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O provided we make sure that [Fol=A), then put xt,=xos. We claim that F, c { xrp: p < A ) ( c E, u F,) holds; this will imply that Et+ is also dosed under f .\n fact,assume,on the contrary, that y < I is the feast ordinal such that y,,4{~,~:/? u { x a s : a < 5 ) is not free or we must have xtS =y,, for some 6 < y. We show that this is absurd by showing that the former possibility can occur only for < A of Ps (this is obviously true for the latter possibility). Indeed, the set E, is closed under f; and so ytu # f (xmP) holds for any a<5 and B
V, i.e., that Y= r, E J , which we wanted to show.
u
i
writing y , = x , and- y, = x d , we have 0 < k < 1 and
so there must be a j s l - 2 such that
Then { y i : j l i < j + r + 2 ; €.Il, which contradicts our assumption.
SOME SPECIAL NEGATIVE PARTITION RELATIONS FOR
Suppose now there is an X G K with
rz 3
[ X l r + l ~ J 1
141
and ( X I = r + 3 . Let
x i , i < r + 3 be an enumeration of the elements of X in increasing order. Then
{ x i : i < r + 2 f € J 1 , and so
, f ( ( x i :l < i < r + I ; ) > , f ( [ x i : 2 < i < r + 2 i ,
and { x i :l i i < r + 3 f € J I ,which implies , f ( ( x i I: i i < r + l ) ) < , f ( { x i2:5 i < r + 2 ) ) .
This is a contradiction, completing the proof of the theorem. We ngw turn to a discussion of the corollaries of the above re@.
COROLLARY 25.2. Let
K>W.
Then 2K74(h-+, 4)3
PROOF. We have 2"74(h-+)$by (19.14), and so the result follows from ( 3 ) . COROLLARY 25.3. Let
h-
be singular. Then h-f*(h-,
4)3.
PROOF. We have ~ 7 4 t h - i ) for ; singular h- according t o Corollary 21.3, and so the result follows from (3). COROLLARY 25.4. Assume h- L w. Then any of the conditions (i)
2K=2y
(ii)
2K==h->cf(z),
.for some
p
(iii) cf (h-)j,(cf( h - ) ) : ensures that
2F74 (h-, 4 ) j
holds.
PROOF. We have 2 a + ( ~ )under i the above conditions by Corollary 21.5, and so the result follows from (3). COROLLARY 25.5. Let
and, furthermore, for any r 2 4 .
h-20 be
such that
IC+(K)~
Then
I42
NEGATIVE ORUINARY PARTITION RELATIONS CH.
PROOF. We have 2 5 + ( ~ ) :for K with K + ( K ) : implies (6), (3) implies (7), and (4) implies 2"(K,
v
according to (21.14), and so (2)
5)4
This confirms (8) in case r = 4. In case r > 4, ( 8 ) follows from here by a direct application of result a ) in Subsection 9.7 (this result is a monotonicity property in the superscript of the ordinary partition relation). We do not know whether (7) can be strengthened to 2"(~, 4)3.The simplest problem that remains open is whether 2 s - +(K,,,4)3 holds if 2'" < 2%> K,,,for every n
THEOREM 25.6. Assume
K 2 w,
C
2"2 cf (K),K =
K,
, ,Ic 2 4 for any ( < T, and
I
%(u
(91
2k3 for any v < cf
(K).
Then
2K%((4):
(10) K
is regular
2"%(3);
Z2' % (4;)
(11)
for any p 2 w by (24.5).If K is regular and the assumptions ofthe above theorem hold, then 2*'22", and so (10) is satisfied. PROOF.
k t IV=(ll: (
be a partition of
[ r . I 3 verifying relation
be a 1-1 mapping. Write
(9), and let
h : Cf (K)-+'2 A=)(V
2h';
SOME SPECIAL NEGATIVE PARTITION RELATIONS FOR r >
clearly, [ A [=2". Let
3
143
< be the lexicographic ordering of A, i.e., let < =lex (<,: v
where < is the natural welldrdering of the cardinal 2",.We are going to define a partition l = ( l : :t < t i p / p )
of CAI3 verifying (lo), or, rather, the equivalent relation
2"%((44< < ( 4 ) [ , + [ ,) 3 . T9
(12)
To this end, consider an arbitraryelement {.fo, f l , 1;) of [ A I 3 , where & < f 1< f 2 . If fonf1 =f1nf2=v(
{.fO(V)?
If
yo = fon . f i
# .f1
=vl,
nf 2
.fl(V), f 2 ( v ) ) E
1:.
then put {.Lo, .f1,. f 2 )
~ 1 z . i
if a < p and i < 2 are such that i=h(v,) (a)#h(vl) ( a ) ( = 1 -i). For any a < p write lr+a=la.O
and
Ir+p+a=la.l.
This completes the definition of the (not necessarily disjoint) partition I , since it is obvious that l t = [ A I 3 . We claim that 1 verifies (12).
u
,
C
Assume the contrary. Then there is a < < r / p + p and an X E A such that [ X ] 3 c 1 5 , and I X 1 > & ( 2 4 ) in case ( < r , and 1x124 otherwise. Consider first the case C
f ng=v
(13)
for any two distinct J g E X . In fact, we have t o show that fng = f'ng' whenever J g, f',g' E X and f # g , f ' f g ' . Let , f o < f l < f 2 <.f3 be elements of X such that {k g9 f ' , 8';s {Yo, f i , f2r f 3 ) . Then ($0, f,,f 2 ) ? { f l , f2, f31. E I : , and so fonfl=flnf2=f2nf3=v for some v. Then f o p v = f l p v = , f 2 p v = f 3 p v and.fo(v)
E
X ) y E 1;
NEGATIVE ORDINARY PARTITION RELATIONS CH.
144
v
As I' verifies relation (9), we have
I (,f( v ) :f~ X I < i..: Noting that f ( v ) # g ( v ) for any two distinct ,I; g E X by (13), this implies that IXI<E.,, which contradicts the assumption IXI>i.c. Consider now the case ( L T . Then [ X ] 3 ~ l , . for i some a
f3) E
means that i = h ( v , ) ( a ) # . . .,
which is a contradiction. The proof is complete.
COROLLARY 25.7. Let
K
2 w and 2, L cf 2"%(K,
(K).
Then
(4),13.
PROOF. If K is regular, then 2 , 1 K , and so 2"+(4); according to ( I 1). Hence we K,, may assume that K is singular; then K > W in particular. Write K =
1
r
where 0 5 K , < K . We have P + ( K +4)3 , by Corollary 25.2, and so we have 2 " " + ( ~ , 4)3 a fortiori. Hence 2 " + ( ~ , 4, (4),)3, i.e., 2 " + ( ~ , (4),J3, follows from the theorem just proved.
REMARK. Wedonot know whether the condition 2,2cf ( K ) , which isequivalent to cf ( ~ ) - + ( 3 ) ; (cf. (17.32) and (19.17)), can be replaced by the weaker condition cf (K)S,(cf ( K ) , (3),)' if K is singular. (This can certainly be done if K is regular in view of (24.4)). This leads to the following unsolved problem:
PROBLEM 25.8. Assume GCH. Does then KJc,,~,+l%(K (4X2,l3 lw+,~
hold? Note that, assuming GCH, we have
K+,%(Kl+b (3kJ2
(15)
by Theorem 20.2, which implies
%"+,%(KJW+,~ (3k,I2
(16)
according to Corollary 21.2, but Lemma 24.1 is not applicable here to deduce (14).
145
THE FINITE CASE OF THE ORDINARY PARTITION RELATION
COROLLARY 25.9. Let cardinal (r< ti. Then
t i 2 ~ and
2P2cf (ti), and assume that 2"<25 for any
22K%( K ,
(4),)3.
(17)
PROOF.Our assumptions imply in view of Theorem 6.10.d) and c) that cf (25)=cf (ti) and there are ti,,< ti (vccf (ti)) such that 25= 2"~.We have 2"+(~,?,
V
(K)
according to (19.14), and so Lemma 24.1.a) implies 22Kyj, %(ti:, K : ) ~ ; therefore we have 22Kv-+(ti, t i ) 3 a fortiori. Using Theorem 25.6 now, we obtain the desired result. To give an illustration for the role of this result, assume GCH now. Then K:)'
K1 % ( K l , K l , (3h;")2 according to (19.14) and so
by Corollary 21.2. The Negative Stepping-up Lemma (Lemma 24.1) is not applicable here, and fails to confirm the expected relation
And yet, this relation follows from (17) with K = K , ~ .This is the state of affairs which we referred to as the pseudo stepping-upcharacter of Theorem 25.6. There still remain gaps, however, which seem to be filled neither by stepping up nor by pseudo stepping up, as illustrated by Problem 25.8 above. 26. THE FINITE CASE OF THE ORDINARY PARTITION RELATION
The aim of this section is to prove the main known results concerning the partition relation n +(rnX, where m, n, r, and s are integers. The reason we do this at this place in the book is that the derivation of some negative results extensively relies on Lemma 24.2, a negative stepping-up lemma for the finite case whose proof heavily exploited the methods described in Sections 23 and 24, while many of the other proofs are elementary and so can be given at any place. For our discussion below, it will be convenient to extend the meaning of the partition relation n (Ci);< s (1) +
for the case when the c;s are allowed t o denote arbitrary positive real numbers.
NEGATIVE ORDINARY PARTITION RELATIONS CH.
I46
v
Define the meaning of this relation as the same as that of the relation n+(C;);<s
where, for each i < s, ci is the least integer L ci.This is a natural way of defining the relation in (1). In fact, with this definition, ( 1 ) holds if and only if given any coloring ,f: [n]'-+s, there is an i < s such that there is a set of cardinality 2 c i that is homogeneous of color i with respect to f. The first result we prove is due to Erdos [1947] and uses a simple probabilistic argument. THEOREM 26.1. There is a positive quantity c , , ~depending only on r and s such that 1
n%*c,., (log fl).-1)Z
( 2)
holds,for any integers n, r, s ~ 2 I n. case r = 2 and s ~ we2 have n+(2 log n/log s)f .for all integers n 2 ss .
(3 1
PROOF. Let k be an integer, and consider the chances for the existence of a k-element homogeneous set. Denoting by N the number of all colorings f: [n]'+s, we obviously have N = s (;
1.
(4 1
Given a fixed k-element set X E and ~ a fixed color i < s , the number of those colorings f; [n]'+s with respect to which X is a homogeneous set of color i is k
Letting X and i run here, we obtain for the number N , ofthose obviously s(:)-('). colorings .f: [nT+s with respect to which there is a k-element homogeneous set that
since
(3
and sare the number of possibilities for X and i, respectively. (We do
not necessarily have equality here since a coloring may have several k-element homogeneous sets.) If N , < N, then there is a coloring f: [n]'+s with respect to which there is no k-element homogeneous set; i.e.,
n%tk): holds then. By (4) and (5), this is certainly the case if
(6)
THE FINITE CASE OF THE ORDINARY PARTITION RELATION
147
This is equivalent to (7) using the estimate d / k ! >
(2)
, we can see that this certainly holds if
((1)
k logn~logk!+
- 1) log s.
What we have proved so faristhat (8)or(9)implies (6).Toshow (2), observe that
(9
- 1 > ci., k' holds for all integers k > r with some positive quantity c:,
depending only on r and s. Assuming k > r , (9) therefore follows from the inequality k log n Ilog k ! + c;,, k' log s , and so a fortiori from k log n 5 c;.s k' log s;
this latter is equivalent to 1
1
k 1(c;., log s)I-r (log n )r-l.
(10)
As we assumed n 2 2, the inequality 1
k 2 c r . , (log n ) r-l
(11)
with a suitable positive cr.&ensures both that (10)holds and that k > r . Hence (9), and so (6), hold for any integer k satisfying (1 1 ). This establishes (2). To prove (3), we go back to (9) again. For r = 2 this means that k log n S l o g k !
+(k(k-
1)/2- 1) log S ,
and we claim that this holds in case s log S
l log
nS
k log s. 2
-
Using the second inequality here, (12) certainly holds if
kZ 2
- log slog k !
10.
+ (k(k - 1)/2 - 1) log
S,
(12)
I48
NEGATIVE OKDINAKY PAKTlTlON KELATIONS CH.
i.e., if
v
(k/2+ 1 ) log S l log k !.
Ass
k
k
i=2
I
C logi> S l o g t d t = k l o g k - k + l ,
and so it is sufficient to show that i.e., that
(k/2 + 1 ) log k / 2 I k log k - k
+
+1,
(k/2 1 ) (log k -log 2 ) I k log k - k
+1
holds for k r 4 . This is equivalent to (k/2- 1 ) (2- log 2 ) s (k/2- 1 ) log k + 2 log 2- 1 , and noting that 2 log 2- 1 =logi.e.,
4 e
> O , it is enough to show for this that
(k/2- 1) (2-log 2 ) s (k/2- 1 ) log k , 2 -log 2 5 log k
holds for k 2 4. This last inequality can simply be checked by computation; hence (13)implies (12),i.e., (9)withr=2,whichmeans that (6)holdswithr=2incase (1 3) is satisfied, i.e., in case k 2 2 log n/log s and n 2 ss. This establishes (3). The proof of the theorem is complete. Using a sieve method due to L. Lovasz, J. Spencer [I19751was able to obtain a somewhat better estimate for Nk in case r = 2 than given in ( 5 ) , and thereby he proved that even the inequality
entails the relation n+(k):. This is a slight improvement upon the implication asymptotically equal (8)*(6) in case r =2. In fact, noting that '-$? and k/e, (14) roughly means that kk+' nl-s 2 , e
THE FINITE CASE OF THE ORDINARY PARTITION RELATION
I49
while (8) requires a little more, since in case r = 2 it means about as much as k k - l
nS--s
e
2
for large k. Relation (2) above is interesting only in case s = 2 since, as is easily seen, the case s > 2 easily follows from this by the Omission Rule (see Subsection 9.3) with c,.~= c,,. As we shall immediately see, the case r 2 4 is again of no interest, since we can derive stronger results then. Therefore we think that it is worth formulating the important particular cases of (2):
COROLLARY 26.2. There ure positive constonts c2 c m r l c 3 such that the relutions und hold ,for ewrj' integer n 2 2. For an integer k let log, denote the k times iterated logarithm, i.&, put
log, x = .Y
and
log, + s = log logis
(17)
for each integer i and every large enough real number .Y. We have the following result, which throws some light upon the situation also in case r 2 4 . THEOREM 26.3 (Erdds- HajnalLRado [1965]). Let r 2 3 . There tire positire r w / numbers c r , tl,, crnd n, depending only oil r such that crnd
n + ( d r log,-
1 11);
holdjitr a e r y integer n >n,.
PROOF. The result easily follows from the above corollary with the aid of Lemma 24.2. As for (18), this is valid for r = 3 according to (15). Using induction on r. assume that we have
. where r > 3 and n' is such that 2 n ' - 1 < ~ 1 1 2 " 'Then
zn'+ (2c, holds according to (24.17). As n > 2'"
,/log,
-
n'
', we have, say,
log, - n ' s 2 log,-, n
NEGATIVE OKDINAKY PARTITION KELATIONS TH.
I50
v
if n is large enough. Since we also have n<2"', we obtain (18) from here with I cr = 2 2 cr - I . As for (19). we first prove this in case r = 3. Assume that t i is large enough, e.g. 17 > 20 ( > e"),and choose 17' such that 2"'- I
+
log 11' + 1 ,j: .
(('2
As 2"' I < 17, we have log ti' < 2 log log 17 for, say, n > 20. Thus ( 19) follows from here for r = 3 with an appropriate t l , >O. Using induction on r as in the proof of (18), it is now easy to establish (19) for an arbitrary r 2 3 with the aid of (24.17). The proof of the theorem is complete. Let E > O be an arbitrary real number. One can easily see from the above proof that there are integers nr,r depending only on r and I: such that ( 1 8 ) and (19) hold with c r = ( l + ~ ) 2 * - ~ c and ~ (/,=(I + 1 ; ) 2 ' - ~ c ~ (20) provided n Z n , , ) .(For c2 and c 3 see ( I 5 ) and (16). cz 5 2/log 2 holds according to (31.) We now turn to a discussion of positive relations. To this end we shall need the concept of end-homogeneous sets, which was introduced in Section 15. We first prove the following LEMMA 26.4. Let n, k , r , trntl s he integers, trnrl usslime thtrt k > I '2 2 tint1 s 2 2. Gicen a coloring ,fi [nIr+s, the existence qf'u k-element etitl-homo~enroussot i s guartrnterd by the condition
i=r-
I
Hence, a jorriori, it is guurtrnteeti by the condition
in
ctrse
k =r
+ I , rrnrl by the condition
in cuse k > r + 1.
