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'" cofinal branches. Thus the full binary tree of height w' is an No-Kurepa tree. Are there any ~1-Kurepa trees? Evidently, the full binary tree of height WI doesn't qualify. EXERCISE 14.8 (G). Convince yourself that the full binary tree of height", is a ",-Kurepa tree if and only if '" is a strong limit cardinal, i.e., 2>' < '" for all A < "'. 0 and allowing any number of dummy variables.
14.
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31
From now on we will call NI-Kurepa trees simply Kurepa trees. The following statement is known as the K urepa Hypothesis. (KH) There exists a K urepa tree. Theorem 22.8 shows that KH is relatively consistent with ZFC. On the other hand, if the existence of an inaccessible cardinal is consistent with ZFC, then so is the negation of KH. In many applications of trees it is crucial to know whether a given tree has at least one cofinal branch. The tree in Example 14.6 has no cofinal branches, but rather large levels (of size h'1). On the other hand, the slender trees of Examples 14.2 and 14.4 have the property that every branch is cofinal. It is natural to conjecture that trees with relatively small levels will always contain cofinal branches. Let us explore this hunch. LEMMA 14.2 (Konig). Let (T, :ST) be a tree of height w such that T(n) is finite for all nEw. Then T contains a cofinal branch. PROOF. We construct recursively a sequence (tn)n<w of elements of T so that for all n < w:
(i)
tn
Let to be any element of T(O) such that I{s E T: to
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(more precisely, the Principle of Dependent Choices) has been used in our proof. Not as a matter of convenience, but out of necessity: Konig's Lemma is not provable in ZF. EXERCISE 14.10 (G). Show that the theory ZF + Konig's Lemma entails that Xn I- 0 for every indexed family {Xn : nEw} of nonempty finite sets.
TInEw
EXERCISE 14.11 (G). If T is as in Lemma 14.2, then each level T(n) is finite and can be wellordered without the help of the Axiom of Choice. Why can't we prove Lemma 14.2 in ZF by always letting tn be the smallest eligible element of T(n)? We are going to show that Konig's Lemma is equivalent (Le., equivalent in ZF) to weak versions of Tychonoff's Theorem and the Compactness Theorem. If you are not curious how such logical niceties illuminate the connection between seemingly disparate notions of compactness, you may want to skip ahead and resume reading right after Exercise 14.23. Consider the following statements: Every product TInEw Xn of finite non empty Hausdorff spaces Xn (WTY) is nonempty and compact. (WCT) Let L be a first-order language without functional symbols, and let (Sn)nEw be an increasing sequence of finite sets of quantifier-free sentences of L. If each Sn has a model, then the theory S = UnEw Sn also has a model. THEOREM 14.4 (ZF). The following statements are equivalent: (i) Konig's Lemma; (ii) WTY; (iii) WCT. Before proving Theorem 14.4, let us draw your attention to some subtleties in the wording of WTY and WCT. For k E w\{O} let D(k) denote the k-element space {O, ... , k - I} with the discrete topology. Now consider the statements: For every sequence (kn)nEw of positive integers (vWTY) the product TInEw D(k n ) is nonempty and compact. (vWCT) Let L be a first-order language without functional symbols, and let ('Pk)kEW be a sequence of quantifier-free sentences of L. If for each nEw the theory {'Pk : k < n} has a model, then the theory {'Pk : k E w} also has a model. It may appear that vWTY says the same as WTY and vWCT says the same as vWCT, but this is not the case. EXERCISE 14.12 (X). Prove that vWTY and vWCT are theorems of ZF. Note also that in stating WCT we used the phrase "S has a model" rather than "S is consistent." Recall that a theory S is consistent if no contradiction can be derived from S. But since formal derivations (i.e., proofs) are finite strings of formulas that can contain only finitely many formulas from S, one does not need the Axiom of Choice to show that if a theory is inconsistent, then it has an inconsistent finite fragment. Of course, by the Completeness Theorem, a theory S is consistent if and only if it has a model. But all known proofs of the Completeness
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33
Theorem use the Axiom of Choice. Thus Theorem 14.4 implies that any proof of the Completeness Theorem must use the Axiom of Choice. PROOF OF THEOREM 14.4. Let us start with the easiest part and show that WTY implies Konig's Lemma. So suppose that (T, :::;T) is a tree of height w with all levels finite. We can treat each T(n) as a discrete topological space. Let X = TInEw T(n). By WTY, this is a nonempty compact space. For k E w, let Uk = {x EX: x(k) ZT x(k + I)}. Each Uk is open in X. CLAIM 14.5. For each nEw, the set Uk::;n Uk does not cover X. PROOF. Fix nEw and let t E T(n + 1) be such that for all m > n there exists r E T(m) with t :::;T r. For each k :::; n + 1 let t(k) be the unique s E T(k) such that s :::;T t. By WTY, the space Y = TIk
14. TREES
34
EXERCISE 14.15 (X). Formulate and prove an analogue of Claim 14.6 for sentences 'P without quantifiers that may contain not only constant and relational symbols, but functional symbols as well. COROLLARY 14.7. If S is a nonempty set of tidy formulas of Land S has any model at all, then S has a model whose size does not exceed the cardinality of the set of constant symbols that occur in the formulas of S. PROOF. Let S be as in the assumptions, and let 2t = (A, (Rf)iEI, (Cff)kEK) be a model of S. Define Ko = {k E K : Ck occurs in some 'P E S}, and let A1 = {Ck : k E Ko}· Since S is nonempty, and since every tidy formula must contain at least one constant symbol, Ko '# 0. Let ko E K o, and let
= ct for all k
= Cko
for all k E K\Ko. By Claim 14.6, the model2t 1 = (A1' (Rfn(At}7'o(i))iEI' (CkhEK) is as required. Ck
E K o,
Ck
0
Now let us assume that Konig's Lemma holds and prove that the theory S = Sn has a model. For each nEw, let Kn = {k E K : Ck occurs in some 'P E Sn} and In = {i E I : ri occurs in some 'P E Sn}. Since Sn ~ Sn+1, we also have that Kn ~ Kn+1 and In ~ I n+1. For technical reasons we will assume that Kn is a proper subset of Kn+1 for all n E w. 3 Let mn = IKnl. Since the nature of the objects that comprise the universe of a model does not matter, Corollary 14.7 implies that every theory Sn has a model with universe {a, 1, ... ,mn -I}. Define: U nEw
Mn
= {M
: M = (mn' (Rf!)iEIn , (C!:)kEKn
)
1\
M ~ Sn}.
Since the sets In, Kn are finite, and since for each nEw and i E In there are only finitely many subsets of m~o(i) and finitely many functions from Kn into m n , the sets Mn are finite. Now let T = UnEw M n , and define a relation ~T on T by letting M ~T N if and only if M is a submodel of N. EXERCISE 14.16 (G). Show that T(n) = Mn for all nEw. Hint: It is here that you will have to use the extra assumption that Kn is a proper subset of K n+1'
Since each Mn is finite and nonempty, Konig's Lemma implies that there exists a cofinal branch B = {Mn : nEw} of T such that Mn E Mn for each nEw. Fix such B and define: M
= (w, (Rf!)iEU"ew In' (C!:)kEUnew K,,),
where Rf! = U{R~" : i E In} and Cr: = that k E Kn. EXERCISE
14.17 (PG). Show that M
c;:n for some (equivalently: all) n such ~
S.
It appears that we have proved WCT. But have we really done that? If I =j:.
In or K =j:. UnEw K n , then the model M we just constructed is not a model of L, but only of a fragment of L. But do we need to construct a model of L? The last sentence of WCT postulates the existence of a model of S, not a "model of L that is also a model of S." So we are done with our proof, aren't we? Well, yes and no. You will probably admit that the phrase "model of L that is also a model U nEw
3This does not lead to loss of generality, since we can expand L to a new language L + by adding new constant symbols {d n : nEw}, extend each Sn to s;t = Sn U {d n = d n }, and prove the existence of a model Ql+ for UnEw s;t. The L-reduct of m+ will then be the required model of S.
14. TREES
35
of S" is a bit awkward. Although mathematicians usually express themselves very explicitly, in polite company it is understood that if a paragraph starts with "Let L be a first-order language" then each subsequent use of the word "model" refers to models of L, unless something to the contrary is said. So let us bow to the rules of etiquette and conclude this part of our argument with one last exercise. EXERCISE 14.18 (PG). Let L - ~ L be two first-order languages, and let S be a set of sentences in L -. Show that if there exists a nonempty model M of Lsuch that M F S, then there exists a model N of L such that N F S. Be careful to avoid using the Axiom of Choice.
Finally, let us assume WCT and derive WTY. Let (Xn)nEw be a sequence of nonempty finite sets. We first show that TInEw Xn =f. 0. Define K = UnEw Xn X {n}. Let us treat each (x, n) E K as a constant symbol in a first-order language L. The only other nonlogical symbol of L will be a relational symbol r of arity 1. All models considered below will be models of this language L. To illustrate our line of reasoning, we will first show you a slightly phony version of the argument. While reading it, you should work on the following: EXERCISE
14.19 (G). Catch us cheating.
For each nEw, let S;; be the following set of formulas:
S;; = {'Pn} U {-'(r((a,n)) 1\ r((b,n))): (a,b) E (Xn)2\{(a,a) : a E X n }}, where 'Pn is the disjunction of all formulas r( (a, n)), i.e., something like r( (a, n)) V r( (b, n)) V ... V r( (z, n)) if Xn happens to have 26 elements. (For larger Xn's, you may have to borrow some characters from the Chinese.) Now let us put Sn = Uk
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be the formula:
r((O"(a),n)) Vr((O"(b),n)) V··· V r((O"(z),n)). Now define: s~ =
{
0"
is a permutation of Xn}
{-,(r( (a, n)) /\ r( (b, n))) : (a, b)
E (Xn)2\ {(a, a)
:a
E X n }},
and let S! = Uk~n S~ for all nEw. EXERCISE 14.20 (PG). Convince yourself that the sequence (S~)nEw does not depend on any particular choice of orderings of the Xn's and can be defined without invoking the Axiom of Choice. EXERCISE 14.21 (G). Use the sequence (S~)nEw and WCT to show that the product I1nEw Xn is nonempty. Now consider each Xn as a discrete topological space, and let X = I1nEw Xn with the product topology. Recall that the family {[I] : 3N E w (f E I1n
s;t = S~ U {-'(r((f(O), 0)) /\ r((f(l), 1)) /\ ... /\ r((f(n), n))) :
I E
II Xn /\ 3U E U ([I] ~ U)}. k~n
EXERCISE 14.22 (PG). (a) Convince yourself that the sequence (S;t)nEw has been defined without reference to the Axiom of Choice. (b) Show that for every nEw the theory s;t has a model if and only if UV -=J. X for every V E [U]<~o. (c) Use WCT to deduce that X is compact. 0 EXERCISE 14.23 (PG). Having seen the proof of Theorem 14.4, try your luck with Exercise 14.12 again. If K is an infinite cardinal, then a K-tree is a tree T of height K such that IT(a)1 < K for every a < K. In particular, a K-Kurepa tree is a K-tree. A KAronszajn tree is a K-tree without co final branches. An N1-Aronszajn tree is simply called an Aronszajn tree. An infinite cardinal K has the tree property if there are no K-Aronszajn trees. Konig's Lemma asserts that No has the tree property. Is the analogue of Lemma 14.2 true for other infinite cardinals K? Let us assume T is a K-tree, and let us try to construct recursively a co final branch (te)e<,.; as in the proof of Lemma 14.2. To get started, we would want to choose to E T(O) such that I{s E T : to
14. TREES
37
Note that ~* is a partial pre-order relation, Le., reflexive and transitive. Hence, the relation =* defined by:
eo =*
el
if and only if
eo~· el
and
el ~.
eo
is an equivalence relation on E. We construct recursively a transfinite sequence (ee)e<wI of elements of E such that dom(ee) = ~ and ee ~* e"rJ for all ~ < "., < WI. To get started, let eo = 0. Given ee, let ee+l = ee U {(~, min (w\rng (ee))) }. If 0 < 6 < WI is a limit ordinal and ee is defined for all ~ < 6, we choose an increasing sequence (~n)nEw of ordinals less than 6 such that 6 = SUP{~n : nEw} and use the following lemma to find a suitable eo. LEMMA 14.9. Suppose (en)nEw is a sequence of functions in E such that en ~* en+! for all nEw. Then there exists an e E E such that dom(e) = UnEw dom(e n ) and en ~* e for all nEw. PROOF. Let (en)nEw be as in the assumptions. By recursion over w we construct functions fn E E and a set {an: nEw} ~ w as follows:
fo
= eo;
ao
= min(w\rng(fo)).
Given fn such that fn =* en and {ao, ... ,an}, let
Bn = {k E dom(en+l)\dom(e n ) : en+!(k) E rng(fn) U {ao, ... , an}}. EXERCISE 14.25 (G). Show that Bn is a finite set. Let dom(Jn+!) = dom(en+d. For k E dom(en+l)\(dom(e n ) UBn), let fn+l(k) be equal to en+l(k), and define fn+l fBn in such a way that fn+! is a one-toone function with range disjoint from the set {ao, ... ,an}. Since w\rng(en+d is infinite, we have plenty of room to define fn+! fBn in this way. Finally, let an+l be the smallest element of the set (w\rng(fn+!)) U {ao, ... an}. Note that fn+! =* en+!j and hence we are ready for the next step of the construction. Now let e = UnEw fn. It is not hard to see that this is a one-to-one function from UnEw dom(en) into w with rng(e) n {an: nEw} = 0. Hence e E E; and a 0 straightforward verification shows that en ~. e for all nEw.
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Given a sequence (ee)e<Wl as above, let T = {e E E : 3~ < WI (e =* ee)}. We show that (T,~) is an Aronszajn tree. First note that if e E T, then e = {ef1/ : 1/ E dom( e) }. This set is wellordered by inclusion and has order type dom( e). Hence T is a tree, and ht(e) = dom(e) for all e E T. It follows that T(~) = {e E E : e =* ed for ~ < WI, and T(~) = 0 for ~ 2:: WI' Thus ht(T) = WI. EXERCISE 14.26 (G). Show that every level of T is countable. It remains to show that T has no uncountable branches. Note that if (e7))7)
EXERCISE 14.27 (G). (a) Show that if 8 E ON and (A7))7)
eo
~*
el iff dom(eo) ~ dom(eI) and I{~ E dom(eo) : eo(~) -=f eI(~)}1 < NI .
Consider the equivalence relation ::::::* defined by:
eo ::::::* el iff eo
~*
el and el
~*
eo.
We would like to construct a sequence (e~)~<Wl of elements of EI such that (a) dom(e~) = ~ for all ~ < W2; (b) e~ ~* e7) for all ~ < 1/ < W2· Given a sequence (e~)~<W2 that satisfies (a) and (b), we'd like to define T = {e E EI : 3~ < W2 (e ::::::* e~)}. The same argument as in the proof of Theorem 14.8 shows that (T,~) is a tree of height W2 without cofinal branches. For ~ < W2, the ~-th level T(~) of this tree is equal to {e E EI : e ::::::* ed. It is not hard to see that IT(~)I = N~o if W ~ ~ < W2. Thus, if CH holds, (T,~) will turn out to be an N2 -Aronszajn tree. But can we construct a sequence (e~)~<W2 that satisfies conditions (a) and (b)? Let us try to proceed as in the proof of Theorem 14.8. We let eo = 0. Given e~, one can construct e~+1 exactly as in the proof of Theorem 14.8. But limit stages pose a new challenge. To illustrate the problem, suppose 8 > 0 is a limit ordinal of countable cofinality, and (~n)nEw is an increasing sequence of ordinals with limnEw ~n = 8. For simplicity, suppose that e~n ~ e~n+l for all nEw. EXERCISE 14.28 (G). Convince yourself that if UnEw rng(e~J = there is no eo EEl such that e~n ~* eo for all nEw.
WI,
then
How can we prevent the situation described in Exercise 14.28 from ever arising? Evidently, the requirement that IWI \rng(e~)1 ~ NI is not sufficient. We can make the requirement more stringent by fixing a proper ideal I of subsets of WI and considering the following subset of E I :
E2 = {e E <W2 WI
:
e is an injection and rng( e) E I}.
14.
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39
What properties should I have so that our construction will work? First of all, we will want I to contain some uncountable sets. Let us require that I has the following property: (i) VA E I3B E [W1]Nl (A n B = 0" A U BE I). Second, in order to get our construction past limit stages of countable cofinality, we will want I to be countably complete; i.e., we want I to satisfy the following:
(ii)
If A E [I] No , then
UA E I.
Note that if I satisfies (ii), 8 < W2 is a limit ordinal of countable cofinality, (~n)nEw is an increasing sequence of ordinals cofinal in 8, and for each nEw we have constructed een E E2 such that een ~ een+l for all nEw, then we can simply define eli = UnEw een' It remains to' consider limit stages 8 of cofinality W1. Let us assume that (~Q)Q<Wl is an increasing sequence of ordinals cofinal in 8, and the ee", 's have already been chosen from the set E 2 . For the sake of simplicity, let us assume that ee", ~ ee13 for all a < f3 < W1· If there exists eli E E2 such that eea :::;* eli, there must exist B E I with Irng(eeJ\BI < ~1 for all a < W1' In other words, we need to require the following property of I: (iii) If A E [I]Nl, then there exists B E I such that IA\BI < ~1 for all A E A. It turns out that this property also suffices. Given B as above, use (i) to find C E [wd N1 that is disjoint from B and still a member of the ideal I. Now define recursively eea as follows: If ee", (1]) E B, then let e~J 1]) = eea (1]). If eea (1]) tJ. B, then let eea (1]) be the smallest ordinal in C that has not yet been used in the construction. Finally, let eli = UQEWl e~a' EXERCISE 14.29 (G). Convince yourself that this construction yields a sequence (ee)e<Wl of elements of E2 that satisfies conditions (a) and (b). But does there exist an ideal I that satisfies conditions (i)-(iii)? In Chapter 21 (Lemma 21.11) we will prove the following: LEMMA 14.10. Let", be a regular uncountable cardinal. Then there exists an ideal N SI< on '" with the following properties: (i) VA E NSI<3B E [",]1< (A n B = 0" AU B E NSI<); (ii) If A E [NSI<]
= "', then
If GCH holds, then 2<1< = '" for every infinite cardinal "'. Hence GCH implies the existence of ",+ -Aronszajn trees for every regular infinite cardinal "'. It follows that if GCH holds, then every uncountable cardinal with the tree property is either the successor of a singular cardinal, or strongly inaccessible. Strongly inaccessible cardinals with the tree property are called weakly compact cardinals. In Chapters 15 and 24 we will give some alternative characterizations of weakly compact cardinals. There are other constructions of Aronszajn trees than the one used in the proof of Theorem 14.8. Let us sketch one of them. Let ]P> be the set of all nonempty
14. TREES
40
compact subsets of the rationals Q, and let
K:::;e L if and only if K
~
:::;e denote end-extension, i.e.,
Land Vp E KVq E L\K(p:::; q).
EXERCISE 14.31 (G). Convince yourself that (lP', :::;e) is a p.o. without uncountable wellordered chains. So if T ~ lP' and (T, :::;e) is an wI-tree, then (T, :::;e) is automatically an Aronszajn tree. But how to construct an wI-subtree of lP'? By transfinite recursion of course, that is, level by level. It is easy to construct T(O), T(l), T(2), ... in such a way that (TC w), :::;e) is a splitting tree of height W with countable levels. Then TCw) has 2No cofinal branches. Since T(w) is supposed to be countable, we will put nodes in T(w) only on top of countably many of these branches. Note that a branch in T(w) will remain a branch in T if and only if it has no extension in T(w). But are there any cofinal branches of TC w ) that can be extended? Not necessarily. Note that an increasing chain Ko :::;e KI :::;e K2 :::;e ... is unbounded in (lP', :::;e) if and only if limn->oo max(Kn) = 00. Thus, had we chosen for example the T(n)'s in such a way that max(K) ;::: n for all nEw and K E T(n), then none of the branches in TC w ) could be extended. The trick is to give ourselves the necessary flexibility by requiring that the following condition (*)", holds for every a < WI:
(*)",
V~
< T/ < aVK
E T(~)Vq E
Q(q > maxK
--+
3L E T(TJ)K:::;e Ll\maxL:::; q).
EXERCISE 14.32 (PG). (a) Suppose that 8 E WI n LIM and (TCo) , :::;e) is constructed as above. Show that if (*)0 holds, then at least some cofinal branches of Tco) can be extended. (b) Construct a T ~ lP' such that (T, :::;e) is an WI-tree. An antichain in a tree (T, S;T) is a subset A ~ T of pairwise incomparable elements of T. This notion of "anti chain" is different from the one in Chapter 13, but there is a close connection between the two concepts. EXERCISE 14.33 (G). Show that if T is a tree and A ~ T, then A consists of pairwise incomparable elements if and only if it consists of elements that are pairwise incompatible in the dual p.o. (T, S;T)' EXAMPLE 14.7. Every level T(a) of a tree T is an antichain in T. The set {nl' ql, tI, yt} is an antichain in the tree of Example 14.1. If an denotes the function with domain n + 1 that takes values an(n) = 1 and an (k) = 0 for all k < n, then {an: nEw} is an antichain in the full binary tree of height w. EXERCISE 14.34 (G). Show that if T is a splitting tree and T has a branch of cardinality /'i" then T also has an antichain of cardinality /'i,. Trees of height WI without uncountable chains or antichains are called Buslin trees. 5 The Buslin Hypothesis, abbreviated SH, is the following statement: (SH) There are no Buslin trees. Note that the Kurepa Hypothesis asserts that there are Kurepa trees, while the Suslin Hypothesis asserts that there are nO'Suslin trees. Was Suslin a nihilist? Obviously, Suslin trees are Aronszajn trees. The converse is not so obvious, since the absence of uncountable antichains of the form T( a) does not automatically imply that a tree T has no antichains whatsoever. In fact, the assertion that every 5 Read:
"Soos\een." One often sees the spelling "Souslin."
14.
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41
Aronszajn tree is Suslin is simply false. To see this, consider the tree (T, ::;e) constructed in Exercise 14.32. Each element of T is a nonempty compact subset of Q and hence has a maximum element. Let f : T -+ Q be the function that assigns to each t E T its maximum element. Then f satisfies the following condition: Vs,t E T(s
(*)
s : a + 1 ~ W for some a E s(max dom(s)).
WI.
Define a function
'Tr :
S
-+ W
by 'Tr(s) =
EXERCISE 14.35 (G). Convince yourself that 'Tr is a specializing function for S, Le., show that 'Tr- 1 {n} is an antichain in S for every nEw. We have shown that S is special. How about To itself? EXERCISE 14.36 (G). Let h : To -+ S be a function that assigns to each t E To(a) a node h(t) E To(a+1) with t ~ h(t). Let 'Tr be as above, and define 'Trl = 'Troh. Show that 'Trl is a specializing function for T. We have shown that To is special. But is To also Q-embeddable? Yes. 14.12. Let T be an Aronszajn tree. Then the following are equivalent: T is special; T is Q-embeddable.
LEMMA
(i) (ii)
PROOF. We have already seen that Q-embeddable Aronszajn trees are special. To prove the converse, let us assume that (T, ::;T) is a special Aronszajn tree, and let 'Tr : T -+ w be a specializing function. Define a function f with domain T by: f(t)
= {n: 3s (s:5Ttl\1r(s)=n)}
EXERCISE 14.37 (G). Show that f is a Q-embedding of T.
o
In Chapter 19 we will show that Martin's Axiom (a statement relatively consistent with ZFC) implies that every Aronszajn tree is special, which in turn implies the Suslin Hypothesis. In Chapter 22 we will prove that the existence of a Suslin tree, Le., the negation of the Suslin Hypothesis, is also relatively consistent with ZFC. CLAIM 14.13. Let,.. be a regular infinite cardinal, and let T be a ,..-tree without cofinal branches. Then there exists a splitting ,..-subtree SofT.
14.
42
PROOF. Let
K,
TREES
T be as in the assumption, and define:
S={sET: l{tET: s
o
Clearly, N1 -subtrees of Aronszajn trees are Aronszajn, and N1 -subtrees of Suslin trees are Suslin. It follows that every Aronszajn tree contains a splitting Aronszajn subtree, and every Suslin tree contains a splitting Suslin subtree. Recall from Chapter 2 that if (A, :::SA) and (B, :::SB) are p.o.'s, then the simple product (A, :::SA) ®S (B, :::SB) is the p.o. (A x B, :::Ss), where
(a, b) :::Ss (c, d) if and only if a:::SA b 1\ C :::SB d. Suslin trees, if they exist, provide examples of c.c.c. partial orders whose simple product does not satisfy the C.C.c.: By Exercise 14.33, if (T "5:.r) is a Suslin tree, then (T, "5:. T ) satisfies the c.c.c. On the other hand, we have the following: LEMMA 14.14. If (T, "5:.r) is an uncountable splitting tree, then (T, "5:.r) ®" (T, "5:.r) does not satisfy the c.c.c. PROOF. Let T be as in the assumption. For every t E T, choose tt, tT E T such that t
0
We have proved the following result: THEOREM 14.15. If SH fails, then there exists a c.c.c. p.o. (1P',:::S) such that (1P',:::S) ®" (1P',:::S) does not satisfy the c.c.c. In particular, if (T, :::sr) is any Buslin tree, then the p.o. (T, :::ST) is as above. Trees can be used to construct various fancy linear orders. We conclude this chapter with a sample of such applications. Let (T, "5:.r) be a tree. If t E T(a) and /3 "5:. a, then we let [(/3) denote the unique s E T(/3) such that s "5:.r t. If s, t E T are incomparable in T, then we define
6.(s, t)
= min{/3:
8(/3) =I i(/3)}.
Now let :::S be a l.o. relation on T. We define a relation "5:.£ on T, called the lexicographical ordering on T induced by :::S, as follows: s "5:.t t iff s"5:.r t or sand t are incomparable in T and 8(6.(S, t)) -< i(6.(s, t)). EXERCISE 14.40 (G). (a) Convince yourself that for every l.o. relation :::S on T the lexicographical ordering on T induced by :::S is a l.o. relation. (b) Consider the tree of Example 14.1 and let :::S be the alphabetical order of its nodes. Find the lexicographical ordering of the tree induced by :::S. Not surprisingly, the structure of the l.o . .(T, "5:.t) is closely related to the structure of the tree (T, "5:.r). For example, every "5:.r-chain is also a "5:.rchain of the same order type. If some levels of (T, "5:.r) are uncountable, then (T, "5:.1.) may have subsets of order type WI even if (T, "5:.r) does not have uncountable branches. However, no such chains can be introduced if (T, "5:.r) is an Aronszajn tree.
14.
TREES
43
LEMMA 14.16. Suppose (T, :ST) is an Aronszajn tree, ::5 is a l.o. relation on T, and :St is the lexicographical ordering of T induced by::5. Then (T, :St) has no subsets of order type WI or wi. PROOF. Let T, :ST, and :St be as in the assumptions. We will prove that there are no chains of order type WI in (T, :St); the same argument works for chains of order type wi. So suppose towards a contradiction that (tf, : ~ < WI) is such that tf,
8(a)
= {s E T(a)
:
I{~
< WI : s
EXERCISE 14.41 (G). Show that 8(a)
tdl = NI }.
#- 0 for all a < WI'
Now we show that 18(a)1 = 1 for all a < WI. Suppose not, and let ao be the least a such that 18(a)1 > 1. Let r, s be two different elements of 8(ao). Pick ~ < WI such that r
-< b --> :3 d E
D (a
::5 d 1\ d ::5 b)).
The real line doesn't contain uncountable wellordered subsets because it has a countable dense subset. Compare this with the following: LEMMA 14.17. Suppose (T, :ST) is an Aronszajn tree, ::5 is a l.o. relation on T, and :Se is the lexicographical ordering of T induced by::5. If 8 is an uncountable subset of T, then (8, :Se) does not contain a countable dense subset. PROOF. Let 8, T, :ST, and:S e be as in the assumption, and let D be a countable subset of 8. Choose a < WI such that D ~ T(o:). Since T(o:+I) is countable, we can pick s, t E 8\T(O:+l) such that s
A linear order (A, ::5A) is called an Aronszajn line if
14.
44
TREES
(1)
A is uncountable;
(2) (3)
(A, jA) doesn't contain chains of order type WI or wi; and A doesn't contain an uncountable subset B such that (B, jA) is isomorphic to a subset of the real line with the natural order.
Lemmas 14.16 and 14.17 entail the following: COROLLARY 14.18. Suppose (T, :::;T) is an Aronszajn tree, j is a l.o. relation on T, and :::;t is the lexicographical ordering ofT induced by j. Then (T, :::;£) is an Aronszajn line. Let K be an infinite cardinal. We say that a l.o. (A, j) satisfies the K-chain condition (abbreviated K-C.C.) if and only if the order topology induced on A by j satisfies the K-chain condition. In other words, (A, j) satisfies the K-C.C. if and only ifthere is no family {( ae, be) : ~ < K} of pairwise disjoint nonempty open intervals in (A, j). LEMMA 14.19. Let K be an infinite cardinal and let (T, :::;T) be a splitting tree. If there exists a l. o. relation j on T such that the lexicographical order (T,:::;£) induced by j satisfies the K-C. c., then (T, :::;:r) also satisfies the K-C. c. PROOF. We prove the contrapositive. Assume (T,:::;T) is a splitting tree such that (T, :::;:r) does not satisfy the K-C.C. By Exercise 14.33, there exists an antichain {re : ~ < K} of size K in T. Since T is splitting, we may choose for every ~ < K nodes se, te, ue E T such that re
14.
TREES
45
LEMMA 14.21. Let It be an infinite cardinal, and let (T, :ST) be a bushy tree such that (T, :s:r) has the It-c.C. Buppose::5 is a l.o. relation on T such that for every t E T the set of immediate successors of t has no ::5-minimal element, and let :Se be the lexicographical ordering of T induced by ::5. Then (T, :Se) has the It-C. c. PROOF. Again we prove the contrapositive. Let (T, :ST) be a bushy tree, and let ::5,:Se be as in the assumptions. Suppose {( ae, be) : ~ < It} is a family of nonempty pairwise disjoint intervals in (T,:Se). For each~, choose te E (ae,be ) such that ae
~
T
PROOF. The idea of the proof is deceptively simple: Let (T,:ST) be an Aronszajn tree. Construct recursively an increasing transfinite sequence (ae : ~ < WI) of countable ordinals such that every node t E T(ae) has infinitely many successors at level ae+I' Let B = Ue<Wl T(ae). Then (B, :ST) is an wI-subtree of (T, :ST), and thus is an Aronszajn tree. If T is Suslin, then so is S. Moreover, the set of immediate successors in S of a node t E T(ae) is the set of successors of t at level T(ae+I)' Thus S is bushy. O.k., if a sequence (ae : ~ < WI) as above can be found, then the lemma is proved. But the problem is: Given ae, how can we find ae+I > ae such that every node in T(ae) has infinitely many successors at level ae+I? 6Not all Suslin lines are Aronszajn lines. It is true though that Suslin lines do not contain subsets of order type WI or wi.
14.
46
TREES
Maybe Claim 14.13 helps. It implies that we may without loss of generality assume that (T, 5:T) is splitting. If T is splitting and o:~ is given, we can use the fact that T( o:~) is countable to find a countable ordinal 0:~,1 > o:~ such that for every s E T(o:~) there exist 5:T-incomparable nodes to(s), tl(S) E T(a{,tl such that s
Then there exists a tall
EXERCISE 14.48 (G). Derive Theorem 14.23 from Lemma 14.24. PROOF OF LEMMA 14.24. We need a claim. CLAIM 14.25. Let (R,5:R) be an Aronszajn tree, and let r E R. If r is high, then for each (3 < WI with (3 > ht(r) there exists a high s E R((3) such that r
DO COROLLARY 14.26. If there exists a Bus lin tree, then there exists a densely ordered Bus lin line.
The technique of the proof of Lemma 14.24 can be used to obtain a more general result. EXERCISE 14.49 (G). Let a tall K+ -Aronszajn subtree.
K ~
w. Show that every
K+ -Aronszajn
tree contains
While we are at it, let us mention an interesting consequence of Claim 14.25. 7This is the reason why in the previous paragraph we took the precaution to write T(a{,nl instead of T(a{,n)'
14.
TREES
47
THEOREM 14.27. There are no Aronszajn trees with all levels finite. PROOF. Suppose towards a contradiction that (T':ST) is an Aronszajn tree with all levels finite. By Claim 14.25, the set S of all high nodes of T forms a tall Aronszajn subtree with all levels finite, and we may without loss of generality assume that T is tall to begin with. The latter assumption implies that if Q < f3 < WI, then IT( Q) I IT(f3) I· It follows that there exist Qo < WI and nEw such that IT(f3) I = n for all countable ordinals f3 ::::: Qo. Pick s E T(Qo). The choice of Qo implies that the tree does not split above level Qo, and thus the set {t E T : t :ST s V s :ST t} is an uncountable branch of T, which contradicts the assumption that T is an Aronszajn tree. 0
:s
If (L, :SL) is a dense, Dedekind complete linear order with smallest and largest elements, then L will be called a linearly ordered continuum. This name suggests a link to topology: Indeed, if (L, :SL) is a linearly ordered continuum, then the order topology induced by :SL is connected and compact.
FACT 14.28. Every dense linear order can be order-isomorphically embedded as a dense subset of a linearly ordered continuum. The proof of Fact 14.28 is an easy EXERCISE, but it is not particularly relevant to the material of this chapter. Much more important is the following: EXERCISE 14.50 (PC). Suppose (L+,~) is a dense linear order, L is a dense subset of L +, and (L, ~) is a Suslin line. Then (L + , ~) is also a Suslin line. The following question is known as the Suslin Problem: (SQ) Let (L, :SL) be a linearly ordered continuum with no uncountable family of pairwise disjoint open intervals. Is (L, :SL) necessarily isomorphic to the unit interval [0,1] if By Fact 14.28 and Exercise 14.50, if a densely ordered Suslin line exists, then the answer to the Suslin problem is negative. Thus it follows from Corollary 14.26 that the negation of the Suslin Hypothesis implies a negative answer to the Suslin Problem. 8 The connection also works the other way round. FACT 14.29. Suppose (L, :SL) is a linearly ordered continuum with no uncountable family of pairwise disjoint open intervals that is not isomorphic to the unit interval [0,1]. Then (L, :SL) is a Suslin line. Fact 14.29 is a well-known theorem of general topology. Let us skip its proof and devote the rest of this chapter to the proof of the following: THEOREM 14.30. Suppose there exists a Suslin line. Then there exists a Sus lin tree.
PROOF. It will be easier to work with dense Suslin lines; so let us first show that doing so does not lead to loss of generality. EXERCISE 14.51 (PC). Prove that if there exists a Suslin line, then there exists a dense one .. Hint: Let (L, :SL) be a Suslin line, and consider the following equivalence relation .-v on L: l<"oJ l' if and only if 1(l,l')1 < NI . Define a relation ~ on L / ~ by: 1/ ~ ~ l' / ~ if and only if 1 :SL l'. Show that ~ is well-defined and that (L/~, ~) is a dense Suslin line. 8The strange negativity of
SH
starts to look much less puzzling, doesn't it?
14. TREES
48
So let (L, '5.L) be a dense Suslin line, and let I(L) be the set of nonempty open intervals of (L, '5.d. We are going to construct T ~ I(L) such that (T,~) is a Suslin tree. By recursion over 0: < WI we construct families T(o:) of pairwise disjoint elements of I(L) as follows: Let T(O) be any maximal family of pairwise disjoint elements of I(L). Suppose T(o:) has been constructed. For each interval (a, b) E T(o:) we choose a family I(a,b) ~ I(L) of pairwise disjoint intervals, maximal with respect to the property that (c,d) is a proper subset of (a,b) for each (c,d) E I(a,b)' Note that since (L, '5.L) is dense, II(a,b) I > 1 for each (a, b) E T(o:). Then we let T(o: + 1) = U{I(a,b): (a, b) E T(o:)}. At limit stages 0 > 0, if T(o:) has already been constructed for all 0: < 0, let T(o) be a family I of pairwise disjoint elements of I(L), maximal with respect to the property that if (c,d) E I, then Vo: < 03(a,b) E T(o:) ((c,d) ~ (a,b)). Finally, let T = U"'<Wl T(o:). EXERCISE 14.52 (G). (a) Convince yourself that (T,~) is a tree whose o:-th level is equal to T( 0:) for each 0: < WI' (b) Show that if (a, b), (c, d) E T are incomparable with respect to ~, then (a,b) n (c,d) = 0.
It follows immediately from the assumption on (L, '5.d and point (b) of the last exercise that (T,~) does not contain uncountable antichains. Moreover, it follows from the remark made in the construction of successor stages that T is splitting. Hence, by Exercise 14.34, T does not contain uncountable chains either. It is somewhat less obvious that T has height WI' SO let 0:0 is the smallest ordinal 0: such that T(o:) = 0. Clearly, 0:0 can be neither a successor ordinal nor O. Thus, 0:0 must be a limit ordinal :2: w. Consider the set E = {c E L: 3(3 < 0:03(a,b) E T((3) (c = a V c = b)}. EXERCISE
14.53 (G). Show that E is dense in (L, '5.L), and conclude that
~=~.
0 Mathographical Remarks
If you are looking for a source to learn more about trees and their applications, we recommend the article Trees and linearly ordered sets, by Stevo Todorcevic in the Handbook of Set-Theoretic Topology, K. Kunen and J. E. Vaughan, eds., North-Holland, 1984,235-293. More information about equivalences of Konig's Lemma and weak versions of the Axiom of Choice can be found in the article by J. Truss, Some cases of Konig's Lemma, in: Set Theory and Hierarchy Theory. A Memorial Tribute to Andrzej Mostowski, Bierutowice, Poland, 1975, edited by W. Marek, M. Srebrny, and A. Zarach, Lecture Notes in Mathematics 537, Springer Verlag, 1976.
CHAPTER 15
A Little Ramsey Theory Here is a bit of common sense: If more than n pigeons occupy n pigeonholes, then at least one pigeonhole must be occupied by at least two pigeons. Lemma 14.3 is a mathematical counterpart of this folk wisdom. Let us reformulate it here. THEOREM 15.1 (Pigeonhole Principle). Assume,.. is an infinite cardinal, A is a set of cardinality,.., and A = Ui
n Pk
=
0. for i, k
E I, i
1= k.
Let fp : [X1P -+ I be the (unique) function f with f-l{i} = Pi for all i E I. We call fp the canonical coloring associated with P. Conversely, there is a canonical partition of [X1P associated with every coloring f : [X1P -+ a. A set Y is homogeneous for the partition P (as well as for the canonical coloring associated with it), if there is an io E I such that [Y1P ~ Pia' Let "', A, p, a be cardinals. The symbol l ,.. -+ (A)~
stands for the following statement: Whenever X is a set of cardinality,.., I is a set of cardinality a, and P = {Pi: i E I} is a partition of [X1P, there exists Y ~ X with WI = A that is homogeneous for p.2
Theorem 15.1 can be expressed in this notation as follows: THEOREM 15.2. For all infinite cardinals,.. and for all a holds.
< cf(,..), the relation
,.. -+ (,..)~
1 Read: ,.. arrows A super p sub 0'. 20f course, in order to verify that,.. a} of [,..]P.
-+ (A)~
holds, it suffices to consider partitions {Pi: i
49
<
15. A LITTLE RAMSEY THEORY
50
EXERCISE 15.2 (G). Let /\',)..,p,a,/\",)..',p',a' be cardinals such that /\,' 2: /\', )'" ~ ).., p' ~ p, and a' ~ a. Show that if /\, --+ ()..)~, then also /\,' --+ ()..')~'" Any true statement of the form /\, --+ ()..)~ is a partition theorem, but as we shall see later in this chapter, not all partition theorems are of this kind. As one might expect, partition theorems for p > 1 are more difficult to prove than Theorem 15.2. If you want to get your feet wet before reading on, we recommend the following two exercises. EXERCISE 15.3 (PG). Show that 6 --+ (3)~. Hint: You may find it easier to prove this theorem if you give it the following interpretation: Among every group 01 six people, there are three that are mutual acquaintances, or there are three that are mutual strangers. In order to indicate that a statement /\,
--+ ()..)~
is false, we write /\, .;.
()..)~.
EXERCISE 15.4 (G). By constructing a suitable partition, show that 5 .;. (3)~. The study of partition theorems is called partition calculus or Ramsey Theory in honor of the author of the following result. THEOREM 15.3 (Ramsey's Theorem). For all positive natural numbers k, £:
w --+ (w)~. PROOF. By induction over k. For k = I, this is just a special case of Theorem 15.2. Now assume the theorem has been proved for k, Le., assume that w --+ (w)~ for all £ E w. Let P = {Pi: i < £} be a partition of [w]k+l, and let 1 be the canonical coloring associated with P. One of the most important tricks in proving partition theorems is reducing the superscript p by defining so-called induced colorings.' The proof of Ramsey's Theorem gives a beautiful illustration of this technique. We shall define a strictly increasing sequence (ni)i<w of positive natural numbers and a decreasing sequence (Ai)i<w of infinite subsets of w so that for all i < w the following hold:
(i) (ii)
"In E Ai (ni < n); Va, b E [Ai] k (f ( { nd U
a) = 1 ({ni} U b)).
Let no =0. For a E [w \ {O}]k let I(O)(a) = I( {O} U a). Note that
1(0) is a coloring of all kelement subsets of w\ {o}. Thus we have reduced the superscript by one, and we can apply the inductive assumption to find an infinite Ao ~ w\{O} that is homogeneous for the induced coloring 1(0). Note that Ao satisfies (i) and (ii). Now assume that ni, Ai have been constructed for i < j. We put
nj = min A j - 1 •
For a E [A j _ 1 \{nj}]k let l(j)(a) = I({nj} Ua). Use the inductive assumption to find an infinite subset Aj of Aj -1 \ {nj} that is homogeneous for the induced coloring I(j). Now consider the set K = {n; : i < w}.
15.
A LITTLE RAMSEY THEORY
51
It follows from the construction of the n/s and the Ai'S that: If n E Kj a, bE [K \ (n
Let 9 : K
--+
+ lW,
then f( {n} U a) = f( {n} U b).
w be the function that assigns to each n E K the i Va E
< f for which
[K \ (n + 1)]k (f( {n} U a) = i).
Now choose io < f such that Y = g-1 (io) is infinite. We show that Y is homogeneous for f. To this end, let a E [y]k+l and m = min a. We have f(a)
= f( {m} U (a \
{m}))
= g(m) = i o,
o
independently of the choice of a. The next theorem is a version of Ramsey's Theorem for finite cardinals.
THEOREM 15.4. Let m, k, f be positive natural numbers. There exists a natural number n such that n --+ (m):, PROOF. Suppose towards a contradiction that m,k,f E w\{O} are such that
n
r (m); for every nEw. This means that for every nEw there exists a coloring
f : [n]k --+ f without a homogeneous subset of size m. Let F = {J : there exists nEw such that f E [n]k f and there is no homogeneous set of size m for f}, and consider the p.o. (F, ~). EXERCISE 15.5 (G). Show that: (a) (F,~) is a tree. (b) The r-th level of the tree (F, ~) consists of all functions
f E F with domain
[k+r-l]k. Our assumption implies that every level of the tree (F, ~) is nonempty. On the other hand, since there are only finitely many functions from [r + k - l]k into f, each level of the tree is finite. Thus Konig's Lemma (Lemma 14.2) implies that the tree (F,~) has an infinite branch. EXERCISE 15.6 (G). Suppose B is an infinite branch of (F, Show that: (a) 9 is a function from [w]k into fj (b) gf[n]k E F for all nEw.
~),
and let 9
=
U B.
If 9 is as in Exercise 15.6, then Ramsey's Theorem implies that there exists an infinite subset A of w that is homogeneous for g. Let ao < ... < am-1 be the first m elements of A, and let n = am-1 + 1. Then {ao, ... ,am-d is a homogeneous set of size m for gf[n]k, which contradicts Exercise 15.6(b) and the definition of the set F. 0
The technique of the proof of Theorem 15.4 allows us to obtain even stronger partition theorems for finite sets. A subset A ~ w is called relatively large if IAI ;::: min A. We write n ~ (m); if for every coloring f : [n]k --+ f there exists a relatively large homogeneous set of size at least m. THEOREM 15.5 (Paris-Harrington). For all m, k, f E w\{O} there exists nEw such that n ~ (m):, EXERCISE 15.7 (G). Prove Theorem 15.5.
15.
52
A LITTLE RAMSEY THEORY
The remarkable feature of Theorems 15.4 and 15.5 is that we used a result about infinite sets to prove a property of natural numbers. Could we obtain the same results without employing the notion of an infinite set? THEOREM 15.6. For all mE w\{O}, 4m
-t
(m + 1)~.
PROOF. We need a lemma. LEMMA 15.7. Let m E w\{O}, and let f : [4mj2 - t 2 be a coloring. Then there exist a tree (T, '5:.T) of height 2m and an injection h : T - t 4m such that f( {h(r), h(s)}) = f( {h(r), h(t)}) for all r, s, t E T with r
By the choice of T and A we have: (15.1) Since the domain of f. has 2m - 1 elements, there exists io E {O, I} such that If;l{io}1 ~ m. Now it follows from 15.1 that f;l{io} U {a2m-d is a homogeneous set of size at least m + 1 for f. It remains to prove Lemma 15.7. PROOF OF LEMMA 15.7. The elements of T will be functions t E <2m2; the p.o. relation ~T will be inclusion. With each t E T we will associate h(t) E 4m and C t ~ 4m as follows: h(0) = 0, = 4m, C0 = 4m\{0}. Given t E Tn <2m- 12, C t ~ 4m, and i E {O, I}, we let C0i = {a E 4m : f( {h(t), a}) = i} n Ct. We put t~i into T if and only if C0i =I 0. In this case we let h(t~i) = minC0i' and we define
ct,
ct
Ct~i = Ct'i\{h(t~i)}.
Note that this construction ensures that h is an injection. Moreover, note that if r
ct
EXERCISE 15.8 (G). Let j < 2m. Show that
U{Ct+ : It I = j} = 4m\{h(s) : sET n <j2}. Conclude that Ct+
=10 for at least one t E j2.
DO
The proof of Theorem 15.6 doesn't use' infinite sets. Moreover, the theorem gives an upper bound for the number n in the first nontrivial instance of Theorem 15.4. Let us introduce some notation to express the last observation more concisely. For m,k,£ E w\{O}, let R(m,k,£) be the smallest natural number n
15. A LITTLE RAMSEY THEORY
53
such that n -+ (m)~ holds. 3 Similarly, define R*(m, k, f) as the smallest natural number n such that n ~ (m)~ holds. Theorems 15.4 and 15.5 tell us that the numbers R(m, k, f) and R*(m, k, f) are well defined, but they give us no idea about their magnitude. Theorem 15.6 asserts that R(m + 1,2,2) :S 4m. Exercises 15.3 and 15.4 combined imply that R(3, 2, 2) = 6. EXERCISE 15.9 (PG). Show that R(m + 1,2,f):S flm for all m,f E w\{O, I}. EXERCISE 15.10 (R). Modify the proof of Theorem 15.6 to find upper bounds for R(m, k, f) for arbitrary positive natural numbers m, k, f. Be sure not to use infinite sets in your proof. It seems unreasonable to expect that we will ever be able to find explicit formulas for R(m, k, f); the best one can hope for is to pinpoint the rate of growth of this function. At the time of this writing, even the following question is open: Does there exist a constant c so that limm-+oo(R(m, 2, 2))1/m = c? If so, what is the value of c? It is known that c, if it exists, must be between v'2 and 4. The upper bound follows from Theorem 15.6. The lower bound can be obtained by using random colorings. We include the proof of the next theorem to illustrate this interesting technique. THEOREM 15.8. R(m, 2, 2) ;::: 2(m-l)/2 for all m ;::: 2. PROOF. Let m be as above, and let n < 2(m-l)/2. We show that there is a partition P of [nj2 into two subsets without a homogeneous subset of size m. Rather than constructing such a partition, we merely show its existence. Let n be the set of all partitions P of [nj2 into two subsets. We consider n as a probability space with the uniform probability distribution. One gets this probability distribution by performing the following experiment: For each a E [nj2, toss a fair coin. If it comes up heads, put a into Po; if it comes up tails, put a into Pl. Let (Ai: i < (n-:)!m!) be an enumeration of all m-element subsets of n. For each i, define a random variable ~i on n by: ~i(P) = 1 if Ai is homogeneous for P, and ~i(P) = 0 otherwise. Note that E(~i)' the expected value of ~i' is equal to the probability that Ai is homogeneous for P. Thus 2 E(~i) = 2(m-l)m/2' Let ~ = Li ~i' Recall that the expected value E(~) is the sum of the expected values E(~i)' The number of m-element subsets of n can be crudely estimated as follows: 4 n! nm nm 2(m-l)m/2
-,-----:-:--, < -
<-
< ----
(n - m)!m! m! - 2 2 The last of these inequalities follows from the assumption that n < 2(m-l)/2. Multiplying this estimate by E(~i) we get: E(~)
< 1.
Since ~ takes only· natural numbers as values, it follows that ~ must take the value 0 for some partition P. Note that ~(P) counts the number of homogeneous subsets 3The numbers R( m, k, i) are often called Ramsey numbers.
4More skillful estimates give the result R(m,2,2) 2: e+:'(1)2(m-l)/2.
15. A LITTLE RAMSEY THEORY
54
of size m of P. We conclude that for some PEn, no homogeneous subset of size m exists. 0 So far, we have given no estimate of the numbers R*(m,k,f). For good reason: THEOREM 15.9 (Ketonen-Solovay). R* (m + 1, m, m) defines a function from w \ {O} into w that grows faster than any function from w\ {O} into w that is provably total in Peano Arithmetic. Let us dissect the meaning of the above theorem. Peano Arithmetic (abbreviated PA) was mentioned in Chapter 6; it is an axiomatization of number theory. Hereditarily finite sets can be coded as natural numbers, and it has been shown that the properties of hereditarily finite sets provable in PA are exactly the properties provable in the theory one obtains from ZFC by replacing the Axiom of Infinity by its negation. One can thus think of the theorems provable in PA as precisely the theorems about hereditarily finite sets that are provable without employing the notion of an infinite set. Note that Theorem 15.5 is a theorem about hereditarily finite sets: All colorings and homogeneous sets mentioned in it are hereditarily finite. It is not hard to write down a crude algorithm for computing R*(m+ 1, m, m) without reference to infinite sets. EXERCISE 15.11 (G). Describe an algorithm for computing R*(m+ l,m,m). Assume that you don't have constraints upon computing time and memory. By Theorem 15.5, the function R*(m + l,m,m) is total, i.e., the algorithm you described in Exercise 15.11 eventually yields an output for every input m. By the Ketonen-Solovay Theorem, this fact cannot be proved in Peano Arithmetic. In other words, every proof of Theorem 15.5 must use infinite sets in one way or another.5 After this excursion into the realm of finite combinatorics, let us return to set theory proper and study some sufficient conditions for the existence of uncountable homogeneous sets. The first questions that come to a set theorist's mind after seeing Ramsey's Theorem are: Does W1 --+ (W1)~? Or, more modestly, is there any cardinal", such that", --+ (W1)~? Let a and {3 be ordinals. Recall that the lexicogmphical order ::;1 on (3a is defined as follows: Let f, g E (3 a. Then f
#- f(3(o)};
and d(a) = min{6.(a,{3) : a
< {3 < ",+}.
5 A similar phenomenon was discussed in the Mathographical Remark at the end of Chapter 10.
15.
A LITTLE RAMSEY THEORY
55
Note that d is a function from K:+ into K:. Now the following claim yields a contradiction. CLAIM 15.11. The function dis nondecreasing and not eventually constant. EXERCISE 15.12 (G). Prove Claim 15.11. THEOREM 15.12. 21<
f+
o
(K:+)~ for each infinite cardinal K:.
PROOF. Let (fa)o<2K be an enumeration of 1<2, and let ~I be the lexicographical order on 1<2. We define a partition P = {Po,Pd as follows:
Po = {{a,,B}: fa
= [21<j2 \
Po.
Thus Po is the set of pairs where the lexicographical order agrees with the enumeration, and PI is the set of pairs where the enumeration reverses the lexicographical order. If Y is homogeneous for P, then Y yields a monotonic sequence of length IYI in 1<2. By Lemma 15.10, IYI < K:+. 0 By Exercise 15.2, Theorem 15.12 implies in particular that WI f+ (Wl)~' This gives a negative answer to our first question. The answer to the second question is positive. In order to state the relevant theorem in its full generality, we need to introduce some notation. Let K: be an arbitrary cardinal. By recursion over nEw we define: (15.2)
expo(K:) = K:j eXPnH (K:) = 2exPn (I<).
THEOREM 15.13 (Erdos-Rado). (exPn(K:))+ ---- (K:+)~H for every infinite cardinal K: and each nEw. Note that for n = 0 the Erdos-Rado Theorem simply says that K:+ ---- (K:+)~, which is a special case of the Pigeonhole Principle. For n = 1, the theorem says that (21<)+ ---- (K:+)!, which in view of Theorem 15.12 is the best possible result for superscript 2, even if the subscript K: is replaced by 2. For n = 2, the theorem says 2K that (2 )+ ---- (K:+)~, and for larger n's we get higher stacks of exponents on the left hand side. PROOF OF THEOREM 15.13. By induction over n. As we already mentioned, for n = 0 the theorem is a special case of the Pigeonhole Principle. So assume n > 0, let K: be an infinite cardinal, and assume that (exPn_l (K:)) + ---(K:+)~. To streamline notation, let ..\ = eXPn_l (K:), JJ. = (exPn(K:))+. Fix a coloring f: (JJ.]nH ---- K:. We have the following analogue of Lemma 15.7: LEMMA 15.14. There exist a tree (T, ~T) of height ..\+ + 1 and an injection h : T ---- JJ. such that for all s, t E T with S
T, h as in the lemma, fix any t at the top level of T, and let A Define an induced coloring f* : [AJn ---- K: by: f*(a) =f(aU{h(t)}).
15.
56
A LITTLE RAMSEY THEORY
Since IAI = >'+, the inductive assumption implies that f* has a homogeneous set D ~ A of size K;+. EXERCISE 15.13 (G). Show that if D is homogeneous for is homogeneous for f.
f *,
then D U {h( t)}
EXERCISE 15.14 (PG). Assume the tree in Lemma 15.14 is only of height >'+, but otherwise the lemma is unchanged. Could we still derive Theorem 15.13 by defining f* as in the proof of Theorem 15.6 (Le., without taking advantage of the existence of a top level of T)? PROOF OF LEMMA 15.14. This argument is a modification of the proof of Lemma 15.7. The elements of T will be functions t : [or ---+ K;, where a ~ >.+. The p.o. relation ~T will be inclusion. With every t E T we will associate h(t) E fL and ct,ct ~ fL such that ct =F 0, h(t) = minCt, and C t = Ct\{h(t)}. Let us now describe the recursive construction of the t's and ct's. If a = 0, then the only function with domain [a]n is 0. We make this the root of the tree 6 and let = fL· Now suppose a = f3 + 1 ~ >. +, t : [a]n ---+ K;, and s = t f[f3]n E T. We let ct = {e E C s : f({h(tfb]n) : 'Y E a} U {e}) = t(a) for all a E [a]n}, and put t into T if and only if ct =F 0. At limit stages a, if t : [a]n ---+ K; and tf[f3]n is defined for all f3 < a, we let ct = n{Ct[[.BJn : f3 < a} and put t into T if and only if ct =F 0.
ct
EXERCISE 15.15. (a) Show that the function h: T ---+ K; is injective. (b) Assume 8, t E T are such that s
= f(aU {h(t)}).
