove for hp, which is defined by ha (x) := fa (x)  x for
+
XER: (3) h[Yp1 C Ep. Indeed, x E Yp 3: E Xp(1,a) for some a E E p fp(x) = x + a , and hp(x) = fp(x)  x = x + a  x = a E Ep. hp is continuous, and fp is no translation (of type x const). Thus ha is not constant, and therefore the image set Hp := hp[R] is a segment of R with more than one element, so that lHpl = 2 N ~ .
*
*
+
10.4. HOMOGENEOUS POSETS
347
Due to lEp1 < 2 N and ~ I HpI = 2 N we ~ have [Hp \ Epl = 2 N ~This, . together with Hp \ Ep C hp[R] \ hp[Yp] h g [ R\ Yp],entails that h p [ R \ Yp] has a cardinality 2 IHp \ Epl = 2 N ~and , this implies also (4) since IR\Yp I 1 Ihp [R\Yp] I (5) Xp or Wp has cardinality 2'0. Since ISp I and IZp I are < 2 N by ~ (1) and (2), then (0) yields R \ Yp Xp U Zp U Sp U Wp and by ( 4 ) 2N0 = IR \YpI = IXp U WpI. This implies (5).
c
We can now perform the construction of sp and t p . If Wp # 0, we choose a t p E Wp. Then, by definition of W p , we have t p 6 XpUYpUZp andso f p ( t p ) # m t p + a for allm E Z and alla E Ep. Now we define sp := f p ( t p ) . If Wp = 0 we put Fp := ~ { z l #n 0 in Z}. (Here is the set Ix E Ep .) Evidently IFDI 5 I Ep 1 holds for infinite ,B and 1 Fp 1 5 No for finite p. Next we prove:
{E
(6) Xp n ( R \ (Sp U Fp)) # 0. Indeed, since in our case Wp is empty, we have lXpl = 2*0 by ( 5 ) , and by (0): ( 7 ) R \ Yp E Xp U Zp U Sp. Here the left side has cardinality 2 N by ~ ( 4 ) ,and because Sp and. Zp have cardinality < 2 N by ~ (1) and (2), also Xp has cardinality 2 N ~Since . also IFp/ 5 IEp 1 . No < 2" holds, (6) follows. Due to (6) we can choose sp E Xp n ( R \ (Sp U Fp)), and we put ti3 := fp(sj3). (8) sp # mtp a for all m E Z and all all a E Ep.
+
If we would have sp = mtp +a for some m E Z, a E Ep, then m must be # 0, since otherwise sp = a would be in E p and Fp, contradicting the choice of sp. On the other hand from sp E Xp we conclude: There exists an n E Z and b E Ep with t p = fp(sp) = nsp b, where In/ # 1. Then sp=m.(nsp+b)+aandsp(1mn)=mb+a,sothat s p = e . ( m  n is # 1 since In1 # 1.) But then sp would be in Fp since a, b, mb are in the group Ep, and this contradicts the choice of sp $ Fp. So (8) is proved. Further we have:
+
(9) tp $ SB.
In the first case (where Wp # 0 ) , we had tp E Wp, and Wp n Sp = 0 by definition of Wp. In the case where Wp = 0, we had tg = fp(sp) = nsp b with In1 # 1, and sp 4 Fp due to the choice of sp. If now tp E Sp would hold, t b then sp = had to be in S p by 4O, but we had sp 4 Sp in the case where Wp = 0. And thus ( 9 ) holds.
+
5
We can resume: For p = a  1 we have defined elements tp and sp such that, together with the induction hypothesis, for all ,O < a we have by (8) and ( 9 ) :
( D l ) tp 4 Sp, (D2) fp(tp) = sp or f i l ( t p ) = sp, ( 0 3 ) for all m E Z and all a E Ep there holds sp # mtp + a. Now we can define the sets E, and S, : E, := {mtp alm E Z, a E Ep ), S,I := {(sp mtp a ) / n Im,n E Z, n # 0, a E Ep), S,:! := { ( s m t p ) / nIm,n E Z, n # 0, s E Sp), Sol := Sal U Sa2. We now have to verify that E, and S, satisfy the conditions l o 6'. Since Ep is a group by induction hypothesis, it follows immediately that also (E,, +) is a group and that E, contains Ep, so that lo and 2O are fulfilled.
+
+ +
+
3O If E, n S, would be # 0, we would have E, nSa1 # 0 or E, nSa2 # 0. Suppose E, n Sal # 0. Then we can choose for each x E E, n Sal elements p, m , n E Z and a, b E Ep such that x = mtp a = (sp +ptp + b)/n.Then mntp na = sp ptp b, so that sp = ( m n  p ) . tg + na  b follows with contradiction to (D3). Suppose now E, f l Sag # 0. Then for each element x E E, r l Sa2 one can choose elements p, m , n E Z, a E Ep, s E Sp such that x = mtp+a = ( s +ptp)/n. Then ( m n p ) .tp = s  nu. Here s  nu is # 0 since s E Sp and na E Ep, and because Ep n Sp = 0 holds by 3'. Then also on the left side ( m n  p) is # 0, and thus tp = ( s  n a ) / ( m n p). Then by 4' tp E Sp, for we have s E Sp and na E Ep. But this contradicts ( D l ) . Now 3' is satisfied for E,, S,.
+
+ +
+
4O For each x E E, there is an m E Z and a E Ep such that x = mtp + a. And for each s E S, there is s E Sal or s E Sa2,SO that we have
10.4. HOMOGENEOUS POSETS
349
s = ( s p + p t p + b ) / n withp,n E Z, b~ E p or s = (sl+ptg)/n with p, n E Z, st E Sg. In the first case for every integer q # 0 we have S+X = (sp ptp b)/qn (nmtp an)/qn = (sp (p mn) tp P na b)/qn E S,I. In the second case for every integer q # 0 there follows = (sl+pta)/qn (nmtp+an)/qn = (st+ (mn+p)tp+na)/nq. 9 Here by 4) we have st na E Sp, and then belongs to SaZ.So 4) is fulfilled for a.