PROOF. We are first going to deal with conditions ( 2 2 )and ( 2 3 ) .I t is obvious that in case k = r + 1 (21) and ( 2 2 )are equivalent. We are going to show that in c a s e k > r + 1 (23)iniplies(2l);infact,writingI=k-2andm=r- 1,wearegoing
151
THE FINITE CASE OF THE ORDINARY PARTITION RELATION
to show that the inequality
,
c s(ni)<2s(;l) I
rn+
(24)
i = ni
holds for every I > m provided ni 2 1 and s 2 2. We use induction on I . As for the case / = n i + l , the above inequality simply says that
rn + s+ sm
i.e., that
+
< 2s"
+
I ,
rn<sm+I-s,
and this does in fact hold, as ,p+ 1 - ,y =
m+ I
f 1
s'
log s dr 2 rns log s 2 rn 2 log 2 > m ,
because we assumed s 1 2 . Assume now that I>m+ 1 and that (24) holds with I - 1 replacing 1. Then we have
i = ni
Hence, to show (24) we only have to verify that
This, however, obviously holds, since in case I - 1 >rn> 1 and s 2 2 we have
Thus (24)is established, and so we have also proved the implication (23)*(21). We now turn t o the proof of the main part of the lemma. For this purpose assume that (21) holds, and let ,j': [ n ] ' - + s be an arbitrary coloring; we shall then have to find an end-homogeneous set of k elements. In order to d o this, we shall in effect perform a tree argument using the canonical partition tree associated with the coloring ,f (see Sections 15 and 18). In practice, however, we shall not construct the whole tree here, as we shall be able to find a long enough branch without bothering about the whole tree. We are going to construct an end-homogeneous set where x,, < . . . < x k- < n . T o this end, we shall also have t o define the sets Sj for
. i l k - 2 ; intuitively, Sj will be the set of the potential successors ofthe sequence ( x i : i<.j). Put xj=,j and S j = { i : , j < i < n ) f o r j s r - 2 . Let r - 1 1 / 1 k - 2
and
I52
NEGATIVE OKUINAKY PARTITION KELATIONS CH.
v
assume that Sj and x i have already been constructed for all i < , j such that x, = min S,
holds, and we have f ( Z u [ s i ) ) = . j ' ([J;) zu
whenever J ' E S ~Z, G [x,: I < ; ; ,
and IZI=r.-I. Assume, further, that
Note that (25) obviously holds in case i < i ' - 1, and (26) holds vacuously in this case.Finally, (27)also holdswith,j=r.-l. Infact.we have 1Sr-,l=n-r.+2and
(::I:)
=0, and so (27) says that
i=r-2
which is true in view of (21). Define the equivalence relation j'. :E S j - I \ [ x - I ; satisfying
jyr-l
=,ion
S,_ I \
(.Y,;- j
by putting
J-=
i z for any
,f'(ZU(.Y)) = . t ' ( Z u{ y } , whenever Z c is,:i < , j ; and I ZI
= i'
-
(28)
1. Observe that this equality always holds if
. x i - , $Zinviewof(26)withi=,j- 1. Hence,notingthat thereareat most
(;.I;) (j-1)
possibilities for Z satisfying : y j - E Z, we can see that there are at most s r - 2 equivalence classes under the relation = j . Thus, using (27),we can see that there is at least one equivalence class of cardinality 2
Choose this equivalence class as Sj.Then (27)holds with,j replacingj - 1. Putting .xi= min S j , (25)and (26)will also hold with i = i. In fact, (25)needs no comment, and (26) follows from the fact that S j is an equivalence class under the relation E,.
T H t FINITF CASE O F THE ORDINARY PARTITION RELATION
I53
So far we have constructed x i and S j for i s k - 2 . Noting that we have IS,_zlL2 in view of (27), let . x , - ~ be arbitrary such that Thisfinishestheconstruction oftheset X = ( s j : i < k ) .I t iseasytoseefrom (25). (26) (both valid for i s k - 2 ) , and (29) that X is an end-homogeneous set with respect to ,f: The proof is complete. To prove now the existence of large homogeneous sets. we need an analogue of Lemma 15.1. That lemma is not directly applicable here since it concerns coJorings ,f': [ G I ] < " ' - + T , which are of no use in the finite case. We have, however, the following variant of that lemma: LEMMA 26.5. Lct 1 1 . r. s, k r i n d / hcpo.sitirr i17togrr.s. r i n d let,f': [n]"-+s hO ( I c'O/OriI7g. Assunw t h t i t t hcro i s ( I k-elcnirnt e r t r l - / ~ o n i o ~ ~ ~ r sot ~ r owui.tsh rc.spccl t o , f ' r i n d /hut MY /lt/W
k-1
+
(1-1)I-l.
(30)
Then therc~i s N I I I-elenwit hon1ogoneous srt,fOr,f:H e n c jf'thc)rr ~ i s (i k - c + m w t ondhonwgeneou.s st't ji)r U I I J - coloring f : [n]"+,s. thrn (30) iinp/ic.s 17 + (1 );.
(31 1
The proof is omitted, since it is an exact copy of the proof of (15.5)in Lemma 15.1 with 11, s, k - 1, and 1-1 replacing GI,T, p. and Y..(<
(331 ,for all integers n L 3.
PROOF. We shall first establish (33). Let j': [ n I 2 - + sbe an arbitrary coloring. Taking the trivial relation (1 - 2) s+ 1 +(1 - 1 ),!
into account, the preceding lemma implies that an end-homogeneous set of
I54
NEGATIVE OKVINAKY PARTITION KELATIONS CH.
v
k = (1-2) s+2 elements ensures the existence of an /-element homogeneous set. Noting that r = 2 in the present case, (23) above becomes 2 2s" - 2 ) 5
(34)
with this value for k ; according to Lemma 26.4, this inequality guarantees the existence of a k-element end-homogeneous set, and so that of an /-element homogeneous set, provided k r 3 ; this latter holds in case 123. Since a twoelement set is always homogeneous, (34) guarantees the existence of an /-element homogeneous set even without the assumption 123. (34) can be written as log n 2 log 2 - s log s + (/ - 1 ) s log s.
AS s 2 2, we have log 2 < 2 log 2 1 s log s, and so this inequality certainly holds if i.e., if
log n 2 ( I - I ) s log s, log n +l. s log s
Is----
This inequality holds with the least integer I not smaller than
log n , and so we s log s
have n - + ( / ) f with this 1. Thus we have established (33). We now turn to the proof of (32). We use induction on r. In case I' = 2, (32) is 1
. Assume r > 2 now, and let f': [ n ] * + s be a log s coloring. In order to have a k-element end-homogeneous set with k > r + 1, it is enough according to (23) in Lemma 26.4 that we have confirmed by (33) with c:,=
s
This certainly holds if n 2 2 ~ ' " ' ~ - ' , and SO, a fortiori, if words, this latter inequality, that is, the inequality
t l 2 $',
as s 2 2 . J n other
k 5 (log n/log s ) " ~ ensures the existence of a k-element end-homogeneous set, provided k > r we have k-1 + ( ~ ~ - ~ , , l O g r - ~ ( k - l ) ) ~ ~ '
(35)
+ 1. As
for k - 1 > n , by the induction hypothesis, the preceding lemma implies that (35) ensures the existence of a set of cardinality 2 ~ , + - ~ , , l o g(k-1)+ ,-~ 1 that is homogeneous with respect to f' provided k ? r + 1 and k > n , - I , s . Taking
155
THE FINITE CASE OF THE ORDINARY PARTITION RELATION
here the largest integer k satisfying (351,we can see that this number is greater than
provided n is greater than an integer nr,., depending on r and
s.
This proves ( 3 2 )
1
with c:,~= iTcF-l,h. I t is easy to see that if E > O then we can' have czS= 2-c in ( 3 2 ) , provided n is large enough depending on r, s, r ! slog s and E . The proof is complete. There is a different approach to the case r = 2 of the above theorem. In fact, we have the following simple result due to P. Erdos and G. Szekeres [1935], which was rediscovered in Greenwood-Gleason [ 19551 : =( 2- c )
c;.,,/r!=
'
THEOREM 26.7. Let k and n he integers 22.Then w e h a w ,
-(k,n)2.
PROOF. In case k or n equals 2 this simply says that n+(2, n ) 2 or k + ( k , 2)', both ofwhich obviously hold. Using induction, assume therefore that k, n > 2, and that ( 3 6 ) holds if k or n is replaced with any k'< k or n ' < n , respectively. Let X be a set of cardinality
("iFy *)
and let ,f: [ X I 2 + 2 be a coloring. Fix z
EX
arbitrarily,
and write x,=(xEX\~Zj:j'(jXZf)=O;
and
x ,={X€x\
~ z ; : j ' ( ~ x . ) ) =1 ) .
In view of the equality
(
n+k-2 k-1
)=(
n+k-3 k-2
)+(
n+k-3 k-1
)
we must have either n+k-3 k-2
or
n+k-3 k-1
(37)
(38)
156
NEGATIVE OKUINARY PARTITION RELATIONS CH.
v
If (37)holds, then the induction hypothesis implies that there is a subset Yo or Yl of X , of cardinality k - 1 or n that is homogeneous of color 0 or 1, respectively, with respect to .f: In the second case we are ready, and so are we also in the first case, since then You I z ) is a homogeneous set of cardinality k and of color 0. Similarly, if (38) holds, then there is a subset Yo or Yl of X, of cardinality k or n- 1 of X , that is homogeneous of color 0 or 1, respectively. In the first case we are ready, and so are we in the second case, since then Yl U{Z} is a homogeneous set of cardinality n and color 1. The proof is complete. In case 11 = k = I + 1 (36) gives that
holds for any integer 12 1. As we have
(equality above holds only in case I= l ) , this means that
holds for any integer I > 1. One can easily derive (33)with s = 2 from here. Note that (40) is not the best estimate for
. I t is well known that
holds for all integers I > 1. Hence (41) can be improved to 4'/d2=+(/+1);.
(42)
This is, however, insufficient to make any improvement in the constant factor 1
___ in (33)with s=2. Note the discrepancy in the constant factors in (3)and 2 log 2 (33). Putting
r$(n)= sup i x : x > 0 real & n + ( x log a);} ,
(3) and (33) with s = 2 gives that
T H E F I N I T E C A S E OF T H E O R D I N A R Y P A R T I T I O N R E L A T I O N
I57
It would be interesting to know how much, if at all. this inequality can be improved. A more intriguing problem concerns the case r = 3 : PROBLEM 26.8. Does there exist a positive quantity c such that n + ( c log log n ) ;
(44)
holds for every large enough integer n'? Compare this problem with (16),and with the case r = 3 of (19) and (32).
CHAPTER VI
THE CANONIZATION LEMMAS
Given a coloring f : [ A ] ' + 5 , the canonization lemmas ensure under certain assumptions that a 'large' subset of A can be represented as a disjoint union in such a way that the color of an r-tuple depends only on the summands to which the elements of this r-tuple belong, and not on the elements themselves. These lemmas will be very useful in establishing positive partition relations for singular cardinals, especially in the case r = 2 of the ordinary partition relation, but they also have important applications in set theoretical topology (see Section 43) and in the theory of set mappings of type > 1 (see Section 45).
27. SHELAH'S CANONIZATION The main purpose of this section is to show that, as mentioned earlier, Corollary 21.5 indeed gives a necessary and sufficient condition for ~ F - - + ( K , K)' to hold in case 2r is singular (see Corollary 27.4 below). We shall actually do something more general here, as will be seen from the explanation given after the following
DEFINITION 27.1. Let ( A : : 5 (9) be a sequence of pairwise disjoint sets, write A = U A:, and let f : [A']'-)r be a coloring, where A ' z A . We say that t <s
( A : : 5 < r ) is canonical with respect to f if f ( u ) = f ( u ) whenever u, u E [A]' and lunA:l=lunA:lforevery5<9. Wesaythat ( A : : 5 < 9 ) i s weaklycanonicalwith respect to f if f ( u ) = f ( u ) whenever u , u ~ [ A ] ' a n d ( u n A , ( = ( u n A , I I l for every 5<: 9. Given a sequence ( A t : (<9) of pairwise disjoint sets and a coloring f : [ Ar]'+r, we shall be able to construct, under certain general conditions, :<s another sequence ( B t : <<9) such that B:G A:, B: is 'fairly large', and ( B t : 5 < 9) is canonical or weakly canonical with respect to f . The construction of such a sequence (B,: ( < 9) will be called canonization. It seems plausible that the canonicity of (B,: <9) with respect to f will facilitate the construction of a large homogeneous subset of B,. We shall soon be able to give an illustration
u
<
u
159
SHELAH'S CANONIZATION
confirming this. The main result of this section is Shelah's Canonization Lemma (cf. Shelah [1975]), which is an improvement for r = 2 of the original canonization lemma of ErdBs, Hajnal, and Rado (Lemma 28.1):
LEMMA 27.2 (Shelah's Canonization Lemma). Let (K , : 5 <8 ) be a strictly increasing sequence of cardinals, ( A , : 5 < 9) a sequence of pairwise disjoint sets, F , c 9 ( A , ) and, writing A = A,, l e t f : [A]'-r be a coloring. Assume that we t<9 have
for any 5<9 andfor any X E A : with IXI=(2".)' there is an Y E F , such that YE X and I Y I = K~:
.
(4 1
Then there are B , E A:, 5 < 9, such that B,E F , ,
IB,I=K~?,
(5)
and ( B , : 5 < 8 ) is weakly canonical with respect to f. To illustrate the role of F: in the above lemma, we mention a corollary; we shall, however, not need this corollary for future references. The real illustration of the role of the above lemma will be Corollary 27.4 below. COROLLARY 27.3. Let K,, A:, a n d f b e as in the above lemma, and assume ( l ) , (2), and (3). Then there are B , c A: with IB,l = K: such that the sequence ( B E :5 < 9) is canonical with respect to f.
PROOF. Use the above lemma with F:= { X E A < :X is homogeneous with respect to ,f;; then (4) holds in view of the relation (2".)+-(~'?),2(note that T I K ~ I cf. K,; (17.32)). We obtain B : c A , with IB:I=K; and B,E F, such that ( B , : t < S ) is weakly canonical with respect to f: But then the relation B, E F , implies that this sequence is canonical with respect to f.
PROOFOF LEMMA27.2. We may assume that lAcl = (2"()+.
160
THE CANONIZATION LEMMAS CH. VI
Let 5 < 9 and Z E for any x, y E A,,
u
A,. Define the equivalence relation
a<,
~ = - , ~ y iff
= ,=
on A, as follows:
Vz~Z[f({zx))=f({zy])].
Call the element x of A: good with respect to Z if there are at least (2"~)+elements y of Ac such that x ~ - , ~Call y . x good if it is good with respect to any ZE A,
u
U<,
with 121IK,. We claim that there is a good x E A, for any 5 < 9. In fact, for a fixed ZG A, with 121IK ~ the , number of equivalence classes under the equivalence
u
a<<
relation =t,z is 52":. An element of A, which is bad (i.e., not good) with respect to Z is contained in an equivalence class of cardinality 5 2 " ~ ;hence, the number of bad elements with respect to Z is I2":. As I A,\ I2": in view of (1) and (6),
u
a<<
there are at most (2",)Kc = 2": possibilities for 2 ; therefore, all but 2": elements of A, are good. For any 5 < 9, let at. E A,. be good. We are going to define the sets B,E A:, ( < 9, by transfinite recursion such that ( 5 ) holds, f ( b , a , I ) = f({y,a,l)
for any 5 ,I <'9 and x,, y,
E B,,
(7)
and
f({&
(8)
)= f({x,a,j)
for any 5 < 3 < 9, x t E B:, and x, E B,. Fix u < 9, and suppose that the sets B, E A, have already been constructed for all ( < u in such a way that (5), (7), and (8) hold for all 5, [
c,={x E A,: x as a, is good and
I
u Br I
I2".
=a,
(J Bc a,} ; :
by (l), we have
<
ICaI>2"Q.
Define the equivalence relation x
-
ay
iff
-,on C , as follows: for any x, y E C, put
VBCa
There are at most zlyl equivalence classes under the relation -,; note that we have d9' <2". I2". according to (2). So, by (9), there is an equivalence class Ch under the relation -,that has cardinality (2"m)'. Let B , s C ; be a set of cardinality :K such that B, E Fa;there is such a B, by (4). (5) clearly holds with 5 =a. (7)also holds with 5 =a, since x,, y, E B, c C, implies that x, d,,and so
-
161
SHELAH'S CANONIZATION
indeed f({x,a,)) = f({y,a,)) whenever a
u ~ + ~ ~
COROLLARY 27.4 (Shelah [1975]). Let K be a singular cardinal such that ~ < 2 5 and 2p < 25 for any p < K , and assume that Cf(K)+
(cf ( K N ; .
(10)
for any T < cf (K). This result confirms that Corollary 21.5 gives a necessary and sufficient condition for 2 5 - + ( ~ ti,)' to hold in case K is singular, namely, that 25> ti and 2P<2? for any p
PROOF. We shall establish ( 1 1 ) under the assumption cf (K)-+(Cf
(K))22+,
(12)
;
it.will follow from Theorems 29.1 and 29.5 below that (12) is equivalent to (10). Assume there is a coloring f : [ 2 5 ] 2 - 2 4 ~ that refutes the relation in (11). By Theorem 6.lO.c(ii), there are cardinals K:, 5
whenever t < g < c f (K). Let ( A z : <
{xoUxl: xo,x,5 A , & IxoI= Ix,1 = K: &f"[X,]'=
&
{0)&f"[x,]2={1)). .
Observe that (4) holds with this F,. In fact, fixing t;
162
THE CANONIZATION LEMMAS c n .
VI
(2"~)'> K , or it has a homogeneous subset of color [ for some 5 with 2 1 [ < 2 i 7 that has cardinality ~ :>cf f ( K ) . This, however, contradicts our assumption that .f refutes the relation in (11). Thus (4)holds, and so we can use Lemma 27.2. We obtain B:, 5
(K)]'
and
f"[X;]'== ( i )
(i=O, 1 ) .