(c) Show that if a ~ >. +, then
U{ct:
t E
U {h(t) : t E Tn [.B]n K;} .
[a]n K;} = fL\
.B
Since >. ~ K; and fL = (2)')+ = (K;>')+ = (>.+ . K;>')+ and since for all a ~ >.+ the set U.B. + . K;>', it follows that for each a ~ >. + there exists some t E [aJn K; with ct =F 0. Thus, our construction implies that the tree (T, ~T) has height >.+ + 1. DO EXERCISE 15.16 (PG). If you passed on Exercise 15.10 when you first saw it, return to it now and give it another try. Can Theorem 15.13 be strengthened to eXPn(K;) ---+ (K;+)~+1 or to (exPn(K;))+ ---+ (K;+):tl? Theorem 15.12 rules out the first possibility for n = 1. Concerning the second possiblity, the following results show limitations for n = 0 and n = 1. EXERCISE 15.17 (G). Show that K;+
f+
(2)~+.
6The attentive reader will have noticed that fQr every a
the only function with domain
[a]n is the empty set. Our successor step stipulates that at the first level of T we put the function with domain [1]n, and if n > 1, then we would put the same node at several levels of the tree. The standard way to overcome this difficulty is to add some decorations to the first n nodes of the tree, but this would only obscure the notation. So we will pretend that the empty functions with domains [o]n, ... , [n - l]n are actually different sets and ask the reader's kind forgiveness for a white lie.
15.
A LITTLE RAMSEY THEORY
57
THEOREM 15.15. 21< f+ (3)~ lor each infinite cardinalI\,. PROOF. Fix
1\,.
For I,g E 1<2 let 6.(1, g) = min{8: 1(8) =I- g(8)}. Define: Pa = {{J,g}: I,g E 1<2 and 6.(1, g) = a}.
The family {Pa : a < I\,} is a partition of a set of cardinality 21< into I\, pieces 0 for which there is no homogeneous set of cardinality 3. What about n > I? The following result, which is called the Negative SteppingUp Lemma, gives an answer. LEMMA 15.16. Suppose n 2': 2, and I\,,>",(J are cardinals such that 1\,,>" are infinite, (J 2': 2, >.. is regular, and I\, f+ (>..)~. Then 21< f+ (>..)~+1. PROOF. We prove how to step up from superscript 2 to superscript 3 in the case when (J 2': w. For the complete proof of Lemma 15.16, as well as some variations on it, we refer the reader to Lemma 24.1 of [EHMRJ (see Mathographical Remarks). So let F : [I\,J2 --+ (J be a coloring without a homogeneous subset of size >... For {J, g} E [1<2F let 6.(1, g) = min{a : I(a) =I- g(a)}. EXERCISE 15.18 (G). Show that if {J,g,h} E [1<2]3 and I
I
br(l, g, h)
=
{a,
1,
if 6.(1, g) < 6.(g, h)j if 6.(1, g) > 6.(g, h).
The function br characterizes the branching pattern of the triple {J,g, h}. Now we can define a function G : [1<2]3 --+ (J x {a, I} as follows:
G( {J, g, h}) = (F( {6.(I, g), 6.(1, h), 6.(g, h)}), br(l, g, h)). Since we are proving the lemma for the case where (J is an infinite cardinal, we have (J . 2 = (J, and it suffices to show that G does not have a homogeneous set of size >... Let A = {6.(I,g) : {J,g} E [BJ2}. We will derive a contradiction by showing that IAI = >.. and that A is homogeneous for F. Case 1: br(l,g,h) = for all {J,g,h} E [Bp. In this case, for each I E B there exists an ordinal af E I\, such that 6.(1, g) = af for all 9 E B such that I ... Now consider {a,,8} E [AP. Pick lo,b,h,h E B such that 6.(10,/1) = a and 6.(hh) =,8. We may without lo~s of generality assume that 10 is the lexicographically smallest of these elements. Then a = 6.(10, b) = afo = 6.(10, h) = 6.(10, h)· It follows that G(lo,h,h) = (F({a,,B}),O), and since B was assumed homogeneous for G, the first coordinate of this function value does not depend
°
15. A
58
LITTLE RAMSEY THEORY
on the choice of {a,,B}. In other words, A is a homogeneous set for F and we have reached a contradiction. Case 2: br(J,g,h) = 1 for all {J,g,h} E [Bj3. We leave this case to the reader.
o
EXERCISE 15.19. Finish the above prooffor Case 2.
f+
COROLLARY 15.17. Let nEw and let K be an infinite cardinal. Then eXPn (K) f+ (K+)::t I .
(K+)~+I. Moreover, if GCH holds, then (exPn(K))+
EXERCISE 15.20 (G). Prove Corollary 15.17. Hint: Apply Lemma 15.16 to Theorem 15.12 and Exercise 15.17. We have seen that WI f+ (WI)~' Naturally, the question arises whether there is any uncountable cardinal K such that K ~ (KH. DEFINITION 15.18. A cardinal K is weakly compact if K is uncountable and K ~ (K)~. EXERCISE 15.21 (G). Show that if K is a strong limit cardinal.
K
is weakly compact, then 2>'
<
K
for all
>. < K, i.e.,
LEMMA 15.19. Weakly compact cardinals are regular. PROOF. Suppose Cf(K) = >. < K. We show that K is not weakly compact by constructing a function f : [KF ~ K without homogeneous subset of size K. Let (Ki)i<>. be an increasing sequence of cardinals with SUp{Ki : i < >.} = K. Define f by: f( {a,,B}) = 0 iff Vi < >. (a < Ki +-+,B < Ki). Now suppose X ~ K is homogeneous for f. If f({a,,B}) = 1 for all {a,,B} E [XF, then IXI ~ >. < K. If f( {a,,B}) = 0 for all {a,,B} E [X)2, then X ~ Ki for some i < K, and again we have IXI < K. 0 COROLLARY 15.20. Every weakly compact cardinal is strongly inaccessible. Thus weakly compact cardinals are another example of large cardinals: Their definition generalizes a property of w, their existence implies the consistency of ZFC,7 and therefore their existence is not provable in ZFC. EXERCISE 15.22 (PG). One can show that the smallest strongly inaccessible cardinal is not weakly compact. Use this information to show that the existence of a weakly compact cardinal implies consistency of the theory ZFC + "there exists a strongly inaccessible cardinal." Hint: Use the same technique as in the proof of Corollary 12.23. Their dubious existence notwithstanding, weakly compact cardinals have some neat properties that make them worth studying. THEOREM 15.21. If K is weakly compact, then K has the tree property. PROOF. Assume K is weakly compact. Let (T, ~T) be a K-tree, i.e., a tree of height K such that every level contains fewer than K elements. Fix a wellorder ~ of the nodes of T of order type K, and let ~l be the lexicographical ordering on T 7See Chapter 12 for a discussion of the same phenomenon for inaccessible cardinals.
15. A LITTLE RAMSEY THEORY
59
induced by::;, as defined in Chapter 14. Now define a partition P = {Po, Pd of [TF by putting the pair {s, t} into Po if and only if the relations ::;t and ::; agree on {s,t}, i.e., if s < t implies s
= {s E T:
I{t EX: s
= I'\,}.
Theorem 15.21 is an immediate consequence of the following exercise. EXERCISE 15.23 (PG). (a) Show that IY n T(a)1 = 1 for all a < (b) Conclude that the set Y is linearly ordered by ::;T.
1'\,.
0
If I'\, is inaccessible, then the converse of Theorem 15.21 also holds. I'\,
-+
THEOREM 15.22. If I'\, is strongly inaccessible and has the tree property, then (I'\,)A for all n E w\{O} and A < 1'\,.
PROOF. By induction over n. For n = 1, the theorem is a special case of the Pigeonhole Principle. So assume n > 1, I'\, is as in the assumptions, A < 1'\" and let f : [I'\,jn -+ A. We have the following analogue of Lemma 15.14 LEMMA 15.23. There exist a I'\,-tree (T, ::;T) and an injection h : T -+ I'\, such that for all s, t E T with S
I'\,
f+
(w)2'.
8Compare this with Exercise 15.14. Of course the right answer was No.
15.
60
A LITTLE RAMSEY THEORY
PROOF: Let::; be an arbitrary wellorder relation on --+ 2 by:
[1\:1 No .
We define a function
f : [l\:lNo
f(X) = 0 if and only if 'v'Y E [X1No (X ::; Y). We show that f has no infinite homogeneous set. Suppose towards a contradiction that X is an infinite homogeneous set for f. Let X* be the ::;-minimal element of [X1No. Then f(X*) = 0, and homogeneity of X implies that f(Y) = 0 for all Y E [xto. Consider a sequence (Yi)i<w of denumerable subsets of X such that Yi is a proper subset of Yi+l for each i E w. Since f(Yi) = 0, the choice of f implies that Yi+l -< Yi for all i E w. Hence (Yi)iEW is a strictly decreasing sequence in ([1\:1 No , ::;), contradicting the assumption that the latter structure is a w.o. 0 Since infinite superscripts don't give positive partition relations, we might try our luck with superscripts of the form < W. DEFINITION 15.26. Let 1\:, A, (7 be cardinals. The symbol I\: --+ (A);W stands for the statement: Whenever Un : n E w\ {o}} is a family of functions such that fn : [l\:ln --+ (7, there exists a set A ~ I\: of cardinality A such that A is homogeneous for all fn's simultaneously. Note the word "simultaneously." The relation than the statement 'v'n E w\{O} (I\: --+ (A)~).
I\: --+
(A);W is much stronger
EXERCISE 15.25 (G). Show that W f'. (w)iw. Let A be an infinite cardinal. The least I\: such that I\: --+ (A)i W is called the Erdos cardinal I\:(A). If I\: --+ (l\:)iW, then I\: is called a Ramsey cardinal. One can show that I\:(w) (if it exists) is larger than the first weakly compact cardinal, and that A < J1. implies I\:(A) < 1\:(J1.). Thus the first Ramsey cardinal is larger than I\:(W) , I\:(WI) , ... , I\:(ww),. . .. Corollary 23.18 shows that measurable cardinals are Ramsey. If P = {Pi : i < (7} is a partition of a set [X1n, then we say that A ~ X is homogeneous of color i, or i-homogeneous if [A1n ~ Pi, i.e., if the induced coloring takes the value i for all a E [A1n. Let us now consider the question: How bad can counterexamples to partition relations get? For example, we have seen that WI f'. (wd~· It follows from Theorem 15.3 that every partition {Po,Pd of [WIJ2 has an infinite homogeneous set. But could it happen that there is neither an uncountable homogeneous set of color 1 nor an infinite homogeneous set of color o? In order to express this type of questions succinctly, let us introduce a new symbol.
DEFINITION 15.27. Let n E w\{O}, (7 EON, 1\:, Ai be cardinals for i < (7. The symbol I\: --+ (Ai)i'
1\:--+
THEOREM 15.28 (Erdos-Dushnik-Miller). (l\:,w)2.
Let I\: be an infinite cardinal. Then
PROOF. Let I\: be an infinite cardinal, and let P = {Po, Pl} be a partition of 1\:. For a E I\: we define B(a) = {,6 < 1\:: {a,,6} E Pd.
IS. A LITTLE RAMSEY THEORY
61
CLAIM 15.29. If for every A E [1I:l'" there exists a E A such that IB(a)nAI = 11:, then there exists an infinite homogeneous set for P of color 1. PROOF. Let F : [11:]'" --+ II: be a function such that IB(F(A)) n AI = II: for each A E dom(F). We define recursively:
C n +1 = {a E Cn : F(Cn ) < a A {F(Cn), a} E PI}.
o
The set {F(Cn ) : nEw} is infinite and homogeneous of color 1.
Now let us return to the proof of Theorem 15.28. We assume that there is no infinite homogeneous set of color 1, and show that there must be a homogeneous set of color 0 and size 11:. Claim 15.29 allows us to find A E [11:]'" such that IB( a) nAI < K, for all a E A. We fix such A for the rest of this argument. Case 1: II: is regular. EXERCISE 15.26 (G). Convince yourself that in this case there exists a strictly increasing sequence (,0)0<'" of elements of A such that /0 > sup Ui3
{a
E C~
: IB(a) n CI
< II:..,}.
Note that C~ = U..,<" C~,..,. Since II:~ is regular and larger than A,9 the Pigeonhole Principle implies that there is some b(~) with IC~,6(~)1 = II:~. Moreover, there is a set L E [A]" such that v~, 7] E
L (~ < 7] --+
b(~) ~ 7]).
Note that the set U~EL C~,6(~) is still of cardinality 11:. Now define:
E =
U(C..,,6(..,) \ U{B(a) : 3~ <
7]
(~ E L A a E C~,6(~))}) .
..,EL
A straightforward verification shows that E has cardinality II: and is O-homogeneous for P. 0 9Didn't we have the right inkling when we chose Ito to be larger than
)..?
15.
62
A LITTLE RAMSEY THEORY
EXERCISE 15.27 (PG). Let K be an infinite cardinal, f : K -+ K. (a) Show that there exists X E [K]'" such that f(o:) f(f3) whenever 0:,13 E X are such that 0: < 13. (b) Show that if K is regular, there exists Y E [K]'" such that frY is either constant or strictly increasing.
:s
So far, we have been concerned only with the cardinality of homogeneous sets. But as long as n-tuples of ordinals are being partitioned, it also makes sense to ask whether homogeneous sets of a given order type exist. DEFINITION 15.30. Let n E w\ {O}, 0:, f3i, (J' E ON, where i < (J'. The symbol (f3i)i<<7 abbreviates the statement: For every partition P = {Pi: i < (J'} of [o:]n there exist i < (J' and a homogeneous set A of color i and order type f3i.
0: -+
The symbols 0: -+ (f3)~ and 0: -+ (f3);;W are defined analoguosly. One might think that symbols like W2 -+ (wd~ are ambiguous. But since every subset of ON of cardinality K has order type at least K, N2 -+ (Nd~ means exactly the same as W2 -+ (wd~· Some partition results for cardinalities can be generalized to order types. For example, one can define the ErdOs cardinal K(O:) for each ordinal 0: 2: wand show that 0: < 13 implies that K( 0:) < K(f3). Other partition relations for cardinals can be somewhat sharpened if one considers the order types ofthe resulting homogeneous sets. The Erdos-Rado Theorem is a prime example. EXERCISE 15.28 (PG). By analyzing the proof of Theorem 15.13, show that (exPn(K))+ -+ (K+ + n)~+l for every infinite cardinal K and n E w\{O}. As long as K is regular and uncountable, one can also replace the w by w + 1 in the Erdos-Dushnik-Miller Theorem, but this requires a separate proof and will not be done here. How about the Pigeonhole Principle? If there are only finitely many pigeonholes, nothing exciting happens. EXERCISE 15.29. (a) Show that if k E w\{O}, then w . w
w· W
-+
-+
(w· w)l and
(w,wH·
(b) What is the largest
0:
such that W· 9 -+ (o:)l?
However, partitions of uncountable cardinals into countably many pieces can be surprisingly short on homogeneous subsets. THEOREM 15.31 (Miller-Rado). Let K be an infinite cardinal and let Then 0: .;.. (K·W)~.
0:
PROOF. Let K be an infinite cardinal. We want to show that for every there exists a partition {An (0:) : nEw} of 0: such that
0:
< K+. < K+
We construct the An(O:)'s by recursion over 0:. For a = 0, we let An(O) = 0 for all nEw. Suppose 0: = 13 + 1 and the An(f3)'s have been constructed and satisfy (*)/3' We let Ao(a) = {f3} and A~+1 = A~ for all nEw. Clearly, (*)", holds.
15.
A LITTLE RAMSEY THEORY
63
e
Now suppose a is a limit ordinal and (*)/1 holds for all (3 < a. Let ((3t;. : < cf(a)) be an increasing sequence of ordinals cofinal in a such that ot([(3t;., (3e+d) = (3t;.+l for all and (36 = Ue<6 (3t;. for all limit ordinals 0 < cf(a). Define:
e
Ao(a) = 0,
An+l(a) =
U {(3t;. + 'Y : 'Y E An((3t;.+I)}'
t;.
UnEw
An(a), and the An(a)'s are pairwise disjoint. Moreover,
ot(An+l(a)) ::; sup{ot(An((3t;.+d) :
e< cf(a)}· cf(a).
By the inductive assumption and since cf(a) ::; K, the right hand side is an ordinal not exceeding K· n . K = K· n + 1 . Hence (*)0: holds and we are done. 0 Can Ramsey's Theorem also be generalized to "reasonable" countable ordinals? For example, can one show that w· 2 -+ (w·2)~? Or can one at least prove w· 2 -+ (w· 2,w)2, in analogy with the Erdos-Dushnik-Miller Theorem? THEOREM 15.32. w· 2 ~ (w + l,w)2. PROOF. Consider w x w with the lexicographical ordering. This is a set of order type w· 2. Define a coloring fo : [w x W]2 -+ 2 by letting fo({(a,b), (c,d)}) = 1 if and only if a < c and b > d, i.e., if the order on the second coordinate reverses the order on the first coordinate. EXERCISE 15.30 (G). Show that there are no O-homogeneous sets for fo of order type w + 1 and every I-homogeneous set for fo is finite. 0 EXERCISE 15.31 (PG). (a) Define colorings fi : [w x wj2 -+ 2 for 1 ::; i < 6 as follows: h( {(a, b), (c, d)}) = 1 if and only if a < c < b < d. Let fo be as in the proof of Theorem 15.32, and let f4 be such that fi 1 {1} = f Ol{I}Uf: l{I}. Now define hi+l for j < 3 by letting f2j~dl} = f2·~;t{I}U{ {(a, c), (b,c)}: a,b,c E wl\a,b < c}. Show that for each i < 6, there are no O-homogeneous sets of order type w . 2 + 1 (in the lexicographical ordering on w x w) and no infinite I-homogeneous sets for any of the fi'S. (b) Let A E [w]N o • Define A6. = {(n,k): n,k E Al\n < k}. Show that A6. is a subset of w x w of order type w· 2 in the lexicographical ordering, and for every i < 6, hr[A6.j2 has I-homogeneous sets of arbitrarily large finite size.
z
The following theorem shows that partitions of w· 2 cannot get much messier than the ones we just constructed. THEOREM 15.33 (Specker). lfm E w, then w· 2 -+ (w· 2,m). One remarkable thing about Specker's Theorem is that it doesn't generalize to w· k for k > 2; in fact, one can show that w· k ~ (w· k , 3) whenever 3 ::; k < w. For k = w, the situation changes again; by a result of Chang, w. w -+ (w' w , m) for all mEw. We will prove Specker's Theorem by showing that, in a sense, the functions fi constructed in Exercise 15.31 are the only counterexamples witnessing that w· 2 ~ (w· 2 , w). More precisely, we will prove the following: LEMMA 15.34. Suppose g : [w x wj2 -+ 2 is a coloring without O-homogeneous set of order type w· 2 (in the lexicographical order on w x w) and without infinite I-homogeneous set. Then there exist A E [wt o and i < 6 such that gf[A6.]2 = fir[A~12.
15. A LITTLE RAMSEY THEORY
64
PROOF. Let 9 be as in the assumptions. The idea of the proof is to whittle down the domain of 9 so that what remains will resemble more and more closely one of the fi's. As a first step, consider the coloring hI : [wP --+ 2 defined by:
h 1 ({a,b,e}) = g((a,b), (a, e)) for a < b < e < w. Let C E [wt o be homogeneous for hI' It follows from the definition of hI that gnC.6.]2 is homogeneous on all vertical sections, and since 9 has no infinite 1homogeneous sets, all these vertical sections must be O-homogeneous. Now consider colorings h2 : [C]4 --+ 2 and h3 : [CP --+ 2 defined by
h 2({a,b,e,d}) = g((a,b), (e, d)) for a < b < e < d < w, h3({a,b,e}) = g((a,b), (b,e)) for a < b < e < w. Applying Ramsey's Theorem twice, we find B E [cto that is a homogeneous set for both h2 and h3. EXERCISE
15.32 (G). Show that B must be homogeneous of color 0 for both
h2 and h 3· We need to consider three more colorings h4' h 5, h6, where h4 : [BP defined by h4({a,b,e}) = g((a,e), (b,e)) for a < b < e < w, and h 5, h6 : [B]4 --+ 2 are defined as follows:
--+
2 is
h 5({a,b,e,d}) = g((a, d), (b, e)) for a < b < e < d < w, h6({a,b,e,d}) = g((a,e), (b,d)) for a < b < e < d < w. Applying Ramsey's theorem three more times, we find a set A E [BP~o such that A is homogeneous for h4, h5, and h6 simultaneously. EXERCISE 15.33 (PG). (a) Show that A is I-homogeneous for at least one of the functions h5, h6. (b) Show that A is I-homogeneous for all three functions h4' h5, h6 if and only if gf[A.6.j2 = hf[A.6.]2. (c) Formulate and prove analogues of (b) for the functions fi where i < 5. 0
The example we constructed to show that 2~o f+ (Wl)~ was a rather unnatural concoction. One might ask whether it is possible to show that every "nice" partition of [lR]2 into two subsets has an uncountable homogeneous set. To make mathematical sense of this question, we have to specify what "nice" means. One possible interpretation would be "topologically uncomplicated." Suppose X is a topological space, and P = {Po, Pd is a partition of [xj2. One can identify Po with the set p;ym = {(x,y) : {x,y} E Po}. We say that P is an open partition if p;ym is an open subset of X x X with the product topology.1O The following statement is known as the Open Coloring Axiom (abbreviated OCA): If X is a separable metrizable space and (OCA) P = {Po, Pd is an open coloring oftxj2, then • either there exists an uncountable O-homogeneous subset of X, • or X is a union of countably many I-homogeneous sets. 10 Note
that P is open if and only if p:ym is a closed subspace of X x X\{(x,x): x EX}.
MATHOGRAPHICAL REMARKS
65
Note that, in particular, OCA implies that no open partition of a set of pairs of reals witnesses the negative partition relation 2No f+ (wd~. The Open Coloring Axiom is not a theorem of ZFC. However, Stevo Todorcevic has shown that if ZFC is a consistent theory, then so is the theory ZFC + OCA. Thus, every statement provable in ZFC + OCA is relatively consistent with ZFC. Let us conclude this chapter with an interesting consequence of OCA. THEOREM 15.35. Suppose OCA holds and f is a function from an uncountable subset of the reals into the reals. Then there exists an uncountable subset D ~ dom(f) such that f D is weakly monotonic. In fact, if there is no uncountable D ~ dom(f) such that f D is strictly increasing, then there exists a sequence (fn)nEw of nonincreasing functions such that f = UnEw fn.
r
r
PROOF. It suffices to prove the second statement, since the first one follows from it. Let f be as in the assumptions, and let X = {(x, f(x)) : x E dom(f)} be the graph of f. Then X is a subspace of R2; hence X is separable metrizable. Let Po
= {{ (xo, f(xo)),
(Xl,
f(xd)} E [XF : Xo <
Xl
1\ f(xo)
< f(xd};
PI = [X]2\PO• EXERCISE 15.34 (PG). (a) Convince yourself that {Po, PI} is an open partition of [X]2. (b) Derive Theorem 15.35. 0 Mathographical Remarks There are several good sources for learning more about Ramsey theory. Neil H. Williams' text Combinatorial Set Theory, North-Holland, 1977, contains a fairly extensive treatment of partition calculus for infinite cardinals and ordinals. The standard reference for partition theorems for infinite cardinals nowadays is [EHMRJ, i.e., the encyclopaedical treatise Combinatorial Set Theory: Partition Relations for Cardinals, by P. Erdos, A. Haj nal , A. Mate, and R. Rado, NorthHolland, 1984. Mathematics of Ramsey Theory, edited by J. Neseti'il and V. Rodl, Springer Verlag, 1990, is a collection of survey articles covering various topics in Ramsey theory and related fields. If you want to study finite Ramsey theory, you may be interested in the text Ramsey Theory, by R. L. Graham, B. L. Rothschild, and J. H. Spencer; John Wiley, 1990 (2nd edition), or in the shorter Rudiments of Ramsey Theory, by Ronald L. Graham, published by the AMS in 1981.
CHAPTER 16
The .6.-System Lemma A family A is called a Ll-system if there is a set r such that an b = r whenever a, b E A and a =f. b. If IAI ~ 2, the set r is uniquely determined by A and is called its root by some authors, and its kernel by others. As a warm-up for this chapter, consider the following. THEOREM 16.1. Every uncountable family of finite sets contains an uncountable Ll-system. PROOF. Let n > 0 be a positive natural number, and consider the following statement: (Ll n )
Every uncountable B ~
[Wlt
contains an uncountable Ll-system.
CLAIM 16.2. (Ll n ) holds for every natural number n > O. EXERCISE 16.1 (G). Derive Theorem 16.1 from Claim 16.2. PROOF OF CLAIM 16.2. By induction over n > O. To get started, note that every B as in (Ll l ) is a Ll-system with root 0. Now suppose (Ll n ) holds, and let B = {be : ~ < Wl} ~ [Wl]nH. In order to show that B contains an uncountable Ll-system, we distinguish two cases. Case 1: For each a E Wl, the set {b E B : a E b} is countable. In this case, for each countable C C Wl the set {~ < Wl : be n U'1 EC b'1 =f. 0} is countable. We define recursively a function h : Wl - Wl as follows: h(O) = OJ h(~) = min{l1 : b'1
n U bh «()
=
0}.
«e
The family A = {bh(e) : ~ < wd is a Ll-system with kernel 0. Case 2: There is an a E Wi such that I{b E B : a E b}1 = Ni . Fix such a. Let C = {b E B : a E b} and C' = {b \ {a} : b E C}. By (Ll n ), there is an uncountable Ll-system A' ~ C'. Let r denote the root of A', and let A = {a U {a}: a E A'}. This is a Ll-system with root r U {a}. 00 EXERCISE 16.2 (R). Let B be an uncountable family of finite subsets of Wi. Show that there exists a countable subset N C Wi such that every b E B which is not a subset of N is a member of some uncountable Ll-system A ~ B with kernel r ~ N. l Theorem 16) can be generalized as follows: THEOREM 16.3 (Ll-System Lemma). Let K, and A be infinite cardinals such that A is regular and the inequality 11<1< < A holds for all 11 < A. If B is a set 1A
solution to Exercise 16.2 will be given in Chapter 24. 67
16. THE
68
of cardinality at least A such that ~ B with IAI = A.
~-SYSTEM
LEMMA
Ibl < K for all bEE,
then there exists all-system
A
PROOF. Let
K,
A, B be as in the assumption.
EXERCISE 16.3 (G). Show that the assumptions imply the inequality
K
< A.
Without loss of generality we may assume that IBI = A. Since every element of B has cardinality less than A, we may also restrict ourselves to the case where UB~A.
Each b E B is a set of ordinals of order type less than K. By regularity of A, and since K < A, there exist an ordinal p < K and a subset C ~ B of cardinality A such that every c E c- has order type p. Fix such p and C. EXERCISE 16.4 (G). Show that Enumerate C = {c€ : element of c.
~
UC is cofinal in A.
< A}. For ( < p and c E C, let c(() denote the (-th
EXERCISE 16.5 (G). Show that there exists a ( < p such that {c(() : c E C} is unbounded in A. Let (0 be the smallest ( as in Exercise 16.5, and define: ao = sup{c(7])
+ 1:
c E C /\ 7]
< (o}.
Then ao < A and c(7]) < ao for all c E C, 7] < (0' We recursively define a function h : A ~ A by h(O) = min{~ : C€((o) > ao}, h(a) = min{~ : cd(o) > U{ch(-y)(8) : "y < a /\ 8 < p}}. Note that h is strictly increasing. Let D = {Ch(et) : a < A}. EXERCISE 16.6 (G). Convince yourself that the intersection of any two different elements of D is a subset of ao. Since laol
16. THE
~-SYSTEM
LEMMA
69
PROOF. Assume towards a contradiction that {Ua a < wd is a family of pairwise disjoint, nonempty open subsets of DiE! Xi. By shrinking the Ua's if necessary, we may without loss of generality assume that they are basic open sets. This means that Ua = DiE! Via, where Via is a nonempty open subset of Xi for each i E I, and Via =f:: Xi only if i E ba, where ba is a finite subset of I. EXERCISE 16.9 (PG). Fix a < (3 < Wl and let F = ba nb,a. Moreover, let 7r be the projection of DiE! Xi onto DiEF Xi· Show that 7r[Ua l and 7r[U,al are disjoint open subsets of DiEF Xi. Show also that F =f:: 0. By Theorem 16.1, there exist a finite F C Wi and an uncountable A ~ Wi such that the set {b a : a E A} is a D.-system with kernel F. By Exercise 16.9, F =f:: 0 and the family {7r[Ua l : a E A} witnesses that DiEF Xi does not satisfy the c.c.c. This contradicts one of our assumptions. D
CHAPTER 17
Applications of the Continuum Hypothesis In this chapter we illustrate how the Continuum Hypothesis can be used in mathematical proofs. In Section 17.1 we concentrate on the impact of CH on the structure of the ideals of meager and null sets of reals. In Section 17.2 we present some typical applications of CH to a variety of problems.
17.1. Applications to Lebesgue measure and Baire category Let A ~ lR x lR. The vertical section of A at x is the set Ax = {y E lR: (x, y) E A}, and the horizontal section of A at y is the set AY = {x E lR: (x,y) E A}. THEOREM 17.1 (CH). There exists A ~ [0,1] x [0,1] such that each horizontal section of A is countable and each vertical section of A contains all but countably many reals. PROOF. Assume CH. Let :; ~ [0,1] x [0,1] be a binary relation on [0,1] such that ([0,1],:;) is a w.o. of order type WI' EXERCISE 17.1 (G). Show that A = :; is as required in Theorem 17.1.
0
EXERCISE 17.2 (PG). Show that the converse of Theorem 17.1 is also true. EXERCISE 17.3 (PG). (a) Show that the set A of Theorem 17.1 is not Lebesgue measurable. Hint: Let XA : [0,1] x [0,1] ---+ lR be defined by: (17.1)
XA(a,b) = {I, if(a,b)EA; 0, otherwise.
If A were measurable, then XA would be Lebesgue-integrable, and by Fubini's Theorem,
11 11
XAdxdy =
11 11
XAdydx
(b) Conclude that if ([0, 1],:;) is a w.o. of order type WI, then:; is not a Borel subset of [0,1]2. A subset L of lR is called a Luzin set if L is uncountable, but for every nowhere dense subset K of lR the intersection K n L is countable. THEOREM 17.2 (CH). There exists a Luzin set. PROOF. Since a set is nowhere dense in a topological space if and only if its closure is nowhere dense, we can equivalently define a Luzin set as an uncountable set of reals that l).as countable intersection with every closed nowhere dense set of reals. It follows from Claim 11.6 that there are exactly 2No nowhere dense closed subsets of lR. Assume CH holds, and let (K~ : ~ < WI) be an enumeration of all closed nowhere dense subsets of lR. Construct L = {x1) : "l < WI} in such a way that 71
i7. APPLICATIONS OF THE CONTINUUM HYPOTHESIS
72
XT/ E IR\ (Ut;
N
A subset S of IR is called a Sierpiriski set, if S is uncountable, but for every IR of Lebesgue measure zero the intersection N n S is countable.
~
EXERCISE 17.4 (G). Show that CH implies the existence of a Sierpifiski set. Note that one gets an equivalent definition of a Luzin set if one replaces the phrase "of Lebesgue measure zero" in the definition of a Sierpinski set by the phrase "of first Baire category." It quite often happens that if rp is a statement about subsets of the real line and rp* is the statement obtained from rp by replacing every occurrence of "Lebesgue measure zero" by "first Baire category" and vice versa, then a proof of rp can be translated mutatis mutandis into a proof of rp*. Apparently, the notions of Lebesgue measure zero and first Baire category are "dual" to each other. Is there a way to give a precise meaning to this apparent duality? If CH holds, then the next theorem explains the observed phenomenon. Since repeating the phrases "first Baire category" and "Lebesgue measure zero" gets tiring after a while, let us speak of meager and null sets from now on. The family of meager sets of reals will be denoted by M, and the family of null sets by N. Both M and N are count ably complete ideals in P(IR). THEOREM 17.3 (Erdos-Sierpinski). Assume CH. Then there exists a bijection f : IR -+ IR such that for all A ~ IR :
(i) (ii)
J[A] E M ~ A EN; J[A] E N ~ A E M.
EXERCISE 17.5 (G). Convince yourself that if f is as in Theorem 17.3, then J[L] is a Sierpinski set whenever L is a Luzin set, and f[S] is a Luzin set for every Sierpinski set S. PROOF OF THEOREM 17.3. We need two lemmas. LEMMA 17.4. There exists a decomposition of IR into disjoint sets Fo E M and Go EN. PROOF. Choose an enumeration of the set of rationals Q = {qi : i E w}. For every nEw, let Un = UiEw(qi - 2-(n+i), qi + 2-(n+i)). Define Go = nnEw Un and Fo = IR\Go. Note that the Lebesgue measure of Un is at most 2- n+2 • Thus Go is a null set. Since each Un is a dense open set of reals, Fo is meager. D EXERCISE 17.6 (PG). (a) Show that in the proof of Lemma 17.4, the sequence (qi)iEW can be chosen in such a way that [0, 1]\U3 is homeomorphic with the Cantor set D. (b) Conclude that the property of being a null set is not a topological property, i.e., is not preserved by homeomorphisms. if B
Let I be an ideal of subsets of a set X. We say that B is a base of the ideal I ~ I and for every A E I there exists B E B such that A ~ B.
17.1. APPLICATIONS TO LEBESGUE MEASURE AND BAIRE CATEGORY
73
LEMMA 17.5 (CH). There exist bases {Fa: a < 2~o} of M and {G a : a < 2No} of N such that Fa ~ F{3, G a ~ G{3, lFa+l \Fal = IGa + 1\Gal = 2~O for all a < f3 < 2~o, and Fa U{3 0 is a limit
=
ordinal. PROOF.
We need the following:
FACT 17.6. IfF is a meager subset of R, then there exists a meager set F+ :::> F such that IF+\FI = 2No. Similarly, if G is a null set, then there exists a null set G+ :::> G such that IG+\GI = 2No.
17.7 (PG). (a) Prove Fact 17.6. (b) Show that both the ideal of meager sets and the ideal of null sets have bases of size 2No. EXERCISE
Now let {Aa : a < 2No} be a base for the ideal of meager sets. Construct Fa by recursion over a such that for all a: (a) (b) (c) (d)
Fa is meager; F a+ 1 ;2 Fa U Aa; lFa+1 \Fal = 2No; and Fa = U{3
Since the union of countably many (and in particular: two) meager sets is meager, conditions (b) and (d) do not force us to violate condition (a). By Fact 17.6, the same can be said of condition (c). Clearly, conditions (a)-(d) imply that {Fa: a < 2No} is as required. The construction of {G a : a < 2No} is analogous. We have proved Lemma 17.5. 0 Wait a minute. Lemma 17.5 was supposed to be a consequence of CH. EXERCISE
17.8 (G). Where in the proof of Lemma 17.5 did we use CH?
Now we are ready to prove Theorem 17.3. Let {Fa: a < 2No} and {G a : a < 2No} be as in Lemma 17.5. Without loss of generality we may assume that the pair (Fo, Go) is a decomposition of lR. as in Lemma 17.4. For each a, choose a bijection fa : Fa+l \Fa -- G a+! \Ga , and let = Ua<2l
1
1 1-1.
1-1
EXERCISE 17.9 (G). Verify that and (ii) of Theorem 17.3.
1
1
f is a bijection that satisfies conditions (i) 0
Naturally, if a theorem has been proved with the help of CH, one should try to prove the same theorem from a weaker assumption, or, even better, in ZFC alone. Let us go over the proof of Theorem 17.3 and see what we really need to make it work. The only place where CH was used is the proof of Lemma 17.5. More specifically; we constructed recursively sequences (Fa: a < 2No) and (G a : a < 2~o) of meager (respectively null) sets. At limit stages a < 2No we defined Fa = U{3
74
17.
APPLICATIONS OF THE CONTINUUM HYPOTHESIS
set, this still gave us meager sets Fa and null sets G a , and we could proceed with the recursive construction. Do we need the full force of CH for the argument? To explore this question, let us introduce some new concepts. If I is an ideal, define: add(I) = min{K : 3{A~ :
€ < K}
~I
(U A~ ~ I)}. ~
The cardinal add(I) is called the additivity of the ideal I. Note that I is countably complete if and only if add(I) ~ N1 . The numbers add(M) and add(N) are our first examples of cardinal invariants of the continuum. Many consequences of the Continuum Hypothesis or Martin's Axiom can be derived from suitable assumptions about cardinal invariants. EXERCISE 17.10 (G). (a) Convince yourself that CH implies that add(M) = add(N) = 2No. (b) Show that Theorem 17.3 remains valid if we replace CH by the assumption that add(M) = add(N) = 2No. In Chapter 19 we will show that the equalities add(M) = add(N) = 2No are consistent with the negation of the Continuum Hypothesis. It is not possible, however, to prove in ZFC that add(M) = add(N) = 2No , nor even that add(M) = add(N). EXERCISE 17.11 (G). Convince yourself that if add(M) =I- add(N), then there is no bijection f : JR --+ JR that satisfies conditions (i) and (ii) of Theorem 17.3. Let us say that L is a generalized Luzin set if L ~ JR, ILl = 2No , and ILnFI < 2No for each meager set F. Similarly, let us say that S is a generalized Sierpinski set if S ~ JR, lSI = 2No , and IS n GI < 2No for each null set G. EXERCISE 17.12 (G). Show that if add(M) = add(N) generalized Luzin sets and generalized Sierpiiiski sets.
= 2No , then there exist
A set X ~ JR has strong measure zero if for every sequence (Cn)nEw of positive reals there exists a sequence (In)nEw of open intervals such that In has length ~ Cn for each nEw and X ~ UnEw In. Clearly, every countable set of reals has strong measure zero, and strong measure zero sets are null sets. But uncountable sets of strong measure zero are difficult to come by.
EXERCISE 17.13 (R). (a) Show that no uncountable closed set of reals has strong measure zero. (b) Deduce from (a) that no uncountable Borel set of reals has strong measure zero. Presumably motivated by the result of Exercise 17.13, E. Borel made the following conjecture: THE BOREL CONJECTURE: Every strong measure zero set of reals is countable. The Borel Conjecture turned out to be relatively consistent with ZFC. But it is not a theorem of ZFC.
17.1.
APPLICATIONS TO LEBESGUE MEASURE AND BAlRE CATEGORY
75
THEOREM 17.7 (CH). There exists an uncountable set of reals of strong measure zero. Here is a clever way to prove Theorem 17.7: EXERCISE 17.14 (PG). Show that every Luzin set has strong measure zero. And here is a more pedestrian approach: PROOF OF THEOREM 17.7. Assume CH and let ((C~)nEw : 1] < WI) be an enumeration of all sequences of positive reals. We want to construct recursively a sequence (x~ : ~ < WI) of real numbers and a double sequence (I~ : nEw, 1] < WI) of open intervals such that (a) (b) (c)
I~ has length ~ c~ for all nEw and 1] < WI; Q U {x~ : ~ < 1]} ~ UnEw I~ for all 1] < WI; and X7) E (n~:57)UnEwI~)\{x~: ~ < 1]} for all 1] < WI'
The construction can be carried out as follows: At stage 1], we are given x~ and I~ for ~ < 1] and nEw. Since countable sets have strong measure zero, we can choose intervals I~ such that (a) and (b) hold. Note that the presence of Q in (b) ensures that UnEw I~ is a dense set of reals for every ~ ~ 1]. Thus the set n~:57) U nEw I~ is a dense Go-set of reals and hence uncountable (see Theorem 26.5 and Exercise 14.6). So we can pick x7) E (n~:57)UnEwI~)\{x~: ~ < 1]} as required in (c). Now suppose x~ and I~ have been constructed for all ~ < WI and nEw so that (a)-(c) hold. Let X = {x~ : ~ < wd. By (c), X is uncountable. Moreover, if ~ < WI, then (b) implies that x~ E UnEw I~ for ~ < 1], and (c) implies that x~ E UnEw I~ for ~ 2: 1]. Thus X ~ U nEw I~ for each 1] < WI. Since every sequence of positive reals (Cn)nEw is listed as (C~)nEw for some 1] < WI, (a) implies that X has strong measure zero. D Can we replace CH in Theorem 17.7 by an assumption about a suitable cardinal invariant of the continuum? An analysis of the "pedestrian" proof of Theorem 17.7 may shed light on this question. The Continuum Hypothesis was used to enumerate all sequences of positive reals by countable ordinals. This of course cannot be done with the help of anything less than CH. But perhaps we do not need to consider all sequences of positive reals? Wouldn't it be enough if for every sequence (8n )nEw of positive reals our enumeration contained a sequence (C~DnEw such that c~ ~ 8n for all nEw? Let us define:
= min{,..;: 3{(C~)nEw: where JR+ = (0,00). i)"
1] <,..;} ~ W(JR+)'v'(8 n )nEw E W(JR+)31] < ,..;Vn(c~ ~ 8n )},
EXERCISE 17.15 (PG). (a) Show that i)" > No. (b) Conclude that CH -+ i)" = N1 . (c) Show that Theorem 17.7 remains valid if CH is replaced by the assumption that
i)"
= NI .
The cardinal invariant i)" is usually defined in a different way. Let f,g E Ww. We write f <* 9 and say that 9 eventually dominates f if 3mVn > m (f(n) < g(n)). A subset G ~ Ww is said to be a dominating family if Vf E Ww3g E G (f <* g). Define () = min{IGI : G ~ Ww 1\ G is a dominating family}.
17. APPLICATIONS OF THE CONTINUUM HYPOTHESIS
76
The cardinal invariant () is called the dominating number. EXERCISE 17.16 (G). Show that ()
= ()o.
We will prove in Chapter 19 that the equality () = ~1 is not a theorem of ZFC. However, it can be shown that this equality is consistent with the negation of CH. A sequence (fa : a < K,) of elements of Ww is called a K,-scale if fa <* f/3 for all a < f3 < K, and {fa : a < K,} is a dominating family. EXERCISE 17.17 (PG). Show that () =
~1
implies the existence of an WI -scale.
Let us examine your solution of Exercise 17.17. (You did complete it before reading on, didn't you?) You had to deal with situations where for a given countable family {fn : nEw} ~ Ww you wanted to find g E Ww such that fn <* 9 for all nEw. Superficially, this may look like an application of Exercises 17.15(a) and 17.16, but let us look a little closer at the problem. The inequality () > ~o only implies that some 9 E Ww is not eventually dominated by any of the fn's, not that there exists 9 E Ww which eventually dominates all fn's. Let us say that a family F ~ Ww is bounded if there exists 9 E W w such that f <* 9 for all f E F, and unbounded if no such 9 exists. Define: b = min{IFI : F ~ Ww /\ F is an unbounded family}.
The cardinal b is called the bounding number. EXERCISE 17.18 (PG). (a) Show that ~o < b ~ (). (b) Show that there exists a sequence (fa : a < b) offunctions in Ww such that fa <* f/3 for all a < f3 < b and the family {fa: a < b} is unbounded. (c) Infer from (a) and (b) that b is a regular uncountable cardinal. Does the notion of strong measure zero sets have a "dual" for Baire category? This is far from obvious, since there is no category analogue for the length of an interval. But how about this: Let us call a set M ~ JR perfectly meager if MnK is meager in K whenever K is a perfect (Le., closed without isolated points) subset of JR. Perfectly meager sets share some properties of strong measure zero sets. For example, every countable set of reals is perfectly meager and there are no uncountable perfectly meager Borel subsets of JR. Moreover, the equality b = ~1 allows us to construct an uncountable perfectly meager set. Here is how. LEMMA 17.8. Suppose X is a Polish space, (Ja : a < WI) is a decreasing sequence of Fa-subsets of X such that na<wl J a = 0, and K is a perfect subset of X. Then there exists a < WI such that K n J a is meager in K. PROOF. Suppose (Ja : a < WI) is as in the assumption. For each a, let J a = UmEw Ja,m, where the Ja,m's are closed in X. Let K be a perfect subset of X. Then K is also a Polish space. In the remainder of this proof, all topological notions (basic open set, meager set, etc.) refer to the topological space K. Since for each a < WI, KnJa has the Baire property, either KnJa is meager for some a, or for every a < WI there exist a nonempty basic open set Ua and n(a) E w such that K n Ja,n(a) n Ua is comeager in Ua . Let's assume towards a contradiction that the second alternative holds. Since K is. second countable, the Pigeonhole Principle implies that there exist a basic open U and a cofinal subset A of WI such that Ua = U for all a E A. But each comeager subset of U is dense in U. Since 1 Another
name for perfectly meager sets is always first category sets.
17.1.
APPLICATIONS TO LEBESGUE MEASURE AND BAIRE CATEGORY
77
K n Jo.,nCo.) is closed, we must have Jo.,nCo.) ;2 U for each a E A. It follows that no. EA Jo. ;2 U, and since the sequence (Jo. : a < WI) was assumed decreasing, we have no.<Wl Jo. ;2 U. This contradicts the assumption that no.<Wl Jo. = 0. 0 LEMMA 17.9. Suppose X is a Polish space without isolated points, (Jo. : a < is a strictly decreasing sequence of Fq-subsets of X such that no.<Wl Jo. = 0, and Yo. E Jo. \Jo.+ 1 for all a < WI. Then the set Y = {Yo. : a < WI} is perfectly meager.
WI)
PROOF. Let X, Y, Jo. be as in the assumptions and let K be a perfect subset of X. By Lemma 17.8, there is an ao < WI such that the set Kn{yo. : ao < a < wI} is meager in K. The set K n {Yo. : a:::; ao} is countable, and hence also meager in K. Thus, the set K n Y is the union of two meager sets in K, and hence meager inK. 0 LEMMA 17.10. Let X = Ww with the product topology. If b = ~1' then there exists a strictly decreasing sequence (Jo. : a < WI) of nonempty Fq-subsets of X such that no.<Wl Jo. = 0. PROOF. Assume b cise 17.18(b). Define:
~1.
Let (fa : a
J 0. = {g E
W
W :
<
WI)
be a sequence as in Exer-
fa <. g}.
EXERCISE 17.19 (G). Convince yourself that the J 0. 's are as required. Hint: To show that Jo. is a countable union of closed sets, consider Jo.,m = {g E Ww : "In> m Uo.(n) < g(n))}. 0 Since W W is homeomorphic to the set of irrationals with the usual topology, it follows from Lemmas 17.8-17.10 that the equality b = ~1 implies the existence of an uncountable perfectly meager set of reals. Have we found category counterparts to all properties of strong measure zero sets discussed earlier in this section? How about Exercise 17.14? EXERCISE 17.20 (PG). Show that every Sierpinski set is perfectly meager. How about the Borel Conjecture? Is it consistent that every perfectly meager set is countable? In Chapter 20 (Theorem 20.2 and Claim 20.3) we will prove in ZFC the existence of a strictly decreasing sequence of length WI of Fq-subsets of the Cantor set with empty intersection. Thus, by Lemma 17.9, the existence of uncountable perfectly meager sets can be proved in ZFC. Oops, down the drain goes our "dual" Borel conjecture. Did we discover a blemish on the face of Mathematica, or did we pick the wrong candidate for the dual notion of "strong measure zero?" To be completely honest with you: Mathematica's beauty is not flawless. Not always does the most aesthetically pleasing among competing conjectures turn out to be true. For example, the inequality add(N) :::; add(M) is provable in ZFC, but it is consistent that add(N) < add(M). The Lebesgue measure side of Mathematica's face is not a perfect mirror iJllage of the Baire category side. But such little blemishes only add spice to the intimate knowledge of Mathematica's features. Much of the time, mathematical beauty is a surprisingly reliable indicator of mathematical truth. In the case discussed here, we simply suspected the wrong kind of duality. This becomes evident if you consider the following notion: A set X <;;; IR is said to have
17.
78
APPLICATIONS OF THE CONTINUUM HYPOTHESIS
universal measure zero if for every Borel measure J.L that vanishes on the singletons2 there exists a Borel set A such that X ~ A and J.L(A) = O. Now this appears to be the right measure-theoretic counterpart of the notion of perfectly meager sets, doesn't it? And vice versa, of course.
EXERCISE 17.21 (R). (a) Show that every strong measure zero set has universal measure zero. (b) Show that no uncountable closed set (and hence no uncountable Borel set) of reals has universal measure zero. In Chapter 20 we will show in ZFC that there exists a strictly decreasing sequence (JOI. : a < WI) of Fu-subsets of IR such that nOl.<Wl JOI. = 0 and, if YOi. E JOI.+ I \JOI. for all a, then the set {YOI. : a < WI} has universal measure zero. Do we need to persuade you further that the notions of strongly meager sets and sets of universal measure zero are dual to each other? Now back to the drawing board: What is the right category analogue of strong measure zero? The following theorem points in a promising direction. It shows that strong measure zero sets can be characterized by their behavior with respect to meager sets. THEOREM 17.11 (Galvin-Mycielski-Solovay). A set X ~ IR has strong measure zero iff for every meager set F there exists a real y such that (y + X) n F = 0. The set y + X of Theorem 17.11 is defined as the shift of X by y units, i.e., y + X = {y + x : x E X}. Upon seeing Theorem 17.11 it is almost impossible to resist the temptation to call a set Y ~ IR strongly meager if for every null set G there exists x E IR such that (Y + x)nG = 0. Moreover, one is led to the following: DUAL BOREL CONJECTURE: Every strongly meager set is countable. Speaking of temptations: Here are some good ones for you. EXERCISE 17.22 (X). Show that no uncountable closed set of reals is strongly meager. EXERCISE 17.23 (XX). Show that every Sierpinski set is strongly meager. EXERCISE 17.24 (XXX). Show that the family of strongly meager sets forms an ideal of subsets of R The status of the Dual Borel Conjecture is the same&<; the status of the Borel Conjecture: It is consistent with, but not provable in ZFC. Let us demonstrate the latter: THEOREM 17.12. Assume add(N) set of reals of size 2No.
= 2No.
Then there exists a strongly meager
PROOF. Let (Gr, : ( < 2No) be an enumeration of all Go-sets of Lebesgue measure zero. We want to construct sequences of reals (xe : < 2No) and (Ye : < 2No) such that if X = {xe : < 2No }, then .
e
(+)
(yr,
+ X) n Gr,
=
e
e
0 for all ( < 2No.
2This is shorthand for saying that I./. is a measure defined at least on the u-field of Borel subsets of lR such that 1./.( {x}) = 0 for each x E R
17.2. MISCELLANEOUS APPLICATIONS OF CH
79
Since {Ge; : ( < 2No} is a base for the ideal of null sets, condition (+) implies that X is strongly meager. But how can we construct x~, Y~ by recursion over ~ < 2No and ensure that (+) holds? At stage 1} of the construction, we will have chosen x~, Y~ for ~ < 1}, and we may assume that
Consider the set HT/ = U,,< r< (Ge;-x~). Since HT/ is a union offewer than 2No .. T/,,,_T/ N null sets, add(N) = 2 o implies that HT/ is a null set. So we may choose YT/ E lR.\HT/" Note that Y E G - x if and only if Y + x E G. Thus our choice of YT/ implies:
Now let us choose xT/' We have already incurred a lot of obligations: For every 1} we must make sure that Ye; + xT/ ~ Ge;. Choosing xT/ E lR.\ Ue;::;T/(Ge; - yc;) allows us to meet all these obligations. Again, add(N) :::: 2No implies that the set lR.\ Ue;::;T/(Ge; - Ye;) is nonempty, and hence a suitable choice for xT/ exists. 0
( S
EXERCISE 17.25 (G). Show that if add(M) =2 No , then there exists a strong measure zero set of reals of size 2No. If you have become a true believer in cardinal invariants by now, here is a good exercise for you. For an ideal I of subsets of lR, define:
(17.2)
cof(I) = min{IBI : B is a base of I}, cav(I) = min{IAI : A ~ II\ UA = lR}.
EXERCISE 17.26 (G). Show that if cof(N) S cav(N) , then there exists a strongly meager set Y ~ lR. of cardinality cof(N) , and if cof(M) S cav(M), then there exists a strong measure zero set X ~ lR. of cardinality cof(M). Those readers who want to learn more about meager sets, null sets, and cardinal invariants connected with these notions will find suggestions for further reading in the Mathographical Remarks at the end of this chapter.
17.2. Miscellaneous applications of CH
Our next example is an application of CH to partition calculus. THEOREM 17.13 (Erdos-Rado). CH implies that there exists a partition ofw x into disjoint sets Hand K such that Va E [w]N°VB E [Wd N1 (ax B <6 H I\a x B <6 K).
WI
PROOF. We 'need a lemma. LEMMA 17.14 (CH). There exists a sequence (b a : a < WI) of subsets of w such that Va E [w] N°3aa < WI Va ~ aa (Ia n bal = la\bal = No). EXERCISE 17.27 (PG). Prove Lemma 17.14.
80
17.
Let (b a
:
APPLICATIONS OF THE CONTINUUM HYPOTHESIS
a < WI) be a sequence as in Lemma 17.14. Define:
H = {(n,a) : a < WI,n E ba }, K = {(n,a) : a < WI,n
i. bal.
Consider a E [W]N o , B E [WI]N 1 • Let aa be as in Lemma 17.14, and pick a E B such that a 2: aa. If mEa n b" and n E a\b", then (n, a) E (a x B)\H and (m,a) E (a x B)\K. 0 Can CR in Theorem 17.13 (or, more precisely, in Lemma 17.14) be replaced by an assumption about a suitable cardinal invariant; preferably one that has already been introduced? By using a clever and useful trick, one can show that Theorem 17.13 remains valid if CR is replaced by the assumption () = l{I' Let f be a function from W into W such that f(n) > n for all nEw. For k E w, define recursively a function fk E Ww as follows: fO(n) = n, fk+l(n) = f(fk(n)) for all nEw. By the assumption about f, the sequence (fk(O))kEW is strictly increasing. Thus we can associate with f a set b(n <;;:; W as follows:
b(n =
U[f
2k
(0), f 2k + I (0)),
kEw
where [i,m) = {n E w: i::; n < m}. For every a E [W]No let us define a function g(a) E Ww by:
g(a)(n) = min{m E a : m> n}. EXERCISE 17.28 (G). Let a E [w]No, and assume that f E Ww is such that f(n) > n for all nEw and f(a) <* f· Show that lanb(f)1 = la\b(nl = l{o. Now suppose () = l{I, and (I" : a < WI) is an WI-scale. Let b" = bU,,) for a < WI. It follows from Exercise 17.28 that the sequence (b" : a < WI) is as postulated in Lemma 17.14. By considering another cardinal invariant of the continuum, one can show that Theorem 17.13 is not provable in ZFC. The splitting number s is defined as the smallest cardinality of a splitting family, where a family S <;;:; P(w) is splitting if for every a E [w]N o there exists s E S such that la n sl = la\sl = l{o. Note that the family {b" : a < wd of Lemma 17.14 is a (rather special) splitting family of cardinality l{I. In Chapter 19 (Theorem 19.20) we will show that MA + --CR implies that s > l{I. THEOREM 17.15. Assume s > l{I. Then for every partition of W x WI into disjoint sets Hand K there exist a E [w]No and B E [WI]Nl such that a x B <;;:; H ora x B <;;:; K. PROOF. Let W x WI = H U K, where H n K = 0. For a < WI, let Sa = {n E (n, a) E H}. If s > l{I, then the family {s" : a < wd is not splitting. This means, we can pick ao E [w]No such that for all a < WI, either lao n s,,1 < l{o or lao\s,,1 < l{o· .' Casel: l{a<wI: laons"I
W :
17.2.