+
+ +
+
+
+ +
+
+
+
6' By the definition of E, we have tp E E,, and by definition of Sal there follows s p E Sal C S,. Further we had fp(tp) =: sp in the case, where Wp # 0, and fp(sp) =: tp if Wp = 0. Then 6) is satisfied for the immediate predecessor /3 of a, and by induction hypothesis then also for all ordinals /3 < a. We still have to supplement the definition of E, and S, for the case where a is a limit ordinal. Here we put E, := ~ { E p l / 3< a) and S, := u{SpI/? < a}. Condition lofollows so: If a,b are in E,, there is also a /3 < a with a, b E Ep. This is a group which contains a b and their groupinverses, and so this also holds for E,. Condition 2O is clearly satisfied. 3O If E, and E p would have an element in common, this would also be in some E p with @ < a, with contradiction. 4' If x E E, and s E S,, then by 2O there is a /3 < a such that x E E p and s E Sp, and 4O follows. 5O For each /3 < a we have that (Epl and ISp[are I max{No, 1/31). By 2O there follows IEal = sup{lEpIIP < 4 5 m={No, lal) and ISaI = sup{lSpllP < a) I max{No, lal},and 5' holds. Also Condition 6' follows immediately with 2O. Finally we can now define E := U{E, la < we) and S := u{S, la < we). These sets satisfy also the conditions lo 6' (putting E := E,, and S := S,,) , and this is proved in the same way as before for limit ordinals. In particular by 6' we have obtained the following: For every continuous function f : R + R, which is <  preserving and unbounded from below and from above, and which is not a translation, so that f is one
+
of the f p with < w,, there exists by '6 an element t p E E for which a) I := f S ( t p ) E S or b) x := f;'(tp) E S holds. In the first case f = f p is no automorphism of E because f p ( t p ) is in S, and thus not in E, which by 3' is disjoint to S. In the second case fa maps x of S on the element t p E E , and then this has no element in E which is mapped on t p by fa. So finally, if a , b E E and thus a (b  a ) = b, the translation T , which puts T ( x ) := x (b  a ) maps a onto b. Of course, it is the only translation which does that. Since E is evidently not isomorphic to Z, the theorem is proved.
+
+
Ohkuma had also determined the cardinality of the set of types of all uniquely homogeneous chains. We mention his theorem referring to this without proof:
4.7 Theorem [131]. There is a set of cardinality 2' (where c = 2 N ~ ) of uniquely homogeneous chains, which are pairwise nonisomorphic. The theory of automorphisms of posets has meanwhile reached a great extent. It has, of course, an intensive contact to the theory of groups. One of its central intentions is to study the structure of the automorphism group of special classes of posets. Without going into the details we mention a few things which concern some characteristic features of this. A relational structure 2, e.g. a poset, is called k  homogeneous if each isomorphism between two k  element substructures of U extends to an automorphism of U. In this context a theorem of Droste and MacPherson states: 4.8 Theorem [27]. If a countable poset (P,5 ) is 1  and 4  homogeneous, then it is k  homogeneous for each k e N . And there are infinitely many examples of certain countable posets showing that here the number 4 may not be replaced by 2 or 3. And Droste, MacPherson and Mekler proved: 4.9 Theorem [28]. Assuming the settheoretic assumption O(N1),it is shown that for each k E N , there exist partially ordered sets of sixeNl which embed each countable partial order and are k  homogeneous, but not ( k 1 )  homogeneous. This is impossible i n the countable case for k 2 4.
+
We further mention a connection of order theory and group theory:
10.4. HOMOGENEOUS POSETS
351
4.10 Definition. A group (G, +), which is equipped with an order
< is called a pogroup, if it satisfies: For each two elements x, y E G we have (1) x < y = + a + x + b L a + y + E f o r a l l a , b ~ G . In short this condidition can be formulated as: Every group translation is isotone. A pogroup, which is at the same time a lattice, is called a latticeordered group, in short an 1  group. This concept is of importance if we study the set of automorphisms of a linearly ordered set: 4.11 Definition. Let (S, <) be a linearly ordered set, Aut (S) the set of its orderautomorphisms. For f , g E Aut(S) we put f (x) 5 g(x) for all x E S. f g It is immediately clear that Aut(S) is partially ordered by there holds more, namely:
< . But
4.12 Theorem. Let S be a linearly ordered set. Then (Aut(S), <) is a lattice, and then an 1group. Proof. Let f , g E Aut(S). Then we define a mapping h : S + S by h(x) := max{f(x),g(x)} for x E S. We have to show that h is an automorphism of S. If x, y E S satisfy x < y, then f (x) < f (Y) and g(x) < g(y) holds, and further h(x) < h(y), so that h preserves <, and is therefore injective. We have to verify that it is also surjective. Let y E S be given. There are uniquely defined elements X I , x2 E S with f (xl) = y and g(x2) = y. If h(xl) = f (xl) = y holds, we are done. In the other case we have h(xl) = g(xl) > f (xl) = y. Then there must hold 2 2 < xl since g is strictly increasing. This implies f (xa)< (f (xl) =) y, and then h(x2) = g(x2) = y follows. Of course, h is now also the least automorphism of S which is f and g.
>
In this context Holland proved the following fundamental theorem, which we mention without proof: 4.13 Theorem [87]. Every 1group is isomorphic to a subgroup of the automorphism group of a linearly ordered set. And since every linearly ordered set can be embedded in an 7,  set, this statement can be supplemented by:
4.14 Theorem. Every 1group is isomorphic to a subgroup of the automorphism group of a linearly ordered set of a normal type 7, (= ha).
The automorphism groups of these sets have been studied in detail by E.Weinberg, see e.g. [177]. For a linearly ordered set S the group Aut(S) is usually not commutative. Consider e.g. R with its usual linear order. Let f : R + R be the mapping which puts f (x) = x 1 for x E R, and g : R + R defined by g(x) := $ for x E R, then g o f maps 1 onto 1, whereas f o g maps 1 onto 1;.
+
Chapter 11 comparability graphs In every poset (P,5 ) the relation 5 defines the corresponding comparability relation, which arises from by taking the symmetric hull of . This concept effects a connection between order theory and graph theory. If we consider only the comparability relation of an order 5, we lose information, but with this simplification we possibly can gain additional insight into the structure of the order. How farreaching this simplification can be is e.g. illustrated by the fact that all linear ordering on a set S produce the same comparability relation on S, namely the allrelation on S. In particular, an order 5 on a set S generates the same comparability relation as its inverse order
<
<
2.
11.1
General remarks
For the discussion of graphtheoretical properties of posets the directed graphs are of particular interest. They arise from the general graphs by equipping their edges with directions. For an edge {a, b) of a graph there are two possibilities to give a direction to {a, b), which means a linear order in {a, b),one in which a < b and one in which b < a holds. In the first case we obtain the oriented or directed edge (a, b), which has a as initial point and b as endpoint, in the second conversely with (b, a). In the illustration of a directed edge (a, b) we use a line segment with b. This ends a and b and insert an arrow in it as follows: a shall give an association to a corresponding order theoretical meaning a < b. We now define: 1.1 Definition. A directed graph, abbreviated digraph, is a pair G = (V, D), where V is a set and D a subset of (V x V)\{(x, x)Ix E V). The elements of V are again called vertices of G, and the elements of D directed (or oriented) edges of G. We emphasize that a digraph can have a directed edge (a, b) and at the same time the directed edge (b, a). In this case it contains, so to speak, the edge {a, b) in both directions.