(15)
- 2 4 ~by putting
for any 5, q with t < q < c f ( ~ )where , X ~ E Band : x , , E B , , ;as ( B : : t < c f ( ~ ) ) is weakly canonical with respect to A the value of , f ( { & ) ) does not depend on the particular choice of xy and x,,. (12) implies that there is a set H s c f ( K ) of cardinality cf ( K ) that is homogeneous with respect to ,f;let 5 be the color of this ([<2$7). If i = O or 1, then putting homogeneous set, i.e., let (5)=f"[H]'
x= (J x;, (EH
it is easy to see by ( 1 5) and (16) that X is a homogeneous set of color [ and I X 1 = = K I ;as Iff I=cf ( K ) , this is equal to K in view of (13) and (14). so 1 X J= K , CE H
which contradicts our assumption that f refutes (1 1). If 2 5 [ < 2 45 then, pick arbitrary elements x: from each E:,
C~ROLLARY 27.5. Let K be a singular cardinal such that 2@< 2' for all p < K, let be an arbitrary cardirml, let v,, a C T ,be ordinals, and assume that
cf(K)+(Cf(K), ( V : ) Z < ~ ) ' .
(17)
PROOF. Let f:[2~]',-+147be an arbitrary coloring, and assume that there is no homogeneous set of order type vF and of color 14t for any 5 < 5 . We have to prove that then there is a homogeneous set of order type (or cardinality) K and of color 0. To this end, observe that by Theorem 6.lO.c(ii) there are cardinals K , ,
163
THE GENERAL CANONIZATION LEMMA
a
(K),
such that
Cf ( K ) < K , < K ,
and, for any a, fl with a < p
f"[B,]*= (0).. Consider the coloring f : [cf ( K ) ] ' - P 1 i r defined as follows: for any two distinct 0: P
u
as
X
is a homogeneous set of color 0 with respect to f. Y has cardinality which completes the proof.
1 K , =K , .EX
28. THE GENERAL CANONIZATION LEMMA In the preceding section we introduced the concept of canonical sequence (see Definition 27.1) and proved a canonization lemma for r = 2 (Lemma 27.2). Here we are going to prove a canonization lemma for an arbitrary integer r, though in case r = 2 this will say somewhat less than Lemma 27.2. The General
164
THE CANONIZATION LEMMAS CH.
VI
Canonization Lemma will be used later in order to obtain some results in settheoretical topology and about set mappings and the square bracket symbol (see Sections 43, 45, and 54, respectively), and not in the context of the oidinary partition relation. The reason for including this lemma here rather than at some later place lies more in its proofthan in its applications, and also in the fact that in the preceding section we discussed a canonization lemma that had important applicationsfor the discussion of the ordinary partition relation. In order to state the lemma in question, we recall that exp, (K), where n is an integer and K a cardinal, is defined by the recursive equations expo ( K ) = K , exp, +
( K ) = exp,
(2") .
(1 1
LEMMA28.1 (General Canonization Lemma; cf. ErdBs-Hajnal-Rado [19653). Let ( K: : 5 < 9) be a strictly increasing sequence of cardinals, (A: : < 9) a sequence of pairwise disjoint sets, write A = A,, and let T 2 2 be a cardinal and r 2 1 an integer. For any i with 1 Ii Ir let
t<s
,A: [ A ] ' + e ~ p , _ ~(T) be a coloring. Assume that exP(;)(k-<)<exP(;)(4,) for any 5 < q < 9 , K O >_ TI9', W
and for any 5<9, Then there are sets
lA t 1 1 (exP(;) (%)I+
q t s A:, 5 < 9, such that
and the sequence ( E : : 5 < 9) is canonical with respect t o the coloring,fi,foreach i with l s i s r . The purpose of having several colorings in the above lemma instead ofjust one is technical; namely, in this way it will be possible to prove the result by a simple induction. Our main concern in applications will be the coloring fi. In case r = 2 we obtain a weaker version of Corollary 27.3; indeed, the assumption in (3) here requires more than that in (27.2).
PROOF. We may assume that T = ~ 1 9 1w 2
;
indeed, the only assumption on'r is that in (3),and this remains valid ifwe replace T by s'=max { T ~ ~ ~ , K , J ' INote I ] . . also that the lemma is trivially true for
THE GENERAL CANONIZATION LEMMA
r = 1 ; in this case
(3
=
165
r(r- 1)/2=0. We use induction on r; let r 2 2 and
suppose that the lemma is valid with r - 1 replacing r. First we construct a sequence ( A : : 5<9> such that A2sA : , IAkI=(exp(rjl)(K:))+, and, writing A=
(7)
u A:,
:<.9
the following holds:
i f u , o c A , lul=IuI=i
(8)
Fix t < 9;and suppose that A,, has already been constructed for any q < 5 such that (7) and (8) are satisfied. To define A : , define colorings gi on [A:]' for any i with l ~ i by~ putting: r
note that, writing
=expr-i
(T)
+ C Isjsr-i
In case i =r this implies
(exp,-i-j
(T))elpy;')
166
THE CANONIZATIOY LEMMAS CH.
VI
and in case i < r we may disregard the first term if we write expr-i instead of expr-i_j. We obtain
since we have T SK,,<
K(
by ( 3 ) and (2),and r - i
-
1 s r 1 2 1( r i 1 ) , a s r z 2 . B y
( I l ) , (12), and Corollary 16.5, there is a subset A ; of A , of cardinality ( K , ) ) + = (exp(r;I) ( K < ) ) +
(~XP(;)-,~-
that is simultaneously homogeneous with respect to the coloringsgi, 1 ~ i s rI n. view of the definition of the colorings g i , this simply means that ( 8 ) holds. This concludes the construction of the sequence ( A , : ( < 9). We are about to construct the sequence ( B i : ( < 9) by using the induction hypothesis.Tothisend,define thecoloringsh,on [Alifor 1 s i s r - 1 asfollows: for any u E [A]'put hi(u)= ( 0 ' , 5 , , 1 i + j ( ~ u ~ ~ ) ) : ~ c A i & I S~ ,j< ~ r[-A i ; :.] j & (13)
(A' was defined in (9)) Observe that, by virtue of (8), . f i + j ( u u ~here ) depends
only on u,,i and t, and not on the particular choice of 1: (note that u determines i). The number of possibilities for j i + j ( u u ~on ) the right-hand side is Hence
In case i = r - 1 this implies Ira
(hr- 1
T191 )I < -
- T = ~ x P , , -1 ) - 1 r -
I,
(T)
t 14)
in virtue of (6). If 1 2 i < r - 1, then we get ra ( h i )I 2191 ' r 'exp.. ,
~
Z(T)
= p p , ~,
~
=e X P ~ r - ~ ~ - i ( T ) ~
(15)
since l S l . r s ~ s e x p ~ (T) - ~ by - ~ (6). In view of (7), (lo), (14), (15), and the induction hypothesis, we can use the lemma to be proved with r - 1; ( A r : 5
I67
THE GENERAL CANONIZATION LEMMA
that each Br is homogeneous with respect to j i , and this is indeed so, as already A iwas homogeneous with respect to each .fi, 1 c k s r , by (8); in case 1 < k s r , the canonicity of (B::(<9) with respect to ,fi follows from the homogeneity of B,,5<9, with respect to h,the canonicity of ( B : : ( < 8 ) with respect to the hi’s, and the remark made after (13), namely that, in (13), , f ; + j (v~) depends u only on u,,j, and (, and not on L:. The proof of the lemma is complete. COROLLARY 28.2. Let K be u strong limit curdinul, i.e., such that 2” < ~ j i ) runy p < K, and assume K is singular. Let r 2 1 be an integer, T < K u curdinul, und f : [ K ] ‘ + T a coloring. Then there w e puirwise disjoint sets Br E K ~ (, < cf ( K ) ,
of cardinality
u
such that
< Cf
B, has cardinality
(K)
( B , : (
PROOF.Writing 9=cf
K
and the sequence
fi
there is a sequence ( K ~ (<9) : of cardinals satisfying the requirements of Lemma 28.1 such that (K),
K =
1 ?<9
Let ( A , : ( < Y )
K< =
1 (exp(;)q)+ .‘<9
be a disjoint partition of
K
such that IAtI=(exp(;)(Kc))+.
Applying Lemma 28.1, we obtain the sets B: with the required properties. The proof is complete.
CHAPTER VII
LARGE CARDINALS
The ordinary partition relation shows remarkably different features for inaccessible and for accessible cardinals. A crucial role is played here by weakly compact inaccessible cardinals, which are the main topic of the next few sections. The chapter is concluded by a section concerning saturated ideals and partition properties of measurable cardinals. 29. THE ORDINARY PARTITION RELATION FOR INACCESSIBLE CARDINALS
The simplest instance of Ramsey's theorem says that o - + ( w ) : holds. The question arises whether an analogous relation holds for any other infinite cardinal replacing o.A simple result in this direction is THEOREM 29.1 ([Erdos-~Hajnal-~Rado [1965]). ff ~ > ( i ) a n dK + ( K ) ; holds, thert K
is inuccussiblr.
PROOF. Observe that K must be strong limit, i.e., 2 p < K must hold for any p < K . In fact, if 2p2 K for some p < K , then K 2 (ti): holds in view of the relation 2'-%@')f (see (19.14)). We have to show that K must be regular as well. This too is confirmed by our earlier results (namely, Corollary 21.3), but it is worth presenting a direct proof. Assume K is singular and let ( c q : 5
u
;<<
S
~
ORDINARY PARTITION RELATION FOR INAC’CESSIBLt CARDINALS
169
K+(K); fails for a large class of inaccessible cardinals, including the first one. Their result shows that if K + ( K ) ; holds, then K must be a very large inaccessible cardinal that cannot be characterized in a ‘constructive’ way. This means, in particular, that it is consistent with the usual axioms ofset theory (plus an axiom saying that there are ‘many’ inaccessible cardinals) that the relation K + ( K ) ; fails for any inaccessible cardinal K, since we can cut down the universe of sets anywhere below the first cardinal K (if there is any) for which K+(K);,holds. This means that the assumption that there are inaccessible cardinals satisfying K + ( K ) ~ can be considered a new axiom of set theory, in effect saying that the universe of sets cannot be cut down below a certain ‘large’ cardinal. Henq the name, introduced by K. Godel [1947], of this type of axiom being a large cardinal axiom. There are many more, and much stronger, large cardinal axioms than,the one we mentioned here. Their primary importance lies partly in the fact they have interesting consequences for ‘small’ sets, e.g. subsets of w. The reader interested in large cardinals should turn to F. Drake’s book [1974], which also has a good bibliography on the subject. Many properties equivalent to K + ( K ) ; of an inaccessible cardinal*Kare given in the paper of Keisler and Tarski [1964]; these properties concern a large number of branches of mathematics such as infinitary logic, model theory, topology, the theory of Boolean algebras, representation theorems, separation principles, and so on. They all show the importance of the ordinary partition relation. It is beyond the scope of this book to reproduce all these properties; we shall nonetheless need a few of them. In the rest of this section we shall prove a number of implications between some of these properties, which will show that all problems concerning the relation K + ( & ) ; < ~ for inaccessible K that cannot be decided with the aid of our earlier results can be reduced to the problem whether K+ (K); holds. All these implications except (iv)=-(v) below are stated in Silver [1971], where complete references are given. Thus when proving these implications, we shall only mention the authors’ name, while omitting exact references. In order to formulate and prove these implications, we shall need some definitions; the first one concerns trees (for the definition of tree and related concepts see Section 13).
DEFINITION 29.2. Let K be an infinite cardinal. An Aronszajn K-tree is a tree of length K that has no branch of length K and is such that each of its levels has cardinality < K . The cardinal K is said to have the free property if there is no Aronszajn k--tree. Konig’s lemma (Lemma 10.3) says that w has the tree property, and it is a wellknown theorem of Aronszajn and Specker that there is an Aronszajn K,-tree (Theorem 52.1 below).
170
LARGE CARDINALS CH.
VII
DEFINITION 29.3. ( i ) A nonempty set (of sets) F is called a ,field of' sets iffor every E F we have x u y E F and U F ' , x ~ (so, f in particular, 0 and U F belong to F). For acardinal K, F is called K-complete if in addition we have U X E F for any set X c F of cardinality < li. (ii) Given an arbitrary set (of sets) G and a cardinal li, a field of set's li-generated by G is a li-complete field of sets F G . ~ ( U Gsuch ) that it includes G and is minimal with respect to inclusion. Instead of '3-generated' one simply sayl 'generated', and this is the same as 'to-generated'. x, y
Observe that in order to see that F is a h--completefield of sets it is enough to show that u F \ u X E F whenever X G F and I X I < K . LEMMA 29.4. For any set G und,fbr an)' regular cardinal li, there is rxuctly one field F K-generated by li, and F hus cardinalit)' I I GI? + 1.
PROOF. I t is obvious that there is a unique field F h--generated by G. In fact, we have F = F' c .Y( UG ) : G c F' and F' is a li-complete field of sets).
0
(note that the set on the right-hand side here is not empty, since : Y ( u G )belongs to it). In order to see that the assertion about the cardinality of F is valid we have to give another definition of F. To this end put F,=G, and if we have already defined F , for any fl
Weclaim that F, is a li-complete field of sets. By our remark after Definition 29.2,
it is enough to show to this end that
UG\
ux E F ,
(2)
whenever X c F , and I X 1 < K. First we show that if y E F, then y E Fi for some ( < K. In fact, y has the form U G \ Y for some Y c F, with 1 Y (< K. By the
u
regularity of K, we have Y G
u F,
a<€
note that X E F , implies X c
u
Q
for some ( < K; but then y E F , by (1). To prove (2),
u F, C
by the assertion just proved. Hence
U G\ U X E F , by (l), and this is what we asserted in (2). Hence our claim that F, is a K-complete field of sets is established. Now, on the one hand, F, obviously includes G =F , and, on the other, any field of sets including G must include all F,, and so F, in particular. Hence F, equals the field F K-generated by G (and F,= F, for all a 2 K, but we shall not need this).
ORDINARY PARTITION RELATION FOR INACCESSIBLE CARDINALS
171
So F = F , , and all we have to do is to estimate the cardinality of F h . We consider the case I G l 2 2 first. Writing i.=lGl. we then have to prove that IF,I(i.h+ 1 = i t . We have IF,I=IG1=/., and, obviously, IF215
(c IFil). O
holds for all a > 0 . Fix o! with 0 < 0 r s ~and , assume that lF,,ls;.F for all B
wherethelast equality holds by our assumption ;.=lC122. HencejF,I,sICI+for all o! in case I G 1 2 2, and so IF 15 I G I?,proving the lemma in this case. If 1 G 1 s1. then thefield Fh--generated by Gconsistsoftheemptyset and theonlyelement (if any) of G, and so I F l S l G l + 1 S I G / ? + 1. This completes the proof.
DEFINITION 29.5. ( i ) Given a field F of sets, an idvuI iu F is a set I S F such that X E 1, Y Z X , and Y E F imply Y E 1, and whenever X , Y E 1 we have X u Y E 1. ( i i ) 1 is an ideal orer F if 1 is an ideal in F and uI=O F . The ideal I in F is proper if U F $ I . (iii) For a cardinal K, the ideal 1 in F is K-cwnplerr if we have U X E 1 for any X c l with IXI
172
L A R G E (‘AKUINALS
CH. VII
This gives several properties of an inaccessible cardinal h- equivalent to K + ( K ) ; . The proof that this property holds for many inaccessible cardinals (the Hanf-Tarski theorem mentioned above), as well as other equivalent properties will be given in Sections 3&33. THEOREM 29.6 (Erd6s-Tarski [ 19611, Hanf [ 1964a1, Hajnal [1964]). Let K be un inaccessible curdinul or K = Q. Then the ,fOllowing properties (ire equivalent : (i) K does not have the prime ideal property. (ii) K does not have the tree property. (iii) There is an ordered set ( A , <) such that I A I = K‘ and K , K* $ tp ( A , <>. (vi) K f l w : . (v) K%(K, 4Y. (vi) There are an integer r 2 2 and a cardinal t < K such that K + ( K ~ . (vii) K + ( K ) ~ holds with some cardinal t < (see ~ (15.2)for the definition ofthe partition symbol used). Note that all these properties fail for K = W . ( i ) by our remark preceding this theorem, (ii), as we remarked above, by Lemma 10.3, (iii) by obvious considerations, and (iv)-(vi) by Ramsey’s theorem (Theorem 10.2), and as for (vii), the second proof of that theorem shows that o-+(w), holds for all integers n. Hence we could disregard the case K = w in the proof below. Seemingly, this is indeed what we shall do; in actual fact, all our arguments below apply to the case K = W as well, and even if we may use a phrase like “since K is an inaccessible cardinal”, this could always be replaced by the longer phrase “since either K = W or K is an inaccessible cardinal”. So our proof below gives a “new” proof of Ramsey’s theorem relying on the fact that K does not have the tree property, essentially identical to the second proof of this theorem given in Section 10, with the difference that the proof here is more economical, since we now have the concept of partition tree at our disposal.
PROOF. We shall establish the following implications: (i)-(ii), (ii)*(i), and (ii)*(iii), (iii)*(iv), . . ., (vii)*(ii). As will be clear in the proofs of these implications below, only the regularity, and not the inaccessibility of K is needed in the proofs of (ii)*(i) and (ii)*(iii), and not even the regularity of K is used in the proofs of (iii)*(iv), (iv)*(v), (v)*(vi), and (vi)*(vii). K ~ isO of course always assumed. (i)*(ii). Assuming that K has the tree property, we have to show that it also has the prime ideal property. To this end, let F be a field of sets with U F = K K-generated by a set of cardinality S K , and let 1 be a proper K-complete ideal over F ; we have to show that 1 can be extended to a K-complete prime ideal over F. Observe that F has cardinality IK in view of the inaccessibility of K by Lemma
ORDINARY PARTITION RELATION FOR INACCESSIBLE CARDINALS
I73
29.4; in case IF1 < K we would have U F = U l E I by the ti-complrtmrss of I ; this is, however, not the case, since I is proper. So I F I = K ; let ( A : : < < K ) be an enumeration of the elements of F . For every < < K let x: E U F be such that
x~$U(A,,:~<&%A,,EI).