MISCELLANEOUS APPLICATIONS OF CH
81
Case 2: Not Case 1. Then I{a < WI : lao\sc>I < No}1 = NI . Reasoning as in Case 1, we find a E [ao]~o and B E [Wd~l such that a x B is a subset of H. 0 In topology, CH has been used to construct many examples of spaces that exhibit various bizarre combinations of topological properties. Let us present here just one of these examples, the so-called Kunen line. One of the counter-intuitive properties of the Kunen line: It is not a linearly ordered space. THEOREM 17.16 (CH). Let (xc> : a < WI) be an enumeration of JR. Then there 7 on JR that refines the usual metric topology 7M and is such that: (i) {xc>: a <,B} E 7 for each,B < WI; (ii) Icl rM (A)\clr(A)1 ~ No for every A E [lR.]No; (iii) (JR,7) is zero-dimensional; (iv) (JR,7) is first countable; (v) (JR, 7) is locally compact.
exists a topology
Note that (i)-(iii) imply that the Kunen line is an S-space: Regularity follows from (iii). The family {{xc> : a <,B} : ,B < wd is an open cover without countable subcover; hence the Kunen line is not Lindelof. IT X is a subspace of lR and A is a countable subset of X dense in (X, 7M), then by a (ii), X\clr(A) is countable, and it follows that (JR,7) is hereditarily separable. Moreover, note that (JR,7) is scattered: This follows from (i), (v), and Theorem 26.4. EXERCISE 17.29 (G). How do we know that (lR,7) is Hausdorff? PROOF OF THEOREM 17.16. Let us begin by enumerating everything relevant. Let (xc> : a < WI) be as in the assumption. For,B < WI, let X{3 denote the set {xc> : a < ,B}. Fix an enumeration (A,,/ : 'Y < WI) of [lR]~o. For,B < WI, let C{3 = {A,,/ : 'Y <,B t\ A,,/ S;;; X{3 t\ x{3 E clrM(A,,/)}, and fix a sequence (C{3,n : nEw) such that every element of C{3 gets listed infinitely often in it. By recursion over ,B < WI, we will construct a sequence of topologies (7{3 : ,B < WI) such that for all ,B < WI: (a) (b) (c) (d) (e) (f) (g)
7{3 is a topology on X{3 that refines the usual metric topology IT 'Y < ,B, then 7,,/ = 7{3 X,,/; IT,B E WI n LIM, then U"/<{3 7,,/ is a base for 7{3; IT A E C{3, then x{3 E clr~+l (A); 7{3 is a zero-dimensional topology on X{3; 7{3 is a first countable topology on X{3; 7{3 is a locally compact topology on X{3.
r
7M
rX{3;
IT we succeed in doing this, then U{3<Wl 7{3 is a base for a topology 7 on lR. It is not hard to see that (a)-(c) imply that 7 is a refinement OhM that satisfies (i), and, moreover, that 7 X (3 = 7{3 for each ,B < WI. This in turn implies that the whole space (JR,7) has such "local" properties as zero-dimensionality, first count ability, and local compactness if and only if each X{3 has these properties. Thus (e) implies (iii), (f) implies (iv), and (g) implies (v).
r
EXERCISE 17.30 (G). Convince yourself that (d) implies that if A E C{3, then X{3
E clr(A). Infer that (JR, T) satisfies (ii).
82
17. APPLICATIONS OF THE CONTINUUM HYPOTHESIS
It remains to construct the sequence (Tj3 : (3 < W1). For (3 < w, the space Xj3 is finite, and the metric topology TM on Xj3 has all the desired properties. At limit stages (3, it suffices to observe that U,),<j3 T')' is a base for a topology, and choose Tj3 as dictated by (c). Repeating the argument preceding Exercise 17.30, one can easily show that if (a)-(g) hold up to stage (3, then this topology Tj3 will also have all the required properties. Now assume Tj3 has been constructed and satisfies conditions (a)-(c) and (e)(f). We show how to define Tj3+l' The basic neighborhoods of Xo: for a < (3 have already been determined at previous stages. Now we need to choose a countable base for xj3 so that (d) holds. Since xj3 E cl TM Cj3,n for each nEw, we can choose a sequence (Yn)nEw of reals in Xj3 such that Yn E C n ,j3 for all nEw and the sequence (IYn - Xj3l)nEw of distances between xj3 and Yn is strictly decreasing and converges to zero. We are going to choose a decreasing sequence (Vj3,k)kEW of subsets of Xj3+l such that {Yn : n ~ k} ~ Vj3,k, Vj3,k n Xj3 is open for all k E w, and nkEw Vj3,k = {Xj3}. Then we will declare Tj3 U {Vj3,k: k E w} a base for Tj3+1. EXERCISE 17.31 (G). Convince yourself that if Tj3+l is defined as indicated above, then (d) holds at stage (3, and conditions (a)-(c) and (f) will be satisfied at stage (3 + 1. To ensure that Tj3+1 is zero-dimensional and locally compact, some extra care in choosing the Vj3,k'S is needed. Let the Yn's be as above, and choose a sequence (In)nEw of pairwise disjoint closed intervals such that Yn E int (In) for all nEw. For each nEw, pick a compact neighborhood Uj3,n E Tj3 of Yn such that Uj3,n ~ In and let Vj3,k = {Xj3} U Un~k Uj3,n' EXERCISE 17.32 (G). Convince yourself that these Vj3,k'S satisfy the specifications described above, and show that the resulting space (Xj3+l, Tj3+1) is locally compact. To show that (Xj3+l, Tj3+l) is zero-dimensional, it suffices to prove that each of the sets Vj3,k is closed. The rest follows from the inductive assumption. So suppose x E X/3+l \ Vj3,k for some k E w. Then x =1= xj3. Since lim n -+ oo Yn = xj3 and the intervals In do not overlap, there exist fEw and W E TM such that W n Un>i In = 0 and x E W. In other words,
W n V/3,k c U/3,k U U/3,k+l U··· U U/3,i' The right hand side of the last equality is compact and hence closed in T/3' Since WE T[3, there exists a T/3-neighborhood Wo ~ W of x such that Wo n Vj3,k = 0.
o
The applications of CH we have seen so far all relied on transfinite recursion. Let us give a couple of examples with a different flavor. Let A be a family of subsets of a fixed set X. We say that A is an almost disjoint/amilll if IAI = IXI and IAnBI < IXI for all {A,B} E [AJ2. While the size of every family of pairwise disjoint subsets of ~ is bounded by lXI, almost disjoint families can be somewhat larger. 3Some authors would insist on using the phrase "family of pairwise almost disjoint sets of cardinality IXI each." The phrase "almost disjoint" is also being used in the literature as a shorthand for "IA n BI < min{IAI, IBI}·"
17.2. MISCELLANEOUS APPLICATIONS OF CH
83
THEOREM 17.17. Let K be a regular infinite cardinal. Then there exists an almost disjoint family A of subsets of K such that IAI = K+. PROOF. Let K be as in the assumption. By recursion over 0 < K+ we will construct sets Aa E [KJ": such that IAa n At3l < K for all 0 < {3 < K. To get started, let f : K -+ K X K be a bijection and define Aa = f- 1 {(o, (3) : {3 < K} for 0 < K. Now suppose K S 'Y < K+, and {Ba : 0 < 'Y} has been constructed and forms an almost disjoint fanlily. Fix a bijection g : K -+ 'Y. For (3 < K, define recursively bt3 = min{At3\Ua
PROOF. Fix a bijection f : <w2 characteristic function of a. Define:
-+
w. For a
~
w, let Xa : W
-+
2 denote the
C(a) = {J(Xarn) : nEw}.
EXERCISE 17.33 (G). (a) Show that if a =I- b, then IC(a) n C(b)1 < ~o· (b) Conclude that the fanlily C = {C(a) : a E P(w)} is an almost disjoint D family of size 2No. It is not possible to prove in ZFC that there exists an almost disjoint fanlilyof
size
2Nl
of subsets of WI. However, CH does imply the existence of such a family.
THEOREM 17.19 (CH). There exists an almost disjoint family of size
2Ml
of
subsets of WI.
PROOF. If CH holds, the set Ua<wl a2 has cardinality
~I'
EXERCISE 17.34 (G). Take a close look at the proof of Theorem 17.18 and then prove Theorem 17.19. D EXERCISE 17.35 (G). Find and prove a generalization of Theorem 17.19 to arbitrary infinite cardinals. Let K be an infinite cardinal. A family A all pairs of disjoint F, G E [AJ
~
nAn n
P(K) is called independent if for
(K\A)
AEF
=I- 0.
AEG
(We assume thatn0 = K.) We say that A is strongly independent if ICF,GI every pair (F, G). as above.
= K for
THEOREM 17.20 (Fichtenholz-Kantorovitch-Hausdorff). Let K be an infinite Then there exists a strongly independent family A ~ P(K) such that
cardinal.
IAI =2K.
17. APPLICATIONS OF THE CONTINUUM HYPOTHESIS
84
PROOF. Let K, be an infinite cardinal, and let K = [K,j
IKI
= K" and we can prove Theorem 17.20 by constructing a strongly independent
family 8 ~ P(K) of size 21<. The desired family A can then be obtained by fixing a bijection f : K -+ K, and defining A = {J[B] : BE 8}. For A E P(K,) let BA = {(s,T) E K: snA E T}.
Let 8 = {BA : A E P(K,)}. It is not hard to see that if Ao =j:. AI, then BAa =j:. BAl' Thus 181 = 21<. It remains to show that 8 is strongly independent. Let (F, G) be a pair of finite disjoint subsets of 8. Enumerate F U G as {B Ai : i < n}. For i < j < n, fix ~i,j E AiD-Aj . Let S={~i,j:
i<j
T
= {s n Ai : B Ai
E F}.
CLAIM 17.21. Let s, T be as above. If R E [[K,j
EXERCISE 17.36 (G). Prove Claim 17.21. Since the set of R's that satisfy the assumption of Claim 17.21 has cardinality K" it follows that 8 is strongly independent. 0 The notions of independence and strong independence can be generalized as follows: Let A :::; K, be infinite cardinals. A family A ~ P(K,) will be called Aindependent if for all pairs of disjoint F, G E [Aj<'\ we have:
CF,G =
n n
(K,\A) =j:.
An
AEF
0.
AEG
We say that A is strongly A-independent if ICF,GI = K, for every pair (P, G) as above. Of course, (strong) No-independence is the same thing as (strong) independence. THEOREM 17.22 (CH). There exists a strongly N1 -independent family A P(WI) such that
~
IAI = 2 Nl .
PROOF. Let K = [wd~No X [[wd~Noj~No. If CH holds, then IKI = N1 , and we can prove Theorem 17.22 by constructing a strongly independent family 8 ~ P(K) of size 2Nl • Sounds familiar? So you know already what comes next. EXERCISE 17.37 (G). Finish the proof of Theorem 17.22.
o
So far, we have not given you an explicit example of an application of the Generalized Continuum Hypothesis. Let us include one for good measure. THEOREM 17.23 (GCH). For every infinite regular cardinal K, there exists a strongly K,-independent family A ~ P(K,) such that IAI = 21<. EXERCISE 17.38 (G). Prove Theorem 17.23.
MATHOGRAPHICAL REMARKS
85
Mathographical Remarks If you would like to see more examples of applications of CH, we recommend W. Sierpinski's classical monograph Hypothese du Continu, Monografie Matematyczne, Warsaw, 1934. This is a collection of 82 different consequences of CH and their proofs. The inequality add(N) ~ add(M) was proved independently by T. Bartoszynski in Additivity of measure implies additivity of category, Transactions of the American Mathematical Society 281 (1984), 209-213, and by J. Raisonnier and J. Stern in Mesurabilite et propriete de Baire, Comptes Rendus, Serie I Mathematique 296 (1983), 323-326. Consistency of the Borel Conjecture was proved by R. Laver in the paper On the consistency of Borel's conjecture, Acta Mathematica 137 (1976), 151-169. Consistency of the Dual Borel Conjecture was proved by T. Carlson. The proof is published in his paper Strong measure zero and strongly meager sets, Proceedings of the American Mathematical Society 118 (1993), 577586. A solution of Exercise 17.22 can also be looked up in this paper (Theorem 5.11). Exercise 17.23 was a famous open problem for many years. It was finally shown by J. Pawlikowski in 1992 that Sierpinski sets are indeed strongly meager. By a recent result of Bartoszyriski and Shelah, it is consistent that the strongly meager sets do not form an ideal. So you should not despair if you were unable to solve Exercise 17.24. If you want to learn more about measure and category, you may be interested in one of the following items: The book Measure and Category, by J. Oxtoby, Springer Verlag, 1971, gives a good elementary introduction to the subject. If you are interested in all sorts of exotic subsets of the real line, we recommend A. W. Miller's article Special subsets of the real line, In: Handbook of Set-Theoretic Topology, K. Kunen and J. E. Vaughan, eds., North-Holland, 1984, 201-235. Among other things, this article contains a proof of Theorem 17.11 and a solution of Exercise 17.14. If you are familiar with forcing and curious to learn more about the ideals M and N, the monograph Set Theory. On the Structure of the Real Line, by T. Bartoszynski and H. Judah, A K Peters, 1995, will give you all the information you could possibly want. For example, Corollary 8.1.4 of that book provides a solution of Exercise 17.13(a).
CHAPTER 18
From the Rasiowa-Sikorski Lemma to Martin's Axiom Martin's Axiom is a powerful tool that has found many applications both inside and outside of set theory. It is not a theorem of ZFC, but it is relativly consistent with ZFC. It can be thought of as a generalization of the Rasiowa-Sikorski Lemma, which is provable in ZFC. This section is organized as follows: First we state and prove the Rasiowa-Sikorski Lemma. Then we show how one can transform, step by step, a proof involving a recursive construction into one using the Rasiowa-Sikorski Lemma. Next we introduce Martin's Axiom and show how it can be applied to derive a more general result from the transformed argument. Finally, we discuss variants of the c.c.c. as well as some consistent and some inconsistent modifications of Martin's Axiom. Let (!P, ~> be a p.o., and let V be a family of dense l subsets of lP. A filter G in lP is V-generic if D n G =f:. 0 for all D E V. LEMMA 18.1 (Rasiowa-Sikorski). Let (!P, ~> be a p.o., and let V be a countable family of dense subsets oflP. Then there exists a V-generic filter in lP. PROOF. Let '0= {Di : i < w}. We construct recursively a sequence (Pi)i<w of elements of lP with the following properties: (i) Pj ~ Pi for all i < j < w; (ii) For each i < w there exists qi E Di with Pi ~ qi. We let Po be an arbitrary element of Do. Assume that 0 < n < w and that the Pi'S have already been constructed for i < n. Since Dn is dense in lP, there exists P E Dn with P ~ Pn-l. We choose some such P as our Pn. The filter generated by {Pi : i < w} is as required. 0 Recursive constructions as in the proof of Lemma 18.1 occur quite frequently in mathematical arguments. In many cases it would be possible to replace such constructions by applications of the Rasiowa-Sikorski Lemma. This is not commonly done though, for at least two reasons, one sound and one regrettable. The sound reason is that since the proof of Lemma 18.1 is so simple, a reference to the Lemma would not usually decrease the overall length of an argument. In fact, as we shall see in a moment, this approach may actually result in a longer proof. The other reason is that many mathematicians find it easier to work with recursive constructions than with dense subsets of p.o. 's. This is unfortunate, because recursive constructions have a more limited scope of application. For example, it is not possible to use a procedure as in the proof of Lemma 18.1 when one is trying IThis notion and other relevant concepts pertaining to p.o.'s were defined in Chapter 13. 87
88
18.
FROM THE RASIOWA-SIKORSKI LEMMA TO MARTIN'S AXIOM
to construct a countable object that satisfies uncountably many conditions. An approach using partial orders may still work in such cases, provided there exists a filter that meets all the uncountably many dense sets involved. Martin's Axiom postulates the existence of such filters under certain circumstances. In order to take full advantage of the opportunities this creates, one needs to be at ease with expressing properties of an object constructed from a filter in terms of dense subsets of a relevant p.o. The aim of the next example is to demonstrate, step by step, how a recursive construction can be translated into an argument involving partial orders. An almost disjoint family A of infinite subsets of w is maximal (or, a mad family) iffor each infinite b ~ w there exists a E A such that lanai = No. There are finite mad families; e.g., {w} is one. The following theorem shows that all infinite mad families are uncountable. THEOREM 18.2. Let A be a denumerable almost disjoint family of infinite subsets of w. Then there exists an infinite d ~ w such that Id n al < No for all a E A. This theorem can be proved by a recursive construction as follows: PROOF 1. Let A = {ai : i < w}. We construct recursively a sequence (Xi)i<w of elements of w so that
(U ak
U {Xk : k < i}) k
Xi
E
w\
Let us analyze the above reasoning. Our aim was to construct an object d that satisfies for each i E w the following condition: (8 i ) dnai~{xO"",Xi}' We are able to formulate the conditions to be met by d in this way because both the Xi'S and the ai's can be enumerated by the same set w. But suppose we were trying to prove the following: STATEMENT 18.3. Let A be an almost disjoint family of size Nl of infinite subsets of w. Then there exists an infinite d ~ w such that Id n al < No for all a E
A.
.
In this case, the index set for the xi's would still be w, but for the ai's we need Nl indices, so we cannot meaningfully express all the conditions on d in the form (8 i ). But note that neither the order of the xi's nor the order of the ai's really matters in the proof of Theorem 18.2. We could rewrite it as follows: PROOF 2. Let {Fi: i E w} be an increasing sequence of finite subsets of A such that UiEw Fi = A. We construct recursively a sequence (Si)i<w of finite subsets of w so that
Si~Si+1
1\
ISil~i
1\
Si+1\Si~W\UFi
for each i E w. Since the set on the right hand side contains all but finitely many elements of any a E A \ Fi , it is infinite, and hence nonempty. Therefore, the rules
18.
FROM THE RASIOWA-SIKORSKI LEMMA TO MARTIN'S AXIOM
89
of the game always allow us to put an extra element into Si+1 if need be. Now let d = UiEW Si. This d is as required, since dna ~ Si for each a E Fi , the sets Si are finite, and A = UiEw Fi . 0 Of course, the difference between Proof 1 and Proof 2 is only notational, but the latter argument allows us to reformulate the requirements on d in a manner independent of their numbering: Let d = UiEW Si. Then the following requirements should be met for each JEW and a E A: There is some Si with ISil ? j; (wa) There is some Fi with a E Fi . Now you may suspect that one should be able to eliminate the subscripts i from the conditions B j and w a . As the next proof of Theorem 18.2 shows, this is indeed the case. (Bj)
PROOF 3. Let lP' = {(s, F) : onlP' by:
S
E [wJ
(S,F):::;(Sl,Ft) iff sl~s,F1~F, and (s\sl)nUF1 =0. Let us convince ourselves that :::; is a partial order relation. Only transitivity requires an argument. Suppose (so,Fo) :::; (sl,F1 ) and (s1,F1) :::; (s2,F2). By transitivity of inclusion, S2 ~ So and F2 ~ Fo. Moreover,
(so \ S2) n UF2 = [(so \ Sl) nUF2J u [(Sl \ S2) n UF2J
~ [(so \ Sl) n U F1J U [(Sl \ S2) n U F2J = 0. Thus (so, Fo) :::; (S2' F2). How do we want to use lP' in our proof? The idea is to specify a countable family 1) of dense subsets of lP' and pick a 'V-generic filter G ~ lP' whose existence is guaranteed by the Rasiowa-Sikorski Lemma. Then we shall put
d
= U{s: 3F((s,F)
E
G)}
and show that d is as required. For j < w let D j = {(s,F) E lP': lsi? j}. It is not hard to see that the Dj's correspond to the requirements (B j precisely, we have:
).
More
n D j =f. 0, then Idl ? j. E G n D j , then lsi? j; and since s ~ d we have Idl ? j.
CLAIM 18.4. If G PROOF. If (s, F)
0
CLAIM 18.5. Each D j is dense in lP'. PROOF. Let (s,F) E lP'. Since Iw \ UFI = ~o, there exists t ~ w \ UF such that It I =j. Then (sUt,F) E D j , and (sUt,F):::; (s,F). 0 For a E A let
D a = {(s, F) E lP' : a E F}. These sets are designed to take care of the requirements (W a ). More precisely: CLAIM 18.6.
If G n Da =f. 0, then
Id n al < ~o.
90
18. FROM THE RASIOWA-SIKORSKI LEMMA TO MARTIN'S AXIOM
PROOF. Let (so, Fo) E tG n Da. Then for every (s, F) E tG there exists an H) E tG such that (Sl, F1 ) ~ (s, F) and (Sl, F1 ) ~ (so, Fo). Since (Sl \ so) n UFo = 0, we also have (Sl \ so) n a = 0, which is equivalent to Sl n a ~ so. Since s ~ Sl, we also have s n a ~ so. But s was the first coordinate of an arbitrary element of tG; therefore dna ~ so, and thus Id n al < w. 0 (Sl,
CLAIM
18.7. For each a E A the set Da is dense in lP'.
PROOF. Let (s, F) E lP'. Then (s, F U {a}) E D a and (s, F U {a}) ~ (s, F).
0
Now let (18.1)
v=
{D j
:
j < w} U {Da : a E A}.
Since V is a countable set, the Rasiowa-Sikorski Lemma, in conjunction with Claims 18.5 and 18.7, tells us that there is a V-generic filter tG ~ lP'. If we construct d ~ w from tG as indicated above, then by Claims 18.4 and 18.6, d is as required. 0 The last proof is much longer than each of the previous ones, but one could easily reuse it as a proof of Statement 18.3 (recycling is smart) if only there were a generalization of the Rasiowa-Sikorski Lemma to uncountable families of dense sets. Martin's Axiom, which we are now going to describe, is such a generalization. Let /'i. be a cardinal. By MA(/'i.) we abbreviate the following statement: Let (lP, ~) be a p.o. that satisfies the c.c.c., and let V be a family of dense subsets of lP' with IVI Then there exists a V-generic filter in lP'.
~
/'i..
Martin's Axiom (abbreviated MA) is the following statement: (MA)
MA(/'i.) holds for all /'i. < 2No.
Since MA(~o) follows from the Rasiowa-Sikorski Lemma, the Continuum Hypothesis implies MA. Many authors use MA as an abbreviation for "MA + ...,CH." The negation of CH does not imply MA, but is consistent with it. Now let us return to Statement 18.3. Note that the definition of lP in Proof 3 of Theorem 18.2 does not depend on the size of A. Neither does the rest of the argument, except for the last two sentences. Since the family V defined in 18.1 will now be uncountable, the Rasiowa-Sikorski Lemma no longer applies. But (lP',~) satisfies the C.C.c. (see Fact 18.15 and Claim 18.16 below). Therefore, MA(~d implies Statement 18.3. The latter observation can be generalized as follows: Let a denote the minimum cardinality of an infinite mad family of subsets of w. Then we have the following: THEOREM 18.8. Let /'i. be an infinite cardinal, and assume that MA(/'i.) holds.
Then a> /'i.. Theorem 18.8 has an important consequence. By Exercise 9.20, there exists a mad family A. Since A is contained in P(w), its cardinality must be less than or equal to 2No. Thus we get the following:
18.
FROM THE RASIOWA-SIKORSKI LEMMA TO MARTIN'S AXIOM
91
COROLLARY 18.9. MA(2 No) fails. of
Now let us show that if we drop the countable chain condition in the formulation then we obtain a false statement.
MA(~l)'
DEFINITION 18.10. Let I, J be nonempty sets. Recall from Chapter 13 that Fn(I, J) denotes the set of all functions p such that Ipi < ~o, dom(p) ~ I, rng(p) ~ J. When we speak of the partial order F(I, J), the partial order relation is understood to be inverse inclusion 2. EXAMPLE 18.1. Consider Fn(w,wl). For a < WI, let Po. = {(O,a)}. Then {Po. : a < WI} is an antichain of cardinality ~l in Fn(w,wd, and it follows that Fn(w, WI) does not satisfy the c.c.c. Now consider the sets Do. = {p E Fn(w,wl) : a E rng(p)}, where a < Wl' If q E Fn(w, WI), then q > qU {(min(w \ dom(q)) , a)} E Do.. Thus, each Do. is a dense subset of Fn(w, WI). Let V = {Do. : a < WI}, and assume that IG is a V-generic filter. EXERCISE 18.1 (G). Show that if IG is as above, then UIG is a function with dom(U IG) ~ w. By V-genericity of IG, if a < Wl, then there exists some p E IG n Do.. This implies that a E rng(UIG). But then rng(UIG) = WI, which is impossible. Now let us discuss some techniques of showing that certain p.o. 's have the c.c.c. Let us start with a trivial case. FACT 18.11. Every countable p.o. has the c.c.c. DEFINITION 18.12. Let (IP',::::S) be a p.o., and let B ~ 1P. Then B is called centered if every finite F ~ B has a lower bound2 in IP'. We call B linked if every two elements of B are compatible in 1P. Centered sets are always linked, but not vice versa. EXERCISE 18.2 (PG). Consider (P(w)\0, 2). Find a subset B of P(w)\0 that is linked but not centered in this p.o. FACT 18.13. If B is linked in a p.o. (IP',::::S) and A is an anti chain in (IP', ::::S), then IAnBI ~ 1. DEFINITION 18.14. A p.o. (1P,::::S) is a-centered (a-linked) if IP' can be represented as IP' = UnEN lPn' where N is a countable set and each of the IP'n's is centered (linked). Note that we do not require that the IP'n's are pairwise disjoint. FACT 18.15. (a) a-centered p.o. 's satisfy the c.c.c. (b) a-linked p.o. 's satisfy the c.c.c. p.o.
Many applications of MA use a-centered p.o.'s. For example, consider the (IP',~) of Proof 3 of Theorem 18.2. CLAIM 18.16.
(IP',~)
is a-centered.
2I.e., there exists a q E lP' such that q ::; p for every p E F.
92
18.
FROM THE RASIOWA-SIKORSKI LEMMA TO MARTIN'S AXIOM
PROOF. Let s E [w]
ILl
be one-
A(c:) = {U s:;; lR: U is open in lR 1\ ILl(U) < c:}. (A(c:), 2) is known among set theorists as amoeba forcing. CLAIM 18.17. For each c: > 0 the p.o. (A(c:),2) is a-linked. PROOF. Fix c: > 0 and let B be a countable open base for the topology on lR. For each F E [B]
A(c:)p = {U E A(c:) :
UF s:;; U 1\
ILl (U\
EXERCISE 18.3 (PG). (a) Show that A(c:) = (b) Show that each A(c:)p is linked. (c) Deduce Claim 18.17.
- 8p UF) < -c: 2 -}.
UPE[Bj
A(c:)p.
EXERCISE 18.4 (PG). Fix c: > O. (a) Show that if B is a centered subset of (A (c:) , 2), then ILl
o UB
is open in lR and
(U B) :S c:.
(b) Show that if {Bn : nEw} is a countable family of centered subsets of A( c:), then there exists a U E A(c:) such that ILl (U\ UnEw Bn) > 0 for every nEw. (c) Conclude that (A (c:) , 2) is not a-centered. One can think of a-linkedness as a particularly neat way of satisfying the c.c.c. Unfortunately, there are also much messier reasons for having the c.c.c. For example, let I, J i- 0, and consider the p.o. Fn(I, J). For which I, J is Fn(I, J) a-centered or a-linked? If J is uncountable, then Fn(I, J) does not even satisfy the c.c.c. (see Example 18.1 for a special case). If IJI = 1, then Fn(I, J) is simply centered, which is not very interesting. So the most interesting case occurs when 1 < IJI :S l{o· For f : I -+ J, let B f = {p E Fn(I, J) : p s:;; J}. FACT 18.18. Let I, J i- 0. (a) p, q E Fn(I, J) are compatible iff p U q E Fn(I, J). (b) Let B s:;; Fn(I, J). Then the following are equivalent:
(b1) (b2) (b3)
B is linked. B is centered. B s:;; B f for some f E IJ.
EXERCISE 18.5 (G). Prove Fact 18.18. Hint: Use Exercise 13.7 Now consider J as a topological space with the discrete topology, note that IJ = I1iEI J, and let T be the product topology. CLAIM 18.19. Let I, J i- 0 and F s:;; IJ. Then Fn(I, J) = U{B f : f E F} if and only if F is a dense subset of J, T).
e
EXERCISE 18.6 (PG). Prove Claim 18.19.
·
18.
FROM THE RASIOWA-SIKORSKI LEMMA TO MARTIN'S AXIOM
93
COROLLARY 18.20. Assume that I¥-0 and 1 < IJI :S ~o. Then Fn(I, J) is rT-centered if and only if III :S 2No. PROOF. If J is as in the assumptions, then J is separable. Therefore, by the Hewitt-Marczewski-Pondiczery Theorem (Theorem 26.6) and by Theorem 26.7, IJ is separable if and only if III :S 2No. 0 So what about Fn(I, J) if I > 2No and J is countable? Is this still a c.c.c. p.o.? Yes, but for a more subtle reason. LEMMA 18.21. Assume I ¥- 0, J is countable, and let A be an uncountable subset of Fn(I, J). Then A contains an uncountable linked subset B. PROOF. Let I,J,A be as in the assumptions, and let D = {dom(p): pEA}. This is a family of finite subsets of I. Since for a given finite subset of I there are only countably many functions from I into J, the Pigeonhole Principle implies that D is uncountable. By the 6.-System Lemma, there is an uncountable 6.-system Dl <;; D. Let r be the root of D l , and for each function p : r -+ J, let Afp = {q E A: dom(q) E Dll\ qfr
= pl.
Since there are only countably many p's in r J, applying the Pigeonhole Principle again, we conclude that some Afp must be uncountable. Now the following observation concludes the proof of Lemma 18.21: FACT 18.22. For each p E Fn(I, J), the set Afp is linked. EXERCISE 18.7 (G). Prove Fact 18.22. (Note that for some p, the set Afp may 0 be empty, but the empty set is linked by default.) The property expressed by Lemma 18.21 deserves its own name. DEFINITION 18.23. (a) A p.o. (IP,::5) has Property K (or the Knaster property) if every uncountable subset A of IP contains an uncountable linked subset. (b) A p.o. (IP,:::5) has precaliber ~l if every uncountable subset A of IP contains an uncountable centered subset. EXERCISE 18.8 (G). (a) Convince yourself that Precaliber =}
~l
=}
Property K
C.C.c.
(b) Convince yourself that if J is countable, then Fn(I, J) has precaliber
~l'
Property K behaves nicely with respect to products.
::;0
LEMMA 18.24. Let (IPo, ::;0), (IPl' :Sl) be p.o. 's. (a) If both (IPo, ::;0) and (IP l , :Sl) have Property K, so does (IPo x IP l ,::;o ®8 :Sl). (b) If (IPo, :So) has Property K and (IPl' :Sl) satisfies the c.c.c., then (!Po x IP l , ®8 :Sl) also satisfies the C.C.c.
PROOF. We prove only point (b). Let (IPo, :So) and (IPl' :Sl) be as in the assumptions of (b), and let A be an uncountable subset of IPo x IP l . We show that A is not an antichain in (IPo x IP l ,:So ®8 :Sl). By Property K, we can choose an uncountable subset B of A such that the set of first coordinates of elements of B is linked in (IPo, :So). Since (IPI> ::; 1) satisfies the c.c.c., the set of second coordinates of elements of B does not form an uncountable antichain in (IPl , :S1). Therefore, we can find (po,P1), ~,pD E B such that PI and p~ are compatible in the sense of :S1' (In particular, this will be the case if PI = p~.) By the choice of B, Po and Po are compatible in the sense of 50. Therefore (PO,Pl) and (P~,p;) are compatible in the sense of :So ®8 :Sl, and we are done. 0
94
18. FROM THE RASIOWA-SIKORSKI LEMMA TO MARTIN'S AXIOM EXERCISE
18.9 (G). Prove Lemma 18.24(a).
Do all c.c.c. p.o.'s have Property K? Not necessarily. If (T, ::;:r) is a Suslin tree with the inverted order, then (T, ::;:r) has the c.c.c., but (T x T, ::;:r 0 8 ::;:r) does not (see Theorem 14.15). Thus, by Lemma 18.24, (T, ::;:r) does not have Property K. On the other hand, as we will show in the next chapter, MA + -,CH implies that every c.c.c. p.o. has Property K.
Mathographical Remarks Various modifications of Martin's Axiom have been proposed. For example, if P is a property of p.o.'s, then MAp is the statement: "If (lP', ::;) is a c.c.c. p.o. that has Property P, and if V is a family of dense subsets of lP' with IVI < 2No, then there exists a V-generic filter in lP'." We have: MA
-+
MAproperty
K -+
MA"'-linked
-+
MA".-centered
-+
MAcountable.
One can show that none of the above implications can be reversed, and that even MAcountable is not a theorem of ZFC. Although Example 18.1 shows that one cannot drop the c.c.c. altogether, it is possible to strengthen MA by extending it to some p.o. 's that do not satisfy the c.c.c. The most important strengthening in this direction is the Proper Forcing Axiom, which is commonly abbreviated by PFA. It is obtained by taking our definition of MA(~l) and replacing the phrase "p.o. that satisfies the c.c.c." by "proper p.o." This is not a proper place to define the notion of a proper p.o.; let us just remark that c.c.c. p.o. 's are proper and, for example, the set of all countable partial functions from WI into W partially ordered by reverse inclusion is proper, but does not satisfy the c.c.c. PFA is equiconsistent with the existence of certain large cardinals. Somewhat surprisingly, PFA implies 2No = ~2' (MA(~d only implies that 2No ~ ~2') J. E. Baumgartner's article Applications of the Proper Forcing Axiom, in the Handbook of Set-Theoretic Topology, K. Kunen and J. E. Vaughan, editors, North-Holland, 1984, is a good place to start learning more about PFA.
CHAPTER 19
Martin's Axiom 19.1. MA essentials
We begin with the historically first l application of Martin's Axiom. THEOREM 19.1. MA(Nd - SH. PROOF. Suppose towards a contradiction that MA(NI) holds and that (S, ::;8) is a Suslin tree. By Exercise 14.33, the dual p.o. (S, ::;s) satisfies the C.c.c. By Lemma 14.24 we may asswne without loss of generality that (S,::;8) is tall. This means that for each 0: < WI the set Da = S \ Sea) is dense in (S, ::;s). Thus, by MA(Nd, there exists a filter IG in (S, ::;s) that intersects each of the sets Da. The following claim yields a contradiction. CLAIM 19.2. IG is an uncountable chain in (S, ::;8). PROOF. If IG were countable, then IG ~ Sea) for some 0: < WI. This is impossible, since IG n Da '" 0. So it remains to show that G is a chain. Let s, t E G. Since G is a filter in (S, ::;s), there is an rES such that r ::;s s, t. In other words, s ::;8 rand t ::;8 r, Le., {s, t} ~ f. Since (S, ::;8) is a tree, f is a chain, and hence sand t are ::;8-comparable. DO EXERCISE 19.1 (PG). Show that CH implies the following two statements: (a) If A, BE [[W]N O]< 2N O are such that for each a E A and each finite subset E of B we have la \ UEI = No, then there exists a set d ~ W with la n dl = No for all a E
A and Ib n dl < No for all b E B. [[w]No]< O are such that la n bl < No for each a E A and b E B,
(b) If A, B E then there exists a set d la\dl
2N
~ W
with
< No for all a
E
A and Ib n dl < No for all
bE B.
In the next chapter we will show that the statement in Exercise 19.1(b) is inconsistent with the negation of CH (Theorem 20.8). But in Exercise 19.1(a), CH can be replaced by MA. 2N
LEMMA 19.3 (Solovay's Lemma). Assume MAu.centered. If A, BE [[W]NO]< O are such that for each a E A and each finite subset E of B we have la \ UEI = No, then there exists a set d la
n dl
~ W
with
= No for all a E
A and Ib n dl < No for all b E B.
lIn a sense, Theorem 19.1 is even older than Martin's Axiom itself. In their paper Iterated Cohen extensions aiJd Souslin's problem, Annals of Mathematics 94 (1971) 201-245, R. Solovay and S. Tennenbaum had given a consistency proof of SH. Upon seeing their proof, D. A. Martin realized that the argument showed something more general, namely the consistency of what would later become known as Martin's Axiom. 95
96
19.
MARTIN'S AXIOM
PROOF. This proof is similar to Proof 3 of Theorem 18.2. Let A,B be as in the assumption, and assume MA".-centered. Define:
lP' = {(8, E) : 8 E [w]
E
[BJ
(8, E) ::; (t, F) if and only if t ~ 8, F ~ E, and (8 \ t) n
UF = 0.
We show that ::; is a p.o. relation on lP'. Only transitivity requires a proof. Let (8, E) ::; (t, F) and (t, F) ::; (u, G). Then u ~ 8 and G ~ E. In order to show that (8\U) nUG = 0, note that (8 \ u) nU G ~ ((8 \ t) nUG) u ((t\ u) nU G). The second term (t\ u) nU G is empty since (t, F) ::; (u, G). Moreover, F :2 G; thus the first term (8 \ t) n U G is contained in (8 \ t) n U F, and since (8, E) ::; (t, F), this is again the empty set. In order to see that (lP',::;) is a-centered, let 8 E [wJ
Dn,a = {(8, E) E lP': la n 81 2: n}, Db = {(8,E) E lP': bEE}. EXERCISE 19.2 (G). Show that the assumptions about A and B imply that Da,n and Db are dense in lP' for all a E A, nEw, and b E B. N ow let V = {Dn,a : nEw, a E A} U {Db : b E B}. This V is a family of dense subsets of lP' with 1'01 < 2N o. Let G be a V-generic filter, and define:
d=U{8: 3E((8,E)EG)}. Let us show that d is as required. Fix a E A, nEw. If (8, E) E G n Dn,a, then dna :2 8 n a and 18 n al 2: n; hence Id n al 2: n. Thus Id n al 2: n for all nEw, i.e., Id n al is infinite. Now let b E B and (8, E) E G n Db. Consider an arbitrary (t, F) E G. There exists a common lower bound (8', E') of (8, E) and (t, F). Then (8' \ 8) n U E = 0, and bEE. Hence (8' \ 8) n b = 0, and thus 8 n b = 8' n b. Since t n b ~ 8' n band (t, F) was assumed an arbitrary element of G, this in turn implies, in view of the definition of d, that d n b = 8 n b. Thus Id n bl < l{o. 0 Solovay's Lemma can be used to code subsets of f), for f), < 2No by subsets of w. This technique allows us to derive an important consequence of Martin's Axiom for cardinal exponentiation. THEOREM 19.4 (MA".-centered). If f), < 2No, then 2'" ::; 2N o. PROOF. Assume MA".-centered and let f), < 2N o. Let C be an almost disjoint family of subsets of w such that ICI = f),. The existence of such a family is an easy consequence of Theorem 17.18. For each B ~ C pick dB ~ w such that IdBnal = l{o for a E C \ Band Id B n al < l{o for a E B. Solovay's Lemma implies the existence of suitable dB'S. EXERCISE 19.3 (G). Convince yourself that B =I- B' implies that dB =I- dB,· Since IP(C)I = 2"', Exercise 19.3 implies that the set {dB: B ~ C} has cardinality 2"'. But the latter set is contained in P( w), and thus has cardinality at most 2No. The inequality 2'" ::; 2No follows. 0
19.1.
MA ESSENTIALS
97
COROLLARY 19.5 (MAr-centered). 2No is a regular cardinal. PROOF. Suppose not and let A < 2No be the cofinality of 2No. By Theorem 19.4, 2>' = 2No , so cf(2)') = A, which contradicts Konig's Theorem (Corollary 11.24). 0 Next we are going to discuss several equivalent formulations of Martin's Axiom. The first of these is topological. Consider the following statement. (MAtop)
No compact Hausdorff space with the C.C.c. is a union of less than 2No nowhere dense subsets.
LEMMA 19.6. MA
---t
MA top .
PROOF. Let X be a compact Hausdorff space, and let JP> be the family of all closed subsets F ~ X such that F has nonempty interior. If X is a c.c.c. topological space, then (JP>,~) also satisfies the c.c.c. Let K. < 2No , and let {NO!: 0: < K.} be a family of nowhere dense subsets of X. For 0: < K., let DO! = {F E JP>: F n NO! = 0}. Then DO! is dense in (JP>, ~). To see this, consider FE JP>. Since NO! is nowhere dense in X, we can pick a nonempty open U ~ F such that Un NO! = 0. Let x E U. By regularity of X, there is a closed neighborhood V of x such that V ~ U. Clearly, V E DO!, and V ~ F. Now assume MA holds, and let G be a filter in (JP>, ~) such that G n DO! =f: 0 for each 0: < K.. Then G is a family of closed subsets of X with the finite intersection property (it is even true that H has nonempty interior for every H E [G] <W). By 0 compactness of X, there exists some x EnG. Clearly, x ~ UO!
n
Recall that cov(M) is the minimum number of meager sets that are needed to cover the real line. Note that IR is homeomorphic to the open interval (0,1), and if A is a covering of (0,1) by meager sets, then Au HO, I}} is a a covering of the compact space [0,1] by meager sets. Thus, by Lemma 19.6, MA implies that cov(M) = 2No. Even more is true. EXERCISE 19.4 (G). Show that MAcountable ---t cov(M) = 2No. Hint: Let JP> = {[a, b] : ~ a < b ~ 1, a, bE Q}, and mimic the proof of Lemma 19.6.
°
Now let us consider two versions of Martin's Axiom for Boolean algebras. (MA Ba ) If B is a Boolean algebra that satisfies the c.c.c., and if V is a family of dense subsets of B with IVI < 2No , then there exists a V-generic filter in B. (MA cBa )
If B is a complete Boolean algebra that satisfies the c.c.c., and if V is a family of dense subsets of B with IVI < 2No , then there exists a V-generic filter in B.
Recall from Section 13.3 that if B is a Boolean algebra, then its Stone space st B is the topological space (Ult B, r), where Ult B is the set of all ultrafilters in Band r is the topology generated by the base U = {Ua : a E B}, where Ua = {F E Ult B': a E F}. We have shown in Section 13.3 that Stone spaces of Boolean algebras are compact Hausdorff spaces. EXERCISE 19.5 (PG). Let B be a Boolean algebra. Show that a set D ~ B+ is dense in B if and only if the set D* = {F E Ult B ; F n D ;f 0} is dense and open in st B.
19.
98
MARTIN'S AXIOM
Now suppose MAtop holds, let B be a Boolean algebra that satisfies the c.c.c., and let V be a family of fewer than 2No dense subsets of B. By Exercise 19.5, for each D E V the set UltB\D* is a nowhere dense subset of stB. Since stB is a compact Hausdorff space that satisfies the c.c.c., there exists F E Ult B such that F E D* for all D E V. Then F n D =I- 0 for all D E V, i.e., F is V-generic. We have proved the following: LEMMA 19.7. MA top
--+
MA Ba .
Clearly, MABa implies MAcBa. The latter statement in turn implies the original version of Martin's Axiom. LEMMA 19.8. MAcBa
--+
MA.
PROOF. Assume MAcBa and let", < 2No. We want to show that MA(",) holds. Consider the following statement: Let (lP,~) be a c.c.c. p.o. of cardinality ~ "', and let V be a family of dense subsets of lP with IVI ~ "'. Then there exists a V-generic filter in lP.
MA -("') appears to be weaker than MA(",) , but in Chapter 21 (Theorem 21.26) we will show that MA( "') and MA - ("') are equivalent statements. Thus, in order to prove MA(",), we can restrict our attention to p.o.'s of cardinality ~ "'. So assume that (lP, ~) is a c.c.c. p.o. such that IlPl ~ "'; let A ~ "', and consider a family V = {Da : Q < A} of dense subsets of lP. By Theorem 13.34, there exist a complete Boolean algebra B and a function i : lP --+ B+ such that (i)
rng(i) is dense in B+;
(ii) Vp,q E lP(p ~ q --+ i(P) ~B i(q)); (iii) Vp, q E lP (p..l q +-d(p) ..lB i(q)). Fix such i, B, and for
Q
< A, let i[Dal be denoted by Ea·
EXERCISE 19.6 (G). Convince yourself that B satisfies the c.c.c. and each Ea is dense in B. By MAcBa , there exists an ultrafilter IF in B such that IF n Ea =I- 0 for each < A. Define G = i-I [lFl. It appears that we have found a V-generic ultrafilter in lP: Clearly, G n Da =I- 0 for each Q < A. If pEG and p ~ q, then i(P) E IF
Q
and by (ii), i(p) ~B i(q). Since IF is a filter, i(q) ElF, and hence q E G. Similarly, (iii) implies that the elements of G are pairwise compatible in lP. But this does not quite suffice: The definition of a filter requires that if p, q E G, then there exists a lower bound for p and q in G, not just in lP. EXERCISE 19.7 (G). Assume that i satisfies the following strengthening of (ii): (ii)+
Vp, q E lP (p ~ q
+-t
i(p) ~B i(q)).
Show that in this case G is a V-generic ultrafilter in lP. Unfortunately, in general i satisfies just condition (ii), not (ii)+. But by using a trick we can make sure that G is as required. For p E lP, let Dp = {r E lP : r ~ p V r..l pl.
19.1. MA ESSENTIALS
99
EXERCISE 19.8 (G). (a) Show that for every p E lP' the set Dp is dense and open in lP'. (Recall that the latter means that Vq, r E lP' (q E Dp /\ r :::; q ---t r E Dp)). (b) Show that for all p, q E lP' the intersection Dp n Dq is dense in lP'. Here comes the trick: Since 1lP'1 :::; /'i., we may assume without loss of generality that for all p, q E lP', the set Dp n Dq is an element of V! Now let G be as above, let p, q E G, and let rEG n Dp n D q. Since G consists of pairwise compatible elements, we must have r :::; p, q. Thus, p and q have a lower bound in G, and we are done. 0 We have proved the following: THEOREM 19.9. Each of the statements MA top , MA Ba , and MAcBa is equivalent to Martin's Axiom. EXERCISE 19.9 (R). Show that MAcountable is equivalent to cov(M) = 2No. Hint: Use Corollary 25.4. Now let us strengthen Theorem 19.1 by showing that MA(N 1) implies that every Aronszajn tree is not only not Suslin, but even special. THEOREM 19.10. If MA(N 1) holds, then every Aronszajn tree is special. PROOF. Assume MA(NI), and let (T, :::;T) be an Aronszajn tree. We want to construct a function f : T -+ w such that f(t) = f(s) implies that s and tare incomparable with respect to :::;T. In order to use MA(N 1), we need to work with a suitable C.c.c. partial order (lP', :::;). The elements of lP' will be approximations to our function f. More precisely, let lP' denote the set of all functions p such that (i) (ii) (iii)
dom(p) E [T]
We partially order lP' by reverse inclusion, i.e., we define: p :::; q if and only if q ~ p.
LEMMA 19.11. (lP',:::;) satisfies the c.c.c. Before we prove the lemma, let us show how it implies Theorem 19.10. For
t E T let D t = {p E lP': t E dom(p)}. Each D t is dense in lP'. To see this, let t E T and consider p E lP'\Dt . Pick n E w\rng(p), and let q = p u {(t, n)}. Then q < p and q EDt. Now consider the family
V = {Dt : t E T}. This is a family of N1 dense subsets of lP', and by MA(N 1), there exists a V-generic filter G. Let
f=UG. Then f is a function from T into wand for each nEw the inverse image f-1{n} is an antichain in T: Otherwise, there would be s, t, n and p, q E lP' with s
100
19.
MARTIN'S AXIOM
PROOF OF LEMMA 19.11. Let D be an uncountable subset of P. By the .60System Lemma, there exist u E [T]
a = {a(O), a(l), ... , a(n - I)}. Our aim is to find an uncountable subset E of A and Se E e for each e E E such that the set {se : e E E} consists of pairwise comparable nodes of T. This will yield a contradiction, since T was assumed to be an Aronszajn tree and thus not to contain uncountable chains. Note that if n = 1, we can simply take E = A. If n > 1, then we will use the following trick: We will find an uncountable subset E ~ A and Se E c for c E E such that if d, e E E, then the set
S(d,e) = {t E T: t is comparable with both Sd and se} is uncountable. Note that S(d, e) ~ {Se, Sd} USe U Sd U {t E T: {Se, Sd} ~ i}. Since the set {se, Sd}US eUSd is countable, it follows that whenever S(e, d) is uncountable, there exists t E T such that {se, Sd} ~ i. Since i is wellordered by 5:.T, the latter implies that Se and Sd are comparable. So let us construct a set E and nodes Se as above. Fix a uniform ultrafilter F on A. For k < n and SET, let
B(s, k) = {a E A :
S
is comparable with a(k)}.
Since every two elements a, b E A contain comparable nodes, we have for each aEA: A = {a} U U{B(s,k) : k < n}. sEa
Thus for each a E A we find Sa E a and ka < n such that B(sa, k a) E F. Moreover, there exists k* < n such that the set E = {a E A : ka = k*}
is uncountable. For d, e E E we have
S(d, e) 2 {a(k*) : a E B(Sd, k*)
n B(se, k*)}.
101
19.1. MA ESSENTIALS
Note that the map that assigns a(k*) to a E A is an injection. Since F is uniform and B(Sd' k*) n B(se, k*) E F, the set on the right hand side is uncountable, as required. DOD THEOREM 19.13 (MA(Nd). Every C.C.c. partial order has precaliber Nl. EXERCISE 19.10 (PG). Theorem 19.13 can be used to give yet another proof of Theorem 19.1. Can you see how? PROOF OF THEOREM 19.13. Assume MA(Nd, let (IP',::;) be a p.o. that satisfies the c.c.c., and let A be an uncountable subset of IP'. We want to show that A contains an uncountable centered subset. Without loss of generality we may assume that IAI = N1 . Note that every subset of a filter in IP' is centered. Thus, it suffices to find a filter IF in IP' such that the intersection IF n A is uncountable. In order to be able to apply MA(Nd, we need to find a family V of ::; Nl dense subsets of IP', or perhaps of a suitable C.C.c. suborder of IP', such that V-genericity of IF ensures that IF n A is uncountable. The most natural impulse is to try working with the p.o. (IP',::;) itself. However, this may prove futile: For example, it could happen that IF contains some p such that p 1. q for every q E A. In this case no amount of genericity of IF will force IF to intersect A, let alone uncountably often. So we need to consider a suitable subset of IP' instead. As a first step towards finding such a subset, define: B = {p E IP': I{q E A : p
t
q} I ::; No V 'v'r ::; p (I {q E A : r
t
q} I ~ Nd }.
By Zorn's Lemma, there exists an antichain C in IP' that is maximal with respect to the property that C ~ B. Let us show that C is actually a maximal antichain in (IP', ::;): Suppose not, and let p E IP' be such that p 1. p' for all p' E C. Then p fI. B; hence there exists r < p such that I{q E A : r t q}1 ::; No. But then rEB and C U {r} is an antichain in IP' that extends C, which is ruled out by the choice of C. Since C is maximal, every q E A must be compatible with some p E C. By the c.c.c., C is countable. On the other hand, A is uncountable, and hence the Pigeonhole Principle implies that there exists Po E C such that Po is compatible with uncountably many q E A. Since Po E B, the latter implies that 'v'r ::; Po (I{q E A: r
t q}1
~ Nd·
Now let Po be as above, and define lP'o = {p E IP' : p ::; Po}· Then (lP'o, ::;) is also a c.c.c. partial order. Enumerate A as {qo : a < wd. For each a < WI define: Do = {r ::; Po : 3/3 > a (r ::; q(3)}. EXERCISE 19.11 (G). Convince yourself that each of the sets Do is dense in (lP'o, ::;). Now apply MA(Nd to find a filter IFo in (lP'o, ::;) that meets Do for every a < WI' Let IF be the filter in IP' generated by IFo, i.e., let IF = {p E IP': 3r E IFo (r ::; p)}. Then for every a < WI there exists /3 > a such that q{3 E IF n A. Hence IF n A is an uncountable centered subset of A, as required. 0 Since precaliber Nl implies Property K, Lemma 18.24 and Theorem 19.13 yield the following: COROLLARY 19.14 (MA(Nd). The simple product of any two C.C.c. p.o. 's satisfies the c. c. c. In particular, the product of any two topological spaces with the C.C.c. satisfies the C.C.c.
102
19. MARTIN'S AXIOM
Corollary 19.14 in turn gives together with Theorem 16.4: COROLLARY 19.15 (MA(l'~l)). The Tychonoff product of any nonempty family of topological spaces with the c.c.c. satisfies the c.c.c. Corollary 19.14 can also be used to prove the following: THEOREM 19.16 (MA(l'~l)). Every c.c.c. partial order of cardinality Nl is centered.
(j-
PROOF. Assume MA(N 1 ) and let (IP',~) be a c.c.c. partial order such that IIP'I = N1 . Let Q = <wIP', and define a p.o. relation ~ on Q as follows: ql ~ qo if and only if dorn(qo) ~ dom(ql) 1\ Vi E dom(qo) (ql(i) ~ qo(i)). EXERCISE 19.12 (G). Show that that the p.o. (Q,~) satisfies the c.c.c. Hint: Note that for every uncountable A ~ Q there must be nEw such that A n nIP' is uncountable. Then use Corollary 19.14 and induction over nEw. Now define for p E IP':
Dp = {q E Q: 3i E dorn(q) (q(i)
~ pH.
It is not hard to see that each of the sets Dp is dense in (Q, ~). Since IIP'I = Nl and Q satisfies the c.c.c., MA(Nl) implies that there exists an ultrafilter IF in Q such that IF meets each of the sets Dp. Fix such IF and define for i E w: IP'i = {q(i) : q ElF}.
o
Then each of the sets IP'i is centered and IP' = UiEW IP'i.
19.2. MA and cardinal invariants of the continuum In this section we shall examine what Martin's Axiom implies about certain cardinal invariants of the continuum. One result in this direction was already proved in Chapter 18. Recall that a
= min{IAI : A
is an infinite maximal almost disjoint family of subsets of w}.
In this new terminology, Theorem 18.2 asserts that a > translates into the inequality a > ~l.