If we have a (nonoriented) graph G = (V,E ) , and if we choose for each edge { a , b) of E exactly one pair f ( { a ,b)) E { ( a ,b), (b,a ) ) , we say that we have oriented G by f, and call D := { f ( { a ,b ) ) 1 { a , b) E E ) the orientation of G (also of E ) , which is generated by f. We call f ( { a ,b)) the orientation of {a, b), given by f. If (V,D ) is a directed graph, and if we put E := {{a, b)l(a, b) or (b,a ) are in D ) , then (V,E ) is the undirected graph corresponding to (V,D ) . We call a directed graph G = (V,D ) transitively directed (or transitively oriented) and D a transitive orientation, if ( a ,b) E D and (b,c) E D ( a ,c) E D. This entails evidently that if we have a directed path, that means a finite sequence of directed edges (al,a2),(a2,a3),..,(anl,an), also ( a l , a n ) is in Dm 1.2 Definition. A graph G = (V,E ) is called a comparability graph, if it is possible to find an order relation on V such that for each {a,b) E V x V we have {a,b) E E +=+ a < b or b < a.
<
A slightly different characterization of comparability graphs is given by: 1.3 Theorem. A graph G = (V,E ) is a comparability graph i f it can be transitively oriented.
Proof. Let G = (V,E ) be a comparability graph of an order relation 5 on V. Then we orient each edge { a , b ) of E by the directed edge ( a ,b) if a < b, resp. by @,a) if b < a holds. This generates evidently a transitive orientation of G. If we have a transitive orientation D of G, we put a < b for all directed edges ( a , b) E D . Then the relation <, augmented by idv, is an order relation on V, whose comparability graph is G. Of course one is now interested to find more characterizations of the comparability graphs. First we mention some examples. 1.4 Example. Let G = (V,E ) be a graph with V = { v l , . . . ,vn), where n is an odd number 2 5, and E = {{vi, vi+l)li = 1,. . . ,n),where vn+l = v l . Then G is no comparability graph. Suppose the contrary. Then we could find an order 5 in V which has G as comparability graph . W.r.0.g. let vl < v2. Then v2 cannot be < v3 since this would imply vl < v3, but { v l , v 3 ) 6f E. And so v3 < v2 must hold. Successively we would obtain vi < for the odd numbers i E ( 1 , . . . ,n) and vi > vi+l for the even numbers i. Since v,
11.1. GENERALREMARKS
355
is comparable with vl we must have v, < vl or vl < v,. In the first case this would yield v, < (vl <) va, in the second vl(< v,) < v,1, which is impossible since v, is not comparable with vz, and vl not with v,1. Contrary to the situation of 1.4 we have a positive result in the case where n is even: 1.5 Example. If the assumption of 1.4 is given with the difference 0 is an even integer, then G is a comparability graph. And that n this has the form of the graph given in 7.3.6.
>
Another simple example can be constructed using bipartite graphs. We define them as follows: 1.6 Definition. A graph G = (V, E) is said to be bipartite if there is a partition of V in two disjoint subsets A and B, such that each edge of E is of type {a, b), where a E A and b E B.
It follows trivially, that each bipartite graph is a comparability graph. For under the above assumptions we only have to ascribe to each edge {a, b} E E with a E A, b E B the directed edge (a, b) (or to every edge {a, b ) the directed edge (b, a)) to obtain a transitive orientation of G. And these are usually not the only possibilities to reach this aim. But if we have in addition to our assumption that G is connected, we can prove more, namely:
1.7 Theorem. Let G = (V, E) be a connected bipartite graph. T h e n there i s a n order relation o n V such that G i s the comparability graph of (V, I). A n d every order relation o n V that generates the comparability . graph G i s o r the reverse 2 of
<
<
<
Proof. As mentioned before there exists an order I in V whose comparability relation generates G. Let A and B be subsets of V satisfying the condition of 1.6 and let w.r.0.g. a E A, b E B be elements with a < b. Then for each x E B that is comparable with a, we must have a < x since x < a would entail x < b, contradicting the fact, that b and x are both in B. For a similar reason for all y E A that are comparable with b we must have y < b. The connectedness of G now yields that for every two comparable elements x E A, y E B we have x < y. So the relation a < b of the beginning determines the whole relation <. If before we would have had b < a instead of a < b, then we would have obtained the reverse relation
>, and the statement
is proved.
The last theorem presents an example of a case where a comparability graph determines a generating poset "nearly uniquely", namely an order and its reverse order. Graphs with this property are called pographs ([6]). In general there are many orders on a set which generate the same comparability graph. To every chain in a poset there corresponds a clique in the comparability graph, that is a set of vertices which are pairwise linked. To the antichains of a poset there correspond independent subsets of the comparability graph  these are sets of pairwise unlinked vertices. Here the correspondence between the order and its comparability graph is more intensive. Before we proceed we introduce two graphtheoretical concepts: 1.8 Definition. Let G = (V,E) be a graph. A subgraph of G is a graph (V', E') where V' C V and E' 5 E holds. For V' E V we define the subgraph which is induced by V' as the pair (V', E'), where E' contains ad1 edges {a, b ) of E, for which a, b E El. For E' E we define the subgraph which is induced by E' as the subgraph which is induced by uE'. It contains all edges of E', but possibly more.
Concerning the behaviour of subgraphs (resp. induced subgraphs) of comparability graphs we have the following: 1.9 Remark. a) Every graph G = (V, E ) is a subgraph of a comparability graph, e.g. of the graph (V, [VI2),in which every two elements of V are linked, and which is the comparability graph of every linear order on V. b) An induced subgraph of a comparability graph G = (V, E ) is also a comparability graph. For if 5 is an order on V that generates the comparability graph G, and if V' E V holds, the restriction 5 V' has the subgraph of G, which is induced by V', as comparability graph. c) A graph G is a comparability graph. iff each connectivity component of G is a comparability graph.
r
The property to be a comparability graph transfers from the finite induced subgraphs of a graph to the whole graph. This can be proved using the compactness theorem of mathematical logic. Here we present a proof using Rado's Selection Lemma 8.1.6 (and therefore denote here the set of edges by I):
357
11.2. C O M P A R A B I L I T Y GRAPHS
1.10 Theorem. Let G = (V,I ) be a graph of which all finite induced subgraphs are comparability graphs. Then G is also a comparability graph.