(31
Note that the set on the right-hand side here belongs to I , and I is proper; so there is such an x:. We define a function ,ji:5+2 by putting ,f;-(q)=O
iff
xi $ A,,
(41
for every q < < . The set of functions T = [,/;." a : a ~ < < ~ ) ordered by inclusion is a tree of length K, and it is obvious by the inaccessibility of Kthat,forany a < ~ , t h e a t h l e v eof l Ti.e.,the set offunctionsin Twhosedomains are a, has cardinality < K . Since we assumed that K has the tree property, this implies that T has a branch of length K. Let b c T be such a branch, and let .f'= U b ; clearly, .f is a function from K into 2, and b= [,f "a: a<.; E T i.e., Va < ~ 3 ( [ I a 5 < K &.f " a = j;. " a ] .
(5)
We claim that the set J = [A,,:q < ~ & , f ( q ) = O )
is a K-complete prime ideal extending I . In fact, J includes I , since if A,, E I then .h(q)=Oforany 5 with q < 5 < K by ( 3 ) and (4),and so , f ( q ) = O by ( 5 ) ,i.e., A,, E J . If A{ E J and A,E A<,then A,, E J . To see this, choose an a< ti with (, q < a ; by ( 5 ) , there is a 5 with a s 5 < K such that .f' " a = .f>" a . We have f > ( ( ) =,f(()=O in view of A , E J ; hence (4) implies that x:$AS. But then x:$ A,,(cA,), and so f ( q ) = . f i ( q ) = O again by (4),i.e., A,, E J , as we wanted to show. J is a ti-complete ideal. Indeed, if X E K, I XI < K, and A,, E J for every q E X , then we have to show that U [ A , : q E X I E J . Note here that the field F is K-complete (since it is K-generated by a set), and so this set belongs to F; i.e., U[A,: q E X } = A , for some y < K. To see that A , E J , let a < K be an ordinal exceeding y and all elements of X ; in view of ( 5 ) , there is a t with a I 5 < ti such that ,f '" a = f: ,"a. Our assumptions imply f&) = f(q) = 0 for all q E X ; hence xy# A , by (4) if q E X . So xs# A, = =(-.(A,,: r] E X ) ; i.e., .f(y)=,fl.(y)=Oagainby (4),and so A ; , E J ,and this is what we wanted to show. J is proper, i.e., U F $ J . To verify this, let q < K be such that A,= U F ( EF ) , choose an tl with q < a < K, and, in accordance with ( 5 ) , let 5 be such that a<(r
I74
LARGE CARDINALS CH.
VII
<
this, choose an a with (, q < CL < K and, by ( 5 ) , a s' with u 5 < K and ,f * u = .fi. 'u. Then either x t $ A; or x 2 $ A,l, i.e., either , f ' ( [ ) = . j i . ( ( ) = O or ,f(q)=,ji.(q)=O; in other words; we indeed have either A ; E J or A,l E J . Thus we have shown that J is a ri-complete prime ideal extending I , that is, ti has the prime ideal property. (ii)*(i). (We have to assume only the regularity of K here.) Assume ti is regular and has the prime ideal property. We have to show that K has the tree property. To this end, let (7; <) be a tree of length K such that each of its levels has cardinality
G = [ ( x ):x E Ti u
it E T : .\-
Ti ,
I = IX E F : ( X I < K )( = ~ X E T : ~ X J < K ) ) . I is a K-complete ideal over F by the regularity of ti, and I is proper, since the set U F = T obviously has cardinality K. As K has the prime ideal property, there is a prime ideal J over F extending I . We claim that the set b=(xET: (tET:x
is a branch of length ti in T As T was an arbitrary tree, this means that K has the tree property, and this is what we want to establish. First we show that b is a chain, i.e., that x s y or y<x holds for any x , y E b. In fact, if x any y are incomparable in the ordering <,then ( t E T :x < t ) and ( t E T: y < t ) are disjoint, and so one of these sets belongs to J , since J is a prime ideal; i.e., either x q! b or y # b. To see that b is a branch of length K it is now sufficient to show that b intersects each level of 7: Denoting, as usual, by T [ 5 ]the 5th level of 7;we have
T=
0 T[B]u
Baa
u
(tE T:x
YEJ[a]
for every CL< K.'There is a union of less than K sets belonging to F on the right hand side here; so at least one of these sets does not belong to J, because otherwise we would have U F = T EJ . Now each T[P] has cardinality < ti, and so T[B] E Z C J ; hence { t E T : x < t f !$ J, i.e., x E b, holds for some x E T[or], as we wanted to show. (ii)*(iii). (Hanf [1964a]. We have to assume only the regularity of K here.) Assume K is regular and does not have the tree property, i.e., there is an Aronszajn K-tree ( A , < A ) . Clearly, A has cardinality K. We are going to define an ordering < of A for which K , K*$ tp ( A , <).To this end, for anv x E A and any 5
ORDINARY PARTITION RELATION FOR INACCESSIBLE CARDINALS
I75
A[(]. Define now the ordering 4 as follows: for any two x, y E A that are incomparable in < A , i.e., which are such that neither x 5 " y nor y < A x holds, put x < y iff XI t<
<
O<<
final segment X' of X. For any We have
YE
T [ t ] put
x;= [ X € x::x I ( = y ) . x'= u X' I €
7"fi
S'
and, moreover, the sets X ; do not interlace in the ordering <; that is, if y, y' E T[53 and y
1 76
L A R G E C A R D I N A L S CH.
VII
partition tree associated with 1; as defined in the beginning of the proof of Lemma 15.2. (Alternatively, one could use the tree ( K , <,) defined in Section 18).Observe that the relations in (15.8k(15.16)remain valid in the present case, since we only used there that E is an ordinal (i.e., wellordered), and the assumption E = (A!)+ of that proof was used only later. We claim that for each a < K the ath level T[a] of the tree T = (7; c ) has cardinality < K . Assume, on the contrary, that this is not the case and let < < K be the least ordinal for which I T [ ~ ] ~ L K Note . that < # O , as T[O] has only one element (the empty set); 5 = a i 1 for some a is not possible, since then (15.16) implies that IT[<]l=
C hET[i]
lims (/I)[I
1
il[lal+'l''"1<~
h~ T [ z ]
by the inaccessibility of K. Assume therefore that 5 is a limit ordinal. Then, noting that each element of T[5] is the least successor of a path of length 5 in 7;and each such path p has at most one least successor (namely, U p or none), we can see that I T [ 53 I is Ithe number of paths of length 5 in 7:As each such path intersects each of the levels T[a], a< t, in exactly one element, their number is I I T[a]l< K
n
n<;
by the inaccessibility of K. This contradicts our assumption that 1 T[<] I L K; thus our claim that I T[a] 1 < K holds for every a < K is established. Note that IR(h)l= 1 for every h E T by (1 5.15), i.e., I TI = K by Lemma 14.2. Hence T must have length 2 K. As K was assumed to have the tree property, TI K cannot be an Aronszajn tree, i.e.,it must havea branch b= [gz:a<.) oflength K, whereg,denotes theath element of b, i.e., g , E T[a]. Putting C,=s(g,) (cf. (15.8)), it follows that X = = [<,: 1 I a < K ) is an end-homogeneous set by arguments similar to the ones used at the end of Lemma 15.5. In fact, (,<(s whenever a
u (<;k
which we wanted to show. Thus the relation K-+ (K), holds for any i.< K, i.e., (vii) fails. This completes the proof of the implication (vii)*(ii), and so that of the theorem.
REMARKS. The investigation of properties (ik(vii)is also of interest in case K is not inaccessible. As for (i), one can show that it holds for all uncountable accessible cardinals, i.e., that K can have the prime ideal property only if it is inaccessible. As for (ii), it trivially holds if A- is singular. Specker proved under GCH that if i.is regular, then (ii)failsfor K = i +(cf.Theorem 52.1 below). It is not known whether the regularity of i. is necessary here, e.g. if the same conclusion
WEAK COMPACTNESS A N D METAMATHEMATICAL APPROACH TO THE HA'NF-TARSKI RESULT
I77
holdswithi=N,,, (seeErdos--Tarski [1961]).J.Silver [I9711 proved that if2"<'is real-valued measurable (which is consistent relatively to ZFC + 'there is a measurable cardinal' by a theorem of Solovay [1971]; see Section 34 for the definitions of measurability and real-valued measurability), then (ii) fails for K = Pii.Noting that K is regular in this case, this shows that (ii) and (iv) are not provably equivalent if we assume only that h- is regular, since (iv) holds for every accessible cardinal by Theorem 29.1.
30. WEAK COMPACTNESS AND A METAMATHEMATICAL APPROACH TO THE HANF-TARSKI RESULT The present section describes a metamathematical method showing that the relation K + ( K ) : fails for many inaccessiblecardinals. As the tools used herediffer radically from the combinatorial tools more common in this book, the combinatoridly oriented reader. might skip this whole section: he will find an alternative, purely combinatorial proof of the main result (Theorem 30.5) concerning the relation K - + ( K ) ~ in Sections 31-32. It would, however, not be entirely fair to suppress that the ideas in the next section have a metamathematical background. The idea that the question whether K + ( h - ) : holds for certain inaccessible cardinals might be approached by metamathematical methods occurred to Tarski; the actual proof, by these methods, that K - + ( K ) : fails for many inaccessible cardinals was obtained by Hanf. Afterwards, Keisler and Tarski, and later ErdBs and Hajnal, found direct combinatorial proofs of this result. As this section is only for readers having some interest in model theory, we omit the definitions of basic model-theoretical concepts. We start with the following
DEFINITION 30.1. Let % be a set of constants and W a set of (finitary) relation symbols. For a cardinal K , the language Lkk(%, 9) is defined as follows: the atomic formulas of I,,,(%, 9) are strings of symbols of the form R ( . , . , . . . , . ), where R E 9, and the dots (whose number is n if R is an n-ary relation symbol) are to be replaced by constants (i.e., elements of%) or variables (we suppose that there are K variables, e.g. u, for each a < K). The set of (well-formed) formulas of Lkk(%,9) are defined as members of the smallest set W such that (i) every atomic formula belongs to W, (ii) if cp belongs to W then so does its negation 1cp; (iii) if a < K and each cp:, 5
I2 Combmatonal
I78
LARGE CARDINALS CH. VII
(iv) if a < K and cp belongs to W then so does the quantific&on ( 3 ~ ~ ) : < ~ c p , where x:, t ) ~ abbreviates <~ 1(3x2):<%-I. $=
:
The formulas of LA,(%, 9)will be interpreted semantically in a fairly straightforward manner: Let % = ( A , c', R'),.,,,,,, be a structure, and let cp be a formula of L,,(%, .@). For given elements a,,, v < y , of A as values of the free variables y,,,v < y , in cp, the truth value [cp[a,,/y,:v < y]J (which is always 0 or 1 ) of cp is defined as follows: (i') if rp is an atomic formula, then [rp[a,/y,: v < a ] J = 1 iff the relation R' corresponding to cp holds with the appropriate arguments, with a , substituted for y,,;
(iii') [ V cpJa,/y,: v < y]]
=1
iff [cpt[av/yv:v <]]y = 1 for some 5 < a ;
:
(if)
B ( ~ x & < . r p [ a , / ~v <~y:l ] =
1 iff Ill(rpCbr/xt:& = a l ) C ~ , / ~v <~ ~: l l J = 1
for some elements b,, 5
DEFINITION 30.2. A cardinal K is called weakly compact if every semantically consistent set of sentences of L,, is consistent.
WEAK COMPACTNESS A N D METAMATHEMATICAL APPROACH TO THE HANF-TARSKI RESULT
I19
w is a weakly compact cardinal by the compactness theorem of finitary first order logic. According to results of Hanf [1964], an uncountable weakly compact cardinal must be a 'large' regular limit cardinal but it is not known whether it must be inaccessible. The importance of weakly compact cardinals for partition relations is shown by the following theorem:
THEOREM 30.3. An inuccrssible cardinal
ti
is weukly compucr ij'.und only ij
ti+ (ti)$.
PROOF. The result is clearly valid also for the case K = w . Our proof does work for this case, and phrases like 'since ti is inaccessible' can always be ieplaced by longer phrases like 'since ti=w or ti is inaccessible' in the arguments below. 'If part. Assume t i e +(K)$.Let S- {(pi: 5 < K ] ( be a semantically consistent set of sentences of LLK(%, 9). As we have to consider only those constants and relations that occur in some (pr, we may obviously assume that Let be a model of the sentences (p.,,, r < < . For each C L < K , and each existential subformula x = ( 3 x V ) , ~ , # of (p3, add Skokm functions f $ *
2I= ( A , c", .f6)1.6u,X.l,
of the structures ( A : , c";, f $*C)(.E5%X,Y where x and v run over the domains described above. That is, A = X , , , A , , c" is the function g c A such that g(<)=c"? for every < < t i , and the function f:: 'zA-+A is such that for any function h E ' Z Aand any ordinal 5 < 9, we have fF(h)(5)=f6,'({(P,h(l0(5)>:11<9,1).
(3)
Let f?=(B, c", f ; ~ f i 3 ~ B ) c E ube . X a, Ysubstructure of 2I that includes the set {c': C E % ) and is closed with respect to all functions .ft. Note that we may 12'
I80
LARGE CARDINALS CH.
VII
assume
IB(
< (1;:w=$~~(o/x'1;,
(5)
where $ runs over all subformulas of the formulas cpz, CI < k-, and 6runs over all possible sequences of elements of B whose length is equal to the length of the sequence x'of free variables in +. Note that there are at most K possibilities for as a formula of L,, has 5 k- subformulas (equality is possible only for singular k - ; this is not the case at present), and by the inaccessibility of k- there are less than kpossibilities for 6, as I B ~ by ~ (4), K and the sequence 6consists of less than kelements of B. Hence the field F is k--generated by k- sets. Put
+,
l=;XEk-:IXI
Then I is a k--complete ideal over F . According to the assumption k - + ( k - ) i , Theorem 29.6 implies that k- satisfies the prime ideal property. Hence there is a proper K-complete prime ideal J over F that includes 1 (cf. Definition 2 9 . N ~ ) ) . We shall define a relation R' for each R E 9 on the structure 52; for each R and go, g , , . . . E B we stipulate that
(R' is a kind of subdirect ultraproduct of the relations RY*.,1; < K . ) Note that, by (5), the set on the right-hand side belongs to the field F provided R(xo,xl, . . . ) with distinct variables xo, x l , . . . is a subformula of some q:, and this is an assumption we can easily afford to make (e.g. by supposing that the formulas 3xoxl . . .[R(x,, x,, . . . )..R(x,, x,, . . .)] occur among the ~0:'s). We claim that for any subformula II/ of each sentence cp:, 5 < K , and for any substitution sequence 6of elements of B for the free variable sequence x'of II/ we have ( B , cv, R')< E'd. R E 9 k II/[6/21
iff
( < : ~ + I I / C ~ ( O$ J/ ~. )
(7)
WEAK COMPACTNESS AND METAMATHEMATICAL APPROACH TO THE HANF-TARSKI RESULT
I81
Note that the set on the right-hand side here belongs to the field F i n view of ( 5 ) . The proof of this claim is like the standard proof of tos's theorem; it proceeds by simple induction on the complexity of the formula $; the argument used in the induction steps depends on which of the forms given in (i)-(iv) of Definition 30.1 the formula $ has. In case (i),i.e., when $ is an atomic formula, the claim holds in view of the definition of the relations Rv in (6); in case (ii) it follows from the induction hypothesis and the fact that J is a prime ideal (i.e., that for any X E F exactly one of the relations X E J and K\ X E J hold); in case (iii) one has to exploit the ri-completeness of J ; finally in case (iv), i.e., if $ is an existential formula, then the Skolem functions ,ft pick the elements of B whose$xistence is claimed by the existential quantifiers (it is here that the fact is used that B is closed with respect to-these Skolem functions). This proves the claim in (7). This claim is valid in particular for the sentences 'p:. But we have {t:i?l+qf;2K\(t+l)
by the choice of the structure a:. The set on the right-hand side does certainly not belong to J , as its complement belongs to I c J ; hence, in viewof (7), the structure ( B , cLL, R"), € % , R E : M is a model for all the sentences q:,t < K. This settles the 'if part.
REMARK. One should note that ifequality occurs among the relation symbols in
9,then it is not necessarily true that equal elements in the obtained model are identical. This is a perfectly sound situation, but it may be inconvenient, since if one defines an isomorphism as a 1-1 mapping satisfying certain requirements, then two models, though essentially the same, may not be isomorphic just because in one of the models there are equal but nonidentical elements. To avoid this inconvenience, one may pick an element from each of the equivalence classes of B under the relation = Q and throw away all the others. 'Only if part. Assume K is weakly compact and inaccessible. We shall then prove that K satisfies the prime ideal property. The model that will be constructed for this purpose will, with minor modifications, be useful in other proofs as well in the remaining part of this section. Let .F be a Kcomplete field of sets with U F = K that has cardinality K (as K is inaccessible, this is the same by virtue of Lemma 29.3 and Theorem 6.3 as saying that F is h--generated by K sets), and let I be a proper K-complete ideal over F. We have to show that 1 can be extended to a complete prime ideal J over F. To this end, consider the language ~==L,,(Icr:crIK)(U(X:XEF) U {?'ti,( = , € ) ) ,
where the first set is the set of constants (intuitively, the constants g and x denote
1 82
LARGE CARDINALS CH.