~o,
and Statement 18.3
EXERCISE 19.13 (G). Convince yourself that Theorem 18.8 remains true if "MACK)" is replaced by "MAa-centered(K)." If a, b ~ w, then we write a ~* b as shorthand for la\bl < ~o. If a ~* b, then we say that a is almost contained in b. We say that a and b are almost equal, and write a =* b, if a ~. b and b ~* a. We write a c· b if a ~. b and a #* b. A subset b of w is called a pseudo-intersection of a family C ~ P(w) if b is infinite and b ~. c for all c E C. We say that a family C ~ P(w) has the strong finite intersection property (sfip) if I nHI = ~o for all H E [C]
EXERCISE 19.14 (G). Suppose C is a nonprincipal ultrafilter on w. Show that C has the sfip but no pseudo-intersection. . Let us define our next cardinal invariant: p = min{ICI : C ~ [wt o 1\ C has the sfip but no pseudo-intersection}.
19.2.
MA AND CARDINAL INVARIANTS OF THE CONTINUUM
103
EXERCISE 19.15 (PG). Show that p > No. THEOREM 19.17. MA,,-centered ~ P = 2No. PROOF. Exercise 19.14 implies that p is well-defined and :S 2 No. So we only need to prove that every family C ~ [wjNo with the sfip and of cardinality less than 2No has a pseudo-intersection. Assume MA,,-centered, and fix C as above. Consider the families A = {w} and B = {w\c : c E C}. Note that the sfip of C implies that for every finite E ~ B we have Iw\ U EI = No· Hence the assumptions of Solovay's Lemma (Lemma 19.3) are satisfied and there exists d ~ w such that d is infinite and Id n bl < No for all b E B. The latter means that d ~* C for all c E C. Hence d is a pseudo-intersection of C, as desired. 0
e e
A /'i,-tower is a sequence (c~ : < /'i,) of elements of [W]N o such that c~ C* cT/ for all TJ < < /'i, and the family {c~ : < /'i,} has no pseudo-intersection. We define:
e
t
= min{/'i,:
there exists a /'i,-tower}.
EXERCISE 19.16 (G). (a) Convince yourself that for some /'i, :S a /'i,-tower. (b) Convince yourself that t is a regular cardinal.
2No
there exists
Let us put down the following observation for later reference. CLAIM 19.18. PSt. One can use Theorem 19.17, Claim 19.18, and the next theorem to give an alternative proof of Theorem 19.4. THEOREM 19.19 (Rothberger). If /'i, < t, then 2'" S 2No. PROOF. Let /'i, < t. For such that:
(i) (ii)
Vl,g VI
E
0:
S /'i, and
1E
0<2 we will construct sets af E [wjNo
E $."'2(f ~ g ~ a g ~* af); <><2 (af~o af~1 = 0).
n
EXERCISE 19.17 (G). Assume that the aI's are constructed in such a way that (i) and (ii) hold. Convince yourself that if {f,g} E ["'2]2, then laf n agl < No and thus af =j:. a g • Infer that 2>< :S 2No. The a f 's can be constructed by recursion over 0:. If af has been constructed, we let {af~o,af~d be a partition of af into infinite disjoint subsets. Now suppose 0: is a limit ordinal S /'i" 1 E 0<2, and the aJ[{3's have been defined for all f3 < 0:. Then (a J[(3 : f3 < 0:) is a ~ *-decreasing sequence of infinite subsets of w of length < t. Thus, it is not a tower, and we can find a pseudo-intersection af of {aJ[{3 : f3 < o:}. This construction yields aI's that satisfy (i) and (ii). 0 and
Now let us show that if MA,,-centered holds, then the cardinal invariants b, il, N 5 that were introduced in Chapter 17 are all equal to 2 o. THEOREM 19.20. MA,,-centered ~ 5
= 2No.
PROOF. Assu~e MA,,-centered, and let A ~ [W]N o be a family of size < 2No such that w\a is infinite for all a E A. We want to show that A is not a splitting family, i.e., we want to construct x E [W]N o such that for all a E A, either Ix n < No or Ix\al < ~o.
al
104
19.
MARTIN'S AXIOM
As in most applications of Martin's Axiom, the crucial step in the proof is finding a suitable p.o. to work with. Let F be a nonprincipal ultrafilter on w, and define: lP = {(s,X): s E [w]
(8', X') j (8, X) iff 8 ~ 8' 1\ X ~ X' 1\ 8'\8 ~ n(X n F)\ U(X\F). The x we are looking for will be defined as: XG
= U{8: 3X((8,X) E Gn,
where G is a V-generic filter in lP for a suitable family V of fewer than 2No dense subsets of lP. The choice of V is dictated by the properties we want x to have: (1) For a E A, we want XG to be almost contained in a or almost disjoint from a, i.e., we want XG ~* a or xGna =* 0. If a E XnF for some (8, X) E G, then x will be almost contained in a. Similarly, if a E X\F for some (8, X) E G, then x will be almost disjoint from a. Thus, for each a E A, we will want the set Va = {(8,X) E lP: a E X} to be in V. (2) We want XG to be infinite. As in our earlier applications of MA, this can be assured by putting the sets En = {(8,X) E lP: 1812: n} into V. This describes the idea of the proof. Now we need to verify that it works. Once you get the knack of it, this part of a proof involving Martin's Axiom will become utterly boring, and you will be tempted to skip it. Don't. Practically all mistakes in proofs involving Martin's Axiom occur because the author skips one of the more obvious steps in the following exercise. EXERCISE 19.18 (G). Let lP, j, Va, En, G, and XG be as above. (a) Show that j is a p.o. relation on lP. (b) Show that (IP, j) is a-centered. (c) Show that for all a E A the set Va is dense in lP. (d) Show that for all nEw the set En is dense in lP. (e) Show that if G is a filter in lP and a E A is such that G n Va =j:. 0, then XG is almost contained in or almost disjoint from a. (f) Show that if G is a filter in lP such that G n En =j:. 0 for all nEw, then IXGI = No. Now let V = {Va: a E A} U {En: nEw}. The assumption on A implies that IVI < 2No. By points (a)-(d) of Exercise 19.18, MA",-centered applies, and there exists a V-generic filter G in lP. By points (e) and (f) of Exercise 19.18, the corresponding XG witnesses that A is not splitting. 0 THEOREM 19.21. MAcountable
-+ ()
= 2No.
PROOF. Assume MAcountable, and let F ~ W w be such that IFI < 2No. Consider the countable set Fn(w, w) partially ordered by reverse inclusion. The idea is to find a family V of fewer than 2No dense subsets of this p.o. such that for every Vgeneric ultrafilter G the set U G is a function from w into w that is not eventually dominated by any function f E F. EXERCISE 19.19 (PG). Finish the proof of Theorem 19.21.
o
It is not possible to find a family V of dense subsets of the p.o. Fn(w, w) such that f <* UG for every f E F and every V-generic filter G ~ Fn(w,w). Thus, if
19.2. MA AND CARDINAL INVARIANTS OF THE CONTINUUM
105
we want to derive the equality b = 2No from Martin's Axiom, we must adorn the elements s E Fn(w,w) with so-called side constraints. THEOREM 19.22. MAer-centered -+ b = 2No. EXERCISE 19.20 (PG). Prove Theorem 19.22. Hint: For F as in the proof of the previous theorem, consider the set lP={(s,F): SE<ww/\FE[F]
(s, F) :::5 (t, G) iff t ~ s /\ G ~ F /\ "In E dom(s)\dom(t)VI E G (s(n) > I(n)). Show that (lP,:::5) is a a-centered p.o., and find a family V of fewer than 2No dense subsets of lP such that for every V-generic filter G in (lP,:::5) the set gIG defined by gIG = U{ s : 3F ((s, F) E G)} is a function from w into w with I <* gIG for all
IE :F. Theorem 19.22 can be used to derive further consequences of Martin's Axiom. THEOREM 19.23. MAer-centered
-+
add(M) = 2No.
PROOF. Assume MAer-centered, and let {Me : ~ < A < 2No} be a family of fewer than 2No meager subsets of lR. Define M = U e<>. Me. We want to show that M is a meager subset of JR.. Since MAer-centered implies MAcountable, it follows from Exercise 19.4 that M i:- JR.; in fact, it even follows that JR.\M is dense in lR. So, let D be a countable dense subset of JR.\M. Enumerate D as (dn)nEw' Since every meager set is contained in a meager Fer-set, we may without loss of generality assume that each Me is an Fer-set. For every ~ < A, choose a sequence (Fe,n)nEW of nowhere dense closed sets such that U nEw Fe,n = Me, and define a function Ie : w -+ W by fe(n) = min{k E w: (dn - 2- k ,dn + 2- k ) n Fe,n = 0}. Since D n Fe,n = 0 for all ~ < A and n < w, these functions are well defined. By Theorem 19.22, there exists a function I : w -+ w such that Ie <* I for all ~ < A. Fix such I, and define for mEw a set Urn = UnEw(dn - 2-!(n)-rn, dn + 2-!(n)-rn). Then each Urn is a dense open subset of JR., and hence the set K = JR.\ nrnEw Urn is meager. EXERCISE 19.21 (G). Show that for each ~ < A and n < w there exists mEw such that Urn n Fe,n = 0. Conclude that M ~ K. 0 The proofs of Theorems 19.20-19.22 presented above gave us opportunities to practice applying Martin's Axiom, but they do not tell us much about how the cardinals (1, b, (),P,s, and t are related to each other. The next theorem fills this gap and also indicates an alternative way of deriving some of the above results. THEOREM 19.24. N1 S pst S min{b,s} S max{b,s} S () S 2No and b S
(1
S
2 No.
PROOF. The inequality N1 S p was established in Exercise 19.15, the inequality PSt is Claim 19.18, the inequality b S () is Exercise 17.18(a), the inequalities min{b,s} S max{b,s}, (1 S 2No , and () S 2 No are obvious. Let us prove the remaining inequalities. t S s: Consider 8 ~ [w]No such that 181 = 1'0 < t. We show that 8 is not a splitting family. Enumerate 8 as (act)ct
106
19. MARTIN'S AXIOM
(1) (2)
b{3 ~* boi and bo ~* a o or Ibo n aol < No·
This can be done as follows: Let bo = ao. If 0 < a < K and b-y has already been constructed for all 'Y < a, then let Co be a pseudo-intersection of {b-y : 'Y < a}. The existence of Co is assured by the assumption that K, and hence a, is less than t. If Co n a o is infinite, we define bo = Co n ao. If not, we define bo = Co. In both cases, condition (2) will be satisfied. Since K < t, condition (1) implies that there exists a pseudo-intersection b of {bo : a < K}. EXERCISE 19.22 (G). Convince yourself that any pseudo-intersection b of {b o a < K} witnesses that S is not a splitting family.
:
t ::; b: With every a E [wP~o we associate an increasing function fa E Ww by letting fa(n) be the (n + l)st element of a.
EXERCISE 19.23 (PG). (a) Convince yourself that if f : w ~ w is increasing, then frng(f) = f· (b) Show that if a E [wP~o and g E w w, then there exits b E [a]No such that g <* fb' (c) Show that if a C* b then fa <* fb. Now assume towards a contradiction that b < t, and let {go : a < b} be unbounded in ww. By recursion over a < b we construct a sequence (ao : a ::; b) of infinite subsets of w such that for all a < f3 ::; b:
(1) (2) (3)
go <* faa+l i la O \a O +ll=Noi a{3 ~* ao·
Exercise 19.23(b) implies that given ao, we can find a subset a o+! of a o such that (1) and (2) hold. At limit stages f3 ::; b, the assumption b < t implies that we can find a pseudo-intersection a{3 of {ao : a < f3}. This, together with taking subsets at successor stages, will take care of requirement (3). Now consider fab' It follows from (1)-(3) and Exercise 19.23(c) that go <* fab for all a < b. But this contradicts the assumption that the family {go: a < b} is unbounded. b ::; a: Let A ~ [w]No be an almost disjoint family of size K, where w ::; K < b. We show that A cannot be maximal. Enumerate A as (ao)o<,,' For each a E [w, K), define a function fo : w ~ w by fo(n) = max(a o n an), where the maximum of the empty set is arbitrarily defined as O. Since K < b, there exists a strictly increasing function f E Ww such that fo <* f for all a E [w, K). Define: a" = {min(a n\(f(n) + 1 U ao U al U ... U an-d) : nEw}. EXERCISE 19.24 (G). Show that a" is an infinite subset of w that is almost disjoint from a o for all a < K. 5 ::; (l:
This is a good exercise.
EXERCISE 19.25 (PG). Finish the proof of Theorem 19.24 by showing that Hint: The same idea as in the proof that in Lemma 17.14 CH can be 0 replaced by () = Nl works here.
5 ::; (l.
19.2. MA AND CARDINAL INVARIANTS OF THE CONTINUUM
107
By Theorem 19.24, all the consequences of MA".-centered for cardinal invariants that we have proved so far already follow from the equality p = 2No. Does this equality perhaps imply MA".-centered itself? THEOREM 19.25 (Bell). MA".-centered is equivalent to the equality p = 2No. PROOF. We have already shown that MA".-centered --+ P = 2No. To prove the converse, assume p = 2No , let (IP,::;) be a a-centered p.o., and let V = {Do: a < It} be a family of dense subsets of IP, where It < 2No. We want to show that there exists a filter IF s:;; IP with IF n Do =f. 0 for all a < It. EXERCISE 19.26 (PC). (a) Convince yourself that we may assume without loss of generality that (1)
IIPI::; It;
(2) (3)
Every Do is open, i.e., Vp, q E IP (p E Do 1\ q ::; P --+ q E Do); For all p,q E IP, the set Dp n Dq E V, where Dp = {r E IP: r::; pVr.l pl.
Hint: Recall the proof of Lemma 19.8.
(b) Conclude that it suffices to show that there is a linked subset L s:;; IP with L
n Do =f. 0 for all a < It. For the remainder of this proof, let IP be as in Exercise 19.26, and fix a partition
IP = UmEw Pm of IP into centered subsets. For p E IP and a < It we define: Ap,o = {m: ::Iq E Pm
n Do (q
::; pH.
EXERCISE 19.27 (C). (a) Show that if a < It and q ::; p, then Aq,o s:;; Ap,o' (b) Show that for every mEw the family Am = {Ap,o : p E Pm 1\ a < It} has the finite intersection property. Now suppose we have a set C E [W]N o and a set L (A) (B)
= {Po: a < It} such that:
Va < It (Po E Do); and Va < 1t3mo E C (Po E Pm" 1\ C\(mo + 1) s:;; Ap",o)'
Then L will be as required: It follows from (A) that L intersects every element of V. If Po,P{3 E L, then by (B), for nEe such that n > max{mo, m{3}, there are qo, q{3 E Pn such that qo ::; Po and q{3 ::; P{3· Since Pn is centered, qo and q{3 are compatible, and so are Po,P{3' It follows that L is linked. Are there sets C and L that satisfy (A) and (B)? Not necessarily. It may happen that the families Am defined in Exercise 19.27(b) only have the finite intersection property, but not the strong finite intersection property. If this is the case, then we cannot expect to find an infinite C as above. Fortunately, in such a situation Theorem 19.25 has a rather simple proof. CLAIM 19.26. Suppose that there exists mEw such that the family Am does not have the sfip. Then there exists nEw such that Pn intersects all Do's.
n
PROOF. Let mEw be as in the assumptions. It suffices to show that Am =f. 0. Suppose this were not the case, and let {Po, ... ,Pk-d s:;; Pm and {ao, ... ,ak-d C It be such that ni
19.
108
MARTIN'S AXIOM
So let us assume from now on that Am has the sfip for each mEw. By recursion over s E <ww we will construct Bs E [wF~o and Ps,a. E IP' as follows: Let B0 be a pseudo-intersection of Ao. Given Bs and nEw, let Bs~n be a pseudo-intersection of An. Such a pseudo-intersection exists since IAnI ~ K < p for all nEw. For Q < K, let P0,a. be an arbitrary element of Po. Suppose {Ps,a. : Q < K} is given. If n E Ap •. a,a., then we let Ps~n,a. E Pn n Da. be such that Ps~n,a. < Ps,a.; otherwise we let Ps~n,a. be an arbitrary element of Pn . Note that for every s E <ww and Q < K the set {n E Bs : -,3q E Pn (q ~ Ps,a.)} is finite. Thus, for each Q < K we can define a function fa. : <ww -T W as follows:
fa.(s) = min{n E Bs : '<1m E Bs \n (m E Ap•. a,a.)}' Since K < p ::; b, there exists a function 9 : <ww -T w such that for each Q < K the set {s : fa.(s) ~ g(s)} is finite. Moreover, we can choose 9 in such a way that g(s) E Bs and g(s) > max(rng(s)) for all s E <ww. Fix such 9 and define recursively an increasing sequence (lkhEw of natural numbers by letting lk = g( (lo, ... ,lk-l)) (in particular, lo = g(0)). Now define C = {lk : k E w}. By the choice of g, for each Q there exists k E w such that fa.((lo, ... ,lj)) < g((lo, ... ,lj)) for all j ~ k E w. Let ma. be the smallest such k, and definepa. =p«(o"",(ma)'a.' Finally, let L = {Pa. : Q < K}. EXERCISE 19.28 (G). Convince yourself that C and L satisfy conditions (A) and (B). 0 Not all implications of Martin's Axiom for cardinal invariants follow from MA,,-centered. For example, in the next theorem, MA,,-linked cannot be weakened to MA,,-centered . THEOREM 19.27. MA,,-linked
-T
add(N) = 2No.
PROOF. Assume MA,,-linked holds, let K < 2No , and let {Aa. : Q < K} ~ N. We want to show that Ua.
~
G such
that W = UG. PROOF. Consider the set i(; = {fa, bj : a, bE Ql\a < M[a, bj ~ W}. Since W is an open subset of JR, we must have W = U IG. Every interval [a, bj E IG is compact, and hence contained in Uo U Ul U ... U Uk, where Ui E G for i ~ k. But since G is a filter in the p.o. (A(€), ~), the union Uo U Ul U··· U Uk is also an element of G.
19.2. MA AND CARDINAL INVARIANTS OF THE CONTINUUM
109
Thus, for each [a, b] E Ie; we can pick U[a,b] E IG such that [a, b] ~ U[a,b]' Since Ie; is countable, so is the family G = {U[a,b] : [a,b] E IG}' And, of course, UG = W. 0 Now let G, W be as above, and assume towards a contradiction that /-L(W) > c:. Then there exists a finite G I ~ G such that /-L(U Gd ~ c:. But since IG is a filter, we would have UG I E IG, which is ruled out by the definition of A(c:). 0 We conclude this section with an example that is not a theorem about cardinal invariants (although it has a similar flavor). Since the p.o. in this example satisfies the c.c.c. for rather unconventional reasons, we will have to use the full force of MA. 19.29 (MA). Let /'i, and A be infinite cardinals such that /'i" A < 2No I and let A = {aa : 0: < /'i,} be a family of countable subsets of A such that laa n a,81 < ~o for all 0: < 13 < /'i,. Then there exists B E [A]A such that IB n aal < ~o for all THEOREM
0:
<
/'i,.
PROOF.
Let us first deal with the easy cases.
EXERCISE 19.30 (G). Show that for /'i, = W Theorem 19.29 can be proved in ZFC, and for the case A = W only MAa-centered is needed.
From now on, let us assume that We define a p.o. (lP,:S) as follows: lP
/'i"
A > w. Let A be as in the assumptions.
= {(s,F) : FE [Aj
The partial order relation is defined by:
(t, G) :S (s, F) EXERCISE
For
0:
<
/'i,
if and only if F ~ G /\ s ~ t /\ (t\s) n
UF = 0.
19.31 (G). Convince yourself that :S is a p.o. relation on lP. and 13 < A let
Da = {(s,F) : aa E F}, E,8 = {(s, F) : s n [13,13 + WI) Define
v=
{Da :
0:
<
/'i,}
U
'"
0}.
{E,8 : 13 < A}.
19.32 (G). (a) Show that Da and E,8 are dense subsets of lP for all and 13 < A. (b) Show that if G is a V-generic filter in 1P', then the set
EXERCISE 0:
<
/'i,
B = U{s: :IF((s,F) E IG)} is as required. It remains to show that (lP,:S) has the c.c.c. Consider an uncountable set {(Sa, Fa) : 0: < WI} ~ lP. We want to show that there are 0: < 13 < WI such that (sa, Fa) and (s,8, F,8) are compatible. Let us first reduce the general case to a more manageable situation. By the 6.-System Lemma, we may assume without loss of generality that there exist s E [>.]
110
19.
MARTIN'S AXIOM
are incompatible, then so are (sa' Fa \F) and (S{3, F{3\F). In other words, we may further simplify the situation by assuming that F = 0. Finally, by the Pigeonhole Principle, we may assume that Isal = n for a fixed nEw and all 0: < WI' Consider the set UiEW U Fi. This set is countable, and by thinning out the sequence if necessary, we may without loss of generality assume that So. n UiEW U Fi = 0 for all 0: with W ::; 0: < WI. Now if i < W ::; 0: < WI, then either (sa' Fa) and (Si' Fi ) are compatible, or SinU Fa =J 0. We show that there are i < W ::; 0: < WI such that the former happens. Consider the sets C k = UFw+k for k ::; n. Since the set D = U{ CknCt : k < l ::; n} is finite, there exists io < W with Sio n D = 0. The latter property of io implies that the family {Ck n Sio : k ::; n} consists of n + 1 pairwise disjoint sets. Since ISio I = n, we can find ko ::; n such that Cko n Sio = 0. This in turn implies that (SiO' Fio) and (Sw+kO' FW+ko) are compatible. 0 19.3. Ultrafilters on
W
In this section, we show how Martin's Axiom or some of its consequences can be used to construct non principal ultrafilters with various interesting properties. Throughout this section, the word "ultrafilter" will be reserved for nonprincipal ultrafilters on w. DEFINITION 19.30. Let F be an ultrafilter. We say that F is selective, if for every partition (ai)iEw of W into pairwise disjoint subsets such that none of the ai's is in F, there exists bE F with Ibnail ::; 1 for all i E w. We say that F is a P-point, if for every partition (adiEW of W into pairwise disjoint subsets such that none of the ai's is in F, there exists b E F with Ib nail < ~o for all i E w. We say that F is a Q-point, if for every partition (ai)iEw of W into pairwise disjoint finite subsets there exists b E F with Ib nail::; 1 for all i E W. EXERCISE 19.33 (G). Convince yourself that an ultrafilter is selective if and only if it is simultaneously a P-point and a Q-point. EXERCISE 19.34 (PG). (a) Show that an ultrafilter F is a P-point if and only if for every sequence (ai)iEW of elements of F there exists b E F such that b ~* ai for all i E w. (b) A point x in a topological space X is called a P-point if x is in the interior of each G6-set that contains x. Consider the set w· of all nonprincipal ultrafilters on w with the topology r generated by the family {Ua : a E [wl~o},whereUa = {F E w* : a E F}. Show that an ultrafilter F is a P-point in the sense of Definition 19.30 if and only if it is a P-point in the topological space (w·, r). THEOREM 19.31. If t = 2~o, then there exists a selective ultrafilter. PROOF. Let {Co. : 0: < 2~o} be an enumeration ofP(w), and let {(af)iEw : {j E nLIM\ {O}} be an enumeration of all partitions of w into pairwise disjoint finite subsets. We are going to construct recursively a transfinite sequence (bo. : 0: < 2~o) of infinite subsets of w such that b{3 ~* bo. for all 0: < f3 < 2~o, and then we are going to define F = {a ~ w : 30: < 2No (bo. ~. a)}. 2~o
EXERCISE 19.35 (G). Show that ifF is constructed as above, then F is a filter. Moreover, show that if F happens to be an ultrafilter, then it is a P-point. Hint: Use the result of Exercise 19.34(a).
19.3.
ULTRAFILTERS
111
ON w
Here is how we construct the bo: 's: We let bo = w. Given bo:, we choose bO:+l E [bo:]No in such a way that bO:+l ~ Co: or, if this is not possible, so that bO:+l nco: = 0, i.e., bo:+ 1 ~ w\co:. This clause ensures that F will be an ultrafilter. Suppose b < 2No is a limit ordinal and the sequence (bo: : a < b) has already been constructed in such a way that bf3 ~* bo: for all a < {3 < b. Since b < t, we can find B6 E [w]No !'luch that B6 ~* bo: for all a < b. Now consider the partition (a~)iEw and define: b6 = {mina~ : i Ed}. EXERCISE 19.36 (PG). Verify that the family {bo: : a < 2No} generates a selective ultrafilter F.
D
A technique similar to the proof of Theorem 19.31 can be used to construct a P-point that is not a Q-point (and hence not selective). We need a lemma. LEMMA 19.32 (MAu-centered). Let (ai)iEw be a partition ofw into pairwise disjoint sets with lail = i for all i E w. Suppose B ~ P(w) is a family such that IBI < 2No , and for every HE [BJ
Define a binary relation j on JP> by:
(8, F)
j
(t, G)
if and only if
t ~ s 1\ G ~ F 1\ s\t ~
n
G.
EXERCISE 19.37 (G). (a) Verify that j is a p.o. relation on JP>. (b) Show that (JP>,~) is a-centered. We want to define c = U{s : 3F((s,F) E G)}, where G is a V-generic filter for a suitable family of dense sets V. "Suitable" means that IVI < 2No , and that V-genericity of G forces c to be as required. EXERCISE 19.38 (G). Find a family Va of cardinality < 2No such that V ogenericity of G implies that c ~* b for all b E B. Don't forget to verify that the elements of Va are really dense in (JP>, j). We have to ensure one more property of c. For each nEw, let
En = {(s, F) E JP>: 3i E w (Iai n sl ~ n)}. EXERCISE 19.39 (G). (a) Verify that each set En is dense in (JP>, ~). (b) Show that if Va is as in Exercise 19.38, and G is Va U {En: nEw }-generic, then c = U{ s : 3F( (s, F) E G)} is as postulated in Lemma 19.32. D THEOREM 19.33. MAu-centered implies that there exists a P-point that is not a Q-point. EXERCISE 19.40 (PG). Prove Theorem 19.33. Hint: Fix a partition (ai)iEw of w into pairwise disjoint sets such that lail = i for all i E w. Then construct recursively a 2No -tower (bo: : a < 2No) as in the proof of Theorem 19.31, but make sure that {Ibo: n ad : i E w} is infinite for all a < 2No. At limit stages, use Lemma 19.32.
19.
112
MARTIN'S AXIOM
An ultrafilter F is called a semi-Q-point if for every partition (ai)iEw of w into pairwise disjoint finite sets there exists bE F such that Ib n ail ~ i for all i E w. EXERCISE 19.41 (R). Use MAa-centered to construct a P-point that is not a semi-Q-point. THEOREM 19.34. MAcountable implies the existence of a Q-point that is not a P-point. PROOF. Assume MAcountable. For the duration of this proof, fix a partition A = (ai)iEw of w into pairwise disjoint infinite sets. A set b ~ w will be called A-large if I{ i E w : Ibnail = ~o}1 = ~o. A family B ~ P(w) will be called A-large if for every finite H ~ B the intersection n H is A-large. In particular, every A-large family has the sfip. We are going to construct an A-large ultrafilter F. EXERCISE 19.42 (G). Convince yourself that an A-large ultrafilter cannot be a P-point. LEMMA 19.35. If B ~ P(w) is an A-large family, then there exists an A-large ultrafilter F with B ~ F. EXERCISE 19.43 (PG). (a) Show that if B is A-large and c ~ w, then at least one of the families B U {c} and B U {w\c} is A-large. (b) Derive Lemma 19.35 from (a). 0 By Lemma 19.35, the proof of Theorem 19.34 boils down to the following: LEMMA 19.36. There exists an A-large family B ~ P(w) such that for every partition (Cn)nEw of w into pairwise disjoint finite sets there exists b E B such that Ibn Cn I ~ 1 for all b E Band nEw. PROOF. Let {(C~)nEw : Q < 2 No} be an enumeration of all partitions of w into pairwise disjoint finite sets. We construct recursively a sequence (ba,)a<2No of subsets of w such that {b a : Q ~ .B} is A-large and Ib,a n ~I ~ 1 for all .B < 2No and nEw. To see that this construction can be carried out, assume that .B < 2No , and let {b a : Q < .B} be an A-large family. Consider the countable set JP = {s E [W]
Dele = {s E JP: Isnail;::: k}. Let
v=
{Df:Ie: HE [{b a
:
Q
< .B}]
EXERCISE 19.44 (G). (a) Show that V consists of fewer than 2No dense subsets of (JP,2). (b) Show that if G is a V-generic filter and b = UG, then b is as required. MAcountable implies that G as in Exercise 19.44(b) exists, and we are done.
DO
EXERCISE 19.45 (R). Show that MAcauntable implies that there exists a semiQ-point that is not a Q-point. Hint: Fix a paitition A = (ai)iEw of w such that lail = i for all i E w, and call a family B ~ P(w) A-large if for every H E [B]
19.3. ULTRAFILTERS ON w
113
By Theorem 19.31, if t = 2No , then P-points exist. One can improve this result a little and show that the existence of P-points already follows from the equality il = 2No. THEOREM 19.37 (Ketonen). Assume il = 2 No; If C ~ P(w) is a family of cardinality < 2No with the sfip, then there exists a P-point F such that C ~ :F. PROOF. We need a lemma. LEMMA 19.38. Suppose C ~ P(w) is a family of cardinality < il with the sfip, and (ai)iEw is a partition of w into pairwise disjoint infinite sets such that for each H E [C]
n
PROOF. For every H E [C]
DD
EXERCISE 19.47 (PG). Show that if il < 2No , then there exists a filter base C such that ICI < 2No and there is no P-point F that contains C. Hint: Let {fo. : a < il} be a dominating family, and let (ai)iEw be a partition of w into pairwise disjoint infinite sets. For a < il, let Co. = UiEw(ai\fo.(i)). Consider the family {w\ai: iEW}U{Co.: a
(ii) For every w.o. (w,~) there exists a E F such that ot( (a, ~)) = w. (iii) F is a P-point. PROOF. The implication (i) --7 (ii) is obvious. To prove the implication (ii) (iii), assume that F is as in (ii). Let (ai)iEw be a partition of w into pairwise disjoint infinite subsets such that ai (j. F for all i E w. Define a relation ~ on w by: --7
n ~m
iff 3i < j < w (( n E ai t\ m E aj) V (n E ai t\ m E ai t\ n
EXERCISE 19.48 (G). (a) Convince yourself that
(w,~)
~
m)).
is a w.o. of order type
W·W.
(b) Show that if a ~ w is such that infinite for at most one i E w. (c) Conclude that F is a P-point.
(a,~)
has order type w, then
an ai
is
For the proof of the implication (iii) -> (i), assume that F is a P-point and is a 1.0. Let L = {n E w : {m E w : n -< m} E F}. We distinguish three cases. Case 1: L = 0. Then there exists a partition (ai)iEw of w into pairwise disjoint sets such that ai iF for all i E w, and Vi < jVn E aNm E aj (m -< n). (w,~)
19.
114
MARTIN'S AXIOM
Case 2: L =f. 0 and L has a largest element ko. Then there exists a partition (ai)iEw of w into pairwise disjoint sets such that ai ~ F for all i E w, ao = L, and and'iO < i < j'in E ai'im E aj (k o -< m -< n). Case 3: L =f. 0 and L has no largest element. Subcase 3.1: w\L ~ :F. Then there exists a partition (ai)iEW of w into pairwise disjoint sets such that ai ~ F for all i E w, ao = w\L, and'iO < i < j'ik E ao'in E ai'im E aj (n
-< m -<
k).
Subcase 3.2: w\L E :F. Let ao = L. As in Case 1, there exists a partition (ai)iEw\{O} of w\L into pairwise disjoint sets such that ai ¢ F for all i E w\ {O}, and 'i0 < i < j'in E ai'im E aj(m -< n). Since F is a P-point, in each case there exists a E F such that ai is finite for each i E w. Since ultrafilters are assumed nonprincipal in this section, we may choose a in such a way that a n ao = 0.
an
EXERCISE 19.49. Let a be as above. Convince yourself that (a,:::s) has order type w~ in Case 1, Case 2, and Case 3.2, and has order type w in Case 3.1. 0 Now suppose F is a P-point, :::S is a 1.0. relation on w, and a E F is such that (a,:::s) has order type w. Can we assume that the relation :::S agrees with the natural order :$ on a? To make this question more precise, consider the following statement: (i)+ for every 1.0. relation :::S on w, there exists a E F such that • either 'in, mEa (n :::S m - n :$ m); • or 'in, mEa (n :::S m - m :$ n). It turns out that Property (i)+ implies more about F than just being a P-point. LEMMA 19.40. If F is an ultrafilter that satisfies (i)+, then F is selective. PROOF. Let F be an ultrafilter that satisfies (i)+, and let (ai)iEw be a partition of w into pairwise disjoint subsets such that ai rf. F for all i E w. Define a relation :::S on w as follows: n :::S m if and only if 3i
< j < w (( n
E ai 1\ n E aj) V (n E ai 1\ m E ai 1\ m :$ n)).
EXERCISE 19.50 (G). Convince yourself that if a E F satisfies condition (i)+, then la nail :$ 1 for all i E w. Conclude that F is selective. 0 Properties (i), (i)+, and (ii) clearly are partition properties of sorts. Now let us define a collection of partition properties that resemble the ones considered in Chapter 15 even more closely. If k, n E w\ {OJ and F is an ultrafilter, then the symbol w --+ (F)~ is an abbreviation of the following statement: For every coloring c : [w]n --+ m there exists a homogeneous set a E F. EXERCISE 19.51 (G). (a) Show that if F is any ultrafilter and m E w\{O}, then w --+ (F);". (b) Convince yourself that w --+ (F)~ implies that F has Property (i)+. An ultrafilter F is called a Ramsey ultrafilter if w --+
(F)~
holds for all n, m E
w\{O}. It follows from Exercise 19.51(b) and Lemma 19.40 that every Ramsey ultrafilter is selective. The converse is also true.
19.3.
ULTRAFILTERS
ON
115
W
LEMMA 19.41. Every selective ultrafilter is a Ramsey ultrafilter. PROOF. Let F be a selective ultrafilter, and let m E w\{O} be fixed. We prove by induction over n that w --+ (F)~. For n = 1 the latter is true by Exercise 19.51(a). Now let us assume w --+ (F)~ and prove that w --+ (F)~+l holds as well. Toward this end, let c : [w]n+l --+ m be an arbitrary coloring of all (n + I)-tuples of natural numbers with m colors. For i E w, we define an induced coloring Ci : [w]n --+ m by letting Ci(X) = c( {i} U x) if x E [w\(i + 1)]n and Ci(X) = 0 if minx:::; i. By the inductive assumption, for each i E w we can find Ai E F such that ci[[Ai]n] = {t'i} for some t'i < m. Since F is a nonprincipal ultrafilter, we can choose the Ai's in such a way that min Ai > i. Now suppose for a moment that we can find b E F such that (diag)
Vi,j E b(i
<j
--+
j E Ai).
EXERCISE 19.52 (G). Suppose b satisfies (diag), and x E [b]n+l. Show that c(x) = t'minx.
bl
If b E F satisfies (diag), then there exists t' < m such that bl E F, where By Exercise 19.52, a = bl is homogeneous for c. It is not hard to find an infinite b such that (diag) holds. But can we find such
= {j E b : t'j = t'}.
a b in:F? Let us call an ultrafilter quasi-normal if for every sequence (Ai)iEW of elements of F there exists b E F such that (diag) holds. The following sublemma provides the missing link in our proof: SUBLEMMA 19.42. Let F be a selective ultrafilter. Then F is quasi-normal. PROOF. Assume F is a selective ultrafilter, and let (A)iEW be a sequence of elements of F. We define a decreasing sequence (bn)nEw of elements of F by letting bo = w, and bn = ni
f(O)
= 0;
f(n
+ 1) =
+ 1. {[f (n), f (n + 1)) :
max{J(n),min(a\bf(n))}
Note that the function f is strictly increasing. Thus, nEw} is a partition of w into pairwise disjoint finite sets. Since F is selective, there exists C E F such that for all nEw, there is at most one mE C with f(n) :::; m < f(n+ 1). Enumerate an C in increasing order as (jm)mEw' Let
do = {hm : mEw},
d 1 = {j2m+1 : mEw}.
There is t' < 2 such that dl E F. We fix this t', and let b = dl. It remains to show that b is as required. So suppose i, j E b are such that i < j. Then there exists nEw such that i < f(n) < f(n+ 1) :::; j. Note that the definition of f implies that if j E a is such that j ~ f(n+l), then j > min(a\bf(n)), and hence j E bf(n). Since bf(n) ~ Ai for each i < f(n), we conclude that j E Ai, and hence b is an element of F that witnesses (diag). DD Let us summarize our findings.
19. MARTIN'S AXIOM
116
19.43. Let F be an ultrafilter. Then the following are equivalent: F is selective. F is Ramsey. w --+ (F)~. F is quasi-normal.
THEOREM
(a) (b)
(c) (d)
Mathographical Remarks
The most comprehensive collection of applications of Martin's Axiom in print at the time of this writing is D. Fremlin's monograph Consequences oEMartin's Axiom, Cambridge Tracts in Mathematics 84, Cambridge University Press, 1984. If you are interested in something less comprehensive, we recommend the following articles: W. Weiss, Versions of Martin's Axiom, In: Handbook of Set-Theoretic Topology, K. Kunen and J. E. Vaughan, eds., North-Holland, 1984,827-886; K. Kunen and F. D. Tall, Between Martin's Axiom and Souslin's Hypothesis, Fundamenta Mathematicae 102 (1979), 173-181. The converse of Theorem 19.16 is also true. In their paper Martin's axiom and partitions, Compositio Mathematica 63 (1987), 391-408, S. Todorcevic and B. Velickovic have shown that for every K, MA(K) is equivalent to the statement: Every c.c.c. partial order of size less than or equal to K is a-centered. Theorem 19.24 represents the current state of knowledge about inequalities between the cardinal invariants mentioned in it. It has been shown that each of the following inequalities is relatively consistent with ZFC: t < min{5, b}, 5 < b, b < 5, max{ 5, b} < il, b < a, a < il, a < 5. It is still open whether the inequalities il < a and p < t are consistent. Concerning the latter, Todorcevic and Velickovic have shown in the paper mentioned above that if p = N1 , then p = t. Neither the existence of P-points nor the existence of Q-points can be proved in ZFC alone. The former was shown by S. Shelah. For an exposition of this result, see E. L. Wimmers, The Shelah P-point Independence Theorem, Israel Journal of Mathematics 43(1) (1982), 28-48. The latter result was proved in A. Miller, There are no Q-points in Laver's model for the Borel Conjecture, Proceedings AMS 78(1) (1980), 103-106. It is still an open question whether there is a model of ZFC in which neither P-points nor Q-points exist.
CHAPTER 20
Hausdorff Gaps After reading the previous chapter you may be under the impression that MA implies that every conceivable cardinal invariant of the continuum is equal to 2No. Things are not quite that simple. In the present chapter, we are going to discuss a cardinal invariant of the continuum that is equal to Nl-no matter how large 2No is and whether or not Martin's Axiom holds. Let 0. = (ae : ~ < "') and b = (b.,., : '" < 'x) be sequences of subsets of w. We say that (0., b) is a ("" ,X*)-pregap if aeo c* ael c* b""l c* b.,.,o for all €o < 6 < '" and < < ,x. A ("" ,X*)-pregap is called a ("" ,X*)-gap if there is no c E P(w) such that ae c* c c* b.,., for all ~ < '" and", < ,x. We have already encountered gaps of one sort: A ,X-tower is the same thing as a (1, ,X*)-gap with ao = 0. Thus, one could redefine t as the smallest ,X such that there exists a (1, ,X*)-gap. The next interesting case to look at is the situation where '" = w. Define:
"'0 "'1
t:b = min {oX: there exists an (w,'x *)-gap }. It turns out that the cardinal t:b is an old acquaintance of ours. THEOREM 20.1 (Rothberger). t:b = b. PROOF. Let ((an: nEw), (b(: « ,X)) be an (w,,X*)-pregap. For each «'x, define a function 1( : w -+ w by letting ldn) = max(an \bd+1, where the maximum of the empty set is arbitrarily defined as O. Suppose that ,X < b. Then there exists 9 E Ww such that 1( <* 9 for all ( < 'x. Define:
c = U(an \g(n)). Clearly, an c* c for all nEw. Moreover, if ( < 'x, then an \g(n) ~ b( for all but finitely many nEw, and it follows that c c* b(. Thus ((an: nEw), (b( : ( < 'x)) is not a gap, and we have shown that t:b ~ b. To prove the inequality t:b :=:; b, let us construct an (w, b*)-gap. Fix a sequence (dk)kEW of pairwise disjoint infinite subsets of w, and an unbounded sequence (Jd«b of elements of Ww such that 1(0 <* k for all (0 < (1 < b. Such a sequence exists by Exercise 17.18(b). Now let an = Uk
20.
ll8
HAUSDORFF GAPS
THEOREM 20.2 (Hausdorff). There exists an (WI,wn-gap. In honor of the author of Theorem 20.2, (WI, wi)-gaps are often called Hausdorff gaps. I Before we prove Theorem 20.2, let us derive two important consequences of it. CLAIM 20.3. Let ((a~ : ~ < WI), (b~ : ~ < WI)) be a Hausdorff gap. For~ < WI, let J~ = {c E P{w) : a~ ~* c ~* bd. Then (J~ : ~ < WI) is a decreasing sequence of nonempty Fa-subsets ofP{w) with n~<wl J~ = 0. PROOF. Recall that the standard topology on P{w) is defined by identifying P{w) via characteristic functions with the Tychonoff product of W copies of the discrete two-element space {O, I}. For every ~ < Wl and nEw, the set K~,n = {c ~ W : a~\n ~ c\n ~ b~\n} is closed, and J~ = UnEwK~,n. Thus, each J~ is an Fa-set. Clearly, J~ =f 0 and J~ ::::> JTJ for all ~ < 'r/ < WI. If c E n~<wl J~, then a~ c* c c" b~ for all ~ < WI, which is ruled out by the assumption that ((a~ : ~ < WI), (b~ : ~ < WI)) is a gap. 0 Claim 20.3 is akin to Lemma 17.10, but the latter uses the extra assumption b = t{I, while the former is a theorem of ZFC. Theorem 20.2, Claim 20.3, and Lemma 17.9 together show that uncountable, perfectly meager subsets of P{w) can be constructed using only the axioms of ZFC. More precisely, we have the following:
set
COROLLARY 20.4. For any Hausdorff gap ((a~ : ~ < WI), (b~ : ~ < WI)), the : ~ < wd U {b~ : ~ < wd is a perfectly meager subset ofP{w).
{a~
Corollary 20.4 has a measure-theoretic dual: {a~
LEMMA 20.5. For any Hausdorff gap ((a~ : ~ < WI)' (b~ : ~ < WI))' the set ~ < wd u {b~ : ~ < wd has universal measure zero.
:
PROOF. Let ((a~ : ~ < WI), (b~ : ~ < WI)) be a Hausdorff gap, and let J.L be a Borel measure on P{w) that vanishes on the singletons. For ~ < WI, let J~ be defined as in Claim 20.3. Since each J~ is J.L-measurable and contains all but countably many elements of {a~ : ~ < wd u {b~ : ~ < wd, it suffices to show that J.L(J~) = 0 for sufficiently large ~. Suppose towards a contradiction that this is not the case. Since the J~ 's form a decreasing transfinite sequence, there must be ~o < WI and co > 0 such that J.L(J~) = co for all ~ > ~o. Define: c = {n E w: J.L{{d E J~o : nEd}) ~ co/2}, and let
6 > ~o be such that c fj.
J~I
.
EXERCISE 20.2 (G). Let c, 6 be as above. Show that in the definition of c, the subscript ~o can be replaced by 6, i.e., show that c = {n E w : J.L( {d E J~I : nEd}) ~ co/2}. As in the proof of Claim 20.3, for nEw, let K~I ,n
= {d ~ w : a~1 \n ~ d\n ~
b~1 \n}.
The K~I,n's form an increasing sequence of J.L-measurable sets with union J~I' Hence, there must exist no E w such that J.L(K~I,no) > co/2. Fix such no. Since c ¢ K~I,no, there must exist nI > no such that either ni E c\b6 or nI E a~1 \c. In the former case, by Exercise 20.2, the set L = {d E J6 : ni E d} has measure lSome authors reserve the term Hausdorff gap only for (Wl,wi}-gaps with special properties (like Property (iv) in the proof of Theorem 20.6).
20.
HAUSDORFF GAPS
119
at least eo/2 and is disjoint from K~l,no' which is impossible, since J.L(K~l U L) ~ eo· In the latter case, the set M = {d E J~l : ni ¢ d} has measure at least eo/2 and is disjoint from K~l ,no, which again leads to a contradiction. 0
J.L(J~1) =
We shall prove a version of Theorem 20.2 that differs from the one originally stated. Let A, B ~ [W]N o • We say that A and B are almost disjoint, and write A .1 B, if la n bl < ~o for all a E A and b E B. We say that A and B can be separated if there exists d ~ W such that a ~* d and Ib n dl < ~o for every a E A and b E B. Of course, if A and B can be separated, then A and B are almost disjoint. However, the converse is not true. EXERCISE 20.3 (G). Let ((a~ : ~ < K.), (bry : TJ < >.)) be a (K., >.*)-pregap, let < >., and let A = {a~ : ~ < K.}, B = {Cry: TJ < >.}. (a) Convince yourself that A .1 B. (b) Show that A and B can be separated if and only if ((a~ : ~ < K.), (b.,., : TJ < >.)) is not a (K.,>.*)-gap.
Cry
= w\b.,., for all TJ
The above exercise implies that Theorem 20.2 is equivalent to the following statement: THEOREM 20.6. There exist transfinite sequences (a~ : ~ < WI) and (b~ : ~ < of infinite subsets of W such that a~ c* ae: and b~ c· be: for all ~ < ( < WI; and the families {a~ : ~ < wt} and {b~ : ~ < wt} are almost disjoint, but cannot be separated.
WI)
PROOF. We shall construct sequences (a~ : ~ < WI) and (b~ : ~ < WI) simultaneously by recursion over WI in such a way that for all ~ < ( < WI: (i) Iw\(a~ U b~)1 = No; (ii) a~ c* ae: and b~ c* be:; (iii) a~ n b~ = 0.
Condition (i) is a technical requirement that allows us to keep going. Conditions (ii) and (iii) together not only ensure that our sequences are increasing in the sense of c*, but also that the families {a~ : ~ < wt} and {b~ : ~ < wt} will be almost disjoint. In order to ensure that these families cannot be separated, Hausdorff devised an ingeneous trick. Let us say that b ~ W is close to A ~ P(w) if for every nEw the set {a E A : a n b ~ n} is finite. EXERCISE 20.4 (G). (a) Suppose that b is close to a set A and b ~* d ~ w. Convince yourself that d is also close to A. (b) Suppose that b is close to a set A and B ~ A. Convince yourself that b is also close to B. We are going to require that for every ( < WI: (iv) ~
be: is close to {a~ : ~ < (}.
If condition (iv) holds for all ~ < ( < WI, then the sets {a~ : ~ < WI} and {b~ : < WI} cannot be separated. This is a consequence of the following observation.
LEMMA 20.7. Let {a~ : ~ < wt} and {b~ : ~ < WI} be separated families of infinite subsets of WI such that a~ n b~ = 0 for all ~ < WI. Then there exists an EX. uncountable set X ~ WI such that (a~ n be:) U (ae: n b~) = 0 for all
e, (
120
20. HAUSDORFF GAPS
PROOF. Let d ~ W be such that a~ ~* d and Ib{ndl < ~o for all ~ < WI. By the Pigeonhole Principle, there exists nEw such that the set Xn = {~< WI : (a~\d) U (b~nd) ~ n} is uncountable. Moreover, using the Pigeonhole Principle again we can see that there are s, t ~ n such that the set X = {~E Xn : a~\d = s /\b~ nd = t} is uncountable. It is straightforward to verify that the above set X is as required. D EXERCISE 20.5 (G). Use Lemma 20.7 to show that if condition (iv) holds for all ~ < ( < WI, then the sets {a~ : ~ < WI} and {b~ : ~ < WI} cannot be separated. Hint: For X as in the lemma, consider the (w + l)st element ( of X and derive a contradiction. Having all ideas in place, let us now describe the construction itself. To get started, let {Co, Cb C2} be a partition of W into three disjoint infinite subsets. Define: ao=co, bO=Cl. Now suppose 0 < ( < Wb and for all ~ < ( the sets a~, b~ have been defined in such a way that conditions (i)-(iv) hold. If ( = 1/ + 1 for some 1/, then we choose a partition {do, d 1, d2 } of w\(ary U bry) into pairwise disjoint infinite sets and define:
a( = ary U do,
b( = bry U d 1 .
A straightforward argument shows that in this case, conditions (i)-(iv) continue to hold at stage (. Now let us consider the case where ( is a limit ordinal. Let us begin by choosing sets x ~ wand d ~ x such that Iw\xl = ~o, a~ ~* d, and b~ ~* x\d for all ~ < WI' This is possible by Theorem 20.1. Think of (x\d, d) as a first approximation to (a(, be:). This approximation satisfies conditions (i)-(iii), but not necessarily (iv). To see to what extent (iv) fails, fix a strictly increasing sequence of ordinals (1/n)nEw such that 1/0 = 0 and ( = SUP{1/n : nEw}, and define for each nEw:
Bn = {1/ : 1/n :::; 1/ < 1/n+1 /\ d n ary ~ n}. Since (iv) holds for every ( = 1/n+1 and since bryn+l ~* d, each of the sets Bn is finite. Thus, if we define B = U nEw Bn, then B either is finite, or has order type w. Let us consider these two cases separately. Case 1: B is finite. Then for all but finitely many n the set = {~ < ( : d n a~ ~ n} is contained in 1/n. Since bryn ~* d and, by the inductive assumption, bryn is close to {a~ : ~ < 1/n}, so is d. Therefore, the set {~ < ( : d n a~ ~ n} is finite. Since this argument applies to all but finitely many nEw, d is close to {a~ : ~ < (l. Case 2: B has order type w. In this case, d may fail to be close to {a~ : ~ < (}. However, d misses the mark of being close just barely.
en
EXERCISE 20.6 (G). Show that d is close to the set {a~ : ~ E (\B}. Let (~n)nEw be an enumeration of B in increasing order. For every nEw, pick Pn E x n a~n \ Ui
b(=dU{Pn:nEw},
. a(=x\b(.
EXERCISE 20.7 (PG). Convince yourself that the pair (a(, be:) satisfies conditions (i)-(iv). Hint: For the proof of (iv), use the results of Exercises 20.4 and 20.6. D
20.
HAUSDORFF GAPS
121
The technique of the proof of Theorem 20.6 can also be used to construct another interesting object. Let A = {a€ : ~ < WI} be an almost disjoint family of infinite subsets of w. We say that A is a Luzin gap if t.here is no partition {B, C} of A into two disjoint uncountable sets such that B can be separated from C. THEOREM 20.8. There exists a Luzin gap. PROOF. We are going to construct recursively a sequence (a€)€<Wl of infinite subsets of W such that for all 0 < ( < w:
(1) (2)
V~
< (Ia€ na(1 < l{o;
a( is close to {a€: ~ <
O.
The construction is somewhat easier than the one in the proof of Theorem 20.6. We let (an)nEw be a sequence of pairwise disjoint infinite subsets of w. At stage ( ~ w, we arrange ( into a sequence (~k)kEW and pick for every k E W a number Pk E a€k \ Ui
«
lt remains to show that the set {a€ : ~
< WI} constructed above is a Luzin gap. Clearly, this set consists of pairwise almost disjoint infinite subsets of w. Let B, C be two disjoint uncountable subsets of A (we do not even need to require that B U C = A). Suppose towards a contradiction that d ~ W is such that a€ ~* d for all a€ E Band laT) n dl < l{o for all aT) E C. By the Pigeonhole Principle, there exists nEw such that the set Bn = {a€ E B : a€ \ d ~ n} is uncountable. Since C is uncountable, we can find a( E C such that the set B~ = {a€ E Bn : ~ < 0 is infinite. Fix such ( and let mEw be such that a( n d ~ m. Then for all a€ E B~ we have a€ n a( ~ max{ n, m}. It follows that a( is not close to the set B~. By Exercise 20.4(b), the latter implies that a( is not 0 close to {a€ : ~ < 0, which contradicts (2). Let us briefly discuss a somewhat different technique for constructing (WI, wi)gaps. Let ((a€ : ~ < WI), (b€ : ~ < WI)) be an (WI,wi)-pregap. For ~ < ( < WI, consider the following property: (v) (a€ n bd U (b€ n ad =1= 0. EXERCISE 20.9 (PG). (a) Show that if ((a€ : ~ < Wl), (b€ : ~ < WI)) satisfies condition (iii) (from the proof of Theorem 20.6) for all ~ < W1 and there is an uncountable Y ~ Wl such that (v) holds for all {~, 0 E [Y]2, then it is an (W1' wi)gap. Hint: Use Lemma 20.7. (b) Show that a Hausdorff gap that satisfies conditions (i)-(iii) (from the proof of Theorem 20.6) and (v) can be constructed by a slight modification of the proof of Theorem 20.6. In a sense, the construction of a Hausdorff gap outlined in Exercise 20.9 is the only way to construct such objects in ZFC. THEOREM 2Q.9. Assume the Open Coloring Axiom, and assume that G = ((a€: ~ < WI), (b€: ~ < WI)) satisfies conditions (i)-(iii) stated in the proof of Theorem 20.6 for all ~ < ( < WI. Then the families {a€ : ~ < WI} and {b€ : ~ < WI} cannot be separated if and only if there exists an uncountable subset Y of WI such that condition (v) holds for all {~, 0 E [Y]2.
20.
122
HAUSDORFF GAPS
e
PROOF. Let G be as in the assumptions, and let X be the set { (ae, be) : < WI} with the topology obtained by treating X as a subspace of P(w) x P(w). Consider the following partition {Po, PI} of [xj2:
Po = {{ (ae, be), (a(, b()} E [X]2 : (ae n bd U (a( n be)
# 0},
PI = [X]2\PO.
EXERCISE 20.10 (G). Recall the definition of an open partition from Chapter 15, and show that the partition {Po, PI} defined above is open. By the OCA, either there is an uncountable set Y such that [Yj2 ~ Po, or X = UnEw X n , where [Xnj2 ~ PI for all nEw. In the former case, condition (v) holds for all {e,(} E [Yj2, and by Exercise 20.9(a), G is a gap. In the latter case, at least one of the Xn's must be uncountable. Without loss of generality we may assume that Xo has this property.· Let d = UeEx o ae. This d separates {ae : < wI} and {be: < wI}.
e
e
EXERCISE 20.11 (G). Show that if d is as above, then ae for all < WI.
e
~. d
and Idnbel < No 0
MathograI>hical Remarks If you would like to learn more about gaps, we recommend the survey article Gaps in w w, by Marion Scheepers, in Israel Mathematical Conference Proceedings 6 (1993), AMS, 439-561.