Proof. For each edge i = {a, b ) E I let Ki be the set { ( a ,b), (b,a ) ) . We put K := ~{&(ili E I ) . For each finite L C I the set L* of edges which link elements of U :=: U L is still a finite subset of I, and by assumption there exists a mapping fL* : L* +=K , which generates a transitive orientation in the subgraph which is generated by U, and which therefore satisfies: { a , b) and {b, c ) E L*, fL* ( { a ,b ) ) = ( a ,b), fL* ({b,c ) ) = (b,c ) =j {a, C ) E L* and fL* ( { a ,c ) ) = ( a ,c). We put f L := fL* L. According to the Rado Selection Lemma there is a function f : I + K, such that for every finite L C I there is a finite M with L* G M C_ I and f ( i )= fM(i)for all i E L*. We define I* to be the set of edges i E I , where each i is oriented by f (i). Now the digraph (V,I * ) is transitively oriented: Let ( a ,b) and (b,c) E I * , and L* the set of all edges of I whose both ends are in { a , b, c). By the preceeding remarks there is a finite M with L* C M E. I for which f ( i )= f M (i)for all i E L*. Since f M induces a transitive orientation in M, it follows: M contains (not only { a , b) and {b, c ) , but also) { a , c ) , which of course is also in L*, and then f ( { a ,c ) ) = f M ( { a ,c ) ) = ( a ,c ) , and thus (V,I * ) is transitively oriented by f and hence a comparability graph.
r
Theorems 1.10 and 1.3 now also imply: 1.11 Theorem. If each finite induced subgraph of a graph G can be transitively oriented, then so also can G.
11.2
A characterization of comparability graphs
In this section we present the characterizations of comparability graphs which were established by GhouilaHouri 1511 resp. Gilmore and Hoffman 1541. In [51] the folllowing concept was introduced: 2.1 Definition. Let G = (V, E ) be a graph which has been oriented by a function f, and let then D be the set of the oriented edges of E . Then G is called quasitransitively directed if ( a ,b) E D and (b, c ) E D ==+ { a , c ) E E. And then D is said to be a quasitransitive orientation.
Then it is trivial that a fortiori transitive implies quasitransitive. On the other hand there holds the following theorem of GhouilaHouri [511: 2.2 Theorem. Let G = (V, E ) be a finite graph, for which we have a quasitransitive orientation Q. Then there also exists a transitive orientation T of G.
Proof. We make the assumption that the assertion is true for IVI < n, where n is a fixed natural number, and consider a graph G = (V, E ) with IVI = n, which has a quasitransitive orientation Q. First we remark: (1) If three vertices a, b, c satisfy (a, b) E Q, (b, c) E Q and (c, a) E Q, each other vertex v of G is linked with 0,2 or 3 of the vertices a, b, c. Indeed, if (w.r.0.g.) e.g. v is linked with b, we have two cases: (v, b) E Q or (b, v) E Q (see Figure 25). In the first case, due to (b, c) E Q we also have {v, c) E E. In the second case (a, b) and (b, v) E Q implies { a , ~E ) E and (1) holds.
Figure 25 We now assume that Q is not transitive, otherwise we would be done. Then there exist three vertices X I , x2, x3 satisfying (2) (XI,x2) E Q, ( ~ 2 ~ xE3 Q ) and ( ~ 3xi) , E Q. By 1.9, c) we can w.r.0.g. assume that G is connected. Then, due to (I), we have to discuss two possible cases: Case 1. Each vertex x of G which is different from X I , ~ 2 ~ xand 3, which is linked with one of these three, is linked with all of them. Now we consider the vertex set V' := V \{x2, x3). If E' is the set of edges of G whose endpoints are in V', we consider the graph G' := (V', El)  it is the subgraph of G, which is induced by V'. The orientation Q of G induces a quasitransitive orientation Q' of GI, and
11.2. COMPARABILITY GRAPHS
359
since G' has fewer than n vertices, it has by induction hypothesis a transitive orientation T I . We extend TI to a transitive orientation T of G as follows: Let x E V' be a vertex which is linked with X I , and then by assumption also with x2,x3. If (x,xl) E TI, we put the pairs (x,x2) and (x,x3) in T. If (xl, x) E TI, we put the pairs (x2,x) and (x3,x) in T. By (2) the vertices XI, x2,x3 are pairwise linked in G, and the pairs (xl, x2), (a2,x3), and (xl ,x3) and all elements of T' become elements of T. Then T is a transitive orientation of G, as can be verified using the following illustration in Figure 26, which considers the two possible cases (x, xi) E T resp. (xi, X) E T for i = 1,2,3.
x
x Figure 26
Case 2. There exists at least one vertex v of G which is different from xi for all i E {1,2,3), and linked with two of these, but not with
the third one. W.r.o.g., due to symmetry, we can assume that there is a vertex v different from these three which is linked with x2 and x3, but not with XI. Then we define A to be the set of all vertices of G that are linked with 2 2 and x3. SO,among others, xl and v E A. Let W denote the set of all twoelement subsets of A that are not edges of G. Then (A, W) is a graph, and let C be the component of xl in this graph. Since v is not linked with X I , v is in C, like XI, and thus we have ICI 2 2. Next we prove: (3) Each element c E C satisfies (c,x2) E Q and
(23, c)
E Q.
If (c, x2) 4 Q would hold, we would have (22, c) E Q. This, together with (x1,x2) E Q (which holds by (2)) entails {xl,c) E E because of the quasitransitivity of Q. But this contradicts the fact that xl and c are not linked in G.
And if (x3,c) E Q would not hold, we would have (c, x3) E Q, and with (23, XI) E Q (which holds by (2)) this would yield {c, X I ) E E with the same contradiction. (4) Each vertex x E V that is not in C is linked with no or with all vertices of C. Indeed, let x 4 C, a, b E C, and let x be linked with a. We have to show that then also x is linked with b. First we see: (5) x is linked with x2 or x3. For if (x, a) E Q, this together with (a,x2) E Q  which holds by (3)  entails {x, $2) E E. And if ( a ,x) E Q, then this with (23, a) E Q  which holds by (3)  , yields 1x3, x) E E. Now we can prove that also {x, b) E E. Due to (5) it suffices to consider the following four cases: a) Let (x, 2 2 ) E Q. This with ( 2 2 , x3) E Q entails {x, 23) E E. b) Let (x2,x) E Q. This with (b, x2) E Q (which holds by (3)) yields {b,x) E E. c) Let (x, x3) E Q. This with (x3,b) E Q (which holds by (3)) entails {b,x) E E. d) Let (x3,x) E Q. This with (x2, $3) E Q yields {x2,x) E E. So, in the cases b) and c) we are done. In the other two cases, a) and d), we have that x is linked with 2 2 and x3 and thus in A. If now x would not be linked with b, we would have {x, b) E W, and because of b E C also x had to be in C, which contradicts the assumption. So (4) is proved. Now we can construct T. We choose a transitive orientation Tc in the induced subgraph C which is possible because it has fewer than n vertices. And we choose a transitive orientation TM in the induced subgraph of G which is generated by M := (V \ C) U {xl). This has fewer than n elements since C has at least two. Then we define T as the union of Tc and TM and the set of all pairs (x, x'), where {x, x') E E with x $ C, x' E C and where {x, x') is oriented in the same way as {x, xl}: (x,xl) E T , and of course, (xl,x) E T (xl,x) E (x,xf) E T T. Together we so obtain a transitive orientation T. To see this we have to check:


(6) (a, b) and (b, c) E T
+ (a, c) E T.