VII
the sets ci and s,and IC i s a new constant symbol), and the second set is the set of relations. Let ZFC,, denote the conjunction of the first ti axioms of Z e r m e b Fraenkel set theory, and, for an integer n to be chosen later, let K' be a cardinal 1K such that ( V ( K ' ) =, , E) FZFC,, where V(K')denotes the set of all sets of rank < K'. The existence of such a K' is well known. (In fact, we could choose K'= K + here and in the next two proofs, since no axiom of set theory will be needed in the model to be constructed that does not hold in ( V ( K +), =, E), but it would be quite tedious to check this fact.) We construct a semantically consistent set of sentences S=S,uS, u S , of Y as follows: a ) Let So be the set of sentences of Y using only the constant symbols g, x (CIS K, x E F ) that are satisfied by ( V ( K ' ) ,=, E) if we interpret g as CI and x- as x. b ) Let S, = ( 7 r E k ' } . c ) Let S,=(n$x:xEI}, where I is the ideal to be extended. I t follows from the inaccessibility of K that ISI=h-. I t is easy to see that S = S o uS , u S , is semantically consistent provided n is large enough (so that the set-theoretical results used to establish this fact are valid in ( V ( K ' ) ,=, E), i.e., they are provable from ZFC,). In fact, if we take a set S ' z S of cardinality < K, then they are satisfied by ( V ( K ' ) ,=, E), provided we interpret g as ol, -5 as x, and the constant 7t as an ordinal X ' < K such that
n'$x holds for any x E I for which x- occurs in S' A S,. There is such a IC'since there are less than K of these x's, and so u { x : x E I and .% occurs in S'nS,) E I , which implies that the set on the left-hand side does not equal K. Hence, by the weak compactness of K, the set S of sentences has a model; denote it by 2I. Put J=(xEF:(11+7t$~x-).
Then I G J by the axioms in S,. J is a Kcomplete ideal over F . In ;act, if x, y E F and x c y , then the sentence x c y (which is the abbreviation of Vz[z E ~ a E y]; z we shall freely use the customary set-theoretical abbreviations in formulas o f ' below) belongs to S o , and so it holds in (11. If now y E J then 2lb n 6 y, and so a fortiori (11 n 6 x,i.e., x E J . If x u , a < t, are elements of F for some c K , then x = x, is also an element of F , and the sentence =
u
+
<
Vz[zExo
v
1<;
ZESJ
W E A K COMPACTNESS A N D METAMATHEMATICAL APPROACH TO THE HANF- TARSKI RESULT
I83
+ +
belongs to S,,, and so it is valid in 2I. If now x, E J, i.e., 2I x $ z3,holds for all u < c, then this sentence implies that 2I n $ z also holds, i.e., we have x E J . This shows that J is a K-complete ideal over F . J is proper, since 91 x E h- holds by the single sentence in S , , and so U F = K $ J . Finally, J is a prime ideal. In fact, if x , y F~ and xny=O, then xny=Q holds in ‘i? byIa sentence of S o . Hence either n $ g or n $ y- holds in %; ii., either x E J or y E J . The proof of the theorem is complete. There are several properties of inaccessible cardinals that are equivalent to weak compactness; the reader interested in them should consult e.g. Silver’s dissertation 119711. Our main purpose here is to prove the Hanf--Tarski result saying that most of the ‘small’inaccessible cardinals are not weakly compact. To thisend we shallneed the Mahlooperation M . (See Mahlo [191l];tbeoperation defined here differs slightly from what’ Keisler and Tarski [1964] call Mahlo operation. )
+
DEFINITION 30.4. Let X be a class of ordinals. We define M ( X ) by putting u e M ( X ) iff either c f ( u ) i o or u\X is not stationary in a. In other words, M ( X ) contains the same limit ordinals as nst (On\X ) , where On is the class of all ordinals, and the operation nst, the nonstationary points of a class of ordinals, was defined right after the announcement of Theorem 5.7. The following theorem is essentially the Hanf-Tarski result (cf. Hanf [ 19641). saying that many inaccessible cardinals (including the ‘small’ones, in particular the first) are n6t weakly compact. We shall prove this theorem again in the Section 32 as Theorem 32.4, and we shall consider its corollaries there. Let AC denote the class of all accessible ordinals (i.e., of all ordinals which are not inaccessible cardinals). We have THEOREM 30.5. Let h- be an inaccessible cardinul, and ussunie thut there are sets Z, E K, u < K, such that, writing
we have K $ X and K E M ( X ) . Then K is not weakly compuct. If each 2, is taken to be the empty set, then this result says that the first inaccessible cardinal is not weakly compact; or, indeed, the class M ( A C ) does not contain any weakly compact inaccessible cardinals.
PROOF. ( K > W is definitely needed now.) Assume, on the contrary, that K is a weakly compact inaccessible cardinal. Let f be a regressive divergent function on the set K\ X ; such a function exists in view of Neumer’s theorem (Theorem 5.3), as K\ X is nonstationary by our assumption K E M ( X ) .Let Z : K - - + Y ( K ) be
I84
LARGE CARDINALS CH.
the function such that Z ( a ) = Z , for every
VII
a
Consider the language
where 7t is a new constant symbol. Let K’ > K be a cardinal as in the second part of the proof of the preceding theorem, i.e., such that ( V ( K ‘ ) ,=. E) kZFC,, for a large enough integer n. Consider the following set S = S o uS , u S , of sentences of 9: a ) So is the set of all sentences of Y not containing n that are satisfied by ( V ( K ’ ) ,=, E) if we interpret g, and f,as a, 2, and J respectively. b) S , contains the single sentence x E-K: c) Put
z,
S,=(gEn:a
As before, it is easy to see from the inaccessibility of K that IS1 = K . (We remark that the theorem being proved has an analogue for the case when K is not inaccessible but only a regular limit cardinal. In that case the set So might have cardinality > K ; it will however turn out that only ‘very few’ of the sentences in So are actually needed.) I t is also easy to see that S is semantically consistent; in fact, ifs’c S hascardinality < K, then all the sentences of S‘ are satisfied by C VKs,=, E) if we interpret a, and f as a, Z , and f, respectively, and II as an ordinal < K exceeding all ordinals a for which a occurs in the sentences of S‘. Hence, by the assumption that K is weakly compact, S has a model, say % = ( A , =’, E’). % satisfies the following sentence:
z,
Vxy[x=y
0 Vz[z
E X 0z E
L’]
(91
and as these sentences, being valid in ( V ( K ’ )E, , = ), belong to SoE S . In fact, the first sentence is the Axiom of Extensionality, and the second one says that the membership relation is well-founded, which is an easy consequence of the Axiom of Regularity. Hence, by Mostowski’s Collapsing Lemma (see Subsection 4.3),% is isomorphic to a transitive model
I!=(&=,€); that is, B is a transitive set (Le., V x d ? [ x c B]),and, furthermore, = and E are the true equalityand membership relations, respectively. Naturally, I!is also a model of s. We claim that we have u2=a
(11)
WEAK COMPACTNESS AND METAMATHEMATICAL APPROACH TO THE HANF-TARSKI RESULT
for every ordinal
tl < ti.
185
In fact, the sentence Vx[x
E
g o
v
x = b]
B
belongs to So, and so it is valid in 2.Hence if we assume that ( 1 1 )is valid for any ordinal S < M , then it is also valid for tl by this formula. On the other hand (12)
tiEk.l'
holds. Indeed, 3 E ~ l by ' the sentence in S , , and a=g2 E X for every c(
$! M'@(g'))n
(ti
i1) = M"(Z, J n(ti 4 1 )
,!
Hence (14)
(15)
holds for any a< ti, where M" denotes the Mahlo operation in the sense of 2. In fact, the equality here holds by (14), and ti E M'(Z,) would mean that ti\Z, was not stationary in the sense of 2. This would in turn mean in view of Neumer's theorem (Theorem 5.3) that, in 2, there was a regressive divergent function on ti\Z,; there is, however, no such function even in the real world by the assumption ti $ X . We can now conclude that K $ p .
(16)
I86
LARtiE CARDINALS CH.
VII
In fact, according to (8), a sentence of So( GS) says that V < [ j E KO(
<
E AC
v
301 < K ( M <
5 & j E M (Z(gJ)))];
now the right-hand side of this equivalence fails for < =ti according to (13)and (15).
It will be easy to see that (16)leads to a contradiction. At the beginning of the proof we picked a function ,f that is regressive and divergent on ti \ X.A sentence of S , c S says that
dom(.f' - )=K\X, and so by (12) and (16) we have K E dom
Now
j '
(f'- ').
V (ti)=S
for some 6 , as a sentence of S,E S says that -f is regressive. This means that the sentences X E d o m ( f X a ~ t & f (<)=dl are valid in 2 for any a < ti. In fact, they are satisfied by
< = ti (cf. ( 1 1 )). For a fixed
a, either this sentence or its negation is true in ( V(K'),=, E) (or, what is the same,
in reality). So either this sentence or its negation must belong to S , z S . But the latter case would mean that this sentence is false in 2, which is not the case. So each of these sentences is true in reality. That is, we have Vcc < ti 3C: E dom (f' ) [ a < 5 & j ' (5 ) = S] .
This contradicts the assumption that j ' is divergent. The proof is complete. Next wetconsider a combinatorial property of weakly compact inaccessible cardinals, due to J . E. Baumgartner, that enables one to prove the above theorem in a direct mathematical way. DEFINITION 30.6. The cardinal ti is said to satisfy Buumgurtner's principle if, given arbitrary regressive functions f,: ti+ti ( a < ti), there are ordinals I], < ti such that
v x s K[lx I < t i 3 3 Y
K[I Y l = t i &
VCi E
xv<E Yji(()=I]a]]
(17)
holds. It is not difficult to prove that a cardinal ti satisfying Baumgartner's principle must be inaccessible, but this fact is irrelevant for purposes. The important fact is expressed by the following
WEAK COMPACTNESS A N D METAMATHEMATICAL APPROACH TO THE HANF TARSKI RESULT
187
THEOREM 30.7 (Baumgartner, unpublished). A n inuccrssihlr curdinul sati.$ries Baumgurtner’s principle if and only ifit i s weuklj~compuct. We shall give a proof of this theorem in the next section by using combinatorial properties equivalent to weak compactness. But it will be instructive to give a metamathematical proof of the ‘if part, since it lucidly tells where the ordinals q3 come from. PROOFOF THE ‘IF‘ PART. Assume that K is weakly compact and ,inaccessible. As in the proof of the preceding theorem, let K ’ > K be a cardinal such that ( V(K‘), =, E) ZFC, for a large enough integer n. We consider the language
9= L,,({g : a I~ } u-{ :fa=< ~ } u { n }{,=, E}) and, analogously as before, we construct a set of sentences S = SouSlUS, of 9 a s follows: a ) So is the set of all sentences of 9 not containing x that are satisfied by ( V(K‘),=, E) if we interpret g and f u as a and f u , respectively. b) S1 consists of the single sentence n E K. c ) We put S2=(gE7C:a
As before, we have IS[= K by the inaccessibility of K, and it is again easy to see that S is semantically consistent by noting that any subset S’ of cardinality < K of S is satisfied by ( V ( K ’ ) = ,E) if we interpret I[ as a large enough ordinal < K. So S has a model CU by the weak compactness of K. As the sentences under (9)and (10) belong to So E S , by Mostowski’s Collapsing Lemma used in the preceding proof, CU is again isomorphic to a transitive model
f?=(B,
=,E);
this 2 is of course also a model of S . By the same arguments as before, the relations in (1 1) and (12) are valid also for this model 2.Put I ] , = f 5 (k.) (18) for any ci < K. Note here that f, is defined on the whole of K ; hence, by a sentence of So,1;is defined on the whole of h2; and by (12), K belongs to this latter set. So the definition of I], is meaningful. Furthermore, we have qa < K by the regressivity of ,f: (confirmed also by a sentence of So). We only have to prove that (17) holds. To this end observe that, given a set X 5 K of cardinality < K, the sentence
x c p < 5 < k. c!k uAc x f,(5,=’1.1 holds in S for every p < K ; in fact, it is satisfied by 5 = K (cf. (18), 1 1 ), and 12);the
I88
LARGE C A R D I N A L S CH.
VII
latter two are, as we pointed out above, also valid in the present model 2 ) .But this is a sentence of 2 not containing n, so either itselfor its negation must belong to SoG S;the latter case is impossible, since this sentence is true in 2. So i! belongs. to So, i.e., it is true in ( V(h-’),=, E), i.e., it is true in reality. That is, we have 35:CP
A
a€
x
.f(O=%l
for any fi < h-. As X was chosen arbitrarily, this means that ( 1 7) is indeed satisfied, i.e., Baumgartner’s principle holds for K. This completes the proof of the ‘if part of the theorem. As mentioned above, the proof of the ‘only if part will be given in the next section (see Theorem 3 1.4). 31. BAUMGARTNER’S PRlNClPLE
In the preceding section we presented a metamathematical approach to the Hanf-Tarski result saying that h - + ( ti); fails for many inaccessible cardinals. As this book is mainly about combinatorics, the reader might prefer to see a combinatorial proof of this result. This will be furnished in this and the next section. For the reader’s convenience, we shall repeat some ofwhat has been said in the preceding section; in particular, we shall give the definition of Baumgartner’s principle again (Definition 31.3 below). But first we define the concept of normal ideals (we recall for this definition that a function .f sending ordinals t o ordinals is regressive if , f ( t < ) 5 holds for every nonzero 5: in its domain; cf. Section 5), and then prove a lengthy lemma about them. A similar, but conceptually simpler, result will be proved below in Section 34 (see Theorem 34.9), and the reader might benefit by studying that theorem as well as Definition 34.1 before reading the following definition and Lemma 31.2.
DEFINITION 3 1 .l. Let F be a h--complete field of sets with U F = K , I an ideal over F, and, finally, H a set of functions mapping h- into K . The ideal I is called nor-mu/ with respect to H if it is h--complete and proper, and, for every regressive function .f E H and every set X E F\ I , there is an ordinal a < K such that [ < E X :. f ( t ) = a ) 6 I .
Although not required in the definition, we should stress that a normal ideal 1 as above is probably useless unless t, C -
, f ( t ) =E~F)
holds for every regressive fundtion ,f E H and for every a<%. Next we give a
189
BAUMGARTNER'S P R I N C I P L E
result asserting existence of normal prime ideals under certain conditions for cardinals h- satisfying h--+(h-)::
LEMMA 3 1.2. Let h- > (IJ he a cardinul satisfying the prime ideal property. Let H he set of' cardindil)~h- of,func~tion.smapping h- into h- that sa1i.sf;e.s the jid/owing properties : (i) H c~ontainsthe idenrit)~jiriicrionid, on h- and the c.onstaiit,futic~tiori,s cI E u j i ~ all u
X = { ~ < K f(()
,/'(t)=g(o;.
(1 1
where f and g run over elements of H . Then there exists a prime ideal over F that is normal with ruspet't t o H . PROOF. Note that the one element sets [ u ; , U < K , belong to F . In fact, the functions id, and c1 belong to H by (i) and we have I ) = (t
and this set belongs to F by ( 1 ). Write I={Xch-:IXl
and let J be a prime ideal over F that extends I ; there is such a J since h- was assumed to satisfy the prime ideal property (note that F is h--generated by h- sets, since (HI = h-, and so there are i h- sets listed in ( 1 ); in fact there are exactly ti of them since, as we have just seen, all the one element subsets of h- are among them). Consider the following equivalence relation on H : given ,f;g E H , write
,.
(21
. f - J g o { < : f ' ( < ) = g ( < )#; J .
Note that the set on the right-hand side of this formula belongs to F, as the sets listed in (1 )all belong to F . It is an easy exercise to verify that (2)indeed defines an equivalence relation. To show e.g. transitivity, assume that ,f g and g h; then f h because we have
-
IC.
-
. ( r < K : . f ( t ) # h ( t ) ) . G ( r < h - : . f ( r ) # g ( r ) ju
u I5 < K :g(5 ) # h( t)}E J (in fact, both sets o n the right-hand side belong to J , since their complements d o
LARGE CARDINALS CH. VII
190
not, and J is a prime ideal), and so the complement ofthe set on the left-hand side cannot belong to J. Let [j’]denote the equivalence class of the function J E H. We define an ordering c Jon the set
n= i [ . f ’ ] : , f ~ H ’ ,
by putting for any J g
C.fI < J kl E
iff
(t
.f(O
(3 1
H . Before discussing the soundness of: this definition, note that
[t< K : .f(O<s(5)’,E F
(4 1
holds for any f ; g E H. indeed, the characteristic function, say h, of the set on the left-hand side belongs to H by (ii); so does the constant function c1 = 1 by (i). Hence [t
-
J ;
f(t)
obviously holds, and the sets on the right-hand side here belong to F by (1) and (4). Hence exactly one of them does not belong to J , since these sets are pairwise disjoint and J is a prime ideal over F. Weclaim that is a wellordering. To see this, assume on the contrary that ([f,]:n < o )is an infinite descending sequence of elements of li in the ordering Then each of the sets { ( < ~ : f ~ ( t ) s f , belongs + , ( ( ) )to J (since its I 1] , <, Jf.3; the fact is used here complement does not, in virtue of the relation [ that J is a prime ideal); hence we have <J
< J .