CHAPTER 21
Closed Unbounded Sets and Stationary Sets 21.1. Closed unbounded and stationary sets of ordinals Let, be a limit ordinal. A set C S;;; , is bounded in , if there exists {3 < , such that C S;;; {3. Otherwise C is unbounded in ,. C is closed in , if Q E C for all limit ordinals Q such that 0 < Q < , and C n Q is unbounded in Q. C is club i in, if C is closed and unbounded in ,. EXERCISE 21.1 (G). Show that C is closed in , iff C is closed in the sense of the order topology on ,. EXAMPLE 21.1. If {3 < , then the set ({3,,) = {Q : {3 < Q < ,} is closed and unbounded in " but, if, is a limit ordinal, then this set is not closed in any ordinal Q > ,. If cfb) = w, then every cofinal subset of order type w is club in ,. The set, n LIM of limit ordinals smaller than, is closed in any ordinal " and, if cfb) > w, then this set is also unbounded in ,. EXERCISE 21. 2 (G). Find limit ordinals " 0 of countable cofinality that are not cardinals so that LIM n, is unbounded in, and LIM n 0 is bounded in o.
Let, E ON. We define: CLUBb)
= {A S;;;,: :3C S;;; A (C is club in ,)}.
EXAMPLE 21.2. If A = w U {{3 < WI : w < {3 and {3 is a limit ordinal}, then A E CLU B(wI), but A is not club since w S;;; A but sup(w) = w ¢ A. EXAMPLE 21.3. Let K be a cardinal of uncountable cofinality, and suppose A is an unbounded subset of K. We let
A'
= {Q < K
:
Q = sup( A n Q)}.
We call A' the derived set of A. Note that if we consider K with the order topology, then this notion coincides with the concept of derived set used by topologists. EXERCISE
in
21.3 (G). (a) Show that A' is club in
K
whenever A is unbounded
K.
(b) Show that if A is club, then A' S;;; A. EXERCISE 21.4 (PG). Let K be an infinite cardinal such that Cf(K) > ~o. Let ). denote Cf(K), and let {co: Q < ).} and {do: Q < ).} be club subsets of K with Co < C{3 and do < d{3 for Q < {3 < ).. Show that the set C = {co: Q < ). 1\ Co = do} is club. Hint: To show that C is unbounded, let {3 < K and define recursively a sequence (Qn)nEw by letting: Qo = min{Q : Co ~ {3}, Q2n+l = min{Q : Q ~ Q2n 1\ do ~ C02n } ; Q2n+2 = min{Q: Q ~ Q2n+ll\co ~ d o2n +J· lSome authors write "cub." Remark 21.29.
The reason why the word club is italic here will appear in
123
124
21. CLOSED UNBOUNDED SETS AND STATIONARY SETS
EXAMPLE 21.4. Let", be a cardinal of uncountable cofinality, and let "'. We say that I is normal if it satisfies the following conditions: (i) If a < 13 < "', then I(a) :S l(f3)j (ii) If I' < '" is a limit ordinal, then Ih) = sup{J(a) : a < I'}j (iii) I(a) ~ a for all a < "'.
I: '" ~
Note that (i) says that I is nondecreasing, whereas (ii) means that I is continuous in the order topology. Note also that if I is strictly increasing, then (iii) is automatically satisfied. EXERCISE 21.5 (G). Suppose F ~ It", is a set. of normal functions such that for all a < "', sup{J(a) : IE F} < '" exists and is smaller than "'. Show that the function 9 : '" ~ '" defined by g(a) = sup{J(a): IE F} is also normal. LEMMA 21.1. II", is a cardinal 01 uncountable cofinality and I E It '" is normal, then the set Fix(f) = {a < "': I(a) = a} is club in "'. PROOF. In order to see that Fix(f) is closed, either recall that the set of fixed points of any continuous function of a Hausdorff space into itself is closed (see Theorem 26.1), or note that if the set {a : I(a) = a} is cofinal in 13, then sup{J(a) : a < f3} = sup{a : a < f3} = 13· We show that Fix(f) is unbounded: Let 0 < "'. We want to show that there exists some a E Fix(f) with 0 < a. To this end, we construct recursively a sequence (f3n)nEw as follows: 130 = o. Given f3n, let f3n+l > l(f3n). This construction yields an increasing sequence with
0=130 :S 1(130) < 131 :S 1(131) < ... < f3n :S l(f3n) < f3n+l < .... It follows that if a w} = a.
= sup {f3n:
nEw}, then a
> 0 and I(a) = sup {J(f3n):
n E
0
EXERCISE 21.6 (PG). Let D = [0, l)nQ, let order on D x WI defined by:
~a
denote the antilexicographical
(d,a) :Sa (e,f3) iff a < 13 V (a = 13/\ d:SQ e) (for the natural orders :SQ on Q and :S on WI respectively), and let A, B ~ WI \ {a}. Prove that ((D x wd\( {a} x A), ~a) is order-isomorphic to ((D x WI)\( {a} x B), ~a) if and only if there exists a club C ~ WI such that An C = B n C. EXAMPLE 21.5. Let", be a regular uncountable cardinal, let nEw \ {O}, and let 9 be any function from ",n into "'. Then the set
Mg
=
{a < "': rng(gfa n )
~
a}
is club in "'. To see this, consider the function Ig : '" ~ '" defined by Ig (a) = sup ( {g(jJ) + 1 : jJ E an} U {a}). Note that Ig is a normal function. Moreover, Ig has been defined so that Fix(f) = Mg. We shall refer to Mg as the set of ordinals a that are closed under g. Note that this use of the word "closed" has nothin~ to do with the order topology. In a certain sense, Examples 21.4 and 21.5 are the only examples of club sets. THEOREM 21.2. II I' is an ordinal and C is club in 1', then there exists a ' normallunction I : I' ~ I' such that C = Fix (f) .
21.1. CLOSED UNBOUNDED AND STATIONARY SETS OF ORDINALS
125
PROOF. Let C be club in 'Y. Define a function f : 'Y --+ 'Y by f(a) = min {C\a}. A straightforward verification shows that f is as required. 0 EXERCISE 21.7 (PG). (a) Show that for every club C ~ function g : WI --+ WI such that C = Mg. (b) Let C = [WI,W2). Then C is club in W2. Show that if g: nEw \ {O}, then C =I- Mg.
WI
there exists a
w2
--+ W2
for some
THEOREM 21.3. Let'Y be an ordinal of uncountable cofinality A. If K < A and {Co: : a < K} is a family of club subsets of 'Y, then the intersection C
=
nCo:
is club in 'Y.
PROOF. For each a < K, let fo: : 'Y --+ 'Y be a normal function such that Fix(fo:). For (3 < 'Y, let f((3) = sup{Jo:((3) : a < K}. Since K < cf("(), f((3) < 'Y for each (3. By Exercise 21.5, f is a normal function from 'Y into 'Y. Now the theorem follows from the next exercise. Co:
=
EXERCISE 21.8 (G). Convince yourself that Fix(f)
= no:<", Co:.
0
COROLLARY 21.4. Let'Y E ON and A = cf("() A-complete filter.
> w. Then CLUB(,,() is a
PROOF. Let K < A, and suppose {Ao: : a < choose a club set Co: ~ Ao:. Then
CLUB(,,(). For each a <
K} ~
Thus no:<",Ao: contains a club, i.e., no:<",Ao: E CLUB(,,().
K,
o
COROLLARY 21.5. If A is a regular uncountable cardinal, v < A, and G = {go: : a < v} is a family of fewer than A functions, where go: maps An(o:) into A for some n(a) E W \ {O}, then Me = no:
126
21.
CLOSED UNBOUNDED SETS AND STATIONARY SETS
THEOREM 21. 7. Let K. > W be a regular cardinal, and let (Co)o<1t be a sequence of club subsets of K.. Then the diagonal intersection D.o
K.,
choose a normal function by:
fa :
K. -+ K.
such that
-+ K.
f(o) = sup{f/3(o) : {3 < o}. By regularity of K.,
f is well defined.
EXERCISE 21.10 (G). (a) Show that f is normal. (b) Convince yourself that f(o) = 0 if and only if V{3 < conclude that Fix(f) = D.o
0
(1/3(0) =
0),
and 0
Now we are going to discuss stationary sets. DEFINITION 21.8. Let 'Y be an ordinal of uncountable cofinality. A set 8 ~ 'Y is called stationary in 'Y (or just stationary if 'Y is implied by the context) if VA E CLUBb) (8 n A =I 0). Thus, 8 is stationary in 'Y if and only if 8 ~ C LUB* ('Y), where C LUB* ('Y) = {T ~ 'Y : 'Y \ T E CLUBb)}· Note that every stationary set must be cofinal in 'Y. EXAMPLE 21.8. Let 8 = {{3 < W2 : cf({3) = w}. We show that 8 is stationary in W2. Let A be club in W2, and let (a O)O<W2 be an enumeration of A in strictly increasing order. Then cf(aw ) = w; thus aw E 8 and 8 n A =I 0. But 8 is not closed: Let (SO)O<W2 be a strictly increasing enumeration of 8, and let 'Y = sup{so : 0 < wd· Then cfb) = WI, and thus 'Y ~ 8. This example can be generalized as follows: Let 'Y be a limit ordinal with K. = cf(K.) < >. = cfb)· Then the set 8 = {{3 < 'Y : cf({J) = K.} is stationary in 'Y.
w
S
o <
COROLLARY 21.9. If K. is weakly inaccessible, then there exists a family {8o : K.} of K. pairwise disjoint stationary subsets of K..
We shall see in Chapter 23 (Corollary 23.4) that the analogous result for successor cardinals also holds. LEMMA 21.10. Let'Y EON, K. < >. = cfb), and let {Ao : 0 < K.} ~ Pb) be such that 8 = Uo<1t Ao is stationary. Then there exists an 0 < K. such that Ao is stationary. EXERCISE 21.11 (G). Prove Lemma 21.10. Now we are ready to prove Lemma 14.10. LEMMA 21.11. Let K. be a regular uncountable cardinal, and let N 81t denote the family of nonstationary subsets of K.. Then N81t is an ideal on K. with the following properties: (i) VA E N81t 3B E [K.]1t (A n B = 0/\ A u BE NSIt ); (ii) If A E [NSIt]
21.1. CLOSED UNBOUNDED AND STATIONARY SETS OF ORDINALS
127
set B = {.8 E K, : 3a < .8 (.8 E Aon. The set B is often called the diagonal union of the sequence (Ao )o<1t and is denoted by 'V o
THEOREM 21.12. Let K, be a regular uncountable cardinal, 8 a stationary subset of K" and f : 8 ~ K, a regressive function. Then there exists a < K, such that f- 1 {a} is stationary in K,. PROOF. Assume towards a contradiction that for each a < K, there exists a club set Co in K, such that Co n f-l{a} = 0. Let D = tlo
an.
°
Let A = {A{3 : .8 < a} be a family of pairwise disjoint sets. A transversal for A is a set X S;;; UA such that IX n A{31 = 1 for all .8 < a. Recall from Chapter 17 that a family B of subsets of an infinite set B is almost disjoint if IXI = IBI and IX n YI < IBI for all {X, Y} E [BF. THEOREM 21.14. Let K, be an infinite cardinal, A = {Ao : a < K,+} a family of pairwise disjoint sets with IAol = K, for each a < K,+. Then there exists a maximal almost disjoint family F of transversals for A with IFI = K,+ . PROOF. The Ao's are pairwise disjoint and have cardinality K,. So we can assume that Ao = {(a,.8) : .8 < max{K"a}}. Thus each transversal for A can be treated as a function from K,+ to K,+ that is regressive on K,+\K,. For .8 < K,+ define:
F{3
= {(a, 0) : as.8} U {(a,.8):
.8 < a < K,+}.
21. CLOSED UNBOUNDED SETS AND STATIONARY SETS
128
Then the family F defined as: F= {Ff3: (3 < K+}
is an almost disjoint family of transversals for A with IFI = K+. Let us show that F is maximal. Let G be a transversal for A such that G £f. F. Then G is a regressive function on K+\K, and Fodor's Lemma implies that there exists'Y < K+ such that G- I h} is stationary. The latter implies that IGnF')' I = K+, and hence {G} U F is not an almost disjoint family. 0 EXERCISE 21.12 (PC). Let K be a regular uncountable cardinal, let 8 ~ K be stationary, and let {8a. : a < K} be a family of nonstationary, pairwise disjoint subsets of K such that 8 = Ua.<1< 8a.. Use Fodor's Lemma to show that {min8a. : a < K} is stationary in K. Hint: Show that the set 8\ {min 8a. : a < K} is not stationary. . EXERCISE 21.13 (C). Let 8 be a stationary subset of WI. Show that the set 8 n 8' = {a E 8 : sup(8 n a) = a} is stationary in WI. If 8 is a stationary subset of WI, then 8 does not need to contain a dub, i.e., a closed subset of order type WI, but 8 does contain closed subsets of order type a for every a < WI. Even more is true.
THEOREM 21.15. Let 8 be a stationary subset B(8)
= {a : V'Y < a:3 A ~ (8 n a)\'Y (A
O/WI'
Then the set
is closed in a!\ ot(A)
= a)}
is club in WI . PROOF. Let 8 be a stationary subset of WI, and suppose towards a contradiction that B(8) is nonstationary. Let 8 1 = 8\B(8), and define 8 2 = 8 1 n 8~. By Lemma 21.10, 8 1 is stationary, and Exercise 21.13 implies that 8 2 is also stationary. Define a function / : 82 ---+ WI by: /(a) = sup{(3 : V'Y < a:3 A ~ a\'Y (ot(A) = (3 !\ sup A = a)}. Clearly, /(a) ::; a for all a E 8 2. Thus 8 2 = 8 3 U 8 4 , where 8 3 = {a E 8 2 : /(a) = a} and 8 4 = {a E 8 2 : /(a) < a}. The following claim implies that 8 2 is not stationary in WI and thus leads to a contradiction. CLAIM 21.16. Neither 8 3 nor 8 4 is stationary. PROOF. By contradiction. Suppose 8 3 is stationary. By Exercise 21.13, there exists a strictly increasing sequence (an )nEw of elements of 8 3 such that sUPnEw an E 8 3 . Pick such a sequence and let a w = sUPnEw an. Let 'Y < Ow, and let no be such that'Y < a no ' For each n E w\no, choose a set An ~ (an+! \an ) n 8 that is closed in an and of order type an. The the set A = UnEw\no An U {an: nEw} is closed in aw, and ot(A) = aw. Thus a w E B(8), and hence Ow £f. 8 3 . Now suppose 8 4 is stationary. Since the restriction of / to 8 4 is regressive, by the Pressing Down Lemma there exist a stationary 8 5 ~ 8 4 and a countable ordinal (3 such that /(a) = (3 for all a E 8 5 , Choose a strictly increasing sequence (an)nEw of elements of 8 4 such that if a w = sUPnEw an, then a w E 8 4 . Let'Y < aw, and proceed as above to produce a closed subset A of a w \'Y of order type ((3+1)·w > (3 such that A is cofinal in a w • 00 THEOREM 21.17. Consider X = WI X (WI + 1) as a topological space with the product 0/ the order topologies on WI and WI + 1. Then X is not normal.
21.1.
CLOSED UNBOUNDED AND STATIONARY SETS OF ORDINALS
129
PROOF. Let
A = {(a, a) : a <
wd,
B = {(a,Wl) : a < wd.
It is not hard to see that A and B are disjoint closed subsets of X. Let U be an open neighborhood of A. Then for every f3 such that 0 < f3 < WI there exists an ordinal f(f3) < f3 such that (f(f3), f3 + 1) x (f(f3), f3 + 1) ~ U. By Fodor's Lemma there exist f30 < WI and a stationary set S ~ WI such that f ('Y) = f30 for all 'Y E S. Since S is unbounded in WI, it follows that
C = (f30, wI) x (f30, WI) ~
u.
Now let V be an open neighborhood of B. There exists an a such that f30 < a < WI and (f3o + 1, a) E V. The latter implies that Un V =f=. 0, and hence X is not oo~.
0
EXERCISE 21.14 (PG). Show with the help of Fodor's Lemma that for every continuous f : WI -+ IR there exist a real number r and an ordinal a < WI such that f(f3) = r for all f3 E (a,wl). Hint: As a first step, show that there exist an integer n and an ao < WI such that f(f3) E (n - 1, n + 1) for all f3 E (ao, WI). Our last application of Fodor's Lemma is the proof of a famous theorem of
J. H. Silver. This theorem was already mentioned in the Mathographical Remarks at the end of Chapter 11. Silver's original proof used model-theoretic methods that are beyond the scope of this book. THEOREM 21.18. If f3 is a limit ordinal of uncountable cofinality such that f3 < NJ3 , and if there exists a stationary subset S ~ f3 such that 2N", = NO +1 for all a E S, then 2N/3 = NJ3H . PROOF. We give the proof for the special case f3 = WI = S and leave the general case as an exercise. So let us assume that Va < WI (2N", = NoH)' First we code every subset of NWI by an wI-sequence of ordinals. To this end, choose for each a < WI an injection rro : P (No) -+ NoH' A set A ~ NWI is coded by the sequence rr(A) = (rro(AnNo))o<Wl' FACT 21.19. If A, B ~ Nwl' then either A = B, or the set {a < WI : rr(A)(a) rr(B)(a)} is bounded in WI.
=
On P (N W1 ) we define a binary relation R as follows:
R(A, B) iff {a < WI : rr(A)(a) ::; rr(B)(a)} is stationary. Since it is impossible to split a set of the form ('Y, WI) into two nonstationary subsets, Fact 21.19 implies that VA,B ~ NWI (R(A,B) V R(B,A)). Fqr B ~ NWI let
D(B) = {A E P(N W1 ): R(A,B)}. LEMMA 21.20. ID(B)I ::; NWI for every B ~ NW1 ' Let us first show how Theorem 21.18 follows from the lemma. For X ~ P (NwJ define Y(X) = U{.T?(A) : A EX}. COROLLARY 21.21. If X E [P (NWI )]N W1 +1, then Y(X) = P(NwJ. PROOF. Note that if Z E P(Nwl)\Y(X) then D(Z) ;;;1 Y(X) ;;;1 X, and apply Lemma 21.20. 0
21.
130
CLOSED UNBOUNDED SETS AND STATIONARY SETS
Finally, note that it follows from Lemma 21.20 that IY(X)I = NW1 +I for every X E [P(NW1WWl+l. It remains to prove the lemma.
PROOF OF LEMMA 21.20. Assume towards a contradiction that there exists B* E P(N W1 ) such that ID(B*)I ~ NW1 + I ' For A E D(B*)\{B*} let SA ::: {a < WI : lI'(A)(a) < lI'(B*)(a)}. Then SA is a stationary subset of WI. Since ID(B*)I ~ NWI +1, the Pigeonhole Principle allows us to pick a stationary set S ~ WI such that I{A E D(B*) : SA::: S}I ~ NW1 + I ' We define: M
ens is stationary. Consider three cases: 2 Case 0: lI'(B*)b) < N1' for alII E ens. For A E M we define a function T A : ens -+ WI by
Let C = WI
n LIM.
= {A E D(B*) : SA ::: S}.
Then
TAb) = min {a < WI : lI'(A)b) < No,}. By the assumption of Case 0, T A is a regressive function. Hence there exists an aA such that the set TA"I{aA} is stationary. Let us assign such aA to each A EM. Since (TA"I{aA},aA) E P(Nt) x NI , the Pigeonhole Principle implies that there exists a pair (T,8) such that the set
N = {A E D(B*) : (TA"I{aA},aA) = (T,8)} has cardinality NWI + 1. Note that if A E N, then lI'(A)fT : T -+ N6. Since N~l < NW1 +I ' there are different A, BEN such that lI'(A)fT = lI'(B)fT. This contradicts Fact 21.19. Case 1: lI'(B*)b) < N1'+l for alII E ens. For each lEe n S fix an injection 11' : lI'(B*)b) -+ N1" Let
TAb) = min {a < WI: 11'(1I'(A)b)) < No,}, and proceed as in Case O. To get the final contradiction, instead of lI'(A) fT consider functions gA : T -+ N6 defined by:
gAb) = 11'(1I'(A)b))·
Case 2: lI'(B*)b) ~ N1'+I for some lEe n S. This is impossible, since rng(lI'1') ~ N1'+I'
DO
EXERCISE 21.15 (G). Prove Theorem 21.18 in its full generality. EXERCISE 21.16 (R). Prove that if 2N" :5 No+2 for all a < WI, then 2Nwl < NWI +2· Hint: GCH below NWI was only used to make Case 2 in the proof of Lemma 21.20 redundant. Under the weaker assumptions, let the 1I'0'S be injections from P(N o) into No+2, and for A, B ~ Nwl' consider the sets D(A, B) ::: {Y ~ NWI : 3 stationary S ~ wI'i/a E S (1I'(Y) (a) < lI'(A) (a) < lI'(B) (a)}. Then prove the following modification of Lemma 21.20: ID(B,A)I :5 NWI for every A,B ~ NW1 ' 20ne could reduce the proof to Case 1. However, we want to highlight the place in the proof that requires modification for the solutions of Exercises 21.16 and 21.17.
21.2.
CLOSED UNBOUNDED AND STATIONARY SUBSETS OF
[XJ<~
131
EXERCISE 21.17 (X). (a) Prove that if 'Y < WI is such that 2ND :s; N"'+I for all WI, then 2Nwl :s; NWI +1" (b) Prove that if 2Nn :s; Na +a for all a < WI, then 2Nwl < NWI +Wl . (c) Formulate and prove the most general version of Theorem 21.18 you can extract from its proof and points (a) and (b). a <
REMARK 21.22. It is possible and sometimes useful to generalize the notions of closed unbounded and stationary sets to classes. A class C of ordinals is unbounded in ON if it is proper. We say that C is closed if C n'Y is closed in 'Y for every ordinal 'Y. A class S ~ ON is stationary in ON if it has nonempty intersection with every class C that is closed and unbounded in ON. Many results of this section have analogues for proper classes, but it can be tricky to formalize these in Ls. Let us consider just two examples. A proper functional class F ~ ON x ON is normal if dom(F) = ON and If a < {J, then F(a) :s; F({J); (ii) If 'Y E LIM, then F("f) = sup{F(a) : a < 'Y}; (iii) F(a) ~ a for all a EON.
(i)
We have the following analogue of Lemma 21.1: CLAIM 21.23. Suppose F is a proper, normal functional class. Then the class Fix(F) = {a EON: F(a) = a} is closed and unbounded in ON. EXERCISE 21.18 (PG). Prove Claim 21.23. Don't forget to verify that Fix(F) is a class in the first place. EXERCISE 21.19 (PG). Show that if Co, ... ,C n are finitely many closed unbounded classes of ordinals, then Co n ... n C n is a closed unbounded class of ordinals. EXERCISE 21.20 (R). Formulate and prove an analogue of Theorem 21.3. Hint: Of course, everybody's first impulse is to say: "Let {C i : i E I} be a set of classes of ordinals such that each C i is closed and unbounded in ON." But this is streng verboten.
21.2. Closed unbounded and stationary subsets of [X]<1t
Suppose that \5 = (WI, *, 0) is a group. Does there exist a countable subgroup (H, *,0) of \5 such that W ~ H? Yes. To see this, consider functions * : WI x WI --+ WI and i : WI --+ WI, where i assigns to each a E WI its inverse in \5. Let M. and Mi be as in Example 21.5, and let Mi*,i} denote M. nM i . Then M{.,i} is club in WI, and if a E M{.,i} \w, then jj = (a,*,O) is as required. This argument relies heavily on the fact that \5 has cardinality NI . If the universe of \5 were W2 instead, the above reasoning would yield a subgroup jj of \5 whose universe is an ordinal a < W2, but not necessarily countable. jj =
EXERCISE 21.21 (G). Let \5 = (W2, *,0) be a group such that 2 * (WI + 7) = 0, and define M{.,i} as above. Show that the smallest a> win M{.,i} is uncountable. But the group \5 of Exercise 21.21 does have a countable subgroup whose
universe contains w. This is a consequence of the following general theorem.
132
21. CLOSED UNBOUNDED SETS AND STATIONARY SETS
THEOREM 21.24. Let X =I- 0, let G = {ge : ~ < T]} be a family of functions g : Xn(e) --+ X, where n(~) E w for all ~ < T], and let A ~ X. Then there exists a smallest B ~ X such that A ~ Band B is closed under all functions in G. Moreover, IBI -:; max{~o, IAI, IT]I}· The set B of Theorem 21.24 will be called the closure of A under G. Theorem 21.24 has an important application to model theory. If
oX = (X, (R;)iE], (Fj)jEJ, (Ck)kEK) is a model of a first-order language L, and if B ~ X, then B is the universe of a submodel of oX if and only if {Ck : E K} ~ B and B is closed under all functions F j • Since one can treat constants as functions of arity zero, Theorem 21.24 implies the following: COROLLARY 21.25. Let oX = (X, (R;)iEI, (Fj)iEJ, (Ck)kEK) be a model of a first-order language L with nonlogical symbols {ri : i E I} U {Ii : j E J} U {Ck : k E K}, where ri is a relational symbol of arity 7o(i), Ii is a functional symbol with arity 71 (j), and Ck is a constant symbol. Then for every A ~ X there exists a smallest set B ;;;> A such that IBI -:; max{IAI, IJ U KI, ~o} and
r
IB = (B, (R; n BTo(i))iEI, (Fj BTl (j))jEJ, (Ck)kEK)
is a model of L.
In the sequel, we will rather informally write R; instead of R; n BTo(i) and Fj instead of F j BTl (j). Also, we shall often say that B (rather than IB) is the substructure (or submodel) of X (rather than of oX) generated by A. For example, in the situation discussed at the beginning of this section, we would be looking for the subgroup of WI generated by w. Before we prove Theorem 21.24, let us derive another important consequence of it. By MA - (I\:) we abbreviate the following statement: (MA -(I\:)) Let (lP', -:;) be a c.c.c. p.o. of cardinality -:; 1\:, and let V be a family of dense subsets of lP' with IVI -:; 1\:. Then there exists a V-generic filter in lP'.
r
THEOREM 21.26. For each infinite cardinalI\:, the statements MA(I\:) and MA - (1\:) are equivalent. PROOF. Since MA - (I\:) is just MA( 1\:) restricted to a narrower class of p.o.'s, only the implication MA-(I\:) --+ MA(I\:) needs a proof. This proof hinges on the following lemma. LEMMA 21.27. Let I\: -:; A be infinite cardinals, let lP' be a set of cardinality A, partially ordered by a relation -:;, and let V be a family of size I\: of dense subsets of (lP', -:;). Then there exists Q ~ lP' such that: (1)
(2) (3)
If p, q E Q and p is compatible with q in lP', then p is compatible with q in Q, i.e., if there is some r E lP' such that r -:; p and r -:; q, then such an r also exists in Q; Q n D is dense in (Q, -:;) for all D E V;
IQI =
1\:.
EXERCISE 21.22 (G). Derive Theorem 21.26 from Lemma 21.27. Hint: For (lP', -:;) and V as in MA(I\:), form Q as in Lemma 21.27. Note that by (1), (Q, S) still satisfies the c.c.c. Then apply MA-(I\:) to (Q, -:;).
21.2.
CLOSED UNBOUNDED AND STATIONARY SUBSETS OF
[Xj
133
PROOF OF LEMMA 21.27. Let 1) = {De : ~ < ~}. For each ~ < ~, let g~ : lP - De be such that ge(p) ::; p. Moreover, pick a function gl<. : lP2 - lP in such a way that if p,q are compatible, then gl<.(p,q) ::; P and gl<.(p,q) ::; q. Let Po be an arbitrary subset of lP of size ~, and let Ql be the closure of lPo under {ge : ~ < ~ + I}. Then Ql satisfies (1)-(3). DO PROOF OF THEOREM 21. 24. Let X, A, G be as in the assumptions. Consider the family: Me = {B ~ X: A ~ B /I. B is closed under G}. Then Me =J 0, since X E Me, and hence, for every B E Me, we have A C nMe ~ B. Therefore, if nMe is closed under G, this must be the smallest element of Me. EXERCISE 21.23 (G). Show that nMe is closed under G. It remains to prove the last sentence of Theorem 21.24. For this, it suffices to produce at least one element of Me whose cardinality does not exceed max{N o, IAI, 1171}· For C ~ X, let G[Cl = U e<'7 gdCn(e)l. Now define recursively a sequence (An)nEw of subsets of X by letting Ao = A and An+1 = An U G[Anl. Consider the set B = UnEw An. The following exercise concludes the proof of Theorem 21.24.
EXERCISE 21.24 (G). Verify the following properties of the An's and of B: (a) An ~ An+1 for all nEw. (b) B is closed illlder all functions in G. (c) For each nEw, IAnl ::; max{N o, IAI, 1171}. (d) IBI ::; max{N o, IAI, 1171}. 0 The next exercise is redundant for the proof of Theorem 21.24, but interesting on its own. EXERCISE 21.25 (PG). Show that the set B constructed in the last part of the proof of Theorem 21.24 actually is equal to (rather than merely a superset of) the closure of A. Hint: This is similar to Exercise 4.7. You will probably have noticed a certain similarity between the proof of Theorem 21.24 and the arguments used in describing Examples 21.4 and 21.5. In order to express the precise nature of this similarity, we need new concepts. DEFINITION 21.28. Let ~ be an uncountable regular cardinal, and let X =J 0. A subset C ~ [Xl<1<. is unbounded in [Xl<1<. if for every Y E [Xl<1< there is Z E C such that Y ~ Z. A set C ~ [Xl<1< is closed if for every increasing sequence (Yo: : a < v) of length v < ~ of elements of C the union UO:<11 Yo: of is also an element of C. A set C that is both closed and unbounded in [Xl<1<. will be called club in [Xl
134
21.
CLOSED UNBOUNDED SETS AND STATIONARY SETS
EXAMPLE 21.9. If A E [X]<", then the set in [X]<". The set A is called the cone of A.
A=
{B E [X]<" : A ~ B} is club
EXAMPLE 21.10. If /'i, is a regular uncountable cardinal, then /'i, itself is a club subset of [/'i,]
Mo = {B E [X]
~
B
1\
B is closed under G}
is unbounded in [X]
Me = {B
E
[X]
1\
B is closed under G}
is club in [X]
21.2. CLOSED UNBOUNDED AND STATIONARY SUBSETS OF [XJ<"
135
The diagonal intersection of a family {Ca : a E I\:} of subsets of an ordinal I\: can be defined as AaEI
AxExCx = {A E [X]
136
21.
CLOSED UNBOUNDED SETS AND STATIONARY SETS
How about these? Let us call a function F : [XJ
:f 0, let
K
be a regular uncountable cardinal, and let
(a) If for every A E [X]<1t there exists a minimum B E C above A (i.e., there exists B E C such that A ~ B and A n C = iJ n C), then there exists a normal function F : [X]<1t --+ [X]<1t such that Fix(F) = C. (b) There exists a normal function F : [XJ<1t --+ [X]
21.2. CLOSED UNBOUNDED AND STATIONARY SUBSETS OF [XJ
137
EXERCISE 21.32 (PG). Prove Corollary 21.38. Hint: Partition A into sets Ln, each of cardinality A. Construct functions Gn as in the proof of Theorem 21.37(b) and define the g~'s in such a way that if a E [x]n, then Gn(a) = {g~(a) : E Ln}.
e
CHAPTER 22
The O-principle In Chapter 14 we promised you a consistent construction of a Suslin tree. Now is the time to make good on this promise. So let us consider the question: What does it take to construct a Sus lin tree? Let us try a recursive construction in the spirit of Chapter 17. Since a Suslin tree has Nl nodes, we may as well try to build one whose set of nodes is Wi' We will construct a partial order relation ~T on WI level by level. Since the levels of our tree are supposed to be countable, we may as well decide at the outset that T(a) = [w·a,w·(a+1)) = {~: w·a~~ <w·a+w} for all a<Wl. The restriction of the p.o. relation ~T to T(o) will be denoted by ~o' The recursive construction boils down to deciding, in a situation where ~o is given, which elements ofT(a) to put on top of which elements ofT(o)' For technical reasons, we want to construct a tall tree such that every node of it has two immediate successors. The latter will ensure that the resulting tree is splitting, and then we will have to worry only about nonexistence of uncountable antichains. By Exercise 14.34, the absence of uncountable chains will follow automatically. To get a tall tree, we need to make sure that the following condition holds at every level a < WI: (tall(a)) V(3 < aVs E T((3)~t E T(a) (s
EXERCISE
tall(a
+
-+
If a > 0 is a countable limit ordinal and (T(o) , ~o) has already been constructed, we need to put the ordinals in [w· a, w . (a + 1)) on top of cofinal branches of T(o) so as to make tall (a) true. EXERCISE 22.2 (G). Suppose a > 0 is a countable limit ordinal and the tree (T(o) , ~o) is such that tall((3) holds for all (3 < a. Show that for every s E T(o) there exists a cofinal branch Bs of T(o) with s E Bs.
So let {Bs : 8 E T(o)} be a family of cofinal branches of T(o) such that s E Bs for each s E T(o). If we take care to put one ts E T(a) on top of each B s , i.e., if we extend the p.o. relation so that r
22. THE <)-PRINCIPLE
140
This describes what we will call the generic blueprint for the construction of T. The generic blueprint yields a tall, splitting tree (T, :::;T) of height W1 with all levels countable. Moreover, we have the following connection between countable ordinals and subtrees of T: EXERCISE 22.3 (G). Convince yourself that the generic blueprint yields a tree T with the property that if C = {a < W1 : T(ct) = a}, then C = {a < WI : W· a = a} = {w'''( : 'Y E W1 n LIM},
and hence C is club in W1. But how can we ensure that the tree T will not have an uncountable antichain? Let us do some brainstorming: First of all, Zorn's Lemma implies that every uncountable antichain in T is contained in a maximal uncountable antichain. So it suffices to make sure that T has no maximal uncountable antichains. Since the set of nodes of T is WI, every uncountable antichain of T would have to be an uncountable subset of W1' Let us consider an arbitrary A E [Wd NI • How can we make sure that A will not become a maximal antichain in T? Suppose we are at stage a E LIM of the construction, Le., suppose that (T(ct), :::;ct) has been defined, and we want to decide which t E T(a) to put on top of which s E T(ct). Clearly, AnT(ct) is an antichain in (T(ct), :::;ct). If AnT(ct) happens to be a maximal antichain in (T(ct), :::;ct), we will be able to make sure that the following holds:
(K)
Vs E T(a)3r E T(ct) nA (r
This suffices to destroy A as a candidate for becoming an uncountable antichain in T: Since A is uncountable, A\T(ct) =I 0. But for each t E A\T(ct) there exist s E T(a) and rEA n T(ct) with r
(<)O(C))
There exists a sequence (Act: a E C\{O}) such that Act is a subset of a for all a E C\ {O}, and for each A E [Wd NI there exists a E C\{O} such that Ana = Act.
The statement <)O(C) does not look obviously inconsistent: If CH holds, there are only ~1 candidates for Ana, where A E [Wl]NI and a < W1, and it is conceivable that a clever arrangement of some of these sets would contain an initial segment A n a for every A E [WI]Nl. But would Property <)O(C) be sufficient for our purposes? At stage a of the construction, we would want to extend :::;ct in such a way that
Vs E T(a)3r E T(ct) n Act (r
22. THE (>-PRINCIPLE
141
We hinted above that this is possible if Act happens to be a maximal antichain in (T(a), ~a). But will an initial segment Aa of a maximal antichain A in T be a maximal antichain in T(a)? The following claim sheds some light on this question. CLAIM 22.1. Suppose (T ~T) is constructed according to the generic blueprint, and A ~ T is a maximal antichain in T. Then the set D(A)
= {a E
C : A n a is a maximal anti chain in (T(a), ~a)}
contains a closed unbounded subset of WI.
PROOF. It is easy to see that if A is an antichain in T, then A n a is an antichain in T(a) for all a < WI. Now assume A is maximal. Then there is a function g : WI ---+ A such that g({3) ~T {3 or (3 ~T g({3) for all {3 < WI' Let Mg = {a E WI : rng(gf a) ~ a}. By Example 21.5, Mg is a club subset of WI' By Exercise 22.3, the set C n Mg is also club. Now it suffices to observe that D(A);2 CnM g. 0 O.k., if OO(C) holds and A is going to be a maximal antichain in T, then once in a while, Ana will be a maximal antichain in T(a) , and at least for one a E C\ {O}, we will have Ana = Aa. Unfortunately, so far we don't have any reason to believe that there is an a E C\{O} such that these two things happen simultaneously. To eliminate this problem, we would like to have a sequence of Aa's such that for every A E [WI]Nl there are stationarily many a's such that Ana = Aa. This leads to the following statement, which is called the O-principle. (0) There exists a sequence (Aa : a < WI) such that Aa is a subset of a for all a < WI, and for each A ~ WI the set {a < WI : A n a = Aa} is stationary in WI. Clearly, 0 ---+ OO(C). The construction outlined above gives the following:
THEOREM 22.2.
0
---+
...,SH.
PROOF. We need a technical lemma. LEMMA 22.3. Suppose a tree T is being constructed according to the generic blueprint, a > 0 is a countable limit ordinal such that T(a) = a, and Aa is a maximal anti chain in (T(a), ~a). Then (T(a+1) , ~a+1) can be constructed in such a way that both tall(a) and Ka hold. PROOF. Choose an increasing sequence ({3n)nEw of ordinals such that a = SUp{{3n : nEw}. For each r E T(a), choose a sequence (S~)nEw of elements of T(a) such that:
(1) (2) (3)
(4)
So =
r;
s~
Note that since tall({3n) holds for all nEw, we always can find s~'s that satisfy conditions (1)-(3). Also, maximality of Aa is exactly the property that makes condition (4) possible: For each r E T(a) , either s ~T r for some s E Aa, in which case condition (4) will hold for any sl as long as condition (2) is also satisfied; or r
142
22. THE
~-PRINCIPLE
Now enumerate T(a) = {tr : r E T(o:)} and define: s
3n
EW
o
EXERCISE 22.4 (PG). Derive Theorem 22.2. THEOREM 22.4.
0
---+
CH.
PROOF. Let (Bo: : a < WI) be a sequence that witnesses 0 (such a sequence is often called a O-sequence). Each subset B of W is also a subset of WI. Thus, for each B E P(w) there exists a(B) < WI such that W S a(B) and Bo:(B) = B n a(B) = B nw. In fact, we have uncountably many choices for a(B), but we need only one. So let a(B) be the least such ordinal. Note that if B =I C, then a(B) =I a(C). In other words, the function that assigns a(B) to B is a one-to-one function from P(w) into WI. The existence of such a function implies CH. 0 It follows from Theorem 22.4 (and also from Theorems 22.2 and 19.1) that 0 is not a theorem of ZFC. However, the statement 0 is relatively consistent with ZFCj in fact, if V = L, then 0 holds. It has also been proved that 0 is not a consequence of CH. Therefore, whenever a theorem is shown to follow from 0, the question whether this same theorem is also a consequence of CH deserves consideration. This question has been settled for the Suslin Hypothesis: R. B. Jensen has shown the relative consistency of CH + SH with ZFC. EXERCISE 22.5 (PG). Prove that each of the following three statements is equivalent to 0: (a) There exists a sequence (Ao: : a < WI) such that Ao: is a subset ofax W for each a < WI, and for all A <;;;; WI X w, the set {a < WI : A n a x W = Ao:} is stationary in WI. (b) There exists a sequence (Ao: : a < WI) such that Ao: is a subset ofax a for all a < WI, and for each A <;;;; WI X WI, the set {a < WI : A n a x a = Ao:} is stationary in WI. (c) There exists a sequence (J a < WI) such that f 0: is a function from a into a for each a < WI and for all f E w1WI the set {a < WI : a = fo:} is stationary in WI. Q
:
n
There are many useful variants of the O-principle. In the remainder of this chapter we will introduce some of these variants. We begin with one that is equivalent to O. (0 -) There exists a sequence (Ao: : a < WI) such that Ao: E [P(a)l~No for all a < WI, and for each A <;;;; WI the set {a < WI : A n a E Ao:} is stationary in WI.
0-. The implication 0
THEOREM 22.5.
0
~
PROOF. ---+ 0- is obvious. To prove the converse, fix a 0- -sequence, i.e., a sequence (Ao: : a < WI) that witnesses 0-. Without loss of generality, we may assume that each Ao: is infinite. For each a, fix an enumeration (A~)nEw of Ao:. Let C be the club set {a < WI : w·a = a}. For each nEw and a E C, let B~ = {,8 < a: w·,8+n E A~}. For nEw and a ~ C, we let B;; = 0. CLAIM 22.6. There exists nEw such that
(B~
: a < WI) is a O-sequence.
22,
THE ¢-PRINCIPLE
143
PROOF. Suppose not. Then for every nEw there exists Bn such that the set Sn = {a : Bn n a = B~} is nonstationary. Consider the set A = {w . /3 + n : n E W " /3 E Bn}. The following exercise leads to a contradiction with the assumption that (Ao : a < WI) is a <:r -sequence.
22.6 (G). Convince yourself that for every a outside the nonstationary set (WI \C) u UnEw Sn, the set A n a is not a member of Ao. DO EXERCISE
Now let us present two strengthenings of O. (0*) There exists a sequence (Ao : a < WI) such that Ao E [P(a)]::5 No for all a < WI, and for each A ~ the set {a < WI : A n a E Ao} is C LU B in WI •
(0 +)
There exists a sequence (Ao : a < WI) such that Ao E [P(a)]::5 No for all a < WI, and for each A ~ there exists a dub subset C of WI such that Va E C(Ana E Ao "Cna E Ao).
WI
WI
It is clear that 0+ -4 O· and 0* -+ O. Jensen has shown that V and also that 0 f+ O· and O· f+ 0+.
=L
-4
0+,
22.7 (G). (a) Show that the following statement is inconsistent with ZFC: There exists a sequence (Ao : a < WI) such that Ao ~ a for all a < WI, and for each A ~ WI, the set {a < WI : A n a = Ao} is CLUB in WI . (b) Show that 0 implies the existence of a sequence (S~ : ~ < 2Nl) of stationary subsets of WI such that S~ n STI is countable whenever ~ < TJ < 2Nl . EXERCISE
Recall from Chapter 14 that a Kurepa tree is an N1-tree with at least N2 cofinal branches, and that the K urepa Hypothesis asserts the existence of a Kurepa tree. A Kurepa family is a family F ~ P(wd such that
IFI
~ N2 and
Va < WI (I{A n a: A
E
F}I ::; No).
22.7. The following statements are equivalent: There exists a Kurepa tree. There exists a Kurepa family.
LEMMA
(i) (ii)
PROOF. To prove the implication (i) -+ (ii), let (T, ::;T) be a Kurepa tree whose set of nodes is WI, and let F be the set of all cofinal branches of T. Then IFI ~ N2. Moreover, if a < WI and B E F, then B n T(o) = i for some t E T(a). Hence I{B n T(o) : B E F}I ::; IT(a)1 ::; No. Since each /3 < WI is contained in T(o) for some a < WI, it follows that V/3 < WI (I{B n /3: BE F}I ::; No), and hence F is a Kurepa family. To prove the implication (ii) -+ (i), let F be a Kurepa family. For each A E F, let XA : WI -+ 2 be the characteristic function of A. Let T = {XA a: A E F " a <
r
wd· EXERCISE
22'.8 (G). Convince yourself that
THEOREM 22.8.
PROOF.
a family
.A
0+
-+
(T,~)
is a Kurepa tree.
0
KH.
We need a few technical preliminaries. Recall from Chapter 17 that ~ [Wd Nl is pairwise almost disjoint if IA n BI ~ ~o for all A, B E .A
144
22. THE ¢-PRINCIPLE
such that A:j:. B. By Exercise 22.7(b), 0 (and hence 0+) implies that there exists a pairwise almost disjoint family A ~ [Wd N1 of stationary subsets of WI such that IAI = 2N1 . For our argument we will not need the sets in A to be stationary; it will suffice to assume that they are uncountable. For A, B ~ WI we let S(A,B) = {a E A: 3{3 E B({3:S al\--,38 E A({3:s 8 < a))}. We call S(A, B) the shadow of B in A. Note that S(A, B) ~ A, and if A, B are uncountable, then so is S(A, B). Moreover, for all 8 < WI we have S(A, B) n 8 = S(An8,Bn8). For the rest of this argument, let us fix a 0+ -sequence (Aa : a < WI) and a pairwise almost disjoint family {Ao : a < 2Nl} ~ [WIJNl of size 2 N1 . A club set C ~ WI will be called an approximation of a set A ~ WI if 'Va E C (A n a E Aa 1\ C n a E Aa). For each a < 2Nl we choose an approximation Co of Ao. Let F = {S(Ao, Co) : a < 2 N1 }. We show that F is a Kurepa family. Let a < {3 < 2N1 . Since lAo n A,el :S No and S(Ao, Co) E [AojNl, we have S(Ao, Co) :j:. S(A,e, C/3), and thus IFI = 2Nl > NI . Now fix 8 < WI and consider the family T(8) = {S(Ao, Co) n8: a < 2Nl} = {S(Ao n 8, Co n 8) : a < 2N1 }. If 8 E Co, then both Aon8 and Co n8 are elements of A o , and hence S(Aon8, Co n8)
is a member of the countable set {S(A, B) : A, BE Ao}. If 8 fj. Co, let 8* = sup(Co n8). Then 8* < 8 and 8* E Co. Again, S(Ao, Co) n 8* E {S(A,B) : A,B E Ao'}' If Ao n [8*,8) = 0, then S(Ao n 8, Co n 8) = S(Ao n 8*, Co n 8*). If Ao n [8*,8) :j:. 0, then 8* casts a shadow on the minimum element of Ao \8*. In the latter case we have S(Ao n 8, Co n 8) = S(Ao n 8*, Co n 8*) U {min(Ao n [8*, 8))}. We conclude that T(8) ~ {S(A, B) : 38* :S 8 (A, BE A o.)} U {S(A, B) U b} : 38* < 8 (A, B E Ao' 1\ 8* :S I < 8)}. Since the right hand side is countable, so is 0 T(8). Now let us discuss two weakenings of O. The first of these is called the .-principle. (.) There exists a sequence (so : a E WI n LIM\ {O}) such that for all a: (i) So ~ a; (ii) ot( so) = W; (iii) sup So = a; and (iv) Each uncountable subset of WI contains an so' A sequence (so: a E WI n LIM\{O}) that witnesses. is called a.-sequence. EXERCISE 22.9 (G). Convince yourself that
0
-+ • .
LEMMA 22.9. Let (so : a E WI n LIM\ {O}) be a .-sequence. Then for every A E [Wd N1 the set {a E WI : So ~ A} is stationary. PROOF. Let (so: a E WI n LIM\{O}) be a 4-sequence, let A E [Wd N1 , and
let C be a club subset of WI. We want to show that there exists a E C such that So ~ A.
22. THE O-PRINCIPLE
145
EXERCISE 22.10 (G). Show that there exists a sequence (ai3)i3<Wl of elements of A such that V{3 E w1 3'Y E C (ai3 < 'Y < ai3 + 1). Fix a sequence (ai3)i3<Wl as in Exercise 22.10, define B = {ai3 : {3 E WI}, and let Q be such that SQ ~ B. Then Q is the limit of a sequence of elements of C, and hence Q E C. Moreover, since B ~ A we have SQ ~ A. 0 S. Shelah has proved that • does not imply CH. Hence. is strictly weaker than O. However, in the presence of CH these two principles are equivalent.
+ .. +-+ O.
THEOREM 22.10. CH
PROOF. We only need to prove the implication from the left to the right. Let (sQ: Q E WI nLIM\{O}) be a 4-sequence. Let (AQ)Q<Wl be an enumeration of all countable subsets of WI such that every A E [wd ::;No is listed cofinally often. Let C be the closed unbounded set WI n LIM\ {O}. For Q E C we define DQ = U{ESo< ~. For Q tJ. C, let DQ = 0. The following claim implies Theorem 22.10. CLAIM 22.11. (DQ:
Q
E WI) is a O-sequence.
PROOF. Let A ~ WI, and let D be club in exists a'Y E enD such that D{ = An 'Y.
WI'
We need to show that there
EXERCISE 22.11 (G). Construct recursively an increasing transfinite sequence of ordinals (013 : {3 < WI) such that for all {3 < WI:
(i) 3'Y E enD (013 < 'Y < 013+1); (ii) A6i3 +1 = An 013' Let B = {oi3 : 0 < {3 < WI}, and let
Q be such that SQ ~ B. Suppose (Oi3JnEw is an enumeration of SQ. Then for each nEw there exists 'Yn E enD such that 0i3n < 'Yn < 0i3n+l' Since Q = sup{ 0i3n : nEw}, we can infer that Q E enD. Moreover, A6i3n ~ An Q and An 0i3n ~ A6i3n+l for each nEw. It follows that DQ = An Q, and we have proved Claim 22.11. DO
The 4-principle can be further weakened to the so-called stick-principle (abbreviated T): (T) There exists a sequence (AQ : Q < WI) of infinite subsets of WI such that for every A E [Wl]Nl there exists Q < WI with AQ ~ A . . EXERCISE 22.12 (G). (a) Convince yourself that. -) (b) Show that CH -)
T.
T.
Since CH ~ 0, it follows from Theorem 22.10 and Exercise 22.12(b) that strictly weaker than ... However, Tis not a theorem of ZFC. THEOREM 22.12. MA(Nl) -)
-l
PROOF. Let .(AQ : Q < WI) be a sequence of infinite subsets of the c.c.c. partial.order (Fn(wl' 2), ;2). For each Q < WI, let
DQ
Tis
= {p E Fn(wl' 2) :
3€ E AQ n dom(p) (p(€)
WI.
Consider
= On.
EXERCISE 22.13 (PG). (a) Show that each Do is a dense subset of Fn(wl' 2). (b) Derive Theorem 22.12. 0
146
22. THE ¢-PRINCIPLE
Let us briefly mention another way of strengthening O. Let E be a stationary subset of WI. Then O(E) stands for the following statement: (O(E)) There exists a sequence (Aa : a E E) such that Aa ~ a for all a E E, and for each A ~ WI the set {a E E : A n a = Aa} is stationary. Of course, 0 is the same as O(wd. Moreover, 0 trivially implies O(E) for every club set E. By Theorem 22.13 below, V = L implies O(E) for every stationary E. On the other hand, Shelah proved that if ZFC is consistent, then there exists a model of ZFC that contains two stationary subsets E, F of WI such that O(E) holds and O(F) fails. We will show in the next chapter (Corollary 23.7) that O(E) implies the existence of two disjoint stationary sets F, GeE such that both O(F) and O(G) hold. One can also consider O-like principles for cardinals larger than ~1' If I>, is an uncountable cardinal and E ~ 1>" then Ot«E) stands for the statement: (Ot«E)) There exists a sequence (Aa : a E E) such that Aa ~ a for all a E E, and for each A ~ I>, the set {a E E : A n a = Aa} is stationary in 1>,. Of course, O(E) is the same as OWl (E). Instead of 0,,,(1),) one usually writes
0",. THEOREM 22.13. Assume V = L. Let I>, be an uncountable regular cardinal, E a stationary subset of 1>,. Then O",(E) holds.
Mathographical Remarks
The O-principle is due to R. B. Jensen. The "principle is also called the Ostaszewski Principle. In his paper On countably compact, perfectly normal spaces, Journal of the London Mathematical Society 14 (1967), 505-516, A. Ostaszewski used" + CH to construct a first countable, perfectly normal, hereditarily separable, locally countable, locally compact, count ably compact Hausdorff space that is not Lindelof (and hence not compact). K. Devlin later noticed that" + CH already implies O. For variations on the T-principle, see S. Broverman, J. Ginsburg, K. Kunen, and F. D. Tall, Topologies determined by G"-ideals on WI, Canadian Journal of Mathematics 30(6) (1978), 1306-1312. The construction of a model for ",CH appeared in S. Shelah, Whitehead groups may not be free even assuming CH, II, Israel J. Math. 35 (1980), 257-285. The proof of Theorem 22.13 can be found in K. Devlin's book Constructibility, Springer Verlag, 1984. The latter reference also contains information on several other useful combinatorial principles that are consequences of V = L.
,,+
CHAPTER 23
Measurable Cardinals DEFINITION 23.1. Let -+ [0,1] such that (1) 1"(0) = OJ
be a cardinal. A probability measure on K is a function
K
I" : P(,..)
(2) (3)
J.L(K) = Ij J.L(UnEw An) = LnEw J.L(An) for every countable family {An: nEw} of pairwise disjoint subsets of
K.
Property (3) has several names in the literature: a-additivity, countable additivity, or Nl -additivity of 1". If I" is such that 1"( {O) = for every ~ E K, then we say that I" vanishes on the singletons. The family II-' = {N ~ K: J.L(N) = O} is called the ideal of null sets of 1".
°
EXERCISE 23.1 (G). Show that II-' is indeed an ideal on EXAMPLE 23.1. Let
~
E
K,
K.
and define:
J.L(A) = Then I" is a probability measure on the singletons.
{01"
K.
if ~ if ~
E Aj If A.
Of course, this measure does not vanish on
EXAMPLE 23.2. Let U be a count ably complete ultrafilter on
J.Lu (
A) = {I, 0,
K.
Define:
if A E Uj if A If U.
Then J.Lu is a measure on K. The measure of Example 23.1 is a special case of J.Lu for the principal ultrafilter U generated by {O. Note that J.Lu vanishes on the singletons if and only if U is a nonprincipal ultrafilter. EXERCISE 23.2 (G). Show that if I" is any measure on a cardinal K, then the family UI-' = {A ~ K: J.L(A) = I} is a count ably complete filter on K. Moreover, show that UI-' is an ultrafilter on K if and only if rng(J.L) = {O, I}. (The latter property will be referred to by saying that I" is a two-valued measure.) How is Definition 23.1 related to more familiar probability measures, like Lebesgue measure on [0, I]? Of course, the latter is not defined on sets of ordinals. But this is only a superficial difference, since any one-to-one correspondence between the cardinal 2No and the unit interval [0,1] allows us to treat a measure defined on P(2 No) as one defined on P([O, 1]) and vice versa. More relevant for the present purpose is the fact that Lebesgue measure is defined only on a certain a-algebra of subsets of [0,1]' whereas the measures considered here are defined for all subsets of a cardinal K. We shall say that Lebesgue measure can be extended to P([O, 1]) if
there exists a probability measure J.L : P([O, 1]) 147
-t
[0,1] such that J.L([a, b]) = b - a
148
23.
MEASURABLE CARDINALS
for all a, b with 0 ~ a ~ b ~ 1. It is clear that each such measure vanishes on the singletons (consider the degenerate intervals [a, a] = {a}). The following question provided the initial motivation for the notions studied in this chapter. QUESTION 23.2. Does there exist a probability measure on P([O, 1]) that extends Lebesgue measure? Note that this is not the same as asking whether there exists a subset of [0,1] that is not Lebesgue measurable. The latter question has been answered in the negative by Theorem 9.23 of Volume 1. Incidentally, the Vitali set constructed in the proof of that theorem also demonstrates that Lebesgue measure cannot be extended to a translation-invariant measure defined on all subsets of JR. THEOREM 23.3 (Ulam). Let K be an infinite cardinal, and let A = K+. Then there exists an indexed family {Aa,e : 0: < K, ~ < A} of subsets of A such that: (1) Aa,e n Ap,e = 0 for all 0: < f3 < K and ~ < Ai (2) Aa,e n A a,7] = 0 for all 0: < K and ~ < "l < Ai (3) Ua
0:
<
K,
~
< A} that satisfies (1)-(3) is called an
PROOF OF THEOREM 23.3. For each "l < A, let f7] be a one-to-one function from "l into K. For 0: < K and ~ < A define:
Aa,e = {( > ~: fd~)
= o:}.