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361
This is trivial if a, b, c are all in C or all in M. For the rest we consider six cases: There are three cases where one of a, b, c is in M\{xl) and the other two in C, and three cases where one of a, b, c is in C and the other two in M\{xl). SOwe have the six cases: 1) a € M\{xl), b , c ~ C2) . b € M \ { x l ) , a , c E C . 3) C E M \ { x l ) ,
~,~EC.~)~,~EM\{X~),CEC.~)~,CEM\{X~),~ \{xl), a E C. See the following illustrations in Figure 27.
1)
Figure 27
In the cases 1) and 3) we immediately see that (6) holds. Case 2) cannot happen. For here (a,b) E T would imply ( x l ,b) E T , and this (c,b) E T , contradicting (b,c) E T . In case 4 ) we have (b,c) E T + ( b , x l ) E T .This with (a,b) E T and the fact that a, b, xl are in M and that T M is transitive implies (a,x l ) E T and (a,c) E T . In case 5) we have (a,b) E T 3 ( a , x l ) E T , and (b,c) E T 3 ( x l ,c) E T , so that by transitivity of TMwe obtain (a,c) E T . In case 6 ) we have (a,b) E T & ( x l ,b) E T , and (b,c) E T & ( x l , c ) E T since T M is transitive. Then also (a,c) E T . So T is a transitive orientation, and 2.2 is proved.
A bit later we shall see that the statement of 2.2 remains valid if the condition that G is finite is dropped.
2.3 Theorem of GhouilaHouri [51]. Let G = (V, E ) be a finite graph, and let G* be the graph (V*,E*) which is defined as follows: V* contains all pairs (a, b), for which {a, b) E E. (Intuitively speaking: V* contains all edges of E in both orientations. And note: These are in G* the vertices, and not the edges!.) E* is given by: Each vertex (a, b) of V* is linked with (b, a ) and with the vertices (b, c), for which {a, c) 4 E. Then G has a transitive orientation if the graph G* is 2chromatic.
Proof. a) Let D be a transitive orientation of G, then D is a subset of V*.
(I) No two elements of D are linked in G*. For if elements (a, b) , (c, d) of D would be linked in G*, we would have 1) b = c and {a, d) 4 E, or 2) d = a and {c, b) 4 E. Each of both is impossible since D is a transitive orientation. And (I) is proved. Let Do := {(x, y)l(y, x) E D), so that Do contains the edges of D with "opposite direction"
. Also we have by symmetry:
(11) No two elements of Do are linked in G*. Now we color all elements of D with the same color, say red, and all elements of Do with a second color, say green. Due to (I) and (11) this yields a Zcoloring of G*. b) Let now G* be 2chromatic, say V*colored with red and green. So for each edge {a, b) E E exactly one of the oriented edges (a, b), (b, a) is colored red, the other green. The redcolored elements of V* now form a transitive orientation of G. For let (a, b) and (b, c) be redcolored elements of V*. They cannot be linked, and therefore {a, c) must be in G. Then the set of those elements of V* which are colored in red is a quasitransitive orientation of G. According to the previous theorem 2.2, then G also has a transitive orientation. 2.3'. Theorem. The statements of 2.2 and 2.3 hold also, if the assumption that G is finite is dropped.
Proof. 2.2: Let G = (V, E) be a graph which has a quasitransitive orientation. Then a fortiori each finite induced subgraph F of G has a quasitransitive orientation, and by 2.2 then also a transitive orientation. According to 1.3 F is a comparability graph and by 1.10 also G, and thus G has a transitive orientation.
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363
2.3: Let the assumptions of 2.3 be fulfilled, where the condition that G is finite is dropped. Let G have a transitive orientation. Then this also holds for each finite induced subgraph F of G, and the corresponding graph F* is 2  chromatic. Then by Theorem 2.5.5 of Erdos/De Bruijn also G* is 2  chromatic. And if G* is 2  chromatic, the part b) of the proof of 2.3 also works for the general case, where G can be infinite. For before we had seen that quasitransitive implies transitive in general. Also Gilmore and Hoffman [54] established a characterization of the comparability graphs, which is essentially the same as that of GhouilaHouri. We describe this in more detail. First we define, following [54]: 2.4 Definition. Let G = (V, E ) be a graph. A cycle in G is a sequence al, . . . ,ar, (and we put as+i := ai for i = 1,2) of k 2: 3 vertices of V such that {ai, ai+i) E E for i = 1,. . . ,k, for which the pairs (ai, ai+i), i = 1,.. . ,k, are pairwise different. (But it can happen that a pair (ai, ai+1) also appears in the reverse order as (aitl, ai).) This cycle is even (resp odd), if k is even (resp. odd). A triangular chord of the above cycle is an edge of G of type {ai, ai+z) with i E {I,. . . ,k). The criterion of Gilmore and Hoffman is now:
2.5 Theorem. [54]. A graph G = (V, E) is a comparability graph iff each odd cycle has a triangular chord. That this condition is necessary follows easily, and without making use of earlier theorems. For let al, . . . ,ak be an odd cycle (k odd) without a triangular chord. And suppose indirectly that we have an orientation of the edges of E which is to partially order the vertices of G. Then this must give any successive pair of edges of the cycle opposite orientations. Both must be directed towards or away from the common vertex of the pairs. In detail: If {a, b) and {b, c) are edges of the cycle while {a, c) is not, then if ( a ,b) is the orientation which {a, b) obtains, (c, b) must be the orientation given to {b, c), for an orientation (b, c) would, due to the transitivity of order, entail the orientation (a, c), and thus {a, c) also had to be in G, with contradiction. The situation, where (b, a) is the orientation given to {a, b), is analogous. It is easily seen that only in an even cycle can all successive pairs of edges have opposite orientations. So the condition is necessary.
Suppose now that each odd cycle of G has at least one triangular chord. Then there follows: (I) The graph G* = (V*,E*)of 2.3 has no odd cycle. If in G* we would have vertices (al, a2),(a2,as), . . . , (anvl,an) with n 2 3 which form an odd cycle, then the sequence al, . . . ,a, forms an odd cycle of G. By assumption this has a triangular chord {ai, ai+2). But then (ai, ai+i) and (ai+i,ai+2) wouldn't be linked in G*. This yields a contradiction, and thus (I) holds. We have obtained that all cycles of G* are even. By the following simple lemma this implies that G* is 2  chromatic, and then by 2.3' G has a transitive orientation and is thus a comparability graph. 2.6 Lemma. Let G be a graph for which i n all cycles al, . . . ,a,, where the ai, i = 1,. . . ,n, are pairwise diferent, n is even. Then G is 2  chromatic.