(t
3n[fn(t)sf,+1(t)1)=
u
(t
n<w
as J is K-complete and K > w.So there is a 5 < K that does not belong to the set on the left-hand side. Then we have
.fO(5)>fI(t)>fi(5)>.
..
;
this is a contradiction since there is no infinite descending sequence of ordinals, verifying our claim that < is a wellordering.
191
BAUMGARTNER'S PRINCIPLE
Let now h E H be a function such that [h] is the least element of fl in the wellordering that exceeds [c,] for a < t i , where c, denotes the constant function = a ( h is often called a minimalfunction for J ) . Note that there is such an h, since, for the identity function id, on ti, [id,] exceeds all the [c,]. This remark also implies that [h]5J[id,]; hence we may assume that < J
h(<)lt
(5)
holds for any (< ti (we can make this true by changing h on a set belonging to J ) . Put J * = ( X E F : h - I ( X )E J J , (6) h - ' ( X ) = { ( < t i : h ( < ) E X ).
where
We claim that J* is a prime ideal over F that is normal with respect to H. To show this, we first show that if X E F, then
K 1 ( X )E F .
(7)
As the sets listed in (1) k--generate F, it is enough to verify this in case & is one of the sets in ( I ) , i.e., in case
X
=
It < k - : f(t)=g(t)j
for some f i g E H. We have
h - ' ( X ) = {(< t i : h ( ( )E X ) = ( ( < t i : f ( h ( t ) ) = g ( h ( t ) )=) =
It < t i : ( f o h ) (0 = (s 0 h ) (0; f
and this last set also belongs to F, as f o h, g o h E H by (iii). This verifies (7). As J is a K-complete prime ideal over F, the same is true for J* by (6)and (7) (note here that the one element sets ( a ; , a < ti, belong to J*, as h - l ( { a ; ) E J holds because [c.] # [h] by the choice of h ; hence u J * = t i = U F , and so J* is indeed over F). We have yet to show that J* is normal with respect to H . To this end, pick a regressive func&m f E H.As f ( h ( ( ) ) < h ( c )for all 5 < t i unless h(t)=O, we:have in fact,
Cfo h l < J [Ihl ; { 5<
t i : (f 0
(8 )
h ) ({)rh(()] G h - l ( { O J ) € J
by the remark made just before in parentheses. (8) implies by the choice of h that holds for some a < ti, i.e., that
{(
(91
192
LARGE CARDINALS CH.
VII
It will now be easy to show that
[ t < t i :J ' ( < ) = Q )
$J*.
(10)
Note first that the set on the left-hand side belongs to F by ( I ) (and (i)).We have
by (9), and so (10)indeed holds in view of (6).(10)almost amounts to saying that J * is normal with respect to H ; but, in actual fact, we have to show somewhat more: given any X E F \J*, the relation ( g E x : ,f ( 5 )= c1 f
f!
J*
also holds. This is, however, obvious by (lo), since ti\X E J * , as J* is a prime ideal. This verifies our claim that J* is a prime ideal over F that is normal with respect to H . The proof of the lemma is complete. We repeat the definition of Baumgartner's principle (with slight notational changes) given in the preceding section (cf. Definition 30.6). DEFINITION 31.3. The cardinal ti is said to satisfy Baumgurfnur's principle if given an arbitrary set S of cardinality 5 ti of regressive functions on ti, there are ordinals for each J' E S such that
I],
V B c S ( l B l < t i 3 3 Y E I[.
Y I = t i & Vf E svg E Y f ( s ' ) = q , ] ) .
(11)
The main result in this section is the following theorem (essentially identical to Theorem 30.7, but here we prove both implications). THEOREM 3 1.4 (Baumgartner, unpublished). An inaccessible cardinal ti sutisjfies Baumgartner's principle f u n d only fi t i - + ( t i ) : . It is not difficult to prove that Baumgartner's principle fails for all accessible cardinals; hence by Theorem 29. I, the above result holds for any ti > w and not just'for inaccessible ti. This fact is, however, of no interest to us here.
PROOF. 'If part. Assume that t i + ( t i ) ; holds. Then, by Theorem 29.6, ti satisfies the prime ideal property. Given a set S of cardinality Iti of regressive functions on ti, let H be the smallest set of functions mapping K into K that includes S and satisfies (ik(iii)of Lemma 3 1.2 (clearly, J HI = t i ) , and let F be the field of sets associated with this H a s in that lemma. Then that lemma implies the existence of a prime ideal over F that is normal with respect to H; denote this by 1. As the functions f E S are regressive and belong to H , we have 15
$1
193
BAUMGARTNER’S PRINCIPLE
with some qr for each .f
E
S. As 1 is a prime ideal, we have
wh-:,f(t)#YI,; €1. Take a set B G S of cardinality < K. Then
(t
u
IEB
(t
E
1.
because 1 is k-complete (this is included in the definition of normality). Hence the complement of this set,
{t< K : V . f E B .f(O= q/.; ,
does not belong to 1;-so it must have cardinality K, as 01 = U F = ti, and I is K-complete. Taking this set as X we can see that (1 1 ) holds, which completes the proof of the ‘if part. ‘Only if’ part. Assume that ti satisfies Baumgartner’s principle. We shall then show that K satisfies the tree property; this will be enough in view of Theorem 29.6. To this end, let (7; <) be a K-tree such that each of its levels has cardinality < K. We have to prove that T has a branch of length ti. As I TI = ti, we may ) stipulating that actually assume that T= ti. Define the set S = {,f,:a < ~ by ,L: ~ - 2is a function such that ,fJt)=l if
tl
(12)
The functions f a are regressive, except that we may have ,f.(O)= 1 or .f,(1 ) = 1, i.e., tl
(13)
We claim that any two elements tl and p of b are comparable in <. In fact, taking B = { f , , f p j in ( l l ) , there is a t < such ~ that f,(t)=fp(t)=l.We have a, by (12), and so tl and p are indeed comparable, since (K, <) is a tree. Next we claim that b has cardinality ti. To see this assume, on the contrary, that Ibl < K , and let X be a level of ( K , <) that is disjoint from b. Then q,=O for all M E Xby (13), and yet Va E X f , ( t ) = O is possible by (12) only if 5 is in X or in a level below X ; as the number of these 5‘s IS < K, (1 1)must fail with B = { f , : tl E X ) (note that ( B ( I 1x1< K , as each level of 13 Combinatorial
I94
LARGE CARDINALS CH.
VII
( K , <) has cardinality
32. A COMBINATORIAL APPROACH TO THE HANF-TARSKI RESULT In this section, using Baumgartner’s principle, we shall show that K + ( K ) $ fails for many inaccessible cardinals. We start with a simple result which will characteristically show one of the ways in which Baumgartner’s principle can be used. The result in itself is of not much interest to us here, because a far more comprehensive one will be proved below. Note that the result in ( 5 ) can be considered as a lemma for Theorem 32.4 below; it would, however, be uninspiring to formulate it as such. THEOREM 32.1.
K+ (K)$
holds for the j r s t inaccessible cardinal
K.
PROOF. Assume the contrary. Then Baumgartner’s principle holds for Theorem 31.4. Define K regressive functions on K as follows: fo(5)=min { ~ : 2 ~ 2 1 5 1 ) if this value is
<
K
by (1)
f o ( t ) = O ; write
(5 ) = cf ( 5 )
(2)
if is a limit ordinal with cf ( 5 )< 5; otherwise put f l (5)= 0. If i < K is a regular cardinal and t < K is an ordinal with cf ( 5 )=A, then let be an increasing sequence of ordinals tending to 5; if 5 < K and cf ( 5 )#I., then put f A m ( < ) = 0 for any a
(4 )
of regressive functions. Let qf0,q,,, and qfA,be the ordinals that correspond to these functions according to Baumgartner s principle (see Definition 31.3). Our first claim is that we must have q,,=O. In fact, for any fixed q with O < q < K the relation f o ( ( ) = q can hold only if q = I qI and 2IV1L 151,and this holds less than K times, as 21V11~.So (31.11) with B = { f , ) implies our claim. Our second claim is that q,, = 0. Assume the contrary; then we must have q,, = A for some regular cardinal A < K . Consider the set B = { f l ) u{fAa: a
+
A COMBINATORIAL APPROACH TO THE HANF- TARSKI RESULT
for any
195
5 E Y Then, according to (2) and (3), the sequence
must tend to 5 for any 5 E Y ; this is a contradiction, since a sequence can tend only to at most one 5. This establishes our claim. We have therefore established that 1J,=1/, =o,
(5)
and note for later reference that no other assumption was used about K than that it satisfies Baumgartner’s principle with the set S in (4) any that it is inaccessible. What we are now going to show is that ( 5 ) must fail for the first inaccessible cardinal. We shall do this by showing that if we take B = f fo, f, in (31.11),then (5)impliesthatfor theset ,Y whoseexistenceisclaimed by(3l.ll)we must have that if
5 E Y and 5 >o,then 5 is inaccessible.
(61
Hence I Y I = K clearly contradicts the assumption that K is the first inaccessible cardinal. To establish (6),choose a t~ Y with t>o. Then (31.11) and ( 5 ) imply that fo(t,=
f,(t)=O
(7 1
holds. We then have 2”<151
(8)
for any cardinal p < ( by (1) and by fo(()=O. Hence (=It[(because 2151>5 certainly holds), i.e., 5 is a cardinal and so, a fortiori, a limit ordinal; therefore by (2) (or, rather, by the clause after (2)) and the second equality in (7), we have
<
t=cf(t).
This and (8) imply that is an inaccessible cardinal (note that we supposed 5 >o).Thus ( 6 ) is established. As mentioned, ( 6 ) leads t o a contradiction, completing the proof of the theorem. We now repeat the definition of the Mahlo operation given in Section 30 (we again stress that the operation defined here slightly differs from what Keisler and Tarski called Mahlo operation in [1964]). DEF~N~TION 32.2. Let X be a class of ordinals. We define M ( X ) by putting u E M ( X ) iff either cf ( u ) ~ or o u\X is not stationary in u. In other words, M ( X ) contains the same limit ordinals as nst (On\ X ) , where On is the class of all ordinals, and the operation nst, the nonstationary points of a class of ordinals, was defined right after the announcement of Theorem 5.7. The 13‘
I96
L A R G E C A R D I N A L S CH
VII
proof of the next theorem, still not our main result, follows some ideas in the proofofTheorem 5.7. Note again that the implication (17) * (19) below can be considered as a lemma for our main result, just as relation (5) above. THEOREM 32.3. Denote b y AC the class of all accessible ordinals (i.e., of all ordinals that are not inaccessible cardinals). Let ti be an inaccessible cardinal, and assume that there is a set Z E ti such that ti
4 AC u M(Z)
and K E
M(ACuM(Z)).
Then t i + ( t i ) : holds.
PROOF. Assume that, on the contrary, t i - + ( t i ) : holds. Then Baumgartner’s principle holds for ti by Theorem 31.4. For each ~ E M ( Z ) \ A Clet , f;. be a regressive function on t \ Z that is divergent in 5 ; as t \ Z is not stationary in 5, there is such a function by Neumer‘s theorem. (Theorem 5.3).The divergence of fc means that, for each y < t , there is a value g,,(t)
(11)
This defines the regressive function g , ( < ) for each y < ti and for each ~ E M ( Z ) \ A Cwith y < r < t i ; otherwise put g,(t)=O. For each a ~ t i \ Z write , if 5 E M(Z)\ AC and a < 5 < ti (i.e., when the right-hand side is defined), and write h a ( 5 ) = 0 otherwise. Finally, let h be a regressive function on ti such that h P ( t i \ ( A C u M ( Z ) ) is divergent in
and
h ( 5 ) = 0 if
( E A C U M ( Z ) and
ti,
<
(13)
There is such an h by Neumer’s theorem, since K\ ( A C u M ( Z ) )is not stationary in ti according to (10).Consider the set S = { f o ’ , u { f , ’ , u { . f ~ mis: ~regular & a < A < t i ) . u u {gy: y < t i ) u { h z :a E ti\Z) u { h ) ,
(14)
where the definitions of these functions were given in (l), (2), (3), (1 I), (12), and (13) in turn. According to Baumgartner’s principle, for each f E S there is an ordinal q S < such ~ that (31.11) holds. First note that we must have qh=O.
A COMBINATORIAL APPROACH TO THE HANF-TARSKI RESULT
197
In fact, if we choose B = ( h ) in (31.11), then this formula says that there is a set Y c K of cardinality K such that h(<)=qh for any ( E Y . According to the divergence of the function in (13) we must have
I Y n (K\
(ACU M ( Z ) ) ) < ( K,
(16)
and so (15) indeed holds by the clause after (13). Note that the set S in (14) includes that in (4); hence (5) and its consequence ( 6 )holds. Taking a set B G S ( = the S in (14))containing 1,.f l , and h, we must have for the Y in (31.11) by (6) and (16) that I (M(Z)\AC)I < k'.
u
As we can freely omit less than
K
elements of K we may assume that
Yc(M(Z)\AC)nK.
(17)
It is this inclusion that we shall use in what follows, and it is of no relevance to US how we derived it. (This is important, because in the proof of the next theorem we shall obtain an analogous relation but in a slightly different way.) Weghall prove that (17) and the fact that the set S includes the functions in S'=ig,.:?i
imply that K E
M ( Z ).
This of course contradicts our assumption (9), and this contradiction will complete the proof of the theorem. To establish (19)we have to prove that K\ Z is not stationary in K . To this end, by Neumer's theorem it will be enough to verify the following claim: The function (a E K \ Z ) r(C0 = Vh, is a regressive function on K \ Z that is divergent in K. First note that r is regressive; in fact, ha(<) 'lg,.To this end, fix a and y, and choose in (31.11) a set B containing h, and g , . Then there is a set YE K of cardinality K satisfying that formula. By making sure that B contains certain other functions as well (namely fo, f,,and h), we can require, as we saw above,
198
LARGE CARDINALS CH.
v11
that (17)be also satisfied. Note now that (1 1 ) and (12)hold for all but less than K elements 5 of the set (M(Z)\ A C ) n K ; hence by (17)they hold for all but less than K elements 5 of the set Y. Thus we have a >g,(O*h,(O>
Y
for a large enough t E Y by (1 1 ) and (12). As g,(5) = qe,, and h,(t) = q h , hold here
by (31.1 1)and the assumptiong,,, h, E B, and r ( a ) = q h m holds by thedefinition ofr, this verifies (20). Hence, as claimed, r is a regressive function on K \ Z that is divergent in K ; so (19) follows, completing the proof of the theorem. Our next result, the main one in the present section, is identical to Theorem 30.5, which we proved by metamathematical methods. All the ideas in the combinatorial proof below are essentially contained in the proofs of the preceding two theorems above. As before, AC denotes the class of all accessible ordinals. THEOREM 32.4. Let K be an inaccessible cardinal, and assume that there are sets Z, E K ( p< K ) such that, writing
we have
K$!
X and
K E
M ( X ) . Then
K+(K):.
REMARK.S. Shelah proved that if V = L then this theorem gives an exact
characterization of cardinals with K + ( K ) : . In the other direction, M. Magidor constructed a model in WhiCh-K+(K): holds for the first inaccessible cardinal for which this is not confirmed by the above theorem.
PROOF. Assume, on the contrary, that K - + ( K ) : holds. Then Baumgartner’s principle holds for K by Theorem 3 1.4. Let h be a regressive function on K such that h P(k\X) is divergent in
K,
(22)
andlet h ( 5 ) = 0 f o r r ~ X n ~ ; t h e r e i s s u c h a n h , a s ~ ~ M ( X )k:K+Kbethe .Let function defined as follows: k(()=min {p:t ~ M ( Z J \ ( p $ l ) j
(23)
if ( E ( X n K)\AC (in this case the set on the right-hand side is not empty), and
put k ( t ) = O otherwise. k is obviously regressive. Let f’, fi, and fAa be the functions defined in (I), (2), and (3), respectively, and let g; and hi be the functions that are given by the same definitions as gy and ha in (11) and (12), respectively, if the set Z there is replaced by Z , (note that 11 there also depends on
A COMBINATORIAL APPROACH TO THE HANF-TARSKI RESULT
I99
Z). Put
S={fo)u{.fi)u{f,A is regular :~ & u
u
c(<~
( { g ; :y < ~ ) u { h g a :€ K\Z,l)u{h]u{k]
(24)
B
(the function h now differs from the one defined in (13)). We shall use Bauhgartner’s principle with this S. According to this, for each f E S there is an q r < ~ s u c hthat (31.11) holds.Takeaset B c S t h a t includes thefunctions fo,fl, h, and k, and consider the set Y whose existence is claimed in (31.1 1) for this B. Firstly, we must have lyn(K\X)l< K
according to (22),and as there is no harm in omitting less than K elements of X we may assume that YGXnK. Secondly, qjo=qf, = O by (5), and so ( Y n A C l s o according to (6),i.e., we may assume Y E K \ AC. Hence Y c (X n K)\ AC ,
(25)
Now k ( t ) = q k holds for each ( E Y by (31.11) because k E B , which means, according to (23) and (25), that
This is the same as (17) above (expect that Z has a subscript here; but this subscript does not depend on the choice of B in (31 . I l ) , provided k E B ) . Hence this relation implies the analogue of (19), i.e., that K E M(ZJ.