EXERCISE 23.3 (PG). Verify that the Aa,e's defined above satisfy conditions (1)-(3). 0 Before we discuss the impact of Theorem 23.3 on the existence of measures on cardinals, let us present an application to stationary sets. COROLLARY 23.4. Let K be an infinite cardinal, and let S be a stationary subset of A = K+. Then there exists a family S of A pairwise disjoint stationary subsets of A such that US ~ S. In particular, there exists a family of A pairwise disjoint stationary subsets of A. PROOF. Let S be as in the assumptions, and let {Aa,e : 0: < K, ~ < A} be an Ulam matrix. For each ~ < A there exists o:(~) < K such that S n AaCe),e is stationary in A: Otherwise, the stationary set S would be the union of a family of . fewer than cf(A) nonstationary sets, namely {S n Aa,e : 0: < K} U {~+ I}, which is impossible by Lemma 21.10. Now the Pigeonhole Principle implies that there exists 0: < K such that the set Xa = {~ < A : o:(~) = o:} has cardinality A. Fix such 0:, and let S = {S n Aa,e : ~ E Xa}. This is a family of size A of stationary subsets of A, and clearly US ~ S. By point (2) of Theorem 23.3, S consists of pairwise disjoint sets. 0 COROLLARY 23.5. There is no probability measure on WI that vanishes on the singletons. In particular, if CH holds, then Lebesgue measure cannot be extended to a measure on P([O, 1]). 1 The
Polish name Ulam is pronounced Oolawm.
MEASURABLE CARDINALS
23.
149
PROOF. Let {Aa,( : Q < W, ~ < wd be an Ulam matrix. Assume towards a contradiction that f..L is a probability measure on WI that vanishes on the singletons. Note that the latter property implies that f..L(~ + 1) = 0 for each countable ordinal~. Let ~ < WI' Since WI = U({A a,(: Q < w} U {~+ 1}), we have Ea<wf..L(Aa,~) = f..L(~ + 1) + Ea<w f..L(Aa,~) = f..L(WI) = 1. Therefore, it cannot be the case that f..L(Aa,~) = 0 for all Q < w. In other words, there must be some Q(~) < W such that f..L(Aa(O,~) > O. Now the Pigeonhole Principle implies that for some Q < W the set
Xa is uncountable. Fix such an
Q.
n(~) =
= {~:
For
~
Q(~)
= Q}
E X a , let
min{n E W : f..L(Aa,~)
> Tn}.
Using the Pigeonhole Principle once more, we find an uncountable subset Y £; Xa and nEw such that n(~) = n for all ~ E Y. This leads to a contradiction: Let {~i : 0::::; i ::::; 2n} be the first 2n + 1 elements of Y. Since Aa'~i nAa,~j = 0 whenever i ¥ j, we have 2n
f..L(Aa,~o U··· U A a,6 n ) =
L f..L(Aa,~.) ~ (2n + 1) . Tn > 1, i=O
contradicting the assumption that f..L is a probability measure.
o
COROLLARY 23.6. There does not exist a nonprincipal, countably complete ultrafilter on WI . PROOF. This follows from Example 23.2 by an argument as in the proof of Corollary 23.5. 0 COROLLARY 23.7. Suppose E is a stationary subset of WI such that <)(E) holds. There exist disjoint F, G £; E such that both <>(F) and <)(G) hold. PROOF. Let E be as in the assumption. Note that if E- = {Q E E : W· Q = Q}, then <)(E-) also holds. So let us assume without loss of generality that E = E-, i.e., that W • Q = Q for all Q E E. Define:
3 = {F
£; E: -,<)(F)}.
CLAIM 23.8. 3 is a countably complete ideal in P(E). PROOF. It is clear that 3 is closed under subsets. Now suppose Fn E 3 for nEw, and let F = UnEw Fn. We want to show that F E 3. Let (Aa : Q E F) be a sequence such that Aa £; Q for all Q E F. We want to show that there exists B £; WI such that the set {Q E F: B n Q = A",} is nonstationary. EXERCISE 23,4 (PG). Show that for every nEw there exists Bn £; WI such that the set Sn =. {Q E Fn : V{3 < Q(w . {3 + n E Aa - (3 E Bn)} is nonstationary. Let (Bn)nEw and (Fn)nEw be as above. Define B = {w·{3+n: n E wl\{3 E Bn} and S = {Q E F : B n Q = Aa}. Then S is contained in UnEw Sn, and since the Sn's are nonstationary, so is S. Thus B is as required. 0
150
23. MEASURABLE CARDINALS
By Claim 23.8, the dual filter :F = :1* = {G ~ E : E\G E .1} is count ably complete. This filter is nonprincipal, and, since E ¢ .1, it is also proper. By Corollary 23.6, :F is not an ultrafilter in P(E). In other words, .1 is not a maximal ideal in P(E). Thus there exists F ~ E such that neither F nor E\F is in J. This is exactly what we wanted to prove. 0 Let K, A be cardinals. An ideal I in P(K) is said to be A-saturated if for every family {Aa : a < A} ~ P(K)\I there exist a < (3 < A such that Aa n Aj3 ¢ I. EXERCISE 23.5 (PG). (a) Show that if J.L : P(K) -+ [O,lJ is a probability measure on K, then the ideal of null sets IlL is NI-saturated. (b) Show that there is no NI-saturated count ably complete ideal in P(wd that contains all the singletons. Hint: Use an Ulam matrix. (c) Show that if K is a regular infinite cardinal, then the family [KJ<1t is a K-complete ideal on K that is not K+ -saturated. DEFINITION 23.9. Let K, A be uncountable cardinals, and let J.L be a measure on A. We say that J.L is K-additive if J.L(UN) = 0 for every family N of subsets of A such that INI < K and J.L( N) = 0 for each N E N. A few changes in the notation of the proof of Corollary 23.5 yield the following result: THEOREM 23.10. If A is an uncountable successor cardinal, then there is no A-additive probability measure on A that vanishes on the singletons. EXERCISE 23.6 (PG). Show that a measure J.L on A is K-additive if and only if for every family M of pairwise disjoint subsets of A of cardinality IMI < K the following equation holds:
J.L(UM) =
L
J.L(M)
M'EM
The right hand side of this equation is a common abbreviation for the number sup{J.L(U:F) : :F E [MJ<W}. Hint: Use the fact that only countably many sets in M have positive measure, DEFINITION 23.11. An uncountable cardinal A is called real-valued measurable if there exists a A-additive probability measure J.L on A that vanishes on all singletons; A is called a measurable cardinal if, in addition, J.L is a two-valued measure. EXERCISE 23.7 (G). (a) Show that A is a measurable cardinal if and only if there exists a A-complete nonprincipal ultrafilter on A. (b) Convince yourself that if U is a A-complete filter on A such that nU = 0, then U is a uniform ultrafilter on A, i.e., IAI = A for every A E U. Theorem 23.10 tells us that real-valued measurable cardinals are a rather uncommon species; the next theorem makes a similar point. In combination, Theorems 23.10 and 23.12 imply that real-valued measurable cardinals are large cardinals. THEOREM 23.12. Every real-valued measurable cardinal is regular, PROOF. Let A be a singular cardinal, and let (~a : a < K) be an increasing sequence of ordinals cofinal in A, where K = cf(A) < A. Suppose towards a contradiction that J.L is a A-additive probability measure on A that vanishes on
23.
MEASURABLE CARDINALS
151
°
all singletons. By Exercise 23.7(b), f.1.(~o) = for every a < K.. It follows that f.1.e>.) = f.1.(U{~o: a < K.}) = 0, which contradicts the assumption that f.1. is a probability measure. D COROLLARY 23.13. Every real-valued measurable cardinal is weakly inaccessible.
For measurable cardinals, we can say even more. THEOREM 23.14. Every measurable cardinal is strongly inaccessible. PROOF. Since every measurable cardinal is obviously real-valued measurable, it suffices to show that if A is a measurable cardinal, then A is a strong limit cardinal, that is, 21< < A for each K. < A. So assume towards a contradiction that A is measurable, K. < A, and 21< 2: A. Let F : A ---+ 1<2 be a one-to-one function, and let U be a A-complete nonprincipal ultrafilter on A. For each a < K. and i E {a, I}, let X~ = {~ E A: F(O(a) = i}. For each A, X~ n X~ = 0 and X~ U X~ = A. Therefore, exactly one of these sets is in U. Define a function 9 : K. ---+ 2 by choosing g(a) such that X~(o) E U. By A-completeness of U, the intersection of any K. members of U is still a member of U; in particular, X = nO:<1< X~(o) E U. But, by our construction, X = F-l {g}. Since F was assumed to be one-to-one, X is a singleton, contradicting the assumption that U is a nonprincipal filter. D The next theorem improves and actually supersedes Theorem 23.14. However, the proof of Theorem 23.14 given above is of independent interest. THEOREM 23.15. Every measurable cardinal is weakly compact. PROOF. Let K. be measurable, and let c : [K.]2 ---+ 2 be a coloring. We want to show that there exists X E [K.]I< so that X is homogeneous for c. Let U be a K.-complete nonprincipal ultrafilter on K.. By Exercise 23.7(b), each set in U has cardinality K.. Thus, for each a E K., the set of ordinals between a and K. is in U. If we define: X~ = {f3: a < f3 < K. 1\ c({a,f3}) = i}, then exactly one of the disjoint sets X~, X~ is in U. This allows us to define an induced coloring d : K. ---+ 2 by choosing d(a) so that X~(o:) E U. Now we define recursively a sequence (a(~) : ~ < K.) by letting a(O) = 0 and a(1]) = min(ne
23. MEASURABLE CARDINALS
152
EXERCISE 23.8 (PG). (a) Convince yourself that every uniform, normal ultrafilter on an uncountable cardinal K. is K.-complete. Hint: If (Aa : 0 < >.) is a sequence of elements of U with>' < K., let Aa = K. for each 0 with>' ::; 0 < K. and consider the diagonal intersection Aa
Then there exists a
Theorem 23.15 can be strengthened as follows: THEOREM 23.17. Let K. be a measurable cardinal, n E w\ {O}, 0- < K., and let U be a uniform, normal ultrafilter on K.. Then for every coloring c : [K.t - f 0- there exists A E U such that A is homogeneous for c. PROOF. For n = 1 the existence of A follows immediately from Exercise 23.8(a) and the assumption that 0- < K.. We will prove the theorem for n = 2 and leave the general case as an exercise. So suppose 0- < K., and let c : [K.]2 - f 0- be a coloring. For 0 < K. and i < 0-, let Aa,i = {/3 : 0 < /3 < K. A. c({o,/3}) = i}. Since U is K.-complete and since Ui 2. Hint: Use induction and induced colorings. (b) Prove the following consequence of Theorem 23.17: COROLLARY 23.18. Every measurable cardinal is a Ramsey cardinal. Corollary 23.13 implies that if ZFC is a consistent theory, then the existence of real-valued measurable cardinals cannot be proved in this theory. Theorems 23.1423.17 and Corollary 23.18 reinforce this point. Moreover, even if the existence of measurable cardinals (and thus the existence of real-valued measurable cardinals) does not contradict any of the axioms of ZFC, we shall never be able to prove this fact in ZFC. Didn't we perhaps create too exotic a notion somewhere along the way? We started out by asking for an extension of Lebesgue measure to P([O, 1]), not necessarily a 2No -additive one. Does the assumption that on some cardinal K. there exists a (not necessarily K.-additive) probability measure that vanishes on the singletons already imply the existence of large cardinals? THEOREM 23.19. The smallest cardinal earrying a probability measure that vanishes on the singletons is real-valued measurable. Similarly, the smallest cardinal carrying a two-valued probability measure that vanishes on the singletons is measurable. 2Theorem 24.42
23. MEASURABLE CARDINALS
153
PROOF. Let A be a cardinal, and suppose JL is a probability measure on A that vanishes on the singletons. Suppose that JL is not A-additive, i.e., that there exists a family .N = {Nt; : ~ < ;;. < A} of subsets of A such that JL( N) = 0 for each N E .N and JL(UN) = c > O. We show that there exists a probability measure 1/ on ;;. that vanishes on the singletons. Moreover, if JL is two-valued, then so is 1/. In either case, this cannot happen if A is the smallest cardinal which carries a (two-valued) probability measure that vanishes on the singletons; therefore the smallest such A must be real-valued measurable, or, if JL is two-valued, measurable. Note that we can assume without loss of generality that the sets in .N are pairwise disjoint: If not, replace each Nt; by Nt; \ U{NI): TJ < ~}. This gives a family of null sets indexed by ordinals less than ;;. and with the same union of measure c. Now define for X S; ;;.:
1
I/(X) = ~JL(U{Nt; : ~ EX}). This is a probability measure on ;;. that vanishes on the singletons: (1) (2)
(3)
1/(0) = C 1 JL(0) = 0; 1/(;;') = C 1 JL(UN) = C 1 c = 1; I/(UnEwAn) = c 1 JL(U{Nt;: ~ E UnEwAn}) = C 1 JL(UnEw U{ Nt;: ~ E An}) ~ C 1 ~nEw JL(U{ Nt; : ~ E ~nEw 1/( An) for every countable family {An: nEw}
(4)
An}) =
of pairwise disjoint subsets of ;;. (equality (i) follows from the assumption that the Nt;'s are pairwise disjoint); I/({O) = c 1 JL(Nt;) = 0 for all ~ E ;;..
Moreover, if JL takes on only the values 0 and 1, then c = 1, and 1/ is also twovalued. D It is very important that the A in the proof of Theorem 23.19 is the first cardinal that carries a (two-valued) measure. In fact, if ;;. carries a (two-valued) measure JL and A > ;;., then A also carries a (two-valued) measure 1/: Just define I/(X) = JL(X n ;;.) for X S; A. Note that A may even be a successor cardinal or singular; as long as we do not require A-additivity of the measure, we do not contradict any of the Theorems 23.10, 23.12, or 23.14. A cardinal ;;. that carries a (not necessarily ;;.-additive) two-valued probability measure JL that vanishes on the singletons is sometimes called Ulam measurable. Now let us present a beautiful theorem of Ulam that delineates the realm of measurable cardinals among all real-valued measurable cardinals. To state it, we need one more concept. DEFINITION 23.20. Let JL be a measure on;;.. A subset X S; ;;. is called an atom of JL if JL(X) > 0, and for each Y S; X, either JL(Y) = 0 or JL(Y) = JL(X). A measure JL is callE.ld atomless if JL has no atoms. We call JL atomic if for every Z S; ;;. with JL( Z) > 0 there exists an atom X of JL such that X S; Z.
Note that being "atomic" is a stricter requirement than "not being atomless." THEOREM 23.21. Suppose /'Ii is a real-valued measurable cardinal. Then exactly one of the following holds:
154
23. MEASURABLE CARDINALS
(1) (2)
K is measurable a.nd every K-additive measure on K is atomic, or K::; 2~o and every probability measure on K that vanishes on the singletons is atomless.
PROOF.
This boils down to the following:
LEMMA 23.22. If there exists a K-additive atomless probability measure on K, then K ::; 2~o. Note that if JL is an atomless probability measure on any cardinal, then JL vanishes on the singletons, so we did not need to make this assumption explicit in Lemma 23.22. Let us show how Theorem 23.21 follows from the lemma. Consider a K-additive measure JL on K that vanishes on the singletons. Case 1: JL has an atom X. Note that X has size K. SO we can enumerate X = {xa : a < K}, and define a function /.I : P(K) ~ {O, 1} by 1 /.I(A) = JL(X)JL({x a : aEA}). As in the proof of Theorem 23.19, one can show that /.I is a two-valued K-additive probability measure on K. Thus K is measurable; and by Theorem 23.14, K > 2~o. Case 2: JL is not atomic. There exists X ~ K such that JL(X) > 0 and there is no atom Y ~ X. Again, since JL is K-additive and vanishes on the singletons {O ~ X, it follows that IXI = K. Thus, we can enumerate X = {xa: a < K}, and define an atomless probability measure /.I : P(K) ~ [0,1] on K by 1 /.I(A) = JL(X)JL({x a : a E A}). Now it follows from Lemma 23.22 that K ::; 2~o . This proves the dichotomy expressed by Theorem 23.21. PROOF OF
LEMMA 23.22. We need a technical fact about atomless measures.
CLAIM 23.23. Let JL be an atomless probability measure on K, and suppose X ~ K is such that JL(X) > O. Then there exists a partition of X into two disjoint sets XO,Xl such that i-JL(X) < JL(Xo) < £JL(X). REMARK 23.24. Note that if X o, Xl are as in the above claim, then JL(X l ) = JL(X) - JL(Xo). Thus the estimate i-JL(X) < JL(X l ) < £JL(X) holds as well. With a little extra effort, one can prove stronger results along the line of Claim 23.23. But since Claim 23.23 suffices for the present purpose, we leave the improvements as an exercise for the interested reader. PROOF OF
CLAIM 23.23. Let JL, X be as in the assumptions, and let
8 = inf{e : 3Xo, Xl (Xo U Xl = X t\ Xo n'Xl = 0 t\ e = JL(Xo) ~ JL(X l ))}. It suffices to show that 8 < £JL(X). Suppose towards a contradiction that 8 ~ £JL(X), and choose a sequence ((Xo\Xr) : nEw) of partitions of X such that JL(Xr)::; JL(Xr;) < 8+2- n . Note that JL(Xr;nX~+l) ~ 8 for each nEw: !fnot, the pair (Xonx~+l, Xr uXr+l) would be a partition of X that contradicts the choice
23.
MEASURABLE CARDINALS
155
of o. Therefore, we may without loss of generality assume that X~+1 ~ Xl) for all nEw. Now let Xo = nnEw Xl) and Xl = X \ Xo. Then J.L(Xo) = o. Since J.L is atomless, there exists X 2 ~ Xo such that 0 < J.L(X2) < J.L(Xo). If J.L(X2) 2: !J.L(X), then the pair (X 2, X \ X 2) contradicts the choice of o. If J.L(X 2) < !J.L(X), then one of the pairs (Xo \ X 2, Xl U X 2) or (Xl U X 2, Xo \ X 2) contradicts the choice ofo. 0 EXERCISE 23.10 (PG). Show that if J.L and X are as in the assumptions of Claim 23.23, then there exists Xo ~ X such that J.L(Xo) = !J.L(X). Now we are ready to prove Lemma 23.22. Let K, J.L be as in the assumptions. Claim 23.23 allows us to construct by recursion over ~ (which is a well-founded relation on <w2) a family {Xs : s E <w2} with the following properties:
(i) X0 = K; (ii) If dom(s)
= n, then we have X s- o n X s~J.L(Xs) < J.L(X s- o) < ~J.L(Xs).
1
= 0,
X s- O U X s-
1
= X s , and
For each f E w2, let Xf = n{sE<w2: sc!} Xs' Then K = UfEW2 Xf, and J.L(Xf ) = 0 for each f E W2. Thus, we have found 2~o sets of measure 0 that add up to a set of measure 1, and K-additivity of J.L implies that K ~ 2~o. This concludes the proof of Lemma 23.22 and simultaneously of Theorem 23.21. DO Let us return to Question 23.2. By Theorem 23.19, the existence of a realvalued measurable cardinal K ~ 2~o is a necessary condition for Lebesgue measure to be extendable to P([O, 1]). It turns out that this condition is also sufficient. THEOREM 23.25. If there exists a real-valued measurable cardinal K ~ 2~o, then Lebesgue measure can be extended to a probability measure on P([O, 1]). PROOF. Let J.L be a K-additive measure on K ~ 2~o, and let {Xs: s E <w2} be a family of subsets of K like the one constructed in the last part of the proof of Lemma 23.22. By recursion over ~, we construct families of reals {as: s E <w2} and {b s : s E <w2} as follows: (1) (2)
(3)
a0 =0, b0 = 1; as-o = as, bs- 1 = bs for all s E <w2; as-1 = as + J.L(Xs-o) = bs- o for all s E <w2.
Define a function H : w2 ---- [0,1] such that H(J) = x if n{sE<w2: sc!} [as, bs] = {x}. By compactness of [0,1], H is well-defined. Let us define v : P([O, 1]) ---- [0,1] by: v(Y) = J.L(H- 1 y). This is a probability measure defined on all subsets of [0,1]. It follows from our definitions that v([a s , bs]) = J.L(Xs ) = bs - as for every s E <w2. In order to see that the same is true for arbitrary subintervals [a, b] of [0, 1], note that for s E n2, each of the intervals [as,b s] has length J.L(Xs ) ~ (0.75)n. Moreover, for s =f. t, if both s, t E n2, then [as, bs ]n [at, btl contains at most one point and thus has Lebesgue measure zero. Also, for any fixed n, there are at most two s E <w2 such that [as, bs] has nonempty intersection with [a, b] but is not contained in [a, b]. It follows that Ib-a-
156
23. MEASURABLE CARDINALS
In other words, we can approximate the interval [a, bl by intervals of the form [as, bsl with any desired accuracy. Therefore, II( [a, b]) = b - a, and it follows that II extends Lebesgue measure. D Not surprisingly, the assumption that 2No is real-valued measurable has numerous implications for the structure of JR, P(w), and Ww. Let us give just one example that illustrates some of the techniques involved in the proofs of such consequences. THEOREM 23.26. If 2No is a real-valued measurable cardinal, then b < 2No. PROOF. Suppose towards a contradiction that J.£ is a 2No -additive probability measure on P(2 No) that vanishes on the singletons, and b = 2No. By Exercise 17.18(b), there exists a sequence (J~ : ~ < 2No) of elements of Ww such that f~ <* f'1 for all ~ < 11 < 2No, and the set {J~ : ~ < 2No} is unbounded in Ww. For each n,k E w let Bn,k = {~< 2No : fdn) ~ k}. Then J.£(UkEwBn,k) = 1 for each nEw, and since (Bn,k)kEw is an increasing sequence of sets, we can find for each nEw a number k n such that J.£(Bn,kJ > 1 - 2-(n+2). Then J.£( Bn,k,,) = 1 - J.£( (2 No \Bn,k,,)) 2: 1 J.£(2 No \Bn,k,,)
n
U
L
nEw
nEw
nEw
= 1 - "T(n+2)
LJ
nEw
= 1 - ~2 = ~2'
Thus nnEw Bn,k" is a set of positive measure, and since J.£ is 2N°-additive and vanishes on the singletons, InnEw Bn,k" I = 2No. It follows that the sequence (J~ : ~ E nnEw Bn,kJ is cofinal in (J~ : ~ < 2No), and hence unbounded in ww. But consider the function 9 E Ww defined by g(n) = k n + 1 . For all ~ E nnEw Bn,k" we have Ie < g, which leads to a contradiction. D We conclude this section with some remarks on consistency. As we already mentioned, one cannot show in ZFC that real-valued measurable cardinals exist. Moreover, if the theory "ZFC + there exists a real-valued measurable cardinal" is consistent, one cannot prove this fact in the theory "ZFC + CON(ZFC)." The reasons for this have been spelled out in Chapter 12 when we were discussing inaccessible cardinals. Since this makes real-valued measurable cardinal somewhat suspicious, the question arises: Should one study these objects at all? Most contemporary set theorists would answer: "Yes, but cautiously." First, we cannot prove the consistency of the existence of a (real-valued) measurable cardinal, but neither can we prove the consistency of ZFC. However, the existence of (real-valued) measurable cardinals might be consistent, and as long as we have no proof to the contrary, we should study the consequences of this possi bili ty. Second, there is at least strong empirical evidence that the existence of these cardinals is consistent with ZFC. For one thing, many seemingly unrelated questions in mathematics turn out to be equivalent t0 3 or equiconsistent with4 the existence of measurable cardinals. This adds weight to the belief that there is more substance to the concept than just being inconsistency in disguise. Moreover, although the consequences of the existence of (real-valued) measurable cardinals have been extensively studied, no inconsistency has been found so far. This resembles the status 3Question 13.15 is a prime example here. 4The notion of equiconsistency was discussed in Chapter 12.
MATHOGRAPHICAL REMARKS
157
of several famous open conjectures in number theory, like the Riemann Hypothesis. There, we know that if a counterexample exists, it must involve big numbers. Here, we know at least that no easy proof of the inconsistency of a (real-valued) measurable cardinal exists. But some caution never hurts. If a mathematical theorem has been proved under the assumption that there exists a (real-valued) measurable cardinal, one should investigate whether and how much this assumption can be weakened. The aim is to either prove the same result in ZFC, or to prove an equiconsistency result. We conclude this chapter with the perhaps most important equiconsistency results involving measurable cardinals. THEOREM 23.27. If anyone of the following four theories is consistent, then so are the other three. ( a) ZFC + ''there exists a measurable cardinal;" (b) ZFC + ''there exists a real-valued measurable cardinal", < 2No;" (c) ZFC + "2 No is a real-valued measurable cardinal;" (d) ZFC + "some uncountable regular cardinal", carries a ",-complete, ",+saturated proper ideal that contains all singletons."
Mathographical Remarks The equivalences between (a), (b), and (c) of Theorem 23.27 were announced by Solovay in 1966 and published in his article Real-valued measurable cardinals, in Axiomatic Set Theory, D. Scott, ed., Proceedings of Symposia in Pure Mathematics, vol. 13, part 1 (1971), AMS, 396-428. The implication (d)---+(a) of Theorem 23.27 was proved by K. Kunen in Some applications of iterated ultrapowers in set theory, Annals of Mathematical Logic 1 (1970), 179-227. For more information about measurable and other large cardinals we recommend the encyclopaedic treatise The Higher Infinite, by Akihiro Kanamori, Springer Verlag, 1994. D. Fremlin's article Real-valued measurable cardinals, in Set Theory of the Reals, H. Judah, ed., Israel Mathematical Conference Proceedings, vol. 6 (1993), AMS, 151-304, is a comprehensive survey of contemporary knowledge about real-valued measurable cardinals:::; 2No.
CHAPTER 24
Elementary Submodels 24.1. Elementary facts about elementary submodels Throughout this section, L denotes a first-order language with nonlogical symbols {ri : i E J}U{/] : j E J}U{ Ck : k E K}, where ri is a relational symbol of arity To(i), Ii is a functional symbol of arity Tl(j), and the Ck'S are constant symbols. Recall from Chapter 6 of Volume I that if ~ = (A, (Rf)iEI, (F/)jEJ, (Ct)kEK) and 2) = (B, (R{l)iEI, (Fl)jEJ, (Cf)kEK) are models of the language L, then a function h : A - B is a homomorphism from ~ into 2) if (i) For each i E J and each it E ATo(i) , if it E Rf, then h(ii) E R{lj (ii) For each j E J and each it E AT1(j) , h(F/(ii)) = FiB(h(it))j (iii) For each k E K, h(Ct) = Also recall that a homomorphism h is an embedding, if h is injective and satisfies the following condition (iv), which is stronger than (i): (iv) For each i E J and each it E ATo(i), it E Rf iff h(it) E Rf. In this chapter, we will consider a special class of embeddings.
cf
DEFINITION 24.1. Let
~ = (A, (Rf)iEI, (F/)jEJ, (Ct)kEK) and
(B, (R{l)iEI, (Fl)jEJ, (Cf)kEK) be models of L, and let h: A - B be an embedding of ~ into 2). We say that his an elementary embedding if (v) For every formula cp of L with free variables among vo, ... ,Vn and arbitrary ao, . .. ,an E A, ~ 1= cp[ao, ... ,an] if and only if 2) 1= cp[h(ao) , ... ,h(an )]. We say that ~ is an elementary sub model of 2), and write ~ -< 2), if A ~ B and the identity function id : A - B is an elementary embedding. 2) =
REMARK 24.2. Clearly,
~ -<~.
It would make more sense to use the symbol
::5 instead of -<, but we decided to adopt the illogical choice of symbols that is used throughout the literature on mathematical logic. EXERCISE 24.1 (R). Recall that a formula is atomic if it is either of the form
"to = tl" or of the form "ri(to, ... ,tTo (i)-I)" where i E J and to, ... ,tTo (i)-1 are terms of L. Show that if ~,2) are as in Definition 24.1, and h : A - B is an injection, then h ~s an embedding if and only if the following holds: (v)- For every atomic formula
24. ELEMENTARY SUBMODELS
160
Hint: Note that if s : w ~ A is a valuation for 2(, then h 0 s : w ~ B is a valuation for ~. Use induction over the length of t to show that for every term t of L, every valuation s : w ~ A, and each a E A we have: t S = a if and only if thos = h(a). Recall that we write 2( == ~, and say that the models 2( and equivalent, if for every sentence cp of L,
2(
F cp
if and only if
~
~
are elementarily
F cpo
The following claim is an immediate consequence of condition (v). CLAIM 24.3. Let 2( and ~ be models of L such that there exists an elementary embedding from 2( into ~. Then 2( == ~.
24.1. Let Lc be the language of group theory. 1 Let JR+ , Z+ be the sets of nonnegative reals and nonnegative integers respectively, and let lE denote the set of even integers. Consider the following models of Lc: EXAMPLE
2(
= (JR+,., 1); ~ = (Z, +, 0); ~ = (Z+, +, 0); 1) = (lE, +, 0).
Note that ~ is a submodel of~, despite the fact that ~ is a group and ~ is not. But this difference implies, by Claim 24.3, that ~ is not an elementary submodel of~.
The injection h : Z ~ JR+ defined by h(n) = 2n is an embedding of ~ into 2(. However, h is not an elementary embedding, since ~ F VVo (1 =f. Vo + vo), whereas 2( F 3vo (h(l) = Vo' vo). It is not hard to see that the models ~ and 1) are isomorphic. So it may come as a surprise that 1) is not an elementary submodel of ~. EXERCISE
24.2 (PG). Show that 1) is not an elementary submodel
of~.
EXAMPLE 24.2. Let Ls. be the language whose only nonlogical symbol is the binary relational symbol :::;, and consider the following models of Ls.:
2(= (Qn(O,oo),:::;); ~=(Q,:::;); ~=(JR,:::;); 1)= (JR+,:::;); ~=(JRn[-l,oo),:::;). Since 2(, ~, and ~ are models of the complete theory DLONE,2 it is perhaps easy to believe that 2( -< ~ and ~ -< ~. This is indeed the case, but a formal proof of this fact is nontrivial. EXERCISE 24.3 (R). Suppose 2( = (A, :::;A) and ~ = (B, :::;B) are dense l.o.'s without endpoints. Show that 2( ~ ~ implies 2( -< ~. Hint: Assume that ao
Exercise 24.3 implies in particular that 2( -< ~, ~ -< can also be deduced using transitivity of the relation -<.
~,
and 2l -<
~.
The latter
ISee the "Notation" section for a description of LG. 2This theory was discussed in Section 6.1. The letters stand for Dense Linear Order with No Endpoints. 31f 2l = IB, then property (+) expresses the fact that the elements of A are indiscernible. Sets of indiscernible elements play an important role in model theory.
24.l.
ELEMENTARY FACTS ABOUT ELEMENTARY SUBMODELS
161
EXERCISE 24.4 (G). Assume that 2l, IB, and It are models of L. Show that if h is an elementary embedding of 2l into IB and j is an elementary embedding of IB into It, then j 0 h is an elementary embedding of 21 into It. Conclude that if 21 -< IB and IB -< It, then 21 -< It. Now consider the models It, 1), and IE of Example 24.2. Since 1) has a smallest element but It doesn't, it follows from Claim 24.3 that 1) is not an elementary submodel of It. Somewhat more interesting is the fact that 1) is not an elementary submodel of IE either. To see this, note that 1) F 'v'Vl(VI ~ 0 - VI = 0),4 whereas IE F 3Vl(VI ~ 01\ -'(Vl = 0)). In more human terms, while 1) does not have any elements to the left of 0, IE does. It is instructive to compare this situation with the difference between 21 and lB. It is also true that IB contains elements to the left of 0 while 21 doesn't. But since A, the universe of the model 2l, is equal to Q n (0, 00), we have 0 ¢ A, and for every q E A, there does exist an rEA with r < q. Thus, from 21's point of view, all of its elements have the same properties as they have from lB's point of view. Of course, IB has far more radical ideas than 2l about how far to the left one can be, but 21 is not bothered by such extremes, since they are simply not expressible in terms of the parameters known to 21. In other words, if one informally defines ''21's language" as all expressions of L with parameters from A, and "1B's language" as the expressions of L with parameters from its universe Q, then the language of 2l is related to the language of IB as Orwellian Newspeak is related to natural language. EXAMPLE 24.3. Let X be a nonempty set, let U be an ultrafilter on X, let 21 be a model for L, and let X21jU be the corresponding ultrapower. For each a E A, let ga E X21jU be the function that takes the value a for all ~ EX. Then the function h: A _ XAjU defined by: h(a) = gajU is an injection of A into X21jU. EXERCISE 24.5 (PG). Show that the function h defined above is an elementary embedding of 21 into X21jU. Hint: Use LoS' Theorem. The following theorem is often called the Tarski- Vaught criterion for elementary submodels. THEOREM 24.4. Let 21 = (A, .. .), IB = (B, ... ) be models for L such that 21 S;;; lB. Then the following are equivalent: (a) 21 -< IB; (b) For every formula 1/J of L with free variables among Vo, ... , Vn-l, for every e < n, and arbitrary parameters ao, ... , ai-I, aHl, ... , an-l E A, if (24.1)
then there exists ai E A such that (24.2) PROOF. For the implication (a) (24.3)
IB
F 3vi1/J[ao, ...
(b), assume that 21-< IB and
, ai-l, ai+1,'" ,an-l]'
By condition (v) of Definition 24.1, (24.4) 4Recall that this is shorthand for D
F V'VI(VI S VO
......
VI
= Vo)[O].
162
24. ELEMENTARY SUBMODELS
and hence there exists at E A such that (24.5) Applying (v) to the formula 'I/J, we infer that equation (24.2) holds. Now assume that !2l is a submodel of ~ and (b) holds. If r.p is a formula of L with all free variables among Vo, ... ,Vn -l, let {v)cp be the statement: {v)cp
"For all n-tuples (ao, . .. ,an-I) E An, !2l F r.p[ao, . .. ,an-I] if and only if ~
F r.p[ao, ... ,an-I]."
By induction over the length of r.p, we show that {v)cp holds for every formula r.p of L. If r.p is an atomic formula, then {v)cp follows from Exercise 24.1. If {v)cp and {v)", hold, then {v)..,cp and {v)cpi\'" follow from the clauses in the definition of the satisfaction relation F that pertain to negation and conjunction. Now suppose that r.p is of the form 3vt'I/J and {v)", holds. If (24.6) then (24.7) for some a~ E A (possibly, a~
:f:. at),
and {v)", implies that also
(24.8) By the existential quantifier clause in the definition of
F,
(24.9) Now assume formula (24.9) holds. Since Vt is not a free variable in 3vt'I/J, this can be written as (24.1O) By (b), there exists
a~ E
A such that
(24.11) By {v)"" this implies (24.12) and from the definition of
F we conclude that
(24.13)
o LEMMA 24.5. Let ~ = (B, {~)iEI' {Fj)jEJ, (Ck)kEK) be a model of L, let 8 be a limit ordinal> 0, and let (A€ : ~ < 8) be a sequence of subsets of B such that A€ ~ A'1 for all ~ < "l < 8, and each A€ is the universe of an elementary submodel of~ (i.e., !2l€ -< ~, where!2l€ = (A€, (A;o(i) nRi)iEl, {A?(j)+1 nFj)jEJ, (Ck)kEK)). Then U€<6 A€ is the universe of an elementary sub model of~·
24.1. ELEMENTARY FACTS ABOUT ELEMENTARY SUBMODELS
163
Ue<6 Ae, and let (A, (U A;o(i) n R;)iEI, (U A? (j)+1 n Fj)jEJ, (Ck)kEK).
PROOF. Let A =
~=
e<6
e<6
We want to show that ~ -< ~. It suffices to verify that condition (b) of Theorem 24.4 is satisfied. Let 1fJ be an arbitrary formula of L with free variables among Vo, ... , Vn-1, let f < n, and let ao, ... , at-1, at+!, ... , an-1 be arbitrary elements of A such that ~
(24.14)
F 3vt1fJ[ao, ...
,at-1,aH1,··· ,an-1].
We want to find at E A such that (24.15) Since 8 is a limit ordinal, we can find ~ < 8 such that {ao, ... , at -1, aH 1, ... , an - d ~ Ae· Since ~e -< ~, condition (b) of Theorem 24.4 holds for ~e and~. Thus there exists at E Ae such that (24.16) Since Ae
~
o
A, this at is also an element of A, and we are done.
Now it is your turn to practice applications of Theorem 24.4. EXERCISE 24.6 (PG). Let ~ and (~e : ~ < 8) be as in the assumptions of Lemma 24.5. Show that ~e -< ~ for all ~ < TJ < 8. EXERCISE 24.7 (PG). (a) Theorem 24.4 can be treated as a criterion that tells us when the identity mapping is an elementary embedding. Formulate and prove a generalization of Theorem 24.4 to a criterion that tells us when an arbitrary embedding of a model Il into a model ~ is elementary. (b) Use the criterion found in point (a) to prove the following: 24.6. If h : A --+ B is an isomorphism of models (B, .. .), then h is an elementary embedding of~ into ~.
CLAIM
~ =
~
== (A, ... ) and
REMARK 24.7. (a) The models 1) ~ ~ of Example 24.1 and 1) ~ (!: of Example 24.2 are isomorphic, and yet the smaller models are not elementary submodels of the larger ones. This does not contradict Claim 24.6 though, since the inclusion maps are different from the isomorphisms. (b) Theorem 6.4 asserts that isomorphic models are elementarily equivalent. This is an immediate consequence of Claims 24.3 and 24.6. By Corollary 21.31, if L is a countable language, and if ~ = (A, ... ) is an uncountable model of L, then ~ has many countable submodelsj in fact, the universes of these models form a club subset of [A]::;N o • Are at least some of these countable submodels elementary? To study this question, let us first assume that ~ has the following property: (SK) For eachn E w\{O}, each formula cp of L with free variables among vo, ... , Vn , and each f S n there exists a functional symbol f
24. ELEMENTARY SUBMODELS
164
If 2( satisfies condition SK, then we call 2( a Skolemized model. Let Form'l denote the set of formulas of L with free variables among Vo, ... ,Vn . The set {F""n,t: 0 < n < w,l ~ n,cp E Form'l} of interpretations of the function symbols f""n,t is called a set of Skolem functions 5 for L. LEMMA 24.8. Suppose ~ is a Skolemized model and 2( is a nonempty submodel Then 2( -< ~.
of~.
PROOF. If ~ = (B, (~)iEI' (Fj)jEJ, (CkhEK) is any model of L and 2( is a submodel of ~, then the universe of 2( is closed under all functions F j . Thus, if ~ is Skolemized, then 2( and ~ satisfies condition (b) of Theorem 24.4, and hence 2(-<~.
0
So, apparently, at least some uncountable models have lots of elementary submodels. But how common is it for a model2( to satisfy condition SK? Unfortunately, models of the most popular first-order languages never satisfy condition SK. If L has a Skolemized model, there must be a functional symbol f ""n,t in L for every cp E Form'l and l ~ n. For example, the language of set theory Ls contains no functional symbols whatsoever, and hence Ls has no Skolemized models. For a similar reason, no model of Le is Skolemized, not even the one-element group C5 1 . For C5 1 to be Skolemized, we would need, among other things, a function symbol f",,3,3 of arity 3 to represent the Skolem function for the formula ":lV3 (V3 = (vo *VI) *V2)," and no such symbol is available in Le. Isn't that silly? After all, the constant e seems to be such a wonderful "all purpose" Skolem function for the one-element group. Let us try to give a precise meaning to this idea. Recall from Chapter 6 that if L+ is a richer first-order language than L, Le., if the nonlogical symbols of L+ are {ri: i E I+}U{Ii: j E J+}U{Ck: k E K+}, with I ~ 1+, J ~ J+, and K ~ K+; and if~+ = (B'(~)iEI+,(Fj)jEJ+,(CkhEK+) is a model of L+, then the L-reduct of ~+ is the model ~ = (B, (~)iEI' (Fj)jEJ, (Ck)kEK). The notion of an elementary sub model behaves nicely with respect to reducts. CLAIM 24.9. Suppose L+ is a richer language than L, 2(+ and ~+ are models for L +, and 2(, ~ are the corresponding L-reducts. (a) If cp is a formula of L with free variables among Vo, ... ,Vn and ao, ... ,an are elements of the universe of 2(, then 2( F cp[ao, ... ,an] if and only if 2(+ F cp[ao, ... ,an], (b) If 2(+ -< ~+, then also 2( -< ~. EXERCISE 24.8 (PG). Prove Claim 24.9. Now let us consider a language L6 that, in addition to the nonlogical symbols * of L e , has functional symbols Ii of arity j for every j E w\ {o}. The language L6 has exactly one one-element model C5t = ({e}, ii<, (Fj)jEw\{O}), where Fj(e, ... ,e) = e for all j E w\{O}. Clearly, C5 1 is the Le-reduct C5t.
e and
EXERCISE 24.9 (G). Convince yourself that C5t satisfies SK. 50f course, if
24.1. ELEMENTARY FACTS ABOUT ELEMENTARY SUBMODELS
165
We have shown that although ~l is not a Skolemized model of La, it is the reduct of a Skolemized model of a richer (but still countable) language Lt;. As it turns out, ~l is by no means unique in this respect. THEOREM 24.10. Let L be a first-order language with nonlogical symbols {r i : i E I} U {lj : j E J} U {Ck : k E K}, and let ~ = (A, (~)iEI' (Fj)jEJ, (Ck)kEK) be a model for L with nonempty universe. Then there exist J+ :;:2 J and a Skolemized model ~+ for the language L+ with nonlogical symbols {ri : i E I} U {lj : j E J+} U {Ck : k E K} such that 21 is the L-reduct of 21+ . Moreover, J+ can be chosen in such a way that IJ+I = III + IJI + IKI + ~o. PROOF. Let !B be a model with nonempty universe of a first-order language Lm with nonlogical symbols {ri : i Elm} U {Ii : j E J m } U {Ck : k E Km}. We construct a language sk(Lm) and a model sk(!B) as follows: For every n E w\{O}, l ~ n, and
F
, ai-I, f<{J,n,t(ao, ... at-I, at+I,··· , an), at+I,'" , an). Hint: Use Claim 24.9(a). 21+
Unfortunately, the language of~+ is L+, and we need functional symbols f<{J,n,t for every formula '
166
24.
ELEMENTARY SUBMODELS
for each mEw. Let J+ = UmEw J m , and let L + be the language with relational symbols {ri : i E I}, functional symbols {Ii : j E J+}, and constant symbols {Ck : k E K}. Let ~+ = (A, (R.)iEI, (Fj)jEJ+, (Ck)kEK), where the interpretation Fj of the functional symbol Ii in ~+ is the same as in ~m(j), with m(j) being the smallest mEw such that Ii E Jm . EXERCISE 24.12 (PG). Convince yourself that L+ and
~+
are as required.
0
Theorem 24.10 has a number of interesting consequences: THEOREM 24.11. Let ~
1'1,
be a regular uncountable cardinal, and let
= (B, (R.)iE], (Fj)jEJ, (Ck)kEK)
be a model for a language L with nonempty universe and fewer than symbols. Then the set
1'1,
nonlogical
{A E [BJ
is club in [BJ
24.2. APPLICATIONS OF ELEMENTARY SUBMODELS IN SET THEORY
167
24.2. Applications of elementary submodels .in set theory Naturally, we are most interested in elementary submodels of models of set theory. All the models we are going to consider in this section will be standard models, i.e., models of the form (M, E), where M is a set and E stands for the standard interpretation of the binary relational symbol E, that is, E denotes the "real" membership relation restricted to elements of M. Most of the time, we will refer to a model (M, E) simply as "the model M." Of course, there is a major problem with models of ZFC: By G6del's Second Incompleteness Theorem, ZFC does not prove their existence. When we speak of "models of set theory" in this section, we don't necessarily mean models of ZFC, but models of a "sufficiently large fragment of ZFC." Let us illustrate this notion with an example. In Chapter 12 we defined for every infinite cardinal A the class H).. of all sets hereditarily of cardinality less than Aj Le., H).. = {x : ITC(x)1 < A}, where TC(x) denotes the transitive closure of x. We showed that each H).. is a transitive set (Le., H).. is not a proper class, and if y E H).. and x E y, then x E H)..), and that the model (H).., E) satisfies quite a few of the axioms of ZFC. Thus, a suitably chosen H).. may be a model for a "sufficiently large fragment of ZFC." But what does "suitably chosen" mean? Our first application of elementary submodels in set theory will be an alternative proof of the ~-System Lemma. In the proof, we will fix an uncountable family of finite sets B, and show that B contains an uncountable ~-system6 A ~ B. We will pick A such that ITC(B)I < A, Le., B E H)... Then we will argue, using an elementary submodel of H).., that
(24.17)
H)..
1= 3A ~ B (A is an uncountable ~-system).
But are H)..'s beliefs about the existence of an uncountable correct? What we really want is the following:7 (24.18)
~-subsystem
of B
v 1= 3A ~ B (A is an uncountable ~-system).
Does (24.17) imply (24.18)? Let us first show how to answer this question the hard way. If (24.17) holds, then there exists A E H).. such that (24.19) (24.20)
H)..
1= A is a ~-systemj
H)..
1= A is uncountable.
and
(24.21)
We want to show that (24.19), (24.20), and (24.21) remain true if H).. is replaced by V. By Claim 12.19(c), (24.19) implies that V 1= A ~ B. Now consider (24.20). This can be translated into: (24.22)
3r E H)..Va,b E AnH).. (H)..I= a =I- b --+ anb = r).
Since H).. is transitive, it is not hard to see that H).. 1= a =I- b --+ an b = r if and only if the implication a =I- b --+ a n b = r is really true. Similarly, transitivity of 6 Actually,
we will show something stronger, but let us keep this illustration simple. 7The satisfaction relation for proper classes was defined' in Chapter 12. Essentially, the expression "V 1=
24. ELEMENTARY SUBMODELS·
168
H).. implies that An H).. == A. Thus, (24.22) implies: (24.23)
3r\la, bE A (a
In other words, A really is a
=/: b -+ an b == r).
~-system.
EXERCISE 24.14 (G). Show that if f E H).. is such that
(24.24) then
H)..
Ff
is an injection from A into w,
f really is an injection from A into w.
Note that in the last step of transforming (24.22) into (24.23) we took advantage of the fact that if there exists a kernel r for A in H).., then obviously such r exists in V. Formula (24.21) is more tricky in this respect. It translates into
(24.25)
-.3f E H).. (H)..
Ff
is an injection from A into w),
which by Exercise 24.14 translates into:
(24.26)
-.3f E H)..(j is an injection from A into w).
The nonexistence of a certain object in H).. does not automatically preclude the existence of such an object in V. However, in the case of f we are lucky: An injection from A into w is a subset of Ax w, and it is not hard to see that H).. is closed under subsets and Cartesian products of finitely many factors. Thus, any injection from A into w must be an element of H).., and we have proved that (24.17) implies (24.18). A similar argument can be used to show that (24.18) implies (24.17); but this is not needed for the present purpose. Let us analyze our reasoning at a more abstract level. In Chapter 12, we associated with every formula rp of Ls and every class X a formula rp IX (called the relativization of rp to X) that is obtained by replacing each quantifier "3v/, occurring in rp by "3Vi E X" and each quantifier "\lvi" occurring in rp by "\lVi EX." Then we defined the relation X F rp as a fancy way of writing rp/x, and remarked (Exercise 12.9) that if X happens to be a set, then rp IX holds if and only if (X, E) F rp in the usual sense (Le., in the sense of the definition given in the "Notation" section). We called a formula rp E Form2~1 absolute for X if rp(xo, ... ,Xn-l) +-+ rp IX(xO,' .. ,Xn-l) for all xo, ... ,Xn-l EX. Our proof that (24.17) implies (24.18) is nothing else but one half of the proof that the formula "vo contains an uncountable ~-system" is absolute for H).. whenever>. is an infinite cardinal. Now let us go back to the abstract situation. Suppose we are given a formula rp E Form2~1 and sets ao, ... ,an-I, and we want to prove, with the help of elementary submodels, that rp(ao, ... ,an-d holds. All we need is an H).. such that ao, ... ,an- l E H).., the formula rp is absolute for H).., and H).. F rp[ao, ... ,an-I]' In many cases, absoluteness of rp for H).. can be verified by arguments similar to the reasoning in our illustration. However, most of the time even this is not necessary. The next theorem provides a convenient shortcut. THEOREM 24.14 (Reflection Principle). Let rp be a formula of Ls with free variables among Vo, ... , Vn-l' Define Ccp == {a : \lao, ... , an-l E Vo: (V F rp[ao, ... ,an-I] +-+ Vo: F rp[ao, ... ,an-I])}' Then Ccp contains a closed unbounded class of ordinals. How can Theorem 24.14 help us in choosing suitable H)..'s for our arguments? LEMMA 24.15. Let E == {>. E Card: V).. == H)..}. Then E is a closed unbounded class of cardinals.
24.2. PROOF.
(24.27)
APPLICATIONS OF ELEMENTARY SUB MODELS IN SET THEORY
169
If A = SUPo
V>. =
U
V>.",;
U
H>.",.
o
H>. =
o
A2n+l = min{a: ~ A2n : H>'2n s:;; Vo}; A2n+2 = min{1I: ~ A2n+l : V>'2n+l s:;; H,,'}. Let Aw = sUPnEw An. Then Aw ~ A. By (24.27) and (24.28), Aw E E. We have D shown that E is unbounded in ON. Now suppose we are given a formula cp E FOTm~~1 and sets ao, ... ,an-I, and we want to prove, with the help of elementary submodels, that cp( ao, . .. ,an-I) holds. Since every closed unbounded class of cardinals is in particular a closed unbounded class of ordinals, and since the intersection of two closed unbounded classes of ordinals is a closed unbounded class of ordinals,S Theorem 24.14 and Lemma 24.15 imply that there exists a cardinal A such that H>. = V>., and H>. F cp[ao, ... ,an-I] if and only if V F cp[ao, ... ,an-I]' In most cases, we may as well start our argument with picking A as above, and then restrict the proof to showing that H>. F cp[ao, ... ,an _I].9 Why did we write "in most cases"? Don't Theorem 24.14 and Lemma 24.15 imply that there is always a suitable A around? True enough, but you have to be careful about what you can and what you cannot assume about A. Note that Theorem 24.14 guarantees that Ccp contains arbitrary large cardinals, even cardinals of arbitrary large cofinality, but Theorem 24.14 does not guarantee that Ccp contains any regular cardinals. As long as you only need to assume that A is sufficiently large, or even that Cf(A) > w, you may safely use Theorem 24.14 and Lemma 24.15. But if for some reason your argument uses the assumption that A is a regular uncountable cardinal, you are back to square one and you must either verify absoluteness of CPH".[ao, ... , an-d directly, or modify your argument. Let us summarize what we have said so far by giving an "official" definition of the phrase "sufficiently large A" that will appear at the beginning of several arguments later in this chapter: If ao, ... , an-l are given and if we want to prove a formula cp(ao, ... , an-I) with the help of elementary submodels, then A is "sufficiently large" if an, ... , an-l E H>., the formula cp is absolute for H>., and H>. satisfies all the (finitely many) axioms of ZFC that will be used (implicitly or explicitly) in the proof. lO Theorem 24.14 and Lemma 24.15 imply that such A exists. However, note that our approach does not guarantee that if A is suitable and II: > A, 8This was shown in Exercise 21.19. 90f course, we might as well show that V",
F c,o[ao, ...
,an-I], but since the structure of H",
is easier to describe, set theorists usually prefer working with H).. IOThis is what we mean when we speak of a "sufficiently large fragment of ZFC."
24.
170
ELEMENTARY SUBMODELS .
then K is also suitable. We shall always put the phrase "sufficiently large" in scarequotes to alert you that it should not be taken too literally. While we are at it, let us mention another important application of Theorem 24.14. Recall from Chapter 12 the definitions of the La-hierarchy and of the class L. EXERCISE 24.15 (PG). Show that the class D = {a E ON : La = L n Va} is closed and unbounded in ON. Hint: Modify the argument of the proof of Lemma 24.15. The following result appeared in Volume 1 as Lemma 12.35. COROLLARY 24.16. Let a E ON and x,ao, ... ,ak-I E La, and let cP E Form£;I. Then there exists j3 > a such that for every z Ex:
(L{3, E)
F cp[z, ao, ...
,ak-I]
iff L
F cp[z, ao, ...
,ak-I].
PROOF. Let a E ON, and let cP, x, ao, ... ,ak-I be as in the assumptions. Let '¢ be a representation of L, i.e., let '¢ be a formula with one free variable such that L = {y : '¢(Yn. By Theorem 24.14 and Exercise 21.19, there exists j3 > a such that for all z E V{3:
V
(24.29)
F ,¢[z]
iff V{3
F ,¢[z];
and (24.30)
V
F cp/dz,ao, ...
,an-d iff V{3
F cp/dz,ao, ...
,an-I].
The left hand side of (24.30) is equivalent to (24.31)
L
F cp[z,ao, ...
,an-I];
and by (24.29), the right hand side of (24.30) is equivalent to (24.32)
V{3 n L
F cp[z, ao, ... , an-I].
By Exercise 24.15, we can choose j3 as above in such a way that V{3 n L = L{3, which allows us to rewrite (24.32) as: (24.33)
(L{3, E)
F cp[z, ao, ... ,an-I].
Since (24.31) is equivalent to (24.33) for all z E L{3, and since x ~ L{3 by transitivity of L{3, Corollary 24.16 follows. 0 Now let us prove Theorem 24.14. PROOF OF THEOREM 24.14. Let cp be as in the assumption. Let CPo, ... ,CPs be a list of all subformulas of cp (including, of course, cp itself). For each Cpj on this list that is of the form 3VijCPi for some f ::::; s, we define a functional class G j as follows: Fix mj > i j such that the free variables of CPi are among va, ... ,vmj . The domain of G j will be the class of all mrtuples (an, ... ,aij-l> aij+l," . ,amJ (we will use the shorthand ii to denote such tuples). If 3Vij CPi(ii), then we define Gj(ii) = min{a : 3aij EVa CPi(aO, ... ,aij-I,·aij,aj;+I,.·. ,amJ}. If -dVij cpf.(ii), then we define Gj(ii) = O. EXERCISE 24.16 (PG). Convince yourself that G j really is a functional class, i.e., convince yourself that there is a formula '¢j of Ls such that G j = {(ii, a) ,¢j(ii, an.