Proof. W.r.0.g. we can assume that G is connected. Let v be an arbitrary vertex of G. For k E N let Sk be the (socalled ksphere of v, that means the) set of all vertices xk, for which k+l is the length of a shortest sequence v = xo, . . . ,xk, where Xi is linked with xi+l for i = 0,. . . ,k  1. Put So := { v ) . Then we color all vertices of sets Sk with even k red and all other vertices green. This furnishes a 2coloring of G. For if two vertices are linked, one must lie in a set Sk and the other one in Sk1 or Sk+1.To see this consider two vertices x, y of the same ksphere of v and two shortest sequences v = xo, . . . ,xk = x and v = yo,. . . ,yk = y of the above type. Let m be the last index for which xl = yl holds. Then x, y cannot be linked in G because otherwise the sequence $1,. . . ,xk, yk, . . . ,y1, diminished by the last element, would define a cycle of 2(k  1 1)  1 vertices, which is impossible since this number is odd.
+
11.3
A characterization of the comparability graphs of trees
Next we study the comparability graphs of ordertheoretical generalized trees, which are those posets P, in which for each x E P the set of its predecessors is linearly ordered. Wolk [I801 gave a characterization of these graphs, and Jung [93] gave a shorter proof of this, which at the same time furnishes a complete overview of all orientations of a given
11.3. COMPARABILITY GRAPHS OF TREES
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comparability graph of this class. The central concept in Jung's proof is an equivalence relation in the set of vertices of a graph: 3.1 Definition. Let G = (V, E) be a graph and e = {v, v') an edge of E. Then v is said to be a Vroot of e, if there exists a v" E V with v" # v', {v, v") E E and {v', v") 4 E. A relation < on V is called a treeorientation of G, iff V with the relation 5, that is <, augmented by the identity relation idv, is a generalized tree. e = {v, v') is called an n  edge, if exactly n of the vertices v,vlare V  roots of e. So n E { O , l , 2).

3.2 Definition. Let G = (V, E ) be a graph. For v, v' E V we put v = v', or {v,vl) is a 0  edge of G. v v1 3.3 Lemma.
N
is an equivalence yelation in V.


Proof. Only transitivity is nontrivial. Let vl v2 and v2 v3. We assume that the vertices vl, v2, v3 are different, otherwise the assertion is trivial. Then {vl,v2) and {v2,v3) are in E. Now v2 is no V  root of the edge {vl, v2) since this is a 0  edge. And therefore vl must be linked with v3, so that {vl, us) E E. We have to show that {vl, v3) is also a 0  edge. Suppose the contrary: If w.r.0.g. vl is a V  root of {v1,v3), then there is a v4 # v3 with {vl, v4) E E and (213, v4) $ E, and therefore we have v4 # v2. ), If {v2,v4) 4 E would hold, vl would be a V  root of { v ~ , v ~ which is impossible since this is a 0  edge. Also {v2,v4) E E would yield a contradiction since then v2 would be a V  root of the 0  edge {v2,us). So the lemma is proved. From the definition it is clear that the vertices of an equivalence class are pairwise linked in G  they form a socalled simplex of G. And before we proceed to the main theorem we mention a property of the equivalence classes, although we don't need it for the proof. But it gives a better oversight. Namely there holds: 3.4 Lemma. If Cl and C2 are two diflerent equivalence classes, then either all elements of Cl are linked with all elements of C2, or each element of C1 is unlinked with each element of C2.
Proof. Suppose that there is an edge e = {vl, v2) with vl E Cl, va E C2. If x is an element of Cl which is different from vl, then x must also be linked with z.2, for otherwise vl would be a V  root of the 0  edge
{vl, x). So all elements of Cl must be linked with v2. Symmetrically also all elements of C2 must be linked with vl. Then the rest easily follows. 3.5 Theorem [93]. Let G = (V, E) be a graph, and let orientation in E. Then the following two properties hold: (i) If C is an equivalence class of linear order on C.
< be a tree
N, then the restriction 5 1 C is a
(ii) If v is a V  root of {v, v') E E, we have v
< v'.
Proof. (i) All elements of C are by definition pairwise linked by 0  edges. Therefore each two elements of C are comparable, so that C is linearly ordered. (ii) If v is a V  root of {v, v'), there exists a vertex v" # v' with {v, v") E E and {v', v") $ E. Suppose that (ii) is false so that v' < v holds. If now we would have v < v", the transitivity of < would yield v' < v", SO that {v', v") E E would follow, with contradiction. In the case v" < v we would obtain that v" and v' are comparable, since then they would be predecessors (in a tree) of the same v. Again {v', v") E E would hold with contradiction. (ii) yields that not both ends v, v' of an edge of E can be a V  root of {v, v') since this would imply v < v' < v. So we obtain the theorem: 3.6 Theorem. A graph G which admits a tree orientation has no 2  edges.
The main theorem now states that 3.5 can be inverted.
3.7 Theorem [93]. Let G = (V, E) be a graph without 2  edges, and let < be an orientation of G. Then < is a treeorientation iff (i) and (ii) are satisfied. Proof. It remains to prove that (i) and (ii) imply that < is transitive and a treeorientation, namely that every principal ideal (v] of (V, 5 ) is linearly ordered. Let elements vl, v2, v3 with vl < v2 and v2 < v3 of V be given. Then {vl, v2) and {vz, v3) are edges of E. We have to prove vl < v3 and make the indirect assumption that this does not hold. v2 is no V  root of {v1,v2), for otherwise (ii) would imply v2 < vl with contradiction. This implies that {vl, v3) E E, and since we made the indirect assumption that vl < v3 does not hold, we now have v3 < vl . At least one of the edges {vl, v2), {v2,v3), {vl, v3) is no 0  edge, for otherwise we would have that vl, v2, v3 are all in one equivalence class,
11.3. COMPARABILITY GRAPHS OF TREES
367
and by (i) this class would be linearly ordered, and then vl < v3 would follow, contradicting our indirect assumption. Together with the indirect assumption we have vl < v2, v2 < v3, v3 < v1. Due to symmetry we can assume w.r.0.g. that {vl,v3) is no 0  edge. And so v3 or vl is a Vroot of {vl,v3). But vl cannot be that, for otherwise we would by (ii) have vl < vs, contradicting the indirect assumption. Thus v3 is a Vroot of {vl, v3). So there further exists a vertex v4 # vl with {vl,v4) $ E and {v3,v4) E E. If {v2,v4) $ E holds, v3 would be a V  root of {v2,vs). This entails v3 < va by (ii), a contradiction. In the case {v2,214) E E the vertex v2 would be a V  root of {vl, v2), and then va < v1, again with contradiction. So in every case our indirect assumption leads to a contradiction, and the transitivity of < is proved. Now we verify the treeproperty: Let v, v', v" be vertices with v' < v, v" < v, and v' # v". Then {v, v') and {v, v") are in E. If neither v' < v" nor v" < v' would hold, then {vl,v") $ E follows, and v would be a V  root of {v, v'), and by (ii) v < v', contradicting v' < v. So all predecessors of a vertex v are comparable, and (v] is linearly ordered. 3.8 Remark. By the above theorem all treeorientations of a graph without 2  edges are determined: First we orient all 1  edges according to (ii). Then we choose an arbitrary linear order in each equivalence class. This induces an orientation of all 0  edges. In particular we have obtained the criterion of Wolk: 3.9 Theorem [180]. A graph G has a treeorientation ifl G has no
2
 edges.