This contradicts the assumption that K # X , completing the proofofthe theorem. Our last theorem above shows that K + ( K ) ~ holds for many inaccessible cardinals; in fact, it holds if K E M ( A C ) (choose Z,=O for all p K in (21)), or if K E M ( M ( A C ) )(assuming that K # M ( A C ) , choose Z , = M ( A C ) n ~ f l o rall p < ~ ) , etc. The next section describes a large class of cardinals for which K + ( K ) : holds by Theorem 32.4.
-=
200
LARGE CARDINALS CH.
VII
33, HANF'S ITERATION SCHEME The aim of this short section is to describe a large class ofinaccessible cardinals about which Theorem 32.4 confirms that K + ( K ) : . This class is obtained by an iteration scheme, due to Hanf, of the Mahlo operation. Hanfs scheme depends on an arbitrary wellordering R of K. In discerning the meaning of the formulas below, it is perhaps best to take R first equal to the natural wellordering 'less than' of K. But a short contemplation of these formulas will convince the reader that by taking R to be a longer wellordering, i.e., by taking tp (K, R ) > K , we obtain a larger class M R . (We shall make a further comment upon this below, after the definition of the iteration scheme.) This is why the parameter R is needed. To describe Hanf's iteration scheme, let R be an arbitrary wellordering of K. Denote by a;" the 7th element in this wellordering (y<tp (K, R)). For a set X C K put M'R.O)(X)=X ; K
if y < t p ( K, R ) and y = v
1 for some ordinal v, then write
M ' R * y ) ( X ) = M ( R * " ) ( X[M(M'R*"'(X))\ )u (Util)] , and if y is a limit ordinal < tp (K, R), then put
u M'R.V)(X).
M'R"'(X)= Finally set and
V < Y
(J
MR(X) =
Y <tp
(K.
M'"-Y'(X), R)
M * ( X ) = (J M R ( X ) , R
where, in the last formula, R runs over all wellorderings of K. To mention an example, if < K denotes the natural wellordering of K and R denotes the wellordering of type K iK of K in which c$+,, = y i2n and aRK +.y + f.l = = y i2n 1 hold for any integer n and for any limit ordinal y < K (and for y = O), then it is easy to see that MR(X)= M<.(M<=(X)) holds for any X G K (~201. For further examples see Hanfs original paper [1964]. As a corollary of Theorem 32.4, we have 33.1 (Hanf [1964]). r f THEOREM E M*(ACn K ) , then K + ( K ) ~ .
K
i s an inaccessible cardinal and
KE
SATURATED IDEALS, MEASURABLE CARDINALS, STRONG PARTITION RELATIONS
20 I
In contrast to the remark after Theorem 32.4, one can show that K + ( K ) $ holds e.g. for the first inaccessible cardinal with K q! M*(ACn K), although we shall not do so. Again, in thinking about the meaning of this theorem, one should first consider the case K E M R ( A C nK) with R equal to the natural wellordering ‘less than’ of K.
PROOF. According to the assumptions, we have K E M R ( A C nK ) for some wellordering R of K. This means that there is an ordinal y with y 4 1 < t p ( K , R) such that and Define the function f : K + f p
For any p < K put
(K,
R ) as follows:
f(p)=v
iff
p=af.
Z , = ( M ( R , f ( p ) ) ( A CK n) ) n r c
i f f @ ) < y, and 2, = 0 otherwise. Defining X with these Z,’s according to (32.21), it is easy to see that xn(K/ l)=kfcR’Y)(ACnK)n(K/ 1)
holds; hence (1) and (2) imply that K $ X and K E M ( X ) .So according to Theorem 32.4. This completes the proof.
K+(K)~
holds
34. SATURATED IDEALS, MEASURABLE CARDINALS, AND STRONG PARTITION RELATIONS In this section we shall consider a large cardinal property stronger than weak compactness, called measurability. Measurability can be defined in terms of certain prime ideals; as these were already defined in Section 29, we could do without repeating the definition. The situation at present is, however, slightly simpler, since all the ideals considered here are ideals in the field P(K) for some cardinal K. So it would not be fair to recall here the more complicated concepts introduced in Section 28, and we therefore give the definitions here in complete detail; in the following definition K is an arbitrary infinite cardinal.
DEFINITION 34.1. (i) A set I is called an ideal on
K
if U I = K and, moreover,
X,Y E I and Z C X u Y imply Z E I for any X, X and Z. In this case it is
customary to say that K carries the ideal I . I is said to be nontrivial if K $ I . Given a
202
LARGE CARDINALS
CH.VII
cardinal A, the ideal I on K is said to be I-complete if U X E I holds whenever X E I and I X I < I . (ii) Given a cardinal I and an ideal I on K, I is said to be I-saturafed if there is no set K cP(K)\I ofcardinality I such that X n Y E I holds for any two distinct X , Y E K. A 2-saturated ideal is called a prime ideal. (iii) A nontrivial ideal I on K is called normal if it is K-complete and, given any set X c K with X $ I and any regressive function f : X + K , there is an ordinal { < K such that { a E X : f ( a ) = { )$ 1 . (1) It is easy to see that an infinite cardinal K carries a nontrivial K-complete ideal if and only if K is regular; the simplest such ideal is the set [ K ] < ~ w . does not carry a normal ideal, as one can easily ascertain; if K > w is regular, then the ideal of sets nonstationary in K is a normal ideal on K according to Fodor's theorem (Theorem 5.5). The notion of normal ideals is due to D. S. Scott, though Fodor proved his theorem several years prior to the birth of this notion. Just as we have two variants of Fodor's theorem for regular K (Theorem 5.5 and Corollary 5.6), condition (1) in the definition of normality can be replaced by the following: given arbitrary sets X : E I, we have
In fact, the function f : X + K defined by f(a)=min { ( : a € x : \ ( ( i 1))
(a E X )
is regressive and so, assuming that ( 2 )fails, i.e., that X $ I, (1) will also fail with this f . Conversely, assuming that f is an arbitrary regressive function on a set X c K, and putting
x i = { a E x:f ( a ) = t; , we have X : = X : \ (5 i1)by the regressivity of S.If now ( 2 )holds and X : E I for all {
then we obtain that X =
u
X : E I , which means that (1) must hold if
:
X $ I. The question whether a cardinal K carries a I-saturated ideal is a difficult large cardinal problem in case A I2" ;the answer is obviously yes in case A > 2". It is worth noting the following two results about saturated ideals. The first one is quite trivial: LEMMA 34.2. Let K 2 w and I be cardinals, R IK , and let I be a I-complete ideal on Then I is ksaturated if and only if there is no set K EP(K)\I of cardinality I whose elements are pairwise disjoint. K.
SATURATED IDEALS, MEASURABLE CARDINALS, STRONG PARTITION RELATIONS
203
PROOF.The 'only if part needs no comment. To see the 'if part, assume that I is not A-saturated, and let M = ( X , : a < A ) ~ 9 ' ( t i ) \ lb e a set such that X,AX,EI holds whenever a < P < 1.Writing Xh=Xa\ (,J Xp=Xg\ ,
u
(XanXp),
,
the set K = {Xi:a < 1; will consist of pairwise disjoint sets none of which belongs to I in view of the A-completeness of this latter. The proof is complete. The second lemma uses a minor trick:
LEMMA 34.3. Let ti > o be a regular cardinal and 1a normal ideal on ti. Then I is ~ ti'+ such that K+-saturated ifand only ifthere is no set K E , ~ ( K o)f \cardinality IX n Y / < t i holds for any two distinct X, YE K. PROOF.The 'only if part follows from the observation that any subset of ti of cardinality < t i belongs to I by the K-completeness of I. To prove the 'if part, assume that I is not K+-saturated, and let M = { X U : a < t i + ) ~ 9 ( t i ) JbeI a set suchthat X,nX,EIholdswhenevera
where yp =
(,J ( ( X , n X , ) \ ( f , ( a ) i l ) ) .
a<,
In view of the normality of I, (2) implies that Y, E I; hence Xb $1. On the other hand,itisobvious that XhnXb'X,nXbE f,(a)r 1 holdswhenevera
THEOREM 34.4. Given any infinite cardinal ti, there is no nontrivial ti+-complete and ti+-saturated ideal on K + . For the proof we need a technical lemma, also due to Ulam. This lemma is very important in itself, because it has many other applications.
LEMMA 34.5 (Ulam [19301).Let K be an infinite cardinal. Then there is a matrix ( U a p : a < t i & P < t i + ) of subsets of K + such that each row consists of pairwise disjoint sets and the union of each column contains all but I K elements of K + , i.e., such that U,, and Uupeare disjoint whenever a < ti and P < 8' < ti+, and we have
204
LARGE CARDINALS CH.
VII
The matrix whose existence is claimed in this lemma is called an Ulam matrix on K + .
r+~
PROOF. For each 5 < K', let f:: be a 1-1 function (not necessarily onto K), and write u,, = { < K + : j?< & A(/?)= ci)
r
r
for ci < K and B < K'. The sets U,,, < K', are pairwise disjoint for a < K fixed by the one-to-oneness of f ; and we clearly have
u u,,=K+\(Bil),
U C X
which verifies (3). The proof is complete.
PROOF OF THEOREM 34.4. Assume I is a nontrivial K+-complete ideal on K + ; we are going to show that I is not K+-saturated. To this end, consider an Ulam matrix(U,,: a
DEF~NITION 34.6. Let K > w be a regular limit cardinal, and let S E K be a set. The matrix ( Uu,: B E S & a < B ) of subsets of K is called a triangular Ulam matrix on K with support S if S is stationary in K , the sets U,,, a < E S, are pairwise disjoing for any fixed a < K , and IK\
u
a<,
Uu,I
(4)
holds for each B E S. Hajnal's lemma on triangular Ulam matrices runs as follows (see Section 5 for the concepts used):
205
SATURATED IDEALS, MEASURABLE CARDINALS, STRONG PARTITION RELATIONS
LEMMA 34.7 (Hajnal[1969]). Let K > w be a regular limit cardinal such that there is a set S stationary in K and a club C in K , both consisting of uncountable cardinals, such that S E C G ( 2 < K : either A > w is regulur (5) and S n l . is nonstationary in 2. or i. i s singular; , where E. runs over cardinals. Then there is a triangular Ulam matrix support S .
011
K
with
PROOF. We first establish the following CLAIM. For each A E C, there is a 1-1 regressive,function,f, on S n l . We construct the functions fn by recursion on i. E C . Assume to this end that f n ) has already been constructed for each l‘< 2 with 2’ E C. If A is the first element of C, then define ,fn as the empty function (i.e., the empty set). Suppose now that this is not the case, i.e., that Cn?. is not empty, and write p = sup ( C n A).
(6)
We have p E C by the definition of club. We distinguish three cases: a ) p < l , b) p = i L and A is regular, and c) p = l and 1, is singular. Ad a). If p S, then we can simply put f n = f, in this case. If p E S, then we can almost do the same, the only problem being that we also have to define f,(p) in a way that .fn be 1-1. But it is easy to make one value free for ,f,(p) e.g. by putting
f b ( 4 = 1 i.f,(d
(7)
for every a ~ S n p then ; we can write f A = . f b u { ( p O ) ) . Ad b). Notein thiscasethat thereisaclubB’cl\Sin Aaccordingto (5). Write B= B n C ; B is also a club in A because p=sup ( C n A ) = l in this case, and B consists of cardinals since it is a subset of C . Let T,, a < E., be an enumeration ofthe elements of B in increasing order; for each a < I , f,, is a 1-1 regressive function fromSnz,intor,.For eacha
) . f n = ,f,,u whenever (T E S n (z,, z ~ + ~Put
u f,,+,.Then
U
fn
is regressive and 1-1,
and its domain is S n I , as none of the T,’S belong to S because 7, E B G B’G A\S
.
206
LARGE C A R D I N A L S CH.
VII
Ad c). We use a similar argument as in the preceding case, although the situation now is slightly more complicated. Let (tz:a
(g)= T o
icf (1.)if,, (0)
whenever D E S n (to,T , ). So as to make the value 0 also free for f ' , define f io according to (7) with T~ replacing p there. It is then clear that f A = =f'uf,,w f,,,, is a 1-1 regressive function on S n a . Thus our claim is
u
l
established. It is now easy to define the matrix whose existence is claimed in the lemma to be proved: Let h be a 1-1 function from C onto K , and for each a and /3 with 3 and p E S put U I P = ( h ( i . ) : / 3 < I . E C & f A ( f i ) = a ), analogously as in the proof of Lemma 34.5. It is easy to show that ( U a P : BE S ) is a triangular Ulam matrix with support S; e.g., we have U,,=h"(C\ ( p i l)), which entails (4) as h is onto K . The proof is complete.
u
@
Let A C be the class of accessible ordinals, N R L be the class of those ordinals that are not uncountable regular limit cardinals, and, for a cardinal K , denote by < r the natural wellordering of K . Then, using the notations of the preceding section, it is easy to check that the assumptions ofthe preceding lemma hold with some S and C whenever K E M'"*<=)(NRL) \ N R L or K E M(<w.")(AC)\AC; i.e., the above lemma confirms the existence of a triangular Ulam matrix on these cardinals. It is not known, however, if there is a triangular Ulam matrix e.g. on the first cardinal not belonging to M'",'" (AC).The application of the above lemma that we aimed at is the following theorem of Hajnal [1969]: THEOREM 34.8. Let K > o be a regular limit cardinal, and assume that there is a triangular Ulam matrix on K . Then K does not carry any nontrivial K-saturated K-complete ideal.
PROOF. Assume that I is a nontrivial K-complete ideal on K ; we are going to show that I is not K-saturated. To this end, consider a triangular Ulam matrix
SATURATED IDEALS, MEASURABLE CARDINALS, STRONG PARTITION RELATIONS
207
( U , , : a < p E S) on K; note here that S is stationary in K according to Definition 34.6. By (4), there is an ordinal f ( p ) < b for each ~ E with S p # O such that Us(,,B$ I. Neumer’s or Fodor’s theorem (the best reference here is Theorem 5 . 5 ) then implies that there is a set S’ of cardinality K such that ,f is constant on S’, say f ( p )= a for each /?E S’. Then the set { U,,: p E S’] c P ( K ) I\ consists of pairwise disjoint sets, showing that I is indeed not K-saturated. The proof is complete.
According to the remark made before the theorem just proved, this theorem shows that there are many inaccessible or regular limit cardinals rc which do not carry nontrivial rc-saturated rc-complete ideals. It does not, however, give all the known results about rc-saturated rc-complete ideals on rc, since it does not confirm that the first cardinal K not belonging to M@“*<*) (AC) does not carry such a nontrivial ideal (cf. the remark made before the last theorem), although this is a well-known result (cf. our remark made after the proof of Theorem 34.4). Kunen [1970] proved that if I/ = L, then there is no rc+-saturated rc-complete ideal on any uncountable cardinal K . The assumption I/ = L seems necessary here even in case of small K , since Kunen also proved recently that a very strong large cardinal axiom ensures that it is consistent that K, carries an &-saturated K,-complete ideal. Normal ideals are often much more convenient to handle than non-normal ones. Their applicability in the theory of saturated ideals is made possible by the following theorem of D. S. Scott: THEOREM 34.9.Let K and A be cardinals, K regular, 3 II S K + , and suppose that K carries a nontrivial A-saturated K-complete ideal. 7hen K also carries a A-saturated normal ideal. The proof is analogous to that of Lemma 31.2, but the present situtaion is simplified by the fact that we here consider ideals in the complete field qrc)of sets, while the fact that we consider &saturated ideals instead of prime ideals does not cause much complication. PROOF. Let I be a nontrivial I-saturated rc-complete ideal on K, and define the relation <, on ‘ K by putting
for any two functions g E K ~ It. is easy to check that <, is a partial ordering. There is no infinite sequence of functions f, E ‘ K , n
,
u
n<w
{t
f+l(t)=f(t);
208
LARGE CARDINALS CH.
does not equal
K.
Picking a
5 < K not
VII
in this set, we would then have
fO(t)>fl(O>.
..