24.2. APPLICATIONS OF ELEMENTARY SUB MODELS IN SET THEORY
171
For every ordinal ,,(, the restriction Gjr(V,)mj is a function. By the Axiom of Replacement, the range of the above restriction is a set, and hence there exists f3 such that rng(Gjr(V'Y)m j ) ~ V/3. Thus, we can define a functional class F j : ON --+ ON as follows: F j (() = max{(,min{'17: rng(Gjr(V<;)m j ) ~ V1j}}. EXERCISE 24.17 (G). Convince yourself that the functional class F j defined above is normal in the sense of the definition preceding Claim 21.23. By Claim 21.23, the class Fix(F j ) defined as {a E ON : Fj(a) = a} is closed and unbounded in ON. Define D = Fix(Fo) n Fix(Fd n ... n Fix(Fs). By Exercise 21.19, the class D is also closed and unbounded in ON. It remains to show that D ~ C"". EXERCISE 24.18 (PG). Show that for all a E D and all ao, ... ,an - l EVa,
V F !p[aO,'" ,an-I] iff Va F !p[aO,." ,an-I]. Hint: Use an argument similar to the proof of Theorem 24.4.
0
Theorem 24.14 implies that for every finite subtheory T of ZFC, there is an abundance of models of T. Indeed, suppose T = {!Po, . .. ,!Pm}, and let !P be the sentence !Po /\ ... /\ !Pm. By Theorem 24.14, there is a proper class of ordinals a such that Va F !p, and hence, Va FT. One is naturally tempted to indulge in the following reasoning: Let us enumerate the axioms of ZFC by {!PO,!Pl,!P2,' .. }, let Tn = {!Po,!pI, ... ,!Pn}, and let C n denote the class {a : Va F Tn}. By Exercise 21.20, C = nnEw C n is a proper class, hence nonempty. Let a = min C. Then Va F Tn for every n; hence Va F ZFC. So, there exists a set model of ZFC, and we have contradicted G6del's Second Incompleteness Theorem! What is wrong here? In Exercise 21.20, we alerted you that forming the intersection n iE1 Di of a set of proper classes is a problematic operation-since there is no such thing as a nonempty set of proper classes! In some cases, the use of the symbol niEI Di may still be justified though. If D is a class of ordered pairs, and Di = {x : (i, x) E C}, then niEI Di can be understood as shorthand for the class {x : Vi E I ( (i, x) ED)}. However, for the classes C n considered above there is no class C such that (n, a) E C if and only if Va F !Pn for all nEw and a EON. Any representation of such a class C would have to be an infinitely long formula. Thus the expression nnEw C n is meaningless, and we get no contradiction. EXERCISE 24.19 (PG). Convince yourself that we have just proved the following: COROLLARY
24.17. The theory ZFC is not finitely axiomatizableY
It is worth comparing these limitations of Theorem 24.14 with the amount of reflection one gets from large cardinals. Let K. be a strongly inaccessible cardinal. In Chapter 12 we showed that VI< F ZFC. By Theorem 24.11, the set A = {A E [VI<]
THEOREM 24.18. Let K. be a strongly inaccessible cardinal. Then the set B = (VI<, E)} is club in K..
{f3 < K.: (V.B, E) -<
UI.e., there is no finite theory T in Ls such that for all sentences 'P of Ls, T f-- <po
r 'P
iff ZFC
172
24. ELEMENTARY SUBMODELS
PROOF. Let /1" B be as above. Then /1, is a club subset of [/1,J<1< (see Example 21.10). Let A be as in the discussion preceding Theorem 24.18. Then B = An/1" and since the intersection of two club sets is club, the theorem follows. 0 REMARK 24.19. There is nothing special about the language Ls here. For example, if R ~ Vl
Lt
Note that Theorem 24.18 implies that the first ordinal Q such that Va F ZFC, if it exists, is not a strongly inaccessible cardinal. Since all club sets contain ordinals of countable cofinality, Theorem 24.18 also implies that it is possible that Va F ZFC even though Q has countable cofinality (compare this with Remark 12.20). Finally, let us look at Theorem 24.18 from the point of view of consistency strength. For nEw, define recursively a theory ZFcn in Ls as follows: ZFCo = ZFC, ZFcn+1 =ZFcn U {CON(ZFcn)}, where CON(T) can be interpreted as "There exists a model of the theory T." EXERCISE 24.20 (PG). Let B be as in Theorem 24.18, and let (!3n)nEw be an enumeration of the first w elements of B in increasing order. Show that for all nEw, V,6" F ZFcn. Conclude that for each nEw, ZFC + "3 a strongly inaccessible cardinal" f- ZFcn. Now let us assume H>. is a model of a "sufficiently large" fragment T of ZFC, and let us assume that M is a countable elementary submodel of H>.. What does M look like? Since H>. F 3xVy (y fJ. x), also M F 3xVy (y fJ. x). Hence, there exists ao E M such that M F Vy (y fJ. ao)· Since M is an elementary submodel of H>., we also have H>. F Vy (y fJ. ao), and since H>. is transitive, ao must be the empty set. Thus we have shown that 0 EM. A similar argument could now be used to show that {0} E M, and by induction one could show that n E M for every nEw. This is important information that will be used later on, but it is only a special case of the following more general observation. Recall that a E A is definable over a model Qt = (A, ... ) with parameters ao, ... ,ak-1, if there exists a formula cp with free variables Vo, ... ,Vk such that Qt F cp[ao, ... ,ak-1,a] and Qt F Vx (cp(ao, .. · ,ak-1, x) --? x = a). If k = 0, then we simply say that a is definable over Qt. LEMMA 24.20. Suppose !2( = (A, ... ,) and 23 = (B, ... ,) are models of a first order language L such that 23 -< Qt. If a is definable over Qt with parameters ao, ... ,anEB, thenaEB. EXERCISE 24.21 (G). Prove Lemma 24.20. One immediate consequence of Lemma 24-.20 is that elementary submodels of H>. are closed under finite subsets. EXERCISE 24.22 (G). Suppose>. is "sufficiently large" and M -< H>.. Show that if x ~ M and x is finite, then x E M. Hint: If x = {ao, ... ,an}, consider a suitable formula with parameters ao, ... ,an,
24.2.
Vw
APPLICATIONS OF ELEMENTARY SUBMODELS IN SET THEORY
173
EXERCISE 24.23 (PG). Show that if A is "sufficiently large" and M -< H>., then M. Hint: Use Lemma 24.20 and induction over the wellfounded relation E.
~
Lemma 24.20 implies that the definable ordinals w, WI, W2, ••• , W W , W W1 ' • •• can all be assumed to be members of M. Of course, if M is countable, not all countable ordinals can be elements of M. So, what does M n WI look like in this case? To settle this question, we need a technical lemma. LEMMA 24.21. Suppose M -< H>., K is a cardinal less than A, and K Then for every x E M, if H>. F Ixl = K, then x ~ M.
~
M.
PROOF. First note that although in general it is not the case that K ~ M implies K E M,12 our assumptions do imply that K E M: If there exists x E M such that H>. F Ixl = K, then in particular, H>. F 3J.L (lxl = J.L). Byelementarity, M F 3J.L (Ixl = J.L); and hence there exists J.L E M such that M F Ixl = J.L. Again by elementarity, H>. F Ixl = J.L, hence J.L = K, and thus K E M. Now M F 3f E "'x f[K] = x; hence there exists f E M such that M F f E "'x /\ f[K] = X. But then H>. F f E "'x /\ f[K] = x; hence f really maps K onto x. If y E x, then there is some E K such that f(O = y. But K ~ M implies that ~ E M, and hence M F f(e) = ii for some y. Since H>. F f(O = y, this ii must be y. 0
e
COROLLARY 24.22. If M
-< H>. and x is a countable set such that x E M, then
x~M.
PROOF. This follows immediately from Exercise 24.23 and Lemma 24.21.
0
Now suppose a E M n WI' Then a is a countable set of ordinals; hence by Corollary 24.22, a ~ M. In other words, f3 E M for every f3 E a. A similar argument applies if a < K+ and K ~ M. We have proved the following: CLAIM 24.23. Suppose M -< H>.. (a) If M is countable, then M nWl = 8 for some countable ordinal 8. (b) If IMI = K and K ~ M, then M n K+ = 8 for some ordinal 8 < K+. Now we are ready to present some applications of elementary submodels. Our first example will be a solution to Exercise 16.2. LEMMA 24.24. Let B be an uncountable family of finite subsets of WI' Then there exists a countable subset N ~ WI such that every b E B which is not contained in N is a member of an uncountable Il-system A ~ B with kernel r ~ N. PROOF. Let B be as in the assumptions. Fix a "sufficiently large" A, and let M be a countable elementary submodel of H>. such that B E M (recall that the existence of such M follows from Corollary 24.13). Let N = M n WI' By Claim 24.23, N is equal to a countable ordinal 8. We show that N is as required. Consider bE B such that b is not contained in N. Let r = bnN. By Exercise 24.22, b is not an element of M, but r is. Now let a be an arbitrary countable ordinal in M. Then a < 8, and thus
H>.
(24.34)
Fr
~
b /\ b\r =I- 0/\ min(b\r) > a.
In particular, (24.35)
H>.
F 3x E B (r ~ x /\ x\r =I- 0/\ min(x\r) > a).
12For example, if M = VI< then
I<
~ M but
I<
i: M.
24. ELEMENTARY SUB MODELS
174
The parameters a and r of (24.35) are all in M; hence by elementarity, (24.36)
Since a
F 3x E B (r ~ x /\ x\r of. 0/\ min(x\r) > a). was assumed to be an arbitrary element of M n WI, we have
(24.37)
M
M F'Va < w I 3x E B (r ~ x /\ x\r
of. 0/\ min(x\r) > a).
H>. F'Va < wax E B (r ~ x /\ x\r
i= 0/\ min(x\r) > a).
By elementarity, (24.38)
Since H>. 2 WI, (24.38) allows us to recursively construct a transfinite sequence (a€)€<Wl of countable ordinals and a sequence (b€)~<Wl such that r is a proper subset of b~ and a < a~ < min(b~\r) :::; maxb~ < a,., for all € < TJ < WI. It is easy to see that the set {b~ : € < wd is a a-system with kernel r, as desired. 0 EXERCISE 24.24 (G). Prove the following generalization of Lemma 24.24: LEMMA 24.25. Let K be an uncountable cardinal, and let B be a family of finite subsets of K+ with IBI = K+. Then there exists a subset N ~ K+ of cardinality K such that every b E B which is not contained in N is a member of a tl-system A E [BJI<+ with kernel r ~ N. In the proof of Lemma 24.24, the assumption that B consists of subsets of WI played a crucial role. Can we get away without using this assumption? LEMMA 24.26. Let B be an uncountable family of finite sets. Then there exists a countable set N such that every b E B which is not contained in N is a member of an uncountable tl-system A ~ B with kernel r ~ N. PROOF. Let B be as in the assumptions. Fix a "sufficiently large" A, and let M be a countable elementary submodel of H>. such that B E M. Let N = M. We show that N is as required. Consider b E B such that b is not contained in M, and let r = b n N. By Exercise 24.22, b is not an element of M, but r is. In the proof of Lemma 24.24, we proceeded by considering countable ordinals
in M. This won't help us here, since b may not be a set of ordinals. We need to consider something more general: Let D be an arbitrary countable set such that DE M and r ~ D. By Corollary 24.22, D ~ M, and thus bnD = rand b\D i= 0. It follows that (24.39)
M
F 'VD 2
H>.
F 'VD 2 r (IDI
r (IDI :::; No
-+
3a E B (a
i= r /\ anD =
r)).
-+
3a E B (a
i= r /\ anD =
r)).
By elementarity, (24.40)
:::; No
EXERCISE 24.25 (G). Complete the proof of Lemma 24.26.
o
Now let us consider a situation where the kernels of tl-systems may be infinite. THEOREM 24.27. Let J.L and v be infinite cardinals such that vI' = v. Let B be a family of sets such that IBI > v and Ibl :::; J.L for all b E B. Then there exists a set N of cardinality v such that every b E B that is not contained in N is a member of a tl-system A ~ B with kernel r ~ N such that IAI > v. EXERCISE 24.26 (G). Convince yourself that Theorem 24.27 implies Theorem 16.3 for the case when both K and A are successor cardinals. Hint: Let K = J.L+ and A = v+.
24.2.
APPLICATIONS OF ELEMENTARY SUBMODELS IN SET THEORY
175
PROOF OF THEOREM 24.27. Let J.I., v, B be as in the assumptions. Fix a "sufficiently large" A, and let M be an elementary submodel of H>. such that J.I., v, B EM. We'd like to show that N = M is as required. Thus, in particular, M must be of cardinality v. Moreover, if bE B is such that b is not contained in M and r = bnN, then b should not be an element of M, but r should be. How can we make sure that this is the case? The first of these two requirements is relatively easy to handle: Let us choose M in such a way that v ~ M. By Corollary 24.13, it is possible to find M -< H>. of cardinality v such that vU{v,B} C M.13 Note that the assumptions of Theorem 24.2 imply J.I. < v, and hence we have {! ~ M and {! EM for every {! :S J.I.. If M is as above and b E M is a set of cardinality {! :S J.I., then by elementarity, M F Ibl = {!. By Lemma 24.21, b ~ M. In other words, if bE B is not a subset of M, then b ~ M. Making sure that r = b n M is an element of M is trickier. In contrast to the proofs of Lemmas 24.24- 24.26, r does not need to be finite, and Exercise 24.22 no longer applies. We need a new concept. Let K, be an infinite cardinal. We say that M is closed under subsets of size less than K, if [MJ
M
F VD 2 r(IDI
:S v
---+
3a E B (a
=1=
r /I. anD
= r».
Byelementarity, (24.42)
H>.
F VD 2 r (IDI :S v ---+ 3a E B (a =1= r /I. anD = r».
As in Exercise 24.25, we can recursively construct transfinite sequences (D~)~
A ~ Mo; [Ma]<1< ~ M/3; IMa I = v; Ma -< H>..
Actually, at this point we only need Jl ~ M, but at a later stage of the argument we will
use the inclusion
II
~
M.
176
24.
ELEMENTARY SUBMODELS
The construction is straightforward: Use Corollary 24.13 to construct Mo such that (i), (iii), and (iv) hold. Having constructed M a , note that since v = v
(ii)O Ma <; M{J. Thus at limit stages {j :::; v, we can define M6 = Ua<6 Ma. By Lemma 24.5, M6 -< H)". Now let us show that Mv is closed under subsets of size less than K. Note that the assumption v<1< = v implies that c/(v) ~ K. Thus, if x E [MvJ
nw(X) = min{INI : N is a network for X} + No. Note that a network for X consisting of open sets is the same thing as a base for X. It follows that nw(X) :::; w(X) for every topological space X. The inequality may be strict: For example, the space (w + 1,TF) of Example 26.8(b) has uncountable weight, but countable network weight, since the set of singletons is obviously a network for (w+ 1, T F ). However, for compact Hausdorff spaces the network weight is always the same as the weight. LEMMA 24.29 (Arhangel'skii). Let (X, TI) be a Hausdorff space and assume nw(X) Then there exists a topology TO <; TI on X such that (X, TO) is Hausdorff and has weight:::; ,...
=,...
COROLLARY 24.30 (Arhangel'skii). For every compact Hausdorff space X, nw(X) = w(X).
TO
PROOF. If (X, TI) is compact Hausdorff and TO is as in Lemma 24.29, then (see Theorem 26.3). 0
= TI
PROOF OF LEMMA 24.29. Let X, TI, ,.. be as in the assumptions, and let N be a network for (X, TI) of size ,... Choose a "sufficiently large" A, and let M -< H)" be such that Nu{N, X, Td <; M and IMI = ,... Let TO be the topology on X generated by the base M n TI. Clearly, TO <; TI, and by definition, (X,TO) has weight:::; ,... Let us show that (X, TO) is Hausdorff. Consider two distinct points xo, Xl E X (not necessarily in X n M). We want to show that there are disjoint Wo, WI E TO such that Xo E Wo and Xl E WI. Since TI is a Hausdorff topology, there exist disjoint
24.2. APPLICATIONS OF ELEMENTARY SUBMODELS IN SET THEORY
177
open sets Uo, U1 E T1 such that Xi E Ui for i E {O, I}. Unfortunately, nothing we have said so far guarantees that Uo and U1 are in TO. But here comes the trick: By the definition of a network, there exist No, N1 EN such that Xi E Ni ~ Ui for i E {O, I}. Thus (24.43)
H).
F 3Vo, VI
E T1 (Vo
n VI
=
0/\ No
~
Vo /\ N1 ~ ~).
Since N C M, the parameters No and N1 of the last formula are in M, and elementarity implies: (24.44) Thus, there are disjoint open W o, WI E MnTl (and hence in TO) such that No ~ Wo and N1 ~ WI. Since Xi E Ni for i E {O, I}, we have shown that Xo and Xl are separated in TO. 0 THEOREM 24.31 (Arhangel'skii). If X is a Lindel6f, first countable Hausdorff space, then IXI ::; 2 No. PROOF. Let (X, T) be as in the assumptions; choose a "sufficiently large" >', and let M -< H). be such that X,T E M, IMI = 2No , and M is closed under countable subsets. Such M exists by Lemma 24.28. Let Y = X n M. Obviously,
WI ::; 2No.
CLAIM 24.32. For every Y E Y, M contains a local base By at y. PROOF. Let y E Y. Then (24.45)
H).
F 3By (By is a countable local base at y).
M
F 3By (By is a countable local base at y).
By elementarity, (24.46)
Thus, there exists By E M such that By is a countable local base at y. By Corol0 lary 24.22, By ~ M. CLAIM 24.33. Y is a closed subspace of X. PROOF. By first count ability, if X E cl Y, then X is the limit of a converging sequence (Yn)nEw of elements of Y. One can treat this sequence as a countable subset of w x Y. Since M is closed under countable subsets, the sequence (Yn)nEw is also an element of M. By Hausdorffness, convergent sequences have unique limits, and thus X is definable over M with parameters X, T, (Yn)nEw. Hence Lemma 24.20 implies that X E M. 0 Now let us wrap up the argument by showing that X = Y. Suppose not, and let X E X\Y. For every Y E Y, let Uy E By be such that X ¢ Uy. Such Uy exists since X is Hausdorff and By is a local base at y. The family U = {Uy : Y E Y} is an open cover of Y by elements of M such that UU misses x, and hence is not a cover of X. So far there is no contradiction since U does not need to be an element of M. But Y is a closed subspace of a Lindelof space, hence is Lindelof itself. Thus there exists a countable U' ~ U such that UU' ;2 Y. Since M is closed under countable subsets, U' is an element of M. Moreover, since U' covers every element of X n M, we have:
(24.47)
M
F U' is an open cover of X.
24.
178
But since x (24.48)
ELEMENTARY SUBMODELS
tt UU', H>..
F U' is not an open cover of X.
This contradicts the choice of M as an elementary submodel of H>...
o
For dessert, let us show that some of the large cardinals you have encountered in this text can be characterized in terms of elementary embeddings. Let us say that a cardinal K, has the Reflection Property14 if for every R ~ VI< there exists a < K, such that (Va, E, R) -< (VI<, E, R). By Remark 24.19, every strongly inaccessible cardinal has the Reflection Property. The converse is also true. THEOREM 24.34. If K, has the Reflection Property, then sible.
K,
is strongly inacces-
PROOF. Assume that K, has the Reflection Property. Since no (Vn, E) is an elementary submodel of (Vm' E) for n < m S w, we know that K, > w. Now let us show that K, is a regular cardinal. If not, there are fL < K, and a function F : fL -+ K, with range unbounded in K,. Let R = {fL} U F, and let a < K, be such that (Va, E, R) -< (VI<, E, R). Since (24.49)
(VI<, E, R)
F fL is the only ordinal in R,
elementarity implies that (24.50)
(Va, E, R)
F fL is the only ordinal in R.
In particular, fL E Va' Since Va is transitive, we also have fL ~ Va. But this implies that for each ~ E fL there exists TJ such that (~, TJ) E F n Va' Thus F ~ Va, which contradicts the assumption that the range of F is unbounded in K,. Finally, let us show that K, is a strong limit cardinal. If not, there is oX < K, such that 2>" ;::: K,. Fix a surjection G : P(oX) -+ K" and let R = {oX + I} U G. Let a < K, be such that (Va,E,R) -< (VI<,E,R). Again, this implies that oX+ 1 E Va, and by definition of the cumulative hierarchy, P(OX) EVa' EXERCISE 24.27 (G). Show that tradiction.
K,
~
Vat and explain why this yields a con-
0
Now let us consider the "dual" of the Reflection Property. Let us say that K, has the Extension Property if for every R ~ VI< there exist a transitive set X =f. VI< and S ~ X such that (VI<, E, R) -< (X, E, S). What does the Extension Property tell us about K,? THEOREM 24.35. An ordinal K, has the Extension Property if and only if it is a weakly compact cardinal. We shall not give a complete proof of Theorem 24.35 here, but only the most interesting part of this proof. LEMMA 24.36. Suppose K, is a strongly inaccessible cardinal with the Extension Property. Then K, is weakly compact. . 14Be apprised that if you see the phrase "Reflection Property" elsewhere in the literature, it probably has a broader meaning than in this text.
24.2.
APPLICATIONS OF ELEMENTARY SUBMODELS IN SET THEORY
179
PROOF. Let K. be as in the assumption. By Corollary 15.24, it suffices to show that K. has the Tree Property. Let (T, ~T) be a K.-tree. We want to show that this tree has a cofinal branch. Without loss of generality, we may assume that T ~ V",. Let R denote the strict p.o. relation
V",
F 'v'x3a E ON (rank(x) = a).
By elementarity, X has the same property. Now consider x E X\ V"" and let a E X be such that (24.52)
X
Fa E ON I'Irank(x) =
a.
By Claim 12.19(b), the formula "a E ON" is absolute for transitive models. 15 Hence a really is an ordinal, and by the definition of the rank function, (24.53)
X
F x E Vo+!.
EXERCISE 24.29 (PG). Show by induction over a that each of the formulas "x E Vo" and "rank(x) = a" is absolute for transitive models.
By Exercise 24.29, formula (24.53) implies that x really is an element of Vo +1 ' Since x (j. V", and Vo+ 1 ~ V", for all a < K., it follows that a ;::: K.. Thus, we have found an element a of X that either is equal to K. or contains K. as an element, and transitivity of X implies that K. EX. Now consider the relation R. For every a < K., the a-th level T(a) of T can be defined as the set of all t such that (i, R) is a strict wellorder of order type a, where i = {s: (s, t) E R}. The expression "t E T(a)" can be written in the language Lt of Remark 24.19 as: 16 "there exists a bijection 1 with domain a such that for all s, (s E rng(f) iff res, t)) and for all~, "I E a (~ E "I iff r(f(~), 1("1)))." Our assumption that T is a K.-tree implies that IT(a)1 < K. for all a < K.. Thus, each T(a) is actually an element of V"'. Since the definition of the T( a)'s can be expressed in L t, the model (X, E, S) also has its version of the T(a)'s, and elementarity implies that if a < K. and x are such that (24.54)
(V"" E,R)
F x = T(a),
(X, E, S)
F x = T(a).
then also (24.55)
Since both V",' and X are transitive, it follows that for any s, (24.56)
(V"', E, R)
F s E T(a)
iff (X, E, S)
F s E T(a).
ISLe., if Y is a transitive class and Q E Y, then Y Fa E ON if and only if V F a EON. we are going to use a few abbreviations rather than in its pristine form.
16 Admittedly,
L't
180
24. ELEMENTARY SUBMODELS
Since ht(T)
= /'i"
and since ON n VI<
= /'i"
we have
(24.57) By elementarity we get:
(X, E, S)
(24.58)
F Va E ON (T(a) ::p 0).
In particular,
(X, E, S)
(24.59)
F T(/'i,) ::p 0.
Let tl< E X be such that (24.60) Note that (VI<, E, R)
(24.61)
F Vf3 E ONVt E T(f3)Va < f3 3!s E T(a) (r(s, t)).
By elementarity,
(X, E, S)
(24.62)
F Vf3 E ONVt E T(f3)Va < f33!s
In particular, for each a <
/'i,
E T(a) (r(s, t)).
there exists exactly one So: such that
(24.63) By (24.56), if (X, E, S) F So: E T(a), then So: really is an element of T(a)j in particular, So: E VI<' Moreover, since i is wellordered by S, for all a < f3 < /'i, we have (24.64)
(i.e., (so:,
S(3)
E S).
By Exercise 24.28, formula (24.64) implies that (so:, S(3) E R. It follows that the set {so: : a < /'i,} is a cofinal branch of T, as desired. 0 Now let us consider VI<+1, where /'i, is a cardinal. By Theorem 24.35, /'i, + 1 does not have the Extension Property. But perhaps /'i, + 1 could "almost" have the Extension Property in the following sense: (* ) There exists an elementary embedding j : VI<+1 -+ M of VI<+1 into a transitive set M ::p VI<+1 such that VI< ~ M and j rVI< is the identity. Perhaps we could even have: (** ) There exists an elementary embedding j : VI<+1 -+ M of VI<+1 into a transitive set M that contains VI<+ 1 as a proper subset such that j WI< is the identity. It turns out that (*) and (**) give alternative characterizations of measurable cardinals. 24.37. For every cardinal/'i, the following are equivalent: has Property (*)j has Property (**)j is measurable.
THEOREM
(a)
/'i,
(b)
/'i,
(c)
/'i,
PROOF. First suppose /'i, has Property (*), and let j, M be an elementary embedding and a transitive set that witness (*). We are going to show that VI<+l is a proper subset of M, which implies that j and M also witness (**). Since
(24.65)
VI<+l
F
/'i,
is the largest ordinal,
24.2. APPLICATIONS OF ELEMENTARY SUBMODELS IN SET THEORY
181
elementarity of j implies that
(24.66)
M
F= j(/\') is the largest ordinal.
By Claim 12.19(b), the formula "x E ON" is absolute for transitive models, and thus j ( /\,) really is an ordinal. By assumption, j(a:) = a: for all ordinals a: < /\'. Since j is one-to-one, j(/\') must be either equal to or larger than /\'. We will show that the former is inconsistent with our assumptions. If X E V",+l, then X ~ V"" and thus (24.67)
F= j(X) ~ Vj(",).
M
Absoluteness of Va implies that (24.68)
V
F= j(X)
~ Vj(",)
for every X in the domain of j.
Now consider x E V",. Byelementarity, (24.69)
(M
F= j(x)
E j(X)) iff x E X.
Since the formula "x E y" is absolute and jfV", is the identity, (24.69) implies that
(24.70)
x E j(X) iff x E X.
We have thus shown that
(24.71)
j(X) n V",
=X
for every X in the domain of j (Le., in V",+l)'
If j(/\,) were equal to /\', then (24.67) and (24.71) together would imply that j is the identity, which contradicts our choice of j. We have shown that
(24.72) Since M is transitive, j(/\,)
(24.73)
V"'+l
~
Mj in particular, /\, E M. Note that
F= Va: E ON 3x (x = Va).
By elementarity, (24.74)
M
F= Va: E ON3x (x =
Va).
In particular, (24.75) By Exercise 24.29, V", E M, and if X E V"'+l, then (24.71) implies that X is the intersection of two elements of M. Since V"'+l is closed under intersections, so is M, and thus X E M. Hence V"'+l ~ M, and we have shown that j witnesses (**). Now let us contemplate how j acts on subsets X of /\'. If
(24.76)
V",+l
F= X
~ /\',
then (24.77)
M
F= j(X) ~ j(/\').
By (24.71), j(X) contains the same elements of /\, as X, but it may also contain elements ~ /\'. Tn particular, if
(24.78)
V"'+l
F= X
is an unbounded subset of /\',
then
(24.79)
M
F j(X) is an unbounded subset of j(/\,).
24. ELEMENTARY SUBMODELS
182
A strong version of the converse of (24.79) is also true:
EXERCISE 24.30 (PG). Show that if X is a bounded subset of K, then j(X) = X. Hint: Consider a < K such that X ~ a. In particular, j(X) mayor may not contain the ordinal the family
U = {X If X, Y
~ K
~ K: K E
K
itself. Let us consider
j(X)}.
are complements of each other, i.e., if
(24.80)
Vr.+1
F XU Y = K A X n Y = 0,
then
M
(24.81)
F j(X) U j(Y) =
j(K) A j(X)
n j(Y)
=
0.
In other words, exactly one of the sets X, Y is in U. One might now suspect that U is an ultrafilter on K, and this is indeed the case. EXERCISE 24.31 (G). Show that U is closed under finite intersections and under supersets. By Exercise 24.30, if X is a bounded subset of K, then K ¢. j(X), and hence X ¢. U. In particular, U is a nonprincipal ultrafilter on K. SO far we have shown nothing unusual about U. But here comes the punchline that proves measurability of K: CLAIM
24.38. The ultrafilter U defined above is K-complete.
PROOF. Let A < K, and let {Xe : ~ definition of U, for each ~ < A: (24.82)
Xe ~
K
< A} be a family of elements of U. By the
A K E j(Xe).
Define: Z = {(~,a): ~ < AAa E Xe}. Note that Z E V",+l. The intersection Y = ne<>. Xe can be defined as follows:
Y = {a <
K :
'
< A (~, a)
j(Y) = {a < j(K) :
V~
< j(A) (~, a)
E
Z}.
By elementarity, E
j(Z)}.
Note that j(Z) consists of pairs of ordinals such that the first coordinate is less than j(A) and the second coordinate is less than j(K). Thus j(Z) may pick up a whole lot of pairs with new second coordinates that were not in Z. In particular, (24.82) implies that (24.83)
But since j(A) = A, the sets of first coordinates of Z and j(Z) are the same. Hence (24.83) implies that K E j(Y), and thus Y E U. 0 Now assume that K is a measurable cardinal. We are going to show that K has Property (*). Let U be a nonprincipal, K-complete ultrafilter on K, and let ("'V",+dU, E) be the corresponding ultrapower of the model (V",+l, E), as defined in Section 13.2. By Theorem 13.14, the relation E is wellfounded. It is not hard to verify that the relation E is also extensional in the sense of Definition 4.38. Thus, by Theorem 4.39, there exists a (unique) isomorphism F : "'V",+dU ~ M
24.2. APPLICATIONS OF ELEMENTARY SUBMODELS IN SET THEORY
183
of ItVIt+l/U onto a transitive set M (the Mostowski collapse of ItVIt+I!U). Let h: VIt + 1 --+ ItVIt+I!U be the elementary embedding of (V",+I, E) into (ItVIt+l/U,E) defined in Example 24.3, and define j : VIt+1 --+ M as j = F 0 h. We show that j and M witness that K has property (*). By Claim 24.6 and Exercise 24.4, j is an elementary embedding. CLAIM
24.39. For all X E VIt , j(X) = X.
PROOF. For each z E VIt + 1 , let gZ denote the function in ItVIt+l that takes the value z for all a in its domain. By the definition of h, Claim 24.39 can be reformulated as follows:
"For all X E V"" F(gfu)
= x."
We will use this reformulation later in the proof. Now let us prove the claim by induction over the wellfounded relation E. By elementarity, j(0) = 0. Now suppose X E VIt and j(x) = x for all x E X. We want to show that j(X) = X. By elementarity, j(x) E j(X) for all x E X. Hence, it suffices to show that if y E j(X), then y = x for some x E X. So, let y E j(X). By transitivity of M, there exists 9 E '" VIt+l such that F (g jU) = y. Fix such a g. Note that j(X) = F(gfu). Since F is an isomorphism, the choice of 9 implies that ItVIt+I!U
(24.84)
F gjU
E
gfu·
This means that the set B = {f3 E K : g(f3) E gX (f3)} is in U. In view of the definition of gX we have B = {f3 E K : g(f3) E X}, and hence B = UXEX{f3 E K: g(f3) = x}. Since K is measurable and hence strongly inaccessible (Theorem 23.14), by Exercise 12.29, VIt = Hit. Thus IXI < K, and K-completeness of U implies that for some x E X the set {f3 E K : g(f3) = x} is in U, i.e., 9 "'U gX. By the inductive assumption, this implies F(gju) = x, and we have proved Claim 24.39. 0 Next we will show that j(K) is an ordinal greater than K. Since VIt +1 does not contain ordinals> K, this implies that M\ V,,+1 =f. 0, which takes care of the other req uirement in (*). CLAIM
24.40. j(K) > K.
By the previous claim, j(a) = a (Le., F(g;u) = a) for all a < it suffices to show that there exists gid E '" V,,+l such that for all a < K: PROOF.
(24.85)
"V",+I!U
K.
Thus
F g;u E g~t E gju'
Define:
ld (f3) = f3 for all f3 < Then {f3: gid(f3) E g"'(f3)} =
K
E U, and for each a
{f3: gO(f3) E gid(f3)}
K.
<
K,
= K\(a + 1) E u. DO
By definition of the ultrapower, (24.85) follows.
If gid is defined as in the proof of Claim 24.40, is F(g~t) actually equal to K? The answer depends on the properties of U. If there exists a regressive function 9 : K --+ K such that g-l{a} ¢. U for each a < K, then
(24.86)
For all a <
K,
"V",+I!U
F g;u E gju,
184
24. ELEMENTARY SUBMODELS
Moreover, by regressiveness of g, {,B Eli: g(,B) E gid(,B)} = Ii\{O} E u,
(24.87) and hence (24.88)
Thus F(g/u) is between all the a's that are less than Ii and F(g;t) , and hence 'd F ( gju) cannot be Ii. On the other hand, if for every regressive function g : Ii -+ Ii there exists A E U such that A is constant, then for every g/u such that
gr
"V,,+l/U
(24.89)
there exists a
F g/u E g;t
< Ii such that
(24.90) and hence F(g;t) is equal to Ii. Let us call an ultrafilter U on Ii normal if for every X E U and for every regressive function g : X -+ Ii there exists A E U such that grA is constant. Note that we have proved the following: THEOREM 24.41. Let U be a nonprincipal, Ii-complete ultrafilter on a measurable cardinal Ii, let M be the Mostowski collapse of the ultrapower "V,,+dU, let F : "V,,+dU -+ M be the collapsing function, and let gid : Ii - Ii be defined by gid(,B) = ,B for all ,B < Ii. Then F(g;t) = Ii if and only if U is normal. In Chapter 23, we also considered "normal" ultrafilters, but they were defined in a different way. It turns out that the two definitions of normal ultrafilters are equivalent. EXERCISE 24.32 (PG). Let Ii be an infinite cardinal. Show that U is a normal ultrafilter on Ii iff U is closed under diagonal intersections, i.e., iff for every sequence (X", : a E Ii) of elements of U the diagonal intersection ~",<"X", E U. Hint: For the "if"-direction, imitate the proof of Fodor's Lemma (Theorem 21.12). By Exercise 23.8(a), if there exists a uniform, normal ultrafilter on Ii, then Ii is measurable. The converse is also true. THEOREM 24.42. Let Ii be a measurable cardinal. Then there exists a uniform, normal ultrafilter on Ii. PROOF. Let U be any Ii-complete nonprincipal ultrafilter on Ii, and let F : "V,,+dU -+ M be as in Theorem 24.41. Then there exists some g : Ii - Ii with F(g/u) = Ii. Pick such g and define:
V = {X ~ Ii : g-l X E U}. EXERCISE 24.33 (PG). Show that V is a M.iform, normal ultrafilter on Ii.
0
Let us conclude this chapter with an interesting consequence of Property (**). THEOREM 24.43. Suppose Ii is a measurable cardinal and 2A = infinite cardinal >. < Ii. Then 2" = Ii+ .
>.+ for each
MATHOGRAPHICAL REMARKS
PROOF.
(24.91)
Let
K
185
be as in the assumptions, and let j,M be as in (**). Since
F V>" < K (>.. is an infinite cardinal
-+
2), = >.. +),
F V>" < j(K) (>.. is an infinite cardinal
-+
2),
VI<+l
by elementarity,
(24.92)
M
= >..+).
How realistic are M's notions of infinite cardinals?
>.. really is an infinite cardinal, Hint: Argue as in Exercise 24.14.
EXERCISE 24.34 (G). Convince your-self that if
then M
F >.. is an infinite cardinal.
By Exercise 24.34,
(24.93)
M
FK
is an infinite cardinal.
Hence (24.92) implies in particular that
(24.94) This means, there exist
M
f
F 21< =
K+.
E M and an ordinal {3 E M such that
M F f is a bijection from P(K) onto {3, and for every ordinal a: such that K < a: < {3, (24.95)
(24.96)
MFa: is not a cardinal.
By Claim 12.19(j), there exists a set x E M such that onto {3 and (24.97)
M
f
really is a bijection from x
FX =P(K).
Since M witnesses (**) and P(K) ~ VI<, also P(K) ~ M. Moreover, since M is transitive, it is not hard to see that y E x if and only if y really is a subset of K. In other words, x really is equal to P(K). Thus, f really is a bijection from P(K) onto {3, and we have shown that 21< = 1{31. By (24.96) and Exercise 24.34, none of the ordinals between K and {3 is a cardinal. Thus {3 = K+, and we have proved 0 Theorem 24.43.
Mathographical Remarks Theorem 24.35 appears as Theorem 4.5 with a complete proof in The Higher Infinite, by Akihiro Kanamori, Springer Verlag, 1994. The latter is also the ideal source for learning more about characterizations of large cardinals in terms of elementary embeddings. If you want to learn more about applications of elementary submodels in topology, we recommend Alan Dow's survey article An introduction to applications of elementary submodels to topology, Topology Proceedings 13(1) (1988), 17-72.
CHAPTER 25
Boolean Algebras In Section 13.3, we introduced Boolean algebras and outlined their connections to totally disconnected compact Hausdorff spaces and partial orders. The present chapter contains a more detailed treatment of selected topics in the theory of Boolean algebras. Throughout this chapter, letters a, b, c, d are reserved for elements of a Boolean algebra. If no confusion is likely, we shall suppress the phrase "Let a, b be elements of a Boolean algebra ~ = (B, +,., -,0,1)." We start with a connection between Boolean algebras and a certain class of algebraic rings. The symmetric difference between two elements of a Boolean algebra is defined as: at::.b = a . -b + -a· b. In a set algebra, this definition coincides with the usual one. EXERCISE 25.1 (G). Show that at::.b = (-a)t::.(-b) in every Boolean algebra. LEMMA
25.1. Let a, b, c be elements of a Boolean algebra. Then
a = b if and only if at::.b = 0; at::.b = bt::.a; (iii) at::.(bt::.c) = (at::.b)t::.c; (iv) a· (bt::.c) = (a· b)t::.(a· c).
(i)
(ii)
PROOF. (i) We have
a=b iff iff a· -b = iff iff
a :S b and b :S a
(Lemma 13.19)
and - a . b =
(Exercise 13.41(iii))
°
° °
a . -b + b . -a = at::.b = 0.
Point (ii) is obvious. EXERCISE 25.2 (PG). Prove points (iii) and (iv).
o
DEFINITION 25.2. Let !:R = (R, EB, ·,0,1) be an algebraic ring with identity, pER. If p2 = p, we say that p is idempotent. !:R is a Boolean ring if all elements of R are idempotent.
°
Note that we do not require that i:- 1. The one-element ring is a Boolean ring. Let ~ = (B, +,., -,0,1) be a Boolean algebra. We define: r(~) =
(B,t::.,·,O,l).
Lemma 25.1 implies that r(23) is a Boolean ring. 187
188
25. BOOLEAN ALGEBRAS
Now let !Jt = (R, EB, ·,0, 1) be a Boolean ring. We want to show that !Jt is a commutative ring. Idempotence and distributivity imply that a EB b = (a EB b) . (a EB b)
= a . a EB a . b EB b . a EB b . b = a EB a . b EB b . a EB b.
. Subtracting a EB b from both sides of this equation and using commutativity of EB, we obtain: a . b EB b . a
(25.1) For a
= O.
= b, equation (25.1) yields:
(25.2)
a EB a
= O.
It follows that every element of a Boolean ring is of order $ 2. Equations (25.1) and (25.2) combined imply that every Boolean ring is commutative. For !Jt as above and a, bE R we define a + b = a EB b EB (a· b) and -a = a EB 1, and we let b(!Jt) = (R, +, ., -,0,1).
EXERCISE 25.3 (PG). Show that b(!Jt) as defined above is a Boolean algebra. THEOREM 25.3. Every Boolean algebra 23 is isomorphic to b(r(23)). Every Boolean ring!Jt is isomorphic to r(b(!Jt)). EXERCISE 25.4 (PG). Prove Theorem 25.3. EXERCISE 25.5 (G). Let 23 = (B, +,·,0,1) be a Boolean algebra, I ~ B. Show that I is an ideal in the Boolean algebra 23 (i.e., an ideal in the p.o. (B\ {I}, $B)) if and only if I is an ideal in the Boolean ring r(23). Let 2t and 23 be Boolean algebras, kernel of 7r as follows:
ker (7r)
7r:
A
= {a E A:
->
B a homomorphism. We define the
7r(a) = O}.
EXERCISE 25.6 (G). (a) Convince yourself that 7r is also a homomorphism from the algebraic ring rU~l) into r(23), and conclude that its kernel ker (7r) is an ideal in the Boolean algebra 2t. (b) Prove directly (i.e., without reference to algebraic rings) that the kernel of every homomorphism of Boolean algebras is an ideal in the domain of the homomorphism . . If I is an ideal in a Boolean. algebra 23, then the quotient algebra induced by I is the algebra 2311 = b(r(23)II), where r(23)II is the quotient ring. More explicitly, 2311 = (BII,+jT,·jI,-jI,!,1*), where BII is the set of equivalence classes of the relation "'I defined by a "'I b if and only if aAb E I; + j I, . j I, and - j I are the operations on BII induced by the corresponding operations in 23, and 1* = {-a : a E I} is the dual filter of I. Every quotient algebra 2311 is a homomorphic image of 23; this is witnessed by the canonical homomorphism 7rT : B -> B I I which is defined by: 7r I
(b) = {a E B : a· "'I b}.
An element a of a Boolean algebra 23 is called an atom if a I- 0 and Vb E B (b $ a -> (b = 0 V b = a)). The set of all atoms of B is denoted by At(B). We say that a Boolean algebra B is atomic, if At(B) is a dense subset of B, Le., if for each bE B+ there is an a E At(B) such that a $ b. If At(B) = 0, then B is atomless.
25. BOOLEAN ALGEBRAS
189
EXERCISE 25.7 (G). Let B be a Boolean algebra, and consider the function -+ P (At(B)) defined by the formula
P(At(B)). EXERCISE 25.8 (PG). (a) Let X be a compact, totally disconnected Hausdorff space. Show that a clop en set a is an atom in the Boolean algebra Clop (X) if and only if a = {x} for some isolated point x EX. (b) Conclude that a Boolean algebra B is atomic if and only if the set of isolated points is dense in st(B), and that B is atomless if and only if st(B) has no isolated points. (c) Let D be the Cantor set defined in Exercise 13.44. Show that Clop(D) is a countable atomless Boolean algebra. (d) Show that every countable, atomless Boolean algebra is isomorphic to Clop(D). Hint: Do a back-and-forth construction as in the proof of Theorem 4.16. The last exercise has an important consequence. COROLLARY 25.4. Let A be a complete atomless Boolean algebra that contains a countable dense subset. Then A is isomorphic to the algebra RO(D) of regular open subsets of the Cantor set. PROOF. Clop(D)\{0} is dense in the complete Boolean algebra RO(D). Thus, by Exercise 25.8(c), RO(D) contains a dense countable atomless subalgebra. Now Exercise 25.8(d) and Corollary 13.35 imply that, up to isomorphism, RO(D) is the only complete Boolean algebra with this property. 0 As we remarked in Exercise 13.44, D is homeomorphic to D(2)w. More generally, for any infinite cardinal K we set DI< = D(2)1< and we let Fr(K) denote the Boolean algebra Clop(DI<)' EXERCISE 25.9 (G). (a) Show that for every infinite cardinal K, the Boolean algebra DI< satisfies the c.c.c. Hint: Use Theorem 16.4. (b) Show that if A is an infinite cardinal and B is a dense sub algebra of a Boolean algebra A, then B satisfies the A-C.C. if and only if A does. For any infinite cardinal K we have Fr(K) ~ RO(DI<)' Instead of RO(Fr(K)) we shall write cFr(K). Since D(2)1< has a base consisting of clopen sets, Fr(K) is dense in cFr(K). Thus, cFr(K) is the completion of Fr(K). Moreover, Exercise 25.9 implies that cFr(K) satisfies the c.c.c. EXERCISE 25.10 (PG). Let K ~ w, let BI< denote the Boolean algebra of all Borel subsets of DI<' and let MI< denote the ideal of all A E BI< that are meager. Show that cFr(K) ~ BI
a set is often called a maximal antichain in E below b.
190
25.
BOOLEAN ALGEBRAS
Let us consider cFr(K) for K ~ w. The Boolean algebra cFr(K) satisfies the c.c.c., and Fr(K) is dense in cFr(K). By Lemma 25.5, this implies that IcFr(K) I S KNo. Thus for every infinite cardinal K with KNo = K there exists a complete Boolean algebra of cardinality K. Now we are going to show that this is also a necessary condition; i.e., for each infinite complete Boolean algebra B the equality IBI = IBINo holds. First we show this for a special class of complete Boolean algebras. In order to define this class, we need to introduce one more bit of notation. Let ~ = (B, +,., -,0,1) be a Boolean algebra, a E B. We define Bla = {b E B : b sa}. The structure
where x +a y = x + y, X 'a Y = X • y, -aX = -x . a, Oa = 0, and la = a, is also a Boolean algebra. ~Ia is called the factor algebra (or simply factor) of ~ relative to a. EXERCISE 25.11 (PG). Prove the following "Cantor-Schroder-Bernstein Theorem" for complete Boolean algebras: If A and B are complete Boolean algebras such that there exist a E A and b E B with A ~ Bib and B ~ Ala, then A ~ B. Let B be an infinite Boolean algebra with IBI = K. We say that B is weakly Khomogeneous (or simply weakly homogeneous), if for each a E B+ the cardinality of Bla is equal to K. Note that each wekly homogeneous Boolean algebra is atomless. LEMMA 25.6. Let B be a complete, weakly homogeneous Boolean algebra with IBI ~ No. Then IBI = IBINo. PROOF. We need a preliminary observation. EXERCISE 25.12 (PG). Suppose B is an infinite Boolean algebra. Then there exists an infinite set of pairwise disjoint, non-zero elements of B. Hint: Use Konig's Lemma. Let B be as in the assumption and let K = IBI. By 25.12, we can choose a set of pairwise disjoint elements {ai : i E w} ~ B+. For every sequence (bi)iEW such that bi S ai for all i E w, let s((bi)iEW) = ~iEW bi' Note that (bdiEW #- (bDiEW implies s((bi)iEw) #- s((bDiEW)' Thus s is an injection of the set TIiEw Bla; into B. By weak K-homogeneity of B, the existence of such an injection implies that IBI ~ KNO, and the lemma follows. 0 The prime examples of Boolean algebras that are not weakly homogeneous are atomic Boolean algebras. Fortunately, for them the analogue of Lemma 25.6 is easy to prove. EXERCISE 25.13 (G). Show that if B is an infinite, complete, atomic Boolean algebra, then IBIN o = IBI. Hint: Use Exercise 25.7. Exercise 25.13 allows us to reduce our problem to the case of atomless Boolean algebras. To do this in an elegant fashion, let us introduce one more notion. Let 2l = (A, +A, . A, -A, 0 A, lA) and ~ = (B, +B, . B, - B, 0 B, IB) be two Boolean algebras. We define the product algebra 2l x ~ = (A x B, +,., -,0,1) as
25. BOOLEAN ALGEBRAS
191
follows:
(ao,b o) + (al,b l ) = (ao +A al,b o +B bl ); (ao,b o)' (al,b l ) = (aO'A al,bo ' B bl ); -(ao, bo) = (- AaO, - Bbo); 0= (OA,OB); 1 = (lA, 1B)' LEMMA 25.7. Let B be a Boolean algebra, a E B. Then
B
~
(Bla) x (BI - a).
EXERCISE 25.14 (G). Prove Lemma 25.7. Hint: Consider the map (Bla) x (BI - a) defined by 'Tr(b) = (b· a, b· -a).
'Tr :
THEOREM 25.8. Let B be a complete, infinite Boolean algebra. Then
IBINo.
B
--+
IBI =
PROOF. Let B be as in the assumption, and let At be the set of all atoms of B. Let a = EAt. Then BI- a is atomless. EXERCISE 25.15 (G). (a) Let a be as above. Show that at least one of the equalities IBlal = IBI, IBI- al = IBI holds. (b) Derive Theorem 25.8 from the following lemma: LEMMA 25.9 (Pierce). Let B be an atomless, complete, infinite Boolean alge-
bra. Then
IBI
=
IBINo.
PROOF. Let B be as in the assumptions. Consider the set J = {O} U {a E B+ : Bla is weakly homogeneous.}.
EXERCISE 25.16 (R). (a) Show that J is a dense ideal in B. (b) Conclude that there exists a maximal pairwise disjoint subset of B that is contained in J\{O}. (c) Derive Lemma 25.9. Hint: Use an argument similar to the one in the proof DO of Lemma 25.6. Let B be a Boolean algebra, X ~ B. Then (X) denotes the smallest subalgebra of B that contains all elements of X. If B is a subalgebra of A and a E A, we write B(a) instead of (B U {a}). Now let A be a Boolean algebra, B a subalgebra of A, and a E A \B. Consider the set M = {b o . a + bi . -a : bo, bi E B}. Clearly, 0,1 E M. We show that M is closed under +, ., and -: Let bo, bl , b2, b3 E B. We have:
(b o . a + bl . -a) + (b2 . a + b3 . -a) = (bo + b2) . a + (b l + b3) . -a; (bo . a + bi . -a) . (b2 . a + b3 . -a) = (bo . b2) . a + (b i . b3) . -a; - (bo . a + bi . -a) = -bo . a + -b l . -a. Thus M is a subaJ.gebra of B. Since M ~ B(a), we have M = B(a). Let B and C be Boolean algebras with B ~ C. We say that C is a finite extension of B if there exists a finite set {ao 1 •• , 1 an-l} such that C = (B U {ao, ... , an-I})' We call C is a simple extension of B if there exists a such
192
25.
BOOLEAN ALGEBRAS
that C = (B U {a}). Let A, B, and C be Boolean algebras with A ~ B, and let 7r : A ~ C be a homomorphism from A into C. We are going to explore under what conditions 7r can be extended to a homomorphism from B into C. Let A be a Boolean algebra. A pair (S, T) of subsets of A will be called a cut if S ~ T, i.e., if s ~ t for all s E Sand t E T. Note that the word "cut" is used here in a different meaning from that of Section 13.3. An element a E A fills the cut if S ~ a ~ T. A cut (S, T) is called a gap if there is no a E A that fills this cut. Let B be a subalgebra of A and let a E A. We define Sa = {b E B : b ~ a} and Ta = {b E B : a ~ b}. Clearly, (Sa, Ta) is a cut. Moreover, if f : B ~ C is a homomorphism from ~ into It, then UrSa], f[Ta]) is a cut in C. LEMMA 25.10. Let A,B,C,a,f be as above. If some c E C fills the cut UrSa], f[Ta]) , then f can be extended to a homomorphism f+ : B(a) ~ C. PROOF. We define f+(bo . a + b1 • -a) = f(b o) . c + f(b 1 ) . -c. EXERCISE 25.17 (G). Show that the above definition of f+ does not depend on the choice of representations of the elements of B(a).
+ and -. We have: f+((b o ' a + b1 . -a) + (b 2 • a + b3 • -a)) = f+((b o + b2 ) • a + (b 1 + b3 ) • -a) = f(b o + b2 ) • a + f(b 1 + b3 )· -a = f+(b o ' a + b1 . -a) + f+(~' a + b3 . -a). Now let us show that f+ preserves the operations
Moreover, -(bo . a + b1 . -a) = -(bo . a) . -(b 1 . -a) = -bo . -b1 + -bo . a + -b 1 . -a = (-bo . -bt) . a + (-bo . -bt) . -a + -bo . a + -b1 . -a = -bo . a + -b1 . -a. Hence, f+(-(b o ' a + b1 · -a)) = f(-b o)' a + f(-bt}·-a = -f(bo) . a + - f(bd . -a = -(J(bo) . a + f(b 1 ) . -a) = - f+(b o . a + b1 • -a). Since f+(1B) = Ie, Exercise 13.40 and duality imply that f+ is a homomorphism of Boolean algebras. 0
EXERCISE 25.18 (G). Let A,B,C be Boolean algebras with B c A, let a E A\B, and let f : B ~ C be an embedding of Boolean algebras. Show that if there is c E C such that (i) c fills the cut U[Sa] , J[Ta]) , (ii) 'v'b E B (b· a =I 0 ~ f(b) . c =I 0), and
(iii)
'v'bEB(b·-a=lO~f(b)·-c=lO),
then f can be extended to an embedding f+ : B(a) ~ C. THEOREM 25.11 (Sikorski's Extension Theorem). Let A, B, C be Boolean algebras such that B is a subalgebra of A and C is complete. Moreover, let f : B ~ C be a homomorphism. Then f can be extended to a homomorphism f+ : A ~ C. PROOF. Let F be the family of all functions 1* such that dom(J*) is a subalgebra of A, B ~ dom(J*), and 1* is a homomorphism from dom(J*) into C. Then (F,~) is a p.o. in which the union UK of any chain K is an upper bound of K. Zorn's Lemma implies that F has a maximal element f+. Lemma 25.10 implies that f+ is as required. 0
25.
BOOLEAN ALGEBRAS
193
EXAMPLE 25.1. A homomorphic image of a complete Boolean algebra does not need to be complete. To see this, consider B = P(w), A = P(w)/Fin, and let 7r : B -+ A be the canonical homomorphism. The image (7r[B),7r[T)) of each Hausdorff gap (B, T) is a gap in A, and thus Theorem 20.2 implies that A is not complete. Note that the above example also shows that a homomorphic image of a Boolean algebra B does not need to be isomorphic to a subalgebra of B. EXAMPLE 25.2. The following exercise shows that a sub algebra of a Boolean algebra B does not need to be a homomorphic image of B. EXERCISE 25.19 (G). Show that Finco (w) is not a homomorphic image of
P(w). THEOREM 25.12. Let 21 = (A,+A,·A,-A,OA,lA) and ~ = (B,+B,·B,-B, OB,l B ) be Boolean algebras, 21- = (A, SA) and ~- = (B, SB) the associated p.o. 'so Then 21 ~ ~ il and only il 21- ~ ~- . PROOF. Since the relation S is definable in Boolean algebras, the "only if" direction is immediate. For the "if" direction, assume that 1 : 21- -+ ~- is an order isomorphism. Since OA is the minimum element of 21- and OB is the minimum element of ~-, we must have I(OA) = OB. For a similar r.eason, l(lA) = lB. For each a, bE A, one can define a + b as the least upper bound of {a, b} in 21-, and a . b as the greatest lower bound of {a, b} in 21-. Similarly, -a is the maximum element c E A such that the greatest lower bound of {a, c} is 0 A. Thus all constants and operations in 21 are uniquely determined by the structure 21-, and it follows that 1 is also an isomorphism of Boolean algebras. 0 THEOREM 25.13. Let A and B be complete Boolean algebras, let X ~ A and B be dense in the associated p.o. 's (A\{OA}, SA) and (B\{OB}, SB), and let h : X -+ Y be an order isomorphism. Then there exists a unique extension h+ : A -+ B 01 h to an isomorphism from the Boolean algebra A onto the Boolean algebra B.
y
~
PROOF. By Theorem 25.12 it suffices to show that there exists a unique extension of h to an order isomorphism h+ from (A, SA) onto (B, SB). For a E A we define: h+(a) = sup {h(x) : x E X,x sa}.