For finite graphs one obtains from 3.7: A graph G = (V, E) without 2  edges has exactly ITiErlCil ! different treeorientations. Here the {Ci, i E I) is the set of all equivalence classes of V. Finally we mention without proof a theorem of Arditti/Jung [6]: 3.10 Theorem. If G = (V, E ) is a comparability graph of two and (V, have the orders Il and 5 2 on V, then the posets (V, 11) same dimension.
s2)
In view of this theorem it makes sense to define the dimension of a comparability graph G = (V, E) as the dimension of an arbitrary poset (V, I ) , which has G as comparability graph.
For the special case where G is a finite graph the statement of 3.10 was also obtained by Trotter/Moore/Sumner [I711 and Gysin [61]. A penetrating analysis of finite comparability graphs is contained in Gallai's paper [47] of 1967, which characterizes these graphs in terms of the minimum list of graphs that are excluded as induced subgraphs. We also refer to the survey article [95] "Comparability graphs" of D. Kelly.
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Index above, 14 accumulation point, 205 adjacent elements, 19 aleph, 8 allrelation, 11 antichain, 19 antiisotone, 35 antisymmetric, 12 Aronszajn tree, 273 ascending chain condition, 191 ascending tower, 6 automorphic mapping, 343 automorphism, 34,343 automorphism group, 343 axiom of choice, 5 bad sequence, 246 base, 1I basis, 148 below, 14 between, 14 bijective, 2 binary relation, 11 bipartite, 355 block, 161 blockdegression, 165 blocksystem, 160 bounded, 21,205 bounded from above, 2 1 bounded from below, 21 branch, 232 branching point, 267 Cantor's normal form, 138 Cantor's normal representation, 138 carrier set, 11 cartesian product, 4 chain, 19 chaincoveringnumber, 57 change number, 172
character of a cut, 78 characteristic function, 8 choice function, 5,50,297 clique, 356 closed final segment 3 1 closed initial segment, 3 1 closed interval, 3 1 closed set, 205 closure operator, 30 cofinal,7 1, 72, 189 cofinality number, 73 cofinality, 9 coinitial, 7 1, 189 compact, 204 comparability graph, 354 comparable, 19 complementary final segment, 32 complementary relation, 11 complete, 22 complete lattice, 24 completely ordered set, 22 composition, 2 concatenation, 2 connected poset, 28 connected open set, 205 connectivity component, 205 constancysegment, 173 constant chain, 298 continuous poset, 47 continuous mapping, 205 continuously ordered, 47 contraction, 306 contractive, 297 convex, 29 convex hull, 30 countable, 3 cut in a poset 40, cut in a linearly ordered set, 46, 111
cycle, 363 cyclic permutation, 294 decomposition, 17 decreasing mapping, 35 decreasing tower, 6 DedekindMacNeille completion,40 degree, 115 dense, 41 dense in, 111 denumerable, 3 descending chain condition, 191 descending tower, 6 diagonal, 11 digraph, 353 dimension, 206 direct product, 86 directed, 2 1 directed cubic graph, 306 directed edge, 306,353 directed from above, 2 1 directed from below, 21 directed graph, 353 directed path, 354 distributive, 62 domain, 1 doubly monotonic chain, 309 down set, 29 dual statement, 47 dualization, 47 dyadic depth, 180 dyadic splitting, 162 edge, 59 edge set, 59 elementcharacter, 78 embeddable, 37 empty relation, 11 empty sequence, 263 empty set, 1
endcharacter, 78 endpoint, 309 equipotent, 3 equivalence class, 16 equivalence relation, 16 equivalent, 55 euclidean nspace, 1 even ordinal, 296 exorbitant, 75 extensionality, 30 extremal graph, 59 family, 3 final order, 90 final segment of a poset, 29 final segment of an ordinal, 335 finer, 159 finite character, 50 finite set, 3 first element, 22 fixed chain, 298 fixed element, 30 free from, 37 full part, 185 fully comparable, 55 function, 1 Galtonfunction, 3 15 Galtonway, 3 15 gap, 47 gapcharacter, 78 generalized axiom of choice, 164 generalized continuum hypothesis, 9 generalized tree, 160 good sequence, 246 graph, 59 graph of a function, 309 greater, 14 greater or equal, 14 greatest element, 22
Index greatest lower bound, 22 ground set, 11 Hausdorff s maximal principle, 50 height, 24, 162,232 heightnset, 24 hereditary, 194 Hessenberg natural sum, 139 homogeneous, 115,343 idempotency, 30 identity mapping, 2 identity relation, 11 IFpair, 143 image, 2 immediate predecessor, 19 immediate successor, 19 incomparable, 19 increasing, 35 increasing tower, G indecomposable ordinal, 115, 137 independent vertex sets, 185 index set, 3 induced subgraph, 356 infcomplete, 22 infimum, 22 initial character, 78 initial number, 8 initial order, 90 initial ordinal, 8 initial segment of a poset, 29 initial segment of an ordinal, 334 injective, 2 intersection, 3 interval, 3 1 intervalhomogeneous, 121 inverse function, 2 inverse relation, 1 1 irreducible, 49 irreflexive, 12
isolated, 125 isomorphic, 4,34,35 isomorphic mapping, 34 isomorphism, 34 isotone, 35 isotonicity, 30 iterate, 2 jump, 47 jumpspot, 320 kbasis, 148 kcolorable, 59 kernel element, 55 kernel function, 298 kgenerated, 148 larger, 14 last element, 22 lattice, 24,62 latticeordered group, 35 1 least element, 22 left closed, 3 1 left neighbor, 19 left open, 3 1 left subsegment, 173 leftneighborrelation, 19 length, 19,232 length of a sequence, 262 less or equal, 3, 14 less than, 3, 14 lexicographic order, 87 lexicographic product, 87 Igroup, 35 1 limit mapping, 8 limit number, 7 limit ordinal, 7 linear continuum, 1 linear extension, 54 linear kernel, 55 linear order, 13
Index linearly ordered, 14 linearly ordered continuum, 204 linked, 59 lower bound, 21 lower linear extension, 226 lower neighbor, 19 lower set, 29 lower shadow, 289 lowerneighborrelation, 19 mapping, 2 maximal, 23 maximum, 22 minimal, 23 minimum, 22 mixed sum of ordinals, 139 monotone kchain, 307 ncontinuum, 204 ndimensional closed interval, 204 neighbor, 19 neighboring elements, 19 neighboring subsets, 77 nlevel, 24 nspecial, 299 octant, 309 odd ordinal, 296 open final segment, 3 1 open initial segment, 3 1 open interval, 32 open segment, 204 open set, 204 order, 13 order by first differences, 87 order relation, 13 order topology, 204 order type, 36 ordered product, 86 ordered set, 13 ordered sum, 85
orderisomorphic, 35 orderisomorphism, 35 ordinal, 6 orientation, 354 oriented edge, 306,353 partial order, 13 partial ordinal, 25 1 partially ordered set, 13 partially wellordered, 2 12,23 1 partition, 17 path, 232,266 perfect subsequence, 245 pographs, 356 pogroup, 35 1 pointcontaining, 174 pointless, 174 poset, 13 position, 146 positionequivalent, 222 Pouzetset, 2 10 power, 4 power set, 1 predecessor, 19 preorder, 13 preserving, 4 principal filter, 3 1 principal ideal, 3 1 product order, 86 product relation, 11 productive, 90,248 proper cut, 78 proper partition, 159 proper initial segment, 91 proper subset, 1 proper superset, 1 pwoset, 2 12 quasiorder, 13 quasiordered, 14
Index quasitransitive orientation, 357 quasitransitively directed, 357 range, 1 rank, 28 rank fimction, 27 Rcomponent, 49 Rconnected, 49 realization, 3 reducible, 49 reflexive, 12 region, 205 regressing, 297 regressive chain, 298 regular ordinal, 73 regular covering, 290 relation, 11 restriction, 2, 11 reversing, 35 right closed, 3 1 right neighbor, 19 right open, 3 1 right subsegment, 173 rigid, 343 ring, 195 ringclosure, 196 same content, 37 same position, 146 saturated chain, 29 scattered, 38, 194 segment, 29 semichaincomplete, 267 similar, 35 singular ordinal, 73 smaller, 14 spanning tree, 3 18 special contractive mapping, 298 Specker chain, 275 splittable, 3 19
splitting, 161 splitting function, 161 splitting tree, 267 standard linear extension, 268 star, 3 18 strict linear order, 13 strict lower bound, 2 1 strict order, 13 strict upper bound, 2 1 strictly ascending tower, 6 strictly contractive, 297 strictly dense in, 111, 146 strictly descending tower, 6 strictly increasing, 35 strictly isotone, 35 strictly linearly ordered, 14 strictly ordered, 14 stump, 232 subdividing function, 164 subgraph, 356 subrelation, 11 subsequence, 263 subset, 1 subtree, 160 subtype, 37 successor, 19 successor number, 7 successor ordinal, 7 sum of a word, 264 sum of order types, 86 sumorder, 85 supcomplete, 22 superfluous element, 254 superset, 1 supremum, 22 surjective, 2 Suslin chain, 279 Suslin tree, 279
Index symmetric relation, 12 symmetric chain, 293 symmetric cut, 78 symmetric element, 78 symmetric hull, 12 symmetric order, 94 symmetric order type, 37,94 symmetric poset, 35 symmetrically ordered set, 94 symmetry, 94 topological space, 203 total order, 13 totally intransitive, 20 totally unordered, 38 transfinite sequence, 6 transitive, 12 transitive hull, 12 transitive orientation, 354 transitively directed, 354 transitively oriented, 354 tree, 160 treeorientation, 365 triangular chord, 363 unattainable, 121 underlying set, 11 union, 3 uniquely homogeneous, 343 unlinked, 59 unsplittable, 3 19 upper bound, 2 1 upper linear extension, 226 upper neighbor, 19 upper shadow, 289 value, 2 vertex, 59 vertex set, 59 Vroot, 365 wellfounded, 38, 191,23 1
wellordered, 4, 38 wellquasiordered, 38,201,23 1 width, 57 within a cut, 115 word, 263 wqoset, 23 1 Zorn's lemma, 50 &dense, 43 qaset, 101 qaaset, 145 Ahook, 38 vcomponent, 9 1 vpredecessor, 266 psegment, 9 1 odense, 43 7free, 9 1 xa universal, 145 Na universally ordered, 145
List of symbols 14 P, 14 >, 14 1, 14 11, 19 a, 19 D, 19 inf, 22 sup, 22 min, 22 max, 22 Mi, 23 Ma, 23 h b ) , 24 h(P), 24 Ln, 24 44,27 rn(P), 28 W ) , 29 I S ( T ) , 30 FS(T), 30 Conv T, 30 Fix(c), 30 Inc(a), 31 ( P < a ) , 31 ( P l a), 31 ( P > a), 31 (P 1a), 31 ( P ll 4 7 31 (.I, 31 [a), 31 [a,% 31 (a,% 31 [a,% 32 (a, b), 32 A < B, 31 $7
A > B, 31 A < z, 31 A > z, 31 e,35 tp, 36 5 for types, 37 < for types, 37 % A , 38 L(T), 40 W T ) ,40 W ) , 40 DM(T), 40 type (1, I ) , 47 type ( m , l ) , 47 type ( L m ) , 47 type ( m , 4, 47 LK(P), 55 W ) , 57 CCN(P), 57 [SI2, 59 A, 62 V, 62 w(lMl), 71 ord(M), 71 cf, 73 coin, 73 C , 85 X, 86 L, 87 P ( ( T > > , 88 S ( ( 4 ) , 88 Ha, 98 to, 99 ha, 101 Vo, 101 Hk, 105
S ( T ) , 111 Tu, T ( 4 113 tu, t ( 4 114 P ( a ) , 118 T ( ( a 122 do),122 G,, 129 C,, 134 @, 139 P ( X ) , 143 U,, 149 om, E,, 152 h ( T ) , 161 w C ~ ( ( ~ ) ) O )
c(d), 172 Cu(d), 173 l ( D ) , 180 a + (b, c ) ~ ,181 S+(P), 189 6(P), 189 d+(P), 189 d(P), 189 ro(P), 189 .rrl(P), 189 n ( P ) , 189 a ( P ) , 189 R ( S ) , 196 dims, 206 D ( S ) , 210 wqo, 231 pwo, 212 a + [A71, 237 a * EP,rl, 237 o ( X ) , 254 (i),286 V R , 289
AR, 289 P ( A , B ) , 298 ~ ( t )299 , T ( P , Q, n), 299 oct(p), 309 Aut(S), 351