EXERCISE 25.20 (G). Show that h+ is an order-preserving extension of h such that h+(OA) = OB. Let a, b E A, a =I b. Without loss of generality, assume that a . -b =I OA. Then there exists x E X such that x sa· -b. Thus OB =I h+(x) S h+(a), but h+(x)·h+(b) = OB. Hence, h+(a) =I h+(b) and we have shown that h+ is one-to-one. Finally, let bE B. Then b = sup {y E Y : y S b}. Consider a E A defined as a = sup {h- 1 (y) : y E Y,y S b}. Then h+(a) = b, and we have shown that h+[A) = B. Thus, h+ is an isomorphism from A onto B. We leave the proof of uniqueness to the reader. 0
194
25. BOOLEAN ALGEBRAS
EXAMPLE 25.3. Let (L, $) be a linear order with a minimum element OL. We extend L to L U {oo}, where 00 is an object that is not an element of L, and we extend $ to L U {oo} by requiring that a < 00 for all a E L. For a, bEL U { oo} let
[a, b)
= {e E L : a $
e < b}.
We call [a, b) a hal/-open interval. In particular, [OL, 00) = L. Let s = (ao, ... ,a2n-l) be an increasing finite sequence of elements of LU{ oo}. We define: i
EXERCISE 25.21 (G). (a) Show that Intalg (L) is closed under unions, intersections, and complements. (b) Let Q+ be the set of nonnegative rational numbers with their natural order. Show that Intalg (Q+) is a countable, atomless Boolean algebra. (c) Show that Intalg (w) ~ Fineo (w). (d) Show that if (L, $) is a linear order with a minimum element 0, and L1 ~ L is such that 0 E L 1, then Intalg (L 1) is a subalgebra of Intalg (L). (e) Conclude that every countable interval algebra can be embedded in the Boolean algebra Intalg (Q+). Now we are going to explore which Boolean algebras are isomorphic to interval algebras. Let B be a Boolean algebra. A subset A of B is called a ramification system if the following conditions are satisfied: (i) 1 E A, 0 ¢ Aj (ii) \la, bE A(a . b = 0 V a $ b V b $ a). A is a generating ramification system (abbreviated gR), if it satisfies (i), (ii), and (iii) (A) = B. Not every Boolean algebra contains a generating ramification system. EXAMPLE 25.4. We show that P(w) does not contain a gR. Assume towards a contradiction that A ~ P(w) satisfies (i)-(iii). Then IAI > w. Moreover, if a, b E A have nonempty intersection, then a and b are comparable. Therefore, every antichain in (A,~) consists of pairwise disjoint sets and thus is countable. EXERCISE 25.22 (G). Use the Erdos-Duschnik-Miller Theorem and Ramsey's Theorem to show that (A, ~) contains an infinite chain L of order type w or w·. Let L be as in Exercise 25.22. If L has order type w, let L = {ai : i < w} with ai ~ ai+1 for all i < w. Let x = Ui<w(a2i+1 \ a2i)' Then x ¢ (A). The case where L has order type w· is analogous. EXERCISE 25.23 (G). Show that every denumerable Boolean algebra B contains a gR. Hint: Let (bi)i<w be an enumeration of the elements of B and define recursively for each s E <w2 an as E B such that a0 = 1, as~o = as . bn , and a s-1 = as . -b n for all s E<w2.
25.
BOOLEAN ALGEBRAS
195
Now we are going to look at the kind of ramification systems that generate interval algebras. Let B be a Boolean algebra and let A ~ B. Then A is a linear ramification system if A is a chain such that 0 f/. A and 1 E A. We say that A is a linear generating ramification system (lgR), if A is a linear ramification sytem with (A) = B. EXAMPLE 25.5. Let (L,:S;) be a l.o. with smallest element O. Then the family {L} U HO, a) : a E L\{O}} is an 19R for Intalg (L). The result of Exercise 25.23 can be strengthened as follows: LEMMA 25.14, Each countable Boolean algebra B has an 19R. PROOF. Let (bi)i<w be an enumeration of B. We define recursively a sequence (Ci)i<w of chains in B such that for all i < w:
(i) (ii)
0
f/. Co,
1 E Co;
(Ci ) = ({b k : k < i}).
We let Co = {I}. Suppose C n has been defined, and C n c k < ck+l for k < j(n) - 2. We let C':1 = 0, and define for k < j(n): .m+I Uk = Cnk -
1
= {cO, ...
,cj\n) -I}' where
+ ck' n b n'
Let C n+1 = C n U ({elk+! : k < j(n)} \ {o}). It follows from the construction that C n +! is a chain. Note that bn
=
I:
elk+
1
.
-Ck-l'
k<j(n)
hence bn E (Cn + I ), and thus (Cn + I ) = ({bk : k
:s;
n}). It follows that Ui<w C i is
an~
0
THEOREM 25.15. Let B be a Boolean algebra, A an 19R for B. For each bE B there exist uniquely determined n < wand ao, ... ,a2n+! E AU {O} such that ao < ... < a2n+l and b = I:i~n -a2i . a2i+I· EXERCISE 25.24 (G). Prove Theorem 25.15. COROLLARY 25.16. Let B be a Boolean algebra. (a) If A is an 19R for B, then B ~ Intalg ((A, 2:B)). (b) B is isomorphic to an interval algebra if and only if B has an 19R. (c) Every countable Boolean algebra is isomorphic to an interval algebra. (d) Every countable Boolean algebra can be embedded in the countable atomless Boolean algebra Intalg (Q+). EXERCISE 25.25 (R). (a) Show that Intalg (w) ~ Intalg (w
+ w*)
and Intalg (WI
(b) Is Intalg (wd isomorphic to Intalg (WI
+ w*)
+ w) ~ Intalg (wd. or to Intalg (WI + wi)?
R. L. Vaught developed a procedure that allows us to test pairs of countable Boolean algebras for isomorphism. Now we are going to describe this procedure.
25. BOOLEAN ALGEBRAS
196
DEFINITION 25.17. A binary relation Ron a set of Boolean algebras is a Vaught relation if it satisfies the following conditions: R is symmetric, i.e., if ARB, then B RAj (VI) (V2) If ARB and A is trivial, then B is also trivialj (V3) (The back-and-forth-property) If ARB and a E A, then there exists bE B such that (Ala) R (Bib) and (AI - a) R (BI - b). For example, being isomorphic is a Vaught relation. THEOREM 25.18 (Vaught). Let A and B be countable Boolean algebras. If there exists a Vaught relation R such that ARB, then A and B are isomorphic. PROOF. We need a technical concept. Let b = (bi)i
Pe(a) =
II Ci, i
where (25.3)
Ci
=
if e(i) = OJ ai, { -ai, otherwise.
Then {Pe(a) : e E n2} is a partition of unity. Now suppose R is a Vaught relation such that ARB, and let (a2n)n<w and (b 2n+1)n<w be enumerations, possibly with repetitions, of A and B respectively. We construct recursively a1, a3, a5 ... E A and bo, b2, b4 , .•• E B such that for each n <w:
(R)n (I)n
(Alp: (ao, a1,· .. , an-1) R (Blp~ (bo, bl, ... , bn- 1)) for each e E n2, and There exists an isomorphism fn: ({ai : i < n}) - ({b i : i < n}) with f(ai) = bi for all i < n.
Note that the isomorphism fn in (I)n is uniquely determined. Moreover, A = Un<w({ai: i < n}) and B = Un<w({b i : i < n}). It follows that Un<w fn will be an isomorphism from A onto B. Now let us carry out the construction. For n = 0 we have 02 = {0}. Since p:(0) = 1A and pf(0) = 1B , (R)o and (1)0 hold. Suppose n is even and bo, al, b2 ,'" , an -1 have been constructed in such a way that (R)n and (I)n hold. For each e E n2 we choose re E BI(p~(bo,b1"" ,bn- 1)) such that (AI (Pe(ao, a1,· .. , an-1) . an) R (Blre) and
(AI(P:(ao,a1, ... ,an-1)' -an )R(BI(Pe(bo,b1, ... ,bn- 1)· -re)). By (V2), such re exists. We define bn = ~{re : e E n2}. For odd n, we reverse the roles of A and B in the above construction. EXERCISE 25.26 (G). Convince yourself that Properties (R)n and (I)n hold after the n-th step of this construction. 0
25.
BOOLEAN ALGEBRAS
197
Let B be a Boolean algebra. We call B superatomic if each of its homomorphic images is atomic. LEMMA 25.19. A Boolean algebra B is superatomic if and only if each homomorphic image of B other than the trivial Boolean algebra has an atom. PROOF. The implication from the left to the right is clear. To prove the converse, assume that B is not superatomic. Then there exist a homomorphic image A of B and a E A \ {O} such that Ala is atomless. But Ala is a homomorphic image of A and hence of B. Since a =f:. 0, the Boolean algebra Ala is nontrivial and contains no atom. 0 THEOREM 25.20. A Boolean Algebra B is superatomic if and only if each subalgebra of B is atomic. PROOF. Assume that B is a superatomic Boolean algebra, and let B' be a subalgebra of B. Without loss of generality we assume that B is nontrivial, i.e., IBI > 1. We show that B' is atomic. Let I be an ideal in B, maximal in the family of all ideals J in B such that J n B' = {O}. Let 'If : B --+ B I I be the canonical homomorphism induced by I. Then 'lffB' is an embedding of B' into BII. Let a E B'\{O}. Then 'If(a) =f:. 0, and since B I I is atomic, there exists an atom b' E B I I such that b' :S 'If ( a). Then 1 'If- {0} U 'If-I {b'} is an ideal in B that contains the ideal I as a proper subset. It follows from the choice of I that 'If- 1 {b'}nB' =f:. 0. Hence there exists abE B' such that 'If(b) = b'j hence B' is atomic. To prove the converse, let B be a Boolean algebra such that all of its subalgebras are atomic. Suppose towards a contradiction that B is not superatomic. Let 'If : B --+ A be a homomorphism onto an atomless Boolean algebra A. Consider a countable atomless subalgebra A' of A. Let {ai : i E w} be a gR for A' such that ao = 1AI. Recursively we choose, for each i E w, a bi E B such that: (i) bo = 1B and 'If(bi ) = ai for all ij (ii) For all i,j E w, if ai :SAl aj, then bi:S B bjj (iii) For all i E w if ai 'AI aj = ON, then bi'B bj = OB. Then the subalgebra of B generated by {b i atomless.
:
i E w} is isomorphic to A' and thus is
0
EXERCISE 25.27 (PG). (a) Show that Intalg (w· 2 ) is superatomic. (b) Show that the class of superatomic Boolean algebras is closed under finite products, subalgebras, and homomorphic images. Now let us investigate the countable superatomic algebras. An important tool in this investigation will be the Cantor-Bendixson derivative. Let A be a Boolean algebra. The ideal generated by the atoms of A will be denoted by I(A). Note that I(A) = A if and only if A is finite. For each ordinal a we define recursively an ideal I",(A), a Boolean algebra A"" and a homomorphism 'If", : A --+ A", as follows: Let Io(A) = {O}. If I",(A) has already been defined, then we define A", = AII",(A) and we let 'If", : A --+ A", be the corresponding canonical homomorphism. Now suppose a > 0, and Ij3(A) has already been defined for all /3 < a. If a = /3 + 1 for some /3, then we let I",(A) =
'If;;l I(Aj3).
198
25.
BOOLEAN ALGEBRAS
If a is a limit ordinal, then we let
I,,(A)
= U Ifj(A). fj<"
For a ~ (3 we have I,,(A) ~ Ifj(A). Thus there exists a smallest a such that I,,(A) = Ifj(A) for all (3 > a. Now suppose that A is superatomic. In this case, each A" is atomic, and hence there exists a such that I,,(A) = A. The smallest such a must be a equal to {3 + 1, where {3 is the smallest ordinal such that Afj is finite. Let n be the number of atoms in Afj. We refer to the ordered pair ({3, n) as the Cantor-Bendixson characteristics of A. In particular, we call {3 the Cantor-Bendixson height of A (or simply the height of A) and write h(A) for (3. Note that n(A) is the number of atoms of Ah(A), and that the equality n(A) = 0 can only hold if A is the trivial Boolean algebra. EXAMPLE 25.6. The Cantor-Bendixson characteristics of the trivial Boolean algebra are (0,0). The characteristics of the interval algebra Intalg(w) are (1,1). More generally, if (a, n) is an arbitrary pair with a E ON and nEw \ {O}, then (a, n) are the Cantor-Bendixson characteristics of the superatomic Boolean algebra Intalg(w'" . n). LEMMA 25.21. Let A and B be superatomic Boolean algebras. (i) If A ~ B, then h(A) ~ h(B); (ii) If h(A) = h(B), then n(A x B) = n(A) + n(B); (iii) If h(A) < h(B), then n(A x B) = n(B); (iv) If h(A) = a, (3 < a, n < w, then there exists a E A with h(Ala) n(Ala) = n.
= (3
and
EXERCISE 25.28 (PG). Prove Lemma 25.21. THEOREM 25.22. Let A and B be countable superatomic Boolean algebras with the same Cantor-Bendixson characteristics (a, n). Then A 9:! B. PROOF. We prove the theorem by induction over a. If a = 0, then A and B are finite Boolean algebras with the same number of atoms, and hence isomorphic. Let a > O. First note that we can restrict ourselves to the case where n = 1: Suppose ao, .. . , an- l E A and bo, ... ,bn- l E B are such that {ao, ... , an-d is a partition of lA, {bo , ... , bn-d is a partition of 1B , and ad I", bd I" are atoms of A", respectively B". Then each of the algebras Alai and Blb i has Cantor-Bendixson characteristics (a, 1), and the isomorphisms between Alai and Blbi can be combined into an isomorphism from A onto B. Now let A = {Ala: a E A}, 13 = {Bib: bE B}, and define a relation R by C RD iff C,D E AU 13, and C,D have the same Cantor-Bendixson characteristics.
We show that R is a Vaught relation: ClearlYl (VI) and (V2) are satisfied. Now suppose C RD. If the height of C is smaller than a, then C 9:! D by the inductive assumption, and (V3) follows. Suppose h( C) ~ a. Then C has Cantor-Bendixson characteristics (a, 1). Let c E C. It follows from (i) and (ii) of Lemma 25.21 that h(CJc) < a or h(CI - c) < a. Without loss of generality, assume h(Clc) < a. Then Lemma 25.21(iv) implies that there exists d E D with h(CJc) = h(Dld) and
25.
BOOLEAN ALGEBRAS
n(Olc) = n(Dld). By (ii) and (iii) of Lemma 25.21, same Cantor-Bendixson characteristics.
199
CI( -c)
and
DI( -d)
have the 0
Let B be a Boolean algebra, a, b E B. We say that a and b are disjoint if a . b = O. Similarly, we say that A ~ B+ is a disjoint subset of B if c . d = 0 for all c, d E A with c :f. d. The cellularity of B is defined as: c (B) = sup {IAI : A is a disjoint subset of B}. Thus, if IBI satisfies the I\:-c.c., then c (B) ~ 1\:. If 0 is a subalgebra of B, then c (0) ~ c (B). We let c (a) stand for c (Bla). EXAMPLE 25.7. Clearly, c(P(w)) = No. Let A = P(w)/Fin, where Fin is the ideal of finite subsets of w, and assume that X is an almost disjoint family of infinite subsets of w such that IXI = 2No. Since the image of X under the canonical homomorphism from B onto A is a disjoint subset of A of the same cardinality as X, we see that c(A) = 2W. Similarly, for each a E A+, c(a) = 2w. We say that the cellularity of B is attained, if there is a disjoint family A ~ B such that IAI = c (B). If c (B) is a successor cardinal, then the cellularity of B is of course attained. Theorem 25.23 below asserts that the same is true if c (B) is singular. If c (B) is a regular limit cardinal, then c (B) mayor may not be attained. EXAMPLE 25.8. Let I\: be a weakly inaccessible cardinal, and let lL be the set of all finite functions with domain contained in I\: x wand such that if p E lL and (a, n) E dom(p), then p(a, n) E a. The p.o. (lL,2) is called Levy's partial order or the Levy collapse. EXERCISE 25.29 (PG). (a) Show that Levy's p.o. is separative, and thus dense in B+ for some (uniquely determined) complete Boolean algebra B. (b) Let B be as in point (a). Convince yourself that for each a E 1\:\ {O} the set {( (a, 0),.8) : .8 < a} is a disjoint subset of B. Conclude that the cellularity of B is equal to 1\:. (c) Show that there is no disjoint subset A ~ B+ of size 1\:. Conclude that the cellularity of B is not attained. Hint: If a counterexample A exists, then there exists one that is a subset of lL. Apply the D.-System Lemma to the domains of elements of such an A to derive a contradiction. THEOREM 25.23 (Erdos-Tarski). Let B be a Boolean algebra such that c(B) is singular. Then c (B) is attained. PROOF. Suppose c (B) = A, cf A = I\: < A, and let (A"')"'
S
= {a E B+ : c (a) < A}.
200
25.
BOOLEAN ALGEBRAS
EXERCISE 25.30 (G). Show that there exists a maximal disjoint subset C ~ B+ such that C ~ S. Fix a set C as in Exercise 25.30. Case 2: sup {c (a) : a E C} = A. We construct recursively a sequence (aa,)n<x: of distinct elements of C such that c( an) > An for each a < "'. This is possible since c (a) < A for each a E C. For each a < '" we choose a disjoint family An ~ Bla n such that IAnl ~ An. Then A = Un<x: An is as required. Case 3: sup {c(a) : a E C} = J.L < A. We prove that in this case Ie! = A. Assume towards a contradiction that Ie! = J.L' < A. Let C = {cn : a < J.L / }, and denote max{J.L,J.L/ } by 1/. Let D be a disjoint family in B with IDI > 1/. For a < J.L' consider Dn = {cn . a: a E D, Cn ' a ~ O},
and let E = U{Dn : a < J.L / }. On the one hand, E is a disjoint family with lEI ~ IDI > 1/. But on the other hand, lEI = 2:n
(an)n
CLAIM 25.25. There exists a disjoint family Y ~ B such that IY(an)1 = ",+ for each a < A.
If Y is as in Claim 25.25, then one can recursively construct a sequence (Yn)n<>' of pairwise distinct elements of Y so that Yn . an ~ O. For all a < A let Xn = Yn . an and X = {Xn : a < A}. The function 71' : A -+ X defined by 7l'(an ) = Xn witnesses that X is a refinement of A. PROOF OF CLAIM 25.25. We construct an increasing sequence (Yn)n<>. of disjoint families such that Va < A!Yn+l(an)1 = ",+ and (ii) Va,j3 < A (Yn (a/3) = {0} V !Yn(a/3)1 =.",+).
(i)
Note that if (i) and (ii) hold, then Y = Un<>. Yn is as required. We let Yo = 0. If a is a limit ordinal < A, then we let Y n = U/3
>. /\ IZ(a"()I ~ II:}).
Note that since >. ~ 11:, the implication IZ(a"() I = 11:+ particular, !Ya(a,e) I = 11:+.
-+
!Ya(a"() I = 11:+ holds. In 00
Let B be a Boolean algebra, X, Y ~ B, and z E B. We write X ..L Y if x ..L y (i.e., x· y = 0) for all x E X and y E Y. Instead of {z} ..L Y we simply write z ..L Y. Similarly, we write X ~ Y if x ~ y for all x E X and y E Y. An element b of B separates X and Y if X ~ band Y ..L b, or vice versa. Note that if some b separates the sets X and Y, then X ..L Y. We say that B has the countable separation property if for all X, Y E [B]~No such that X ..L Y there exists b E B that separates X and Y. EXERCISE 25.32 (G). Let B be a Boolean algebra. Show that the following are equivalent: (i) B has the countable separation property; (ii) For all disjoint sets X, Y E [B]~No such that X ..L Y there exists b E B separating X and Y; (iii) '
25.
202
BOOLEAN ALGEBRAS
(X)23 into !l, and b is an element of the universe of 'B, then h can be extended to an embedding of the structure (X U {b})23 into 2L EXERCISE 25.34 (R). Suppose 21 = (A, ... ) and 'B = (B, ... ) are models of a first-order theory T, IAI = IBI = K, and both !l and 'B have the K-extension property. Show that !l ~ 'B. Now we want to show that P(w)f Fin is N1-universal for the theory of Boolean algebras. We need some preliminaries. Let B be a Boolean algebra. We say that B has the strong countable separation property if B is infinite and for all X, Y E [Bj~No such that VF E [Xj
EbEX
EXERCISE 25.35 (PG). Prove that if B is a Boolean algebra with the strong countable separation property, then (i)
B has no countable limits; and
(ii) Every infinite maximal disjoint family in B is uncountable. LEMMA 25.27. The Boolean algebra P(w)f Fin has the strong countable separation property. PROOF. Let X, Y E [P (w)j~No be such that VF E [Xj
n G).
n<w i
i
This c is as required.
o
THEOREM 25.28. Every Boolean algebra with the strong countable separation property is N1-universal. PROOF. In order to prove Theorem 25.28,.we need the following: LEMMA 25.29. Let A, B be Boolean algebras such that A is countable and B has the strong countable separation property. If 7r : A - B is a homomorphic embedding and A(a) is a simple extension of A, then 7r can be extended to a homomorphic embedding 7r+ : A(a) - B.
25.
BOOLEAN ALGEBRAS
203
PROOF. Let A(a) be a simple extension of A. If a E A there is nothing to prove, so assume that a ~ A. Consider the cut (Sa, Ta)· Since a ~ A, L:bEF b < fIbEG b for all finite F ~ Sa and G ~ Ta. Since 11" is a homomorphic embedding, it follows that (11"[SaJ, 11" [Ta]) is a cut in B, and L:bEF b < fIbEG b for all finite F ~ 11"[Sa] and
G
~
11"[Ta].
= {s + (-t) : s E Sa, t ETa}. Then J is an ideal in A. For b E A \ J let Y'(b) = {b· - j : j E J}. Then Y'(b) is closed under· and OA ~ Y'(b). Let X = {OB} and Y(b) = 11"[Y'(b)]. Then X < Y(b) and there exists Cb E B such that 0 < Cb < Y(b). Let C = {Cb : b E A \ J}. By the Balcar-Vojtas Theorem and Exercise 25.35(ii), we may assume without loss of generality that the Cb'S are pairwise disjoint. For each b E A \ J we choose disjoint elements db, eb such that o < db, eb < Cb. The sets Let J
U = 11"[Sa] U {db: b E A \ J},
V = 11"({ -t : t ETa}] U {eb : b E A \ J} are countable and U l- V. Hence there exists x E B that separates U and V. Let 11"+ : A --+ B be the extension of 11" given by the formula 11"+(a) = x. By Exercise 25.18, this 11"+ is a homomorphic embedding. 0 EXERCISE 25.36 (PG). Derive Theorem 25.28 from Lemma 25.29.
0
COROLLARY 25.30 (CH). Every Boolean algebra B of cardinality Nl that has the strong countable separation property is isomorphic to P(w)/Fin. PROOF. Since the Boolean algebra P(w)/Fin has the strong countable separation property, Lemma 25.29 applies. In other words, P( w) / Fin is a model of the theory of Boolean algebras that has the N1-extension property. By CH, Exercise 25.34 applies, and Corollary 25.30 follows. 0
a
~
Let B be a Boolean algebra, X (X \ {a}) for all a EX.
~
B. The set X is called irredundant if
THEOREM 25.31 (McKenzie). Every maximal irredundant subset of a Boolean algebra generates a dense subalgebra. PROOF. Let B be a Boolean algebra, X
~
B an irredundant subset.
Let
a E B+ be such that Bla n (X) = {O}. We show that X U {a} is irredundant: Suppose XU{a} is not irredundant. Then there is b E X with b E ((XU{ a}) \ {b}). Let D = (X), Y = X \ {b}. Since b E (Y)(a), there are y,z E (Y) such that b = y. a + z· -a. Then bAz :S: a (since b· -a = z· -a), and since b,z E D, we also have bAz ED. Moreover, z E (Y), but b ~ (Y), since X was assumed irredundant. Thus b #- Zj hence bAz #- 0 and BlanD #- {O}, which contradicts the choice of a. 0 Let B be a Boolean algebra. The 11"-weight of B, denoted by 11" B, is the smallest cardinal K such that there exists a dense subset A ~ B+ of cardinality K. A subset X ~ B is called incomparable if its elements are pairwise incomparable. We define IncB = sup{IXI·: X is an incomparable subset of B}. EXERCISE 25.37 (G). Show that the 11"-weight of a Boolean algebra is equal to the 11"-weight of its Stone space. We conclude this chapter with a theorem that establishes a connection between 11" B and Inc B. First we need a lemma.
204
25.
BOOLEAN ALGEBRAS
LEMMA 25.32. Let B be a Boolean algebra with embedded in P(w).
1l' B
= w.
Then B can be
PROOF. It is not hard to see 2 that Intalg (Q+) can be embedded in P(w). It follows from Corollary 25.16(d) that every countable Boolean algebra can be embedded in P(w). Now let A be a countable dense subalgebra of B, and let f : A -+ P(w) be an embedding. By Sikorski's Theorem there exists an extension to a homomorphism f+ : B -+ P(w). Since f is injective and A is dense in B, f+ is also injective, and thus f+ is a homomorphic embedding of B into P(w). 0 THEOREM 25.33 (Baumgartner und Komjath). Assume B is a Boolean algebra with IncB ~ ~o. Then 1l' B ~ ~o, and hence B can be embedded in P(w). PROOF. Suppose towards a contradiction that B is a Boolean algebra with IncB ~ ~o and 1l' B 2: ~I. Then there exists a sequence (bO:)O:<Wl of elements of B+ such that for each Q < WI there is no d E Bo: with 0 < d < bo:, where Bo: = ({b/3 : (3 < Q}). Let Z
= {Q < WI : 3b E Bo: (b·
bo:
=I 01\ -b· bo: =I O)}.
Case 1: Z is stationary in WI. Let (CO:)O:<Wl be an enumeration of B W1 ' For each Q E Z let f(Q) be the smallest ordinal {3 such that bo: . c/3 =I 0 =I bo: . -C/3. Let
C = {Q < WI : Bo: = {c/3 : {3 < Q}}. Then C is club, and thus Z n C is stationary. By the Pressing Down Lemma (Theorem 21.12), f is constant on some stationary S ~ Z n C. Let c = cJ(o:) for Q E S. For Q E S let do: = c· bo: + -c· -bo:o We show that {do: : Q E S} is an incomparable set, which contradicts our assumption about B. Let Q, {3 E S be such that Q < {3. If do: ~ d/3, then c· bo: ~ b/3, which contradicts the choice of b/3' If d/3 ~ do:, then -c· -b/3 ~ -c· -bo: and thus -c· bo: ~ -c· b/3 ~ b/3' Again we get a contradiction with the choice of b/3. Case 2: Z is not stationary. Let Y = WI \ Z. Then IYI = WI, and for Q, {3 E Y with Q < {3 either bo:· b/3 = 0 or b/3 ~ bo:. Let W = {Q E Y : 1{{3 E Y : b/3 ~ bo:}1 = ~d· Suppose IWI ~ ~o· Then we can recursively construct a sequence (bo:,)i<Wl such that Qi E Y \ Wand bo:, 1:. bO: k for i < k < WI· But then {bo:, : i < wd is an uncountable incomparable set, contradicting the choice of B. Thus IWI = ~I' Suppose that b/3 ~ bo: for all Q, {3 E W with Q < {3. Then {bo: : Q E W} is a chain. Let (Qi)i<Wl be an enumeration of W in strictly increasing order. For i < WI let di = bo:, . -bO: i +1 _ Then {di : i < WI} is an uncountable incomparable set, contradicting the choice of B. Thus there exist Q, {3 E W with Q =I {3 such that bo: . b/3 = O. We pick functions
f : WI
-+
{'Y E Y : b,
~ bo:},
9 : WI -+ bEY: b, ~ b{3} such that f(i),g(i) < f(k),g(k) for i < k < WI- For i <
di = bJ(i)
+ b/3 . -bg(i)'
2Do we still have to remind you that here is an
EXERCISE?
WI
let
MATHOGRAPHICAL REMARK
205
If i < k, then di and dk are incomparable; hence {d i : i < WI} is an uncountable incomparable set. This contradiction shows that B must contain a countable dense subalgebra. 0
Mathographical Remark The three-volume Handbook of Boolean Algebras, edited by J. D. Monk and R. Bonnet, Elsevier Science Publishers, 1989, gives a comprehensive account of the current state of knowledge about Boolean algebras.
CHAPTER 26
Appendix: Some General Topology The main purpose of this appendix is to provide the reader with a convenient source for looking up the concepts and theorems from general topology that are used elsewhere in this book. We also include some proofs that we consider particularly good illustrations of certain set-theoretic techniques. For a serious study of general topology we recommend the encyclopaedic treatise General Topology, by R. Engelking, Heldermann Verlag, Berlin, 1989. In particular, all the proofs that are skipped in this appendix can be found in Engelking's book. Let X be a set, T ~ P(X). We say that T is a topology on X, if 0,X E T and T is closed under finite intersections and arbitrary unions (Le., if F E [T]
ENx 3U E B(x E U ~ N). 207
208
26. APPENDIX: SOME GENERAL TOPOLOGY
Equivalently, B is a base for riff r = {UA: A ~ B}. Similarly, if we are given a family of neighborhood bases {Bx : x E X}, then U E r iff "Ix E U 3B E Bx (B ~ U). Thus topologies can be defined by specifying a base or a family of neighborhood bases. Now let us look at some important examples. EXAMPLE 26.1. Let X be an arbitrary nonempty set, r ogy is called the antidiscrete topology on X.
= {0, X}.
This topol-
EXAMPLE 26.2. Let X be an arbitrary nonempty set, r = P(X). This topology is called the discrete topology on X. In a discrete space, all points are isolated. Clearly, two discrete spaces X and Y are homeomorphic if and only if IXI = WI. If '" is a cardinal, then D(",) denotes the discrete space with", points. EXAMPLE 26.3. Let (L, <) be a strict 1.0., and let a, bEL. We define:
(a, b) = {c E L: a < c < b}; (-00, a) = {c E L: c < a}; (b,oo)={CEL: b
[0,00)
26. APPENDIX: SOME GENERAL TOPOLOGY
209
Given a metric d on X, a point x EX, and a positive real c, one defines the open ball with center x and radius c by: B(x,c) = {y EX: d(x,y)
< c}.
The diameter of a set A ~ X is defined as diam(A) = sup{d(x,y) : {x,y} E [A]2}. Note that (iii) implies that the diameter of an open ball B(x, c) is at most 2c. For x E X, let Bx = {B(x,c) : c > O}. The family {Bx : x E X} is a family of neighborhood bases for a topology Td on X, called the topology induced by the metric d. A topological space (X, T) is metrizable if there exists a metric d on X such that T = Td. Let us look at some specific instances. The function M : JR2 --+ [0,00) defined by M(x, y) = Ix-yl is a metric that induces the usual topology on JR. This shows that the space (JR, T( <)) of Example 26.3(a) is metrizable. Similarly, if K is a cardinal and the function dd : K x K --+ [0,00) is defined by: dd(o:, f3) =
{o,
1,
~f 0: = If 0:
f3;
i= f3,
then dd is a metric such that Tdd is the discrete topology on K. It follows that all spaces D(K) are metrizable. Given a metric d on a set X, we say that a sequence (Xn)nEw of points in X converges (in the sense of d) to x E X iflimn-+oo d(xn'x) = 0. We say that (Xn)nEw is a (d-) Cauchy-sequence if limn-+ oo sUPm>n d( x n , x m ) = 0. Every convergent sequence is a Cauchy sequence, but not necessarily vice versa. We say that a metric d on a set X is complete if every d-Cauchy sequence converges with respect to d to some x E X. A topological space (X, T) is completely metrizable if T is induced by a complete metric on X. For example, the metric dd is complete by default since there are no nontrivial dd-Cauchy sequences. The metric M is complete on JR, but neither the restriction of M to (0,1) nor its restriction to Q is complete. However, the space ((0,1), TMrCO,I)2) is homeomorphic to (JR, TM) and thus is completely metrizable. In contrast, one can show that the space (Q, TMrIQ2) is not completely metrizable. There are several standard constructions of new spaces from old ones. In this text we need only two of them, namely subspaces and product spaces. Let (X,T) be a topological space, Y ~ X. Define TfY = {U n Y : U E T}. The topology T fY is called the subspace topology on Y induced by T, and (Y, T fY) is called a subspace of X. 26.6. (a) (Q,TMrIQ2) is a subspace of (JR,TM)' (b) If 0: E f3 E ON, then (o:,T(E)) is a subspace of (f3,T(E)). (c) If K < .x, :F is a nonprincipal filter on K, and g is a nonprincipal filter on ,x, then (K + 1, T:F) is not a subspace of (,x + 1, T g ). (d) Let w* be the set of all nonprincipal ultrafilters on w. Then (w*, T w*) is a subspace without isolated points of the space (f3w, T) defined in Example 26.5. EXAMPLE
r
Let {(Xi, Ti) : i E I} be a family of topological spaces. The product of these spaces is the space (TIiEI Xi, T), where the product topology T is generated by the base of all sets of the form DiEl Ui such that Ui E Ti for all i E I and Ui i= Xi for only finitely many i E I.
212
26.
APPENDIX: SOME GENERAL TOPOLOGY
For metrizable spaces, most of the countability properties discussed above coincide. In particular, we have the following. THEOREM 26.2. Let X be a metrizable space. Then the following are equivalent: (i) X is second countable; (ii) X is Lindelof; (iii) X is hereditarily Lindelof; (iv) X is separable; (v) X is hereditarily separable. As Examples 26.8(c) and (d) show, the equivalence between (i), (ii), and (iv) of Theorem 26.2 does not generalize to the class of all topological spaces, not even to the class of all regular spaces. The question whether the equivalence between (iii) and (v) generalizes is more complex. A space X is called an S-space if it is regular, hereditarily separable, but not Lindelof, and an L-space if it is regular, hereditarily Lindelof, but not separable. There exist various consistent constructions of S-spaces and L-spaces; for example, the Kunen line constructed in Chapter 17 with the help of CH is an S-space. However, it is relatively consistent with ZFC that there are no S-spaces. At the time of this writing it is still an open problem whether ZFC proves the existence of an L-space. Let us return to Example 26.8(d). The spaces f3w and w* satisfy a very strong version of Lindel6fness, namely: (26.2)
Every open cover of X has a finite subcover.
Topological spaces that satisfy 26.2 are called compact spaces. Compact Hausdorff topologies are minimal topologies in the following sense: THEOREM 26.3. Suppose (X,Tl) is a compact Hausdorff space, and TO C;;; a topology on X such that the space (X, TO) is Hausdorff. Then TO = 1'1'
1'1
is
For a proof of Theorem 26.3, see Corollary 3.1.14 of Engelking's book. If TO, 1'1 are two topologies on the same set X such that TO C;;; 1'1, then we say that 1'1 is a refinement of TO. EXAMPLE 26.9. (a) An ordinal 0: > 0 is compact iff it is a successor ordinal. (b) The real line R is not compact, but its subspace [0,1] is. More generally, every closed interval [a, b] of reals is compact. A Hausdorff space X is called locally compact if every point x of X has a neighborhood that is a compact subspace of X. Example 26.9(b) shows that the real line is a locally compact, noncompact space. Similarly, every ordinal is a locally compact space. The most conspicuous difference between intervals in the real line and ordinals is that the former contain no isolated points, whereas in the latter spaces nonisolated points are few and far between. Let us capture this intuition in a formal definition. A subset A of a topological space X is perfect If it is closed and every point of A is an accumulation point of A; i.e., A is perfect iff A = A'. For example, each closed nontrivial interval is a perfect subset of the real line. The set X' of all nonisolated points of a space X is called the Cantor-Bendixson derivative of X. For example, if 0: is an ordinal with the order topology, then 0:'
26. APPENDIX: SOME GENERAL TOPOLOGY
213
is the set of all nonzero limit ordinals below a. The Cantor-Bendixson derivative of (3w is w·. As in calculus, the operation of taking derivatives can be iterated. By recursion over ~ E ON we define: XCO) = Xj
XCeH) = (XCe))'j X( 6) =
n
XW if 6 is a limit ordinal > O.
e<6 If X is Hausdorff, then the set neEoN XW is either perfect or empty. In the latter case the space X is said to be scattered. All ordinals are scattered spaces. The next theorem gives a sufficient condition for being scattered. We say that a space X is locally countable if every point x E X has a countable neighborhood.
THEOREM 26.4. Every locally countable, locally compact Hausdorff space is scattered. A space is connected if it is not possible to decompose X into two disjoint nonempty open subsets. A continuum is a connected compact Hausdorff space. The unit interval [0, 1] is an example of a continuum. If a Hausdorff space with more than one point contains an isolated point, then X is not connected. In particular, no scattered Hausdorff space with more than one point is connected. The space W2 is an example of a disconnected compact Hausdorff space without isolated points. Compact metrizable spaces are Polish spaces, i.e., they are completely metrizable and separable. The spaces D(~o), P(w), w w, and R. are examples of noncompact Polish spaces. Recall the definition of meager sets from Chapter 13. Let us call a subset C of a topological space X comeager if X\C is meager. THEOREM 26.5 (Baire Category Theorem). In a Polish space, every comeager set is dense. If Y is a dense subspace of a compact Hausdorff space X, then X is called a compactijication of Y and the subspace X\Y is called the remainder of this compactification. Each noncompact, locally compact Hausdorff space Y has a (unique up to homeomorphism) one-point compactijication, i.e., a compactification X such that the remainder X\ Y is a singleton. For example, if It is an infinite cardinal and F is the filter of cofinite subsets of It, then (It + 1, r:F) is the one-point compactification of D(It). In particular, (w + 1, r( E)) can be regarded as the one-point compactification of D(~o). One-point compactifications are the smallest possible ones. On the other end of the spectrum are Stone-Cech compactifications. The Stone-Cech compactijication {3Y of a space Y is a compactification of Y such that for every compactification X of Y there exists a continuous surjection f : {3Y ~ X such that f fY is the identity. The Stone-Cech compactification of Y, if it exists2 , is unique up to homeomorphism. If we identify each natural number n with the principal ultrafilter Fn on w, then the space (3w of Example 26.5 is the Stone-Cech compactification of the discrete space D(~o), and w· is the remainder of this compactification. 20ne can show that Y has a Stone-Cech compactification if and only if Y is a Tychonoff space, where the Tychonoff property is a slightly stronger separation property than regularity.
214
26. APPENDIX: SOME GENERAL TOPOLOGY
Let us conclude this chapter with some proofs that illustrate how set-theoretic arguments can be used as tools in general topology. THEOREM 26.6 (Hewitt-Marczewski-Pondiczery). Let K, be an infinite cardinal. If {Xi: i E I} is a family of topological spaces such that d(Xi ) ~ K, for all i E I, and if III ~ 2"', then d(iliEI Xi) ~ K,. In particular, a product of at most 2No separable spaces is separable. PROOF. Let K" I be as in the assumptions. Since taking the product of a space X with any number of copies of the one-point space results in a homeomorphic space, we may without loss of generality assume that III = 2"', or even more specifically, that I = "'2. So, let {Xf : f E "'2} be a family of nonempty topological spaces of density ~ K, each. For each f E "'2, let Df be a dense subset, and enumerate Df, possibly with repetitions, as {d£ : 0: < K,}. For H E [K,J
r
r
In a certain sense, the result of Theorem 26.5 is the best possible one. THEOREM 26.7. Suppose K, is an infinite cardinal, III > 2"', and {Xi: i E I} is a family of Hausdorff spaces such that IXil > 1 for all i E I. Then d(iliEI Xi) > K,. PROOF. Let {Xi: i E I} be as in the assumptions, and let D be a subset of iliEI Xi of cardinality ~ K,. We show that D is not dense in iliEI Xi' For i E I, pick {Xi, Yi} E [xj2 and two disjoint nonempty open sets Ui, Vi C Xi such that Xi E Ui and Yi E Vi, Let D = {d" : 0: < A}, where A :::; K,. For each i E I, define a function fi : A --+ 2 as follows: fi(O:) =
{I,0,
~f da(~) EUi ;
If d,,(1) ¢ Ui.
Since III > 2'\, the Pigeonhole Principle implies that there exist io, il E I such that io #- i1 and fio = fi Consider the set W = iliEI Wi, where Wio = Ui o , Wil = Vi, , and Wi = Xi for i "E I\{io,id. Then W is a nonempty open subset of iliEIXi with W n D = 0. 0 The last two theorems of this appendix illustrate the use of cardinal invariants of the continuum in topology. More material on this topic can be found in the articles The integers and topology, by E. K. van Douwen, in Handbook of Set-Theoretical Topology, K. Kunen and J. E. Vaughan, eds., North-Holland, 1984, 111-167; and
26. APPENDIX: SOME GENERAL TOPOLOGY
215
Small Uncountable Cardinals and Topology, by J. E. Vaughan, in Open Problems in Topology, J. van Mill and G. M. Reed, eds., North-Holland, 1990, 195-216. A topological space X is called sequentially compact if every sequence of points in X has a convergent subsequence. Every compact metrizable space is sequentially compact. In particular, the space D(2) is sequentially compact. THEOREM 26.8. Suppose K. < t and {(Xo, To) : a < K.} is a family of sequentially compact topological spaces. Then the space I1o<1< X Q is also sequentially compact. PROOF. Let K. and {(Xo,To) : a < K.} be as in the assumptions, and let (fn)nEw be a sequence of elements of I1o<1< Xo. By recursion over a, we construct Xo E Xo and bo E [w]No such that for all a < (J < K.:
(i) VU E To (Xo E U (ii) b{3 ~* boo
--+
I{n E b
Q
:
fn(a) ¢ U}I <
~o);
Sequential compactness of Xo can be used to assure (i), and at limit stages (J one uses the assumption that K. < t to assure (ii). The latter assumption also implies that there exists b E [w]No such that b ~* bo for all a < K.. Fix such a b, and let I E I1o<1< Xo be such that I(a) = XQ for all a < K.. A straightforward 0 verification shows that the sequence (fn)nEb converges to I. THEOREM 26.9. The topological space (D(2))5 is not sequentially compact. PROOF. Let S = {b o : a < s} ~ P(w) be a splitting family. Define a sequence (fn)nEw of elements of 52 as follows:
In(a)
= {I, ~f n Ebo ; O,lfn¢b Q •
We show that (fn)nEw has no convergent subsequences. Let b E [w]N o. Since S is a splitting family, there exists a < 5 such that Ib n bol = ~o = Ib\bol. Fix such a, and let U = {f E 52 : I (a) = I} and V = {f E 52 : I (a) = O}. Then U and V are disjoint open sets in the product space 52, and the sets {n E b : In E U} and {n E b : In E V} are both infinite. Thus the sequence (fn)nEb cannot be convergent. 0
Index
Fq-set, 207 Go-set, 207 Q-embedding of a tree, 41 T2-space, 210 T3-space, 210 ","-principle, 144 ","-sequence, 144 ~-principle, 141 ~-sequence, 142
absolute formula, 168 accumulation point, 207 additivity of a measure, 147 of an ideal, 74 almost contained in, 102 almost disjoint families, 119 almost disjoint family, 82, 127 always first category set, 76 amoeba forcing, 92 antichain in a p.o., 6 in a tree, 40 antidiscrete topology, 208 Aronszajn line, 43 Aronszajn tree, 36 atom of a Boolean algebra, 188 of a measure, 153 atomic formula, 159 atomic measure, 153 atomless measure, 153
lprinciple, 145 Nl-separation property, 201 ~-system, 67 It-Aronszajn tree, 36 It-Kurepa tree, 30 It-chain condition in a Boolean algebra, 19 in a 1.0.,44 in a p.o., 7 in a topological space, 211 It-closed p.o., 6 It-directed closed p.o., 6 It-extension property, 201 It-scale, 76 It-tree, 36 It-universal model, 201 (It, A" )-gap, 117 (It, A")-pregap, 117 A-independent family, 84 7T-base, 211 7T-weight of a Boolean algebra, 203 of a topological space, 211 a-centered p.o., 91 . a-field, 13 a-linked p.o., 91
Baire Category Theorem, 213 Baire property, 14 Balcar-Vojtas Theorem, 200 base of a topology, 207 of an ideal, 72 Baumgartner-Komjath Theorem, 204 Boolean algebra, 14 atomic, 188 atomless, 188 canonical homomorphism, 188 complement, 16
Los' Theorem, 9
complete, 19 217
218 disjoint elements, 199 disjoint subset, 199 dual,16 incomparable subset, 203 K-complete, 19 separated subsets, 201 superatomic, 197 trivial, 14 weakly homogeneous, 190 weakly K-homogeneous, 190 Boolean combination, 33 Boolean order, 15 Boolean ring, 187 Boolean space, 18 Borel Conjecture, 74 Borel set, 13 bounded subset of an ordinal, 123 of w w , 76 bounding number, 76 branch,29 canonical coloring, 49 Cantor discontinuum, 18 Cantor set, 18 Cantor space, 18 Cantor-Bendixson characteristics, 198 Cantor-Bendixson derivative, 197, 212 Cantor-Bendixson height, 198 cardinal function, 210 cardinal invariant, 74 Cauchy sequence, 209 cellularity, 199 centered set, 5, 91 character of a point, 210 of a space, 211 characteristic function, 210 clopen set, 14, 207 closed in ON, 131 under a function, 124 closed subset of a topological space, 207 of an ordinal, 123 of [X)<", 133 closure, 207 closure under functions, 132 CLUB,123 CLUB,133 club, 123 club, 133 cocountable set, 1 cofinal branch, 29 comeager set, 213 compact space, 212 compactification, 213 compactness argument, 31 compatible elements, 3
26. INDEX complete metric, 209 completely metrizable space, 209 completion of a Boolean algebra, 22 of a 1.0., 19 cone, 134 connected space, 213 continuous function, 207 continuum, 213 convergent sequence, 209 countability property, 211 countable chain condition in a p.o., 7 in a topological space, 211 countable limit, 202 countable separation property, 201 cover, 211 cut, 21, 192 de Morgan's Laws, 16 Dedekind cut, 19 definabiIity, 172 dense set in a Boolaen algebra, 19 in a 1.0.,43 in a p.o., 7 in a topological space, 211 density, 211 derived set, 123, 207 diagonal intersection, 125 diagonal union, 127 diameter, 209 discrete space, 18 discrete topology, 208 disjoint refinement, 200 dominating number, 76 dual ideal, 1 elementary embedding, 24, 159 elementary equivalence, 160 elementary submodel, 159 embedding, 159 elementary, 159 Erdos cardinal, 60 Erdlis-Dushnik-Miller Theorem, 60 Erdlis-Rado Theorem, 55 Erdos-SierpiIiski Theorem, 72 Erdos-Tarski Theorem, 199 eventual dominance, 75 extension finite, 191 . simple, 191 Extension Property, 178 factor algebra, 190 Fichtenholz-Kantorovitch-Hausdorff Theorem, 83 field of sets, 12 u-complete, 13
26. INDEX
filter countably complete, 3 generated by a family, 2 in a Boolean algebra, 17 in a p.o., 3 K-closed, 6 generated by a subset, 5 K-closed,3 K-complete, 3 of closed sets, 3 on a set, 1 proper, 1 uniform, 1 filter base, 2 in a p.o., 4 finite intersection property, 2 fip, 2 first countable space, 211 Fodor's Lemma, 127 forest, 28 full binary tree, 28 Fundamental Theorem on Ultraproducts, 9 Galvin-Mycielski-Solovay Theorem, 78 gap, 19, 117, 192 generic blueprint, 140 greatest lower bound, 5 Hausdorff gap, 118 Hausdorff space, 210 height of a node, 28 of a tree, 28 hereditary property, 211 Hewitt-Marczewski-Pondiczery Theorem, 214 homeomorphic spaces, 207 homeomorphism, 207 homogeneous set, 49 homomorphism, 159 ideal in a p.o., 4 of null sets, 147 on a set, 2 proper, 2 idempotent, 187 incompatible elements, 3 independent family, 83 indiscernible elements, 160 induced coloring, 50 initial part of a tree, 29 interior, 207 interval half-open, 194 interval algebra, 194 irredundant set, 203 isolated point, 207
219
kernel of a t::. -system, 67 of a homomorphism, 188 Knaster Property, 93 Konig's Lemma, 31 Kunen line, 81 Kurepa Hypothesis, 31, 143 Kurepa tree, 31, 143 L-space, 212 lattice, 14 level of a tree, 29 Levy collapse, 199 Levy's partial order, 199 lexicographical order, 54 Lindelof degree, 211 Lindelof space, 211 linearly ordered continuum, 47 linearly ordered space, 208 linked set, 91 local base, 207 locally compact space, 212 locally countable space, 213 lower bound, 3 lower semilattice, 5 Luzin gap, 121 Luzin set, 71 generalized, 74 mad family, 88 maximal almost disjoint family, 88 maximal antichain, 7 below an element, 189 McKenzie's Theorem, 203 meager set, 14, 72 measurable cardinal, 3, 12, 150 measure atomic, 153 atomless, 153 K-additive, 150 two-valued, 147 vanishing on the singletons, 147 meet, 5 metric, 208 metrizable space, 209 Miller-Rado Theorem, 62 Negative Stepping-Up Lemma, 57 neighborhood, 207 neighborhood base, 207 neighborhood filter, 1, 207 neighborhood system, 1, 207 network, 176 network weight, 176 node, 27 high,46
immediate successor, 44 splitting, 28 nonstandard natural number, 12
220
normal function on an ordinal, 124 on [Xj<"', 136 normal functional class, 131 normal ultrafilter, 151, 184 nowhere dense set, 13 nuIl set, 72 OCA,64 one-point compactification, 213 open ball, 209 Open Coloring Axiom, 64 open cover, 211 open interval, 208 open partition, 64 open set, 207 open subset of a p.o., 4 order topology, 208 Ostaszewski principle, 146 P-point, 3, 110 Paris-Harrington Theorem, 51 partition, 49 partition calculus, 50 partition of unity, 196 path,29 perfect set, 212 perfectly meager set, 76 Polish space, 213 power set algebra, 12 predense set, 7 pregap,117 Pressing Down Lemma, 127 Prime Ideal Theorem, 6 probability measure, 147 product of Boolean algebras, 190 product space, 209 product topology, 209 Property K, 93 pseudo-intersection, 102 Q-point, 110 quasi-normal ultrafilter, 115 quotient Boolean algebra, 188 ramification system, 194 generating, 194 linear, 195 linear generating, 195 Ramsey number, 53 Ramsey Theory, 50 Ramsey ultrafilter, 114 Ramsey's Theorem, 50 Rasiowa-Sikorski Lemma, 87 real-valued measurable cardinal, 150 reduced product, 8 reduct, 164 refinement, 212 Reflection Property, 178
26. INDEX
regressive function on an ordinal, 127 on [X]
26. INDEX
strongly ,X-independent family, 84 strongly meager set, 78 Bubcover, 211 subfield, 13 submodel elementary, 159 generated by a set, 132 subspace topology, 209 substructure generated by a set, 132 subtree, 27 Suslin Hypothesis, 40 Suslin line, 45 Suslin number, 211 Suslin Problem, 47 Suslin tree, 40 symmetric difference, 187 Tarski-Vaught criterion, 161 tidy formula, 33 topological space, 207 topology, 207 topology induced by a metric, 209 totally disconnected space, 18 transversal, 127 tree, 27 bushy, 44 full binary, 28 lexicographical ordering, 42 splitting, 28 tall, 46 tree property, 36 Tychonoff's Theorem, 6 Ulam matrix, 148 Ulam measurable, 153 ultrafilter in a p.o., 4 fixed,4 free, 4 principal, 4 normal, 151, 184 on a set, 1 principal, 1 quasi-normal, 115 selective, 110 uniform, 150 ultrapower, 11 ultraproduct, 8 unbounded in ON, 131 unbounded subset of an ordinal, 123 of w w , 76 of [X]
weakly compact cardinal, 39, 58 weight, 211 well-met p.o., 5 zero-dimensional space, 210
221
Index of Symbols
.,l.,3
!Bla, 190 fJ. x !B, 190 (X},191 B(a), 191 "'I, 188 [a, b), 194 (X)!2I,201 I1iEI Xi, 209 [s], 210
.1, 3, 119, 201 I1~ExfJ.dU, 8 I1~ExAdU, 8
E,l1 -X, 13 ~~, 13 II~, 13 ~B, 15
I1X,19 LX, 19 ~., 37, 102 =·,37,102 ~e, 40 £,20,27 £({3), 42 181',42 K -+ (>')~, 49 K f+ (>')~, 50 n ~ (m);, 51 K -+ (>,);;W, 60 K -+ (>'i)kO" 60 >. -+ ({3i)kO" 62 >. -+ ({3) 'd, 62 >. -+ ({3);;W, 62 w -+ (F)~, 114 <·,75 y+X,78 C·, 102 A', 123,207 A"<'1 C ",125 A:cExC",,135 \7"
a, 90 add(I),74 A(e), 92 At(B),188 b,76 B+,16 !B., 16 {3w,208 {3Y,213 b(!Jt) , 188 B:c, 207 B(x,e),209 c (a), 199 c(B), 199 cB,22 c.c.c., 7, 211 cFr(K),189 Xa, 210 X(x,X),210 X(X),211 cl B, cl.,.(B) 207 Clop(X) , 14 club, 123 CLUBb),123 club, 133 CLUB([Xj
D,18 il,75 dd,209
il(s, t), 42 diam(A), 209
223
224
26. INDEX OF SYMBOLS
D(K), 18 d(X),211
M g ,124 Mo,125 lJ.u,147 M(x,y),209
F·, 1 F"" 1
N,72 n(A),198 neg U,23
Fe., 4
F(B),4 Finco (A), 13 fip,2
Fix(J),124 Fix(F),131 F< .. (A), 13 flt(A),2 Fn(F, J), 4, 91 FormL,164 fp,49 glb, 13.3 g/u, 8 gid, 183 gZ,183 gR,194
h(A), 198 ht(t), 28 ht(T),28
r,2 I(A), 197 I,,(A), 197 II" 147 IncB,203 Intalg(L),194 intB,207 intr(B), 207 ju,24 K(a), 62 K-C.C., 7, 19, 211 KH,31 K(A),60 ker (7r), 188 .qX),3
IgR,195 lub,15 L(X),211 MA,90 MABa, 97 MAcBa, 97 MA(K),90 MAProperty K, 94 MA,,_linked,94 MA,,-centered,94 MAtop, 97 MA-(K), 98, 132
M,n
Nrr;, 1,207 nw(X),176 OCA,64
w·,209 p,102
PA,54 7r B, 203 7r[, 188 7rw(X), 211 rA,22 rank(x),179 r(~), 187 R(m, k, l), 52 R·(m,k,l),53 RO(X),22 5,80 sfip, 102 SH,40 SK,163
sk(Lm),165 165
sk(~),
SQ,47
stB,18 S(X),211 t,103
T(a), 29 Tea),29 r,207 rd,209 r:F,208 r M ,176
rry,
209
r«); 208 r(E),208
U"" 17 UltB,17 U/i, 147 vWCT,32 vWTY, 32
wct, 32 WTY,32 w(X),211 xV.. /U, l1
xV/U, 24