7
which is impossible, as there is no infinite descending sequence of ordinals. Let now fo be the identity function on K (i.e., let = 5 for each t < K) and, having defined f,for some integer n, choose, if possible, a function ,f,+ E ‘ K with f.+ f, such that { 5 - = K : f,+,(t,=.lEI
fo(c)
<,
for all a < K (note that this clearly holds with fo replacing f,+l). According to the remark just made, there is a least integer n for which it is impossible to choose such a function f,+ 1 . Write h = f,. The function h obviously has the following important properties : h(01 t (9) holds for any 5 < K, since h I,fo, and fo was chosen to be the identity function; for any ordinal a < K we have (t
and whenever f is a regressive function on
K,
(10)
then we have
{ t < K : , f ( h ( t ) ) = a $) 1
(11)
for some a < K . To see this, one only has to observe that (10) with a=O implies fo h < , h, as f ( h ( t ) ) < h ( 5 )whenever h ( t ) # O ; if (1l)were false, then we could choose f,+ = f o h, a contradiction. (A function h with properties (9k(11) is called a minimal function with respect to 1.) We are now in a position to define a %-saturatednormal ideal J on K. For any X C K put X E J if and only if there is a regressive function f on X such that ( ~ C K h : (t)EX&f(h(t))=a)EZ
(12)
holds for any a < K. It is easy to see that J is a K-complete ideal on K ; in fact, to see K-completeness, assume that X , E J, v < p, are pairwise disjoint sets for some p< K. Then there is a regressive function f, on X , for each v < p such that (12) will be satisfied with X, and f v replacing X and respectively, and (12) with f= f y E J will confirm that X = X,= J. (11) shows that J is nontrivial.
u
Y
u
V
J is also a normal ideal; this can be Seen by arguments similar to those used in the proof of Fodor’s theorem (Theorem 5.4) as follows: Assume that J is not normal. Then there is an X c K with X $ J and a regressive function g on X such that x,= (t E x :g (< )= v) E J
S A T U R A T E D IDEALS, MEASURABLE C A R D I N A L S . STRONG PARTITION RELATIONS
209
holds for any v < ti; i.e., by ( 12) there are regressive functions .fv on X , , , v < K, such that ( q E K : h ( l l ) E X & g ( h ( q ) ) = v & , f ; . ( h ( r l ) ) = a€ j 1 (13) holds for any u < ti. Define the regressive function f on X . f ( t ) = m ax tv, whenever
(J X,. by putting
=
fx);
\'
t E X , , ; then . f ( ( ) > g ( ( ) for all ( E X . Given any u < t i ,
we have
{ q < ~h: ( q ) rX & , f ( h ( q ) ) < a ) =
={'I
=
h ( q )E X & g ( h ( q ) ) i a &, f ( h ( q ) ) s a= )
(J ( r l < t i : h ( r l ) E X & g ( h ( l l ) ) = v & f , ( h ( q ) ) = p :€ 1 , fiU5.I
since there are less than ti sets after the last union sign, and each of these belongs to 1according to (13). This, however, means ifwe compare the two extreme sides ofthe above formula that X E J . This is a contradiction, proving that J is normal. Finally, we are going t o show that the ideal J is ;.-saturated. Assume the contrary; then Lemmas 34.2 and 34.3 imply in view ofthe assumption I5 ti+ that there are sets X , E ~ ( K ) \ J p; < A , such that IX, rl X
V I
(14)
holds whenever fi < v < i.Writing
Y,=((
= { 4
: h( g ) E X , n X , ; E I
whenever p < v < i in view of (lo), (14), and the ti-completeness of I . This contradicts our assumption that I is I-saturated. Thus the I-saturatedness of J is also established. This completes the proof. O u r main concern in what follows is to prove a strong partition relation for measurable cardinals, to be defined below. We shall also briefly mention realvalued measurable cardinal since they are often related to combinatorial 14
Combinatonal
210
LARGE CARDINALS CH.VII
problems (in fact, they were mentioned in Section 1 1 while discussing ramifications of the Erdas-Dushnik-Miller theorem); the reader not interested in them may skip parts (ii)-(iv) of the following definition, as well as our short comments about real-valued measurable cardinals made afterwards. DEFINITION 34.10. (i)An uncountable cardinal K is called measurable if it carries a nontrivial k--complete prime (i.e., 2-saturated) ideal. (ii) If X is an indexed set of nonnegative real numbers, then their sum E X is defined as sup(ZY: Y c X & Y is finite), where the sup here is either a real number or + co ;the sum of finitely many real numbers above is to be understood in the usual sense. (iii) Given an infinite cardinal K, a %-additivereal-valued probability measure on K is a function m: 9 ' ( ~ ) + [ 011, , this latter denoting an interval of real numbers, such that m ( K ) = 1 and we have m(UX)=C(m(x):xEX)
for any set X of cardinality < K of pairwise disjoint subsets of K. m is called nontriviaI if m ( { t ] ) = Ofor any f E K. (iv) An uncountable cardinal K is called real-valued measurable if there is a nontrivial h--additive real-valued probability measure on K ; in this case one also says that K carries such a measure. It is easy to see that a measurable cardinal is also real-valued measurable. In fact, if I is a nontrivial li-complete prime ideal on K > O , then the function m defined as m ( X ) = O whenever X E I and m(X)=1 for any other X E K is a nontrivial real-valued probability measure on K. On the other hand, if m is a nontrivial real-valued probability measure on K, then the set
I = (XEK:m(X)=O) is a nontrivial K,-saturated K-complete ideal on K. In fact, to see this, one only has to observe that the sum of K , positive real numbers is +a; this latter assertion can easily be shown, since given K , positive real numbers, there is a real number E > O such that K, among them are > E . We shall soon see that a measurable cardinal must be inaccessible (see our remark immediately after the proof of Theorem 34.1 1). This is not so for real-valued measurable cardinals; in fact, a result of Solovay [1971] says that if ZFC + 'there is a measurable cardinal' is consistent, then ZFC +'2"0 is real-valued measurable' is also consistent. A recent result of K. Prikry says that 2'O is real-valued measurable, then 21 = 2'O for each j. with K,,I~..<~"o; Theorems 34.4 and 34.8 say that a real-valued
SATURATED IDEALS. MEASURABLE CARDINALS. STRONG PARTITION RELATIONS
211
measurable cardinal must be a 'very large' regular limit cardinal. A result of ErdGs and Hajnal [1958] is the following:
THEOREM 34.11. If
ti
is a measurable cardinal, then
(15)
ti-+(ti)A:o
holds for any cardinal , i< K. For the definition of the partition relation in (15) see Subsection 8.6. As pointed out by F. Rowbottom, if there is a measurable cardinal, then there must be an uncountable cardinal K < the first measurable cardinal satisfying the relation in (15); this can be seen by a simple argument involving indescribability. Hint: a measurable cardinal is ll:-indescribable by a theorem of W. P..Hanf and D.. S. Scott [1965].(see Silver [1971] for a proof), while saying that K is the first cardinal > o satisfying (1 5) with all 1 K can be expressed by a I l k , or even by a ll:, sentence. In order to prove the above theorem, we shall first prove the following lemma of F. Rowbottom [1971].
-=
LEMMA 34.12. Let K > O be measurable, I a normal prime ideal on ti,jet n be an integer, and assume t h a r f : [ t i \ {O)]"-+ti is a,function such that . f ( ~ ) < m i nb) holds f o r any u E [ti \ {O;]". Then there is an X [XI".
E ; P ( t i )\
(16) I such that .f is constant on
PROOF.We use induction on n; the assertion holds vacuously in case n, and let .f:[ti\(O)]"+ti be a function satisfying the requirements of the theorem. For each 5 < ~ define , the functionf,: [ti\(5+1)ln-'-+~ by putting n =0, Assume that n > 0 and the assertion holds with n - 1 replacing
.&)=f(mJ4
(17)
whenever u E [K \ (5 4 l)]"- ;the induction hypothesis implies that for each 5 < ti there is an X : E ~ ( K ) and \ I a g ( 5 ) such that
f:[Ix,In-
=
MOj
(18)
ho1ds;we haveg(<)
u (K\(X:U(5il))EI
:
by the normality of I (cf. (2)), i.e.,
14'
212
LARGE ('ARDINALS CH
VfI
Using the normality of I again (cf. ( 1 I), there is an a < K and an X such that g " X = (a;. We claim that
GX'
with X $ I (19)
./ ( u 1 = a
(20)
holds for any u E [XT. In fact, let t be the least element of u, and write u = =.\it).Then we have U C X GX'E (5; 1 )uX : ; the definition ofu implies that o n ( t ~ l ) = O ; h e n c e w e a c t u a l l yh a v e i i s X , . A s t E X , (17),(18),and(19)imply that
f b )= ,h(Ll)=dt)= a
3
which proves (20).This means that the conclusion of the theorem is valid, i.e., the proof is complete.
PROOFOF THEOREM 34.11. There is a normal prime ideal I on IC in view of Theorem 34.9. Fix 1 < K and let f :[ K \ ~ ] ' ~ + ; L be a coloring; it is enough to prove that there is a set X E ~ ( I c ) I\ such that the function f [XI' is constant for each n < o.According to the theorem just proved, there is an X , E Y ( K\ )I for X, 41 each integer n such that f [X,]" is constant; then we have X =
0
A
n
because I is a K-complete prime ideal. This X satisfies our requirements; the proof is complete. Note that (15) with 1 = 2 obviously implies I C + ( I C ) ~ ; hence a measurable cardinal is inaccessible in view of Theorem 29.1. Note also that we have
Infact, put f(x)=Oforanyfiniteset x ~ w i f x , - x , ~ ~ x ~ , w h e r e x , a n darethe x, first two elements of x, and put . f ( x )= 1 otherwise. Then it is easy to check that f verities the relation in (22). Observe that the Axiom of Choice is not used in establishing (22); hence it is interesting to compare this relation with the relation m+(w)?,
(23)
which was shown by A. R. D. Mathias to be consistent with ZF+ the Axiom of Dependent Choices (DC), which is weak form of the Axiom of Choice (see e.g. Jech [1971; p. 793). This seems an anomaly, since one would expect the relation o+(o)y to be stronger than O - + ( W ) , < ~ ,but, obviously, the proof of (22) breaks down if one wants to use it to disprove (23). Of course, the relation in (23) is false in ZFC according to Theorem 12.1. If we have K+(OJ):"' for an infinite cardinal K , then the method of the proof of (24.2)with assumption d ) in Lemma 24.1 implies that 2"+ ( o ) ;holds, ~ and if K is
SATURATED IDEALS. MEASURABLE CARDINALS. STRONG PARTITION RELATIONS
213
singular and A f , ( o ) ; m holds for all 1 < K, then similarly, the method of the proof of (22.5)in Theorem 22.1 shows that K+(o,);" also holds. In this way it follows from (22) that if, (W g l J (241 holds for every cardinal derive from here that
A less than the first inaccessible cardinal K " . It is easy to l);(u
ti"+(Wi
(25)
holds for this K " . In fact, according to (24) there is a coloring j i : [5]'"+2 for each ordinal 5 < K~ verifying the relation t f , ( ~ ) Putting ; ~ . .f(x)+ ff,;(x\ ( 5 ) ) whenever x s K~ is a finite set with t = max x, it is easy'to see that .f berifies the relation in (25). (25) implies in view of Theorem 34.1 l'that the first inaccessible cardinal is not measurable. One can push this argument further so as to prove that the first measurable cardinal K must be the K t h inaccessible cardinal. Of course, we knew this already e.g. in view of Theorem 3p.8, or in view of the results on the relation K + ( K ) : in Sections 30-33. Finally we prove a simple lemma due to Silver [1966], which will be needed in Section 45:
LEMMA 34.13. Let
K
be an infinite curdinul and a be a limit ordinal. If K+(U);O
,:'
holds, then we also have
K +( a
PROOF. Assume (26). Given a coloring f :[K]<w-+E202,
we have to prove that there is a set X of order type a such that f is constant on [XI" for each integer m. For each integer m and every set x E [K]"", denote by xlm the set formed by the first m elements of x. For any two integers m, n and for any set x E [ K ] ~ " . ~ "put (f(xlm)) (n)
(28)
(note here Lhat f ( x l m ) is a function from K O into 2, and so the value on the righthand side here is 0 or 1); if this stipulation does not defineg(x), where x E [ K ] < ~ , then put g(x)=O or 1 arbitrarily (e.g. put g(x)=O always). According to (26), there is a set X c K of order type ct and there are integers ik < 2 ( k < o)such that
g"[X]k=
{ik)
(29)
214
LARGE CARDINALS CH.
VII
holds for each k < w . It is easy to see then that f is constant on [ X I r nfor each integer m.In fact, ifx belongs to this set, then let X ' E [ X I " be such that x'Im=x (here we use the assumption that the order type a of x is a limit ordinal); then we have f ( x ) = (g(x'12" .3"):n < w ) = ( i 2 m . y :n
CHAPTER V l l l
DISCUSSION OF THE ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2
In the first two sections of this chapter we collect the results obtained for the case r = 2 of the ordinary partition relation, and an attempt is made 40 state them in a unified form. Understandably, this is a very unrewarding task, and the reader looking for interesting ideas might be disappointed by studying these two sections. Perhaps it is best for him to skip the detailed proofs here, or even to skip these sections entirely. We do not mean by saying this that these sections are superfluous. Their importance lies in the fact that they put the disarray of the large amount of information for the case r = 2 contained in the previous three chapters in order. They give a practically usable test to see whether gn ordinary partition relation with r =2 holds or fails if this can be decided with the aid of the known results; and, moreover, they help to locate the open problems, the most important of which is probably Problem 35.5. In Section 37, the third and last one in this chapter (which, by the way, the reader should not be afraid to consult, since it is not of the same character as the first two sections here) we shall show that this problem cannot be decided negatively with the aid ofthe methods used in this book. Since a positive solution also seems hopeless to us with present methods, it is quite possible that the solution of this problem will have a really stimulating effect on the development of combinatorial set theory.
35. DISCUSSION OF THE ORDINARY PARTITION SYMBOL IN CASE r = 2 In this section we are going to collect our results and the open problems concerning the case r = 2 of the ordinary partition symbol. So as to make the rather complicated discussion more intelligible, we are not going to formulate the theorems below in the most concise forms possible. We shall study the relation (1)
K4(2:):<~
under the conditions K ~ W 9 , 22,
and
3 1 1 , s ~ for
(<8,
(2)
216
ORDINARY PARTITION RELATION WITH SIJPERSCRIPT
2 CH. VIIl
which will be assumed throughout this section. To this end, we shall need the following
z=(&:
DEFINITTON 35.1. Let I=(>.::t<S) and 5<9') be two sequences of cardinals. We write 142 (read: 2 dominates A', if there is a one-to-one function .f from 9 into 9' such that i: II.>,:,+ holds for any <<9. It is obvious that if ?, dominates I. and we have K + ( & ) ; < ~ , then we also have ~ + ( 2 ~ ) : Our < ~ .main goal would be to show that, whenever relation ( 1 ) holds, there is a sequence (1.;: 5 < 8 ' ) of a certain specified form dominating (2: : 5 < 9) such that ~+(i.:,):,~. holds. In this way, one would always be able t o decide whether ( 1 ) holds o r fails by simply checking whether or not such a dominating sequence exists. Unfortunately, we shall not be able t o achieve this goal completely, though we shall come quite near to doing so. We start by a lemma Listing the cases in which (1 ) fails by our earlier results. The use of the logarithm operation L defined in Section 7 will simplify our considerations. LEMMA 35.2. (1 ) ,fuils prooided at leusr one of the,following conditions (i)-(ix) hold: (i) I.,, 4>L;.,(d ; (ii)
A, = K and I., > min ( L k ( ~Lcrc,, ) , (cf ( K ) ) ) ;
(iii)
j.,=h;
(iv)
9 > L3(K);
(v)
l{<<9: i.:20).l>L;.,(K) ;
(vi)
& = K and 9 2 L 3 ( c f ( K ) ) ;
(vii)
i , = K and I [t;: 2: 2 0: I> LCllk) (cf ( K ) ) .
i.2, cf
(K),and
cf (k.)+(cf
(h-));;
We can strengthen (v) and (vii) in case GCH holds to
(v')
~ > L , , ( Kand )
(vii')
1, = K and 9 2 Lcf(p,(cf ( K ) ) , respectiuely.
(viii)
9 2 c f L , ( K ) ) and there are cardinals K~ < L,(K)for 5
(ix)
2Lzl") = K and at least one of the following conditions holds:
) i. >,L 3 ( ~; ) a ) i . , > L 3 ( ~ and
DISCUSSION OF O R D I N A R Y PARTITION SYMBOL I N CASE r =
2
217
PROOF. We simply proceed by directly applying results established earlier. The monotonicity properties of the ordinary partition relation (see Subsection 9.6) will often be used without any explicit reference. Ad (i). Writing p = L&), the definition of the logarithm function in Section 7 implies that there is a a < A, such that a”> ti. By a change of notation we obtain a 0 + ( ( c r . p ) + , p + ) ’ from (19.12). As O ” ? K and, moreover, a . p < A , and p<;Zl by our assumptions, we obtain K+(&, Hence (1) indeed fails.. Ad (ii). Suppose first that i1> L , ( K ) .We may assume here (as we did in (2)) that I15 ti. But then we have I,, E., > L;Jti) since E., = ti, which shows that this case is covered by (i). Suppose now I , >Lcr(k.,(cf (ti)). Then the result established in the preceding paragraph before with cf (ti) replacing ti shows that cf (ti)+(cf (ti), 1 1 ). Now Corollary 21.2 implies t i + ( t i , 11)2, which we wanted to show. Ad (iii). We have cf (ti)+(cf (ti), by our assumptions; again, Corollary 21.2 implies the desired.relation K + ( K , Ad (iv). We have ~ + ( 3 ) ; ~ according ,~) to (19.18). Hence 9>L,(ti) implies K + ( & ) ~ < $ , because we assumed in (2) that 4 2 3 holds for each 5 5 9 . Ad (v). Writing p = L;,,](K), it is enough to show that ti+().,, (w),,)’ holds (note that p is infinite by Theorem 7.2, and so p - 1 = p). To see this, observe that crp 2 K holds with some a<>., by the definition of the logarithm. Suppose first that ~22,. Then a < p , and so 2”=0”2ti. We have 2”+(3); according to (19.17), which implies that t i + ( E . , , ( w ) , ) ~ . Suppose now p < & . By an alphabetic variant of (19.13) we have a”+((a.p)+, As aY>tiand cr, p < i . , , this means that ti+(;.,, (w),)’, which we wanted to show. Ad (v’).We mentioned after (vii) that in case G C H holds we can strengthen (v) to (v’). To see this, we have to show that x-+(IO,(3)”)’ holds under the assumption of GCH, where p = LjJti) again. First suppose p 2 I,. We mentioned above, in the proof of (v), that 2”ti and 2”+(3), hold then, and these imply the desired result. Suppose now p<j.,. By the definition of the logarithm, there is a a < I , such that ap>ti. O n the other hand, GCH implies that a P l m a x ( a + , p + ) O L , ; a s I , ~ t i btheassumptionsmadein y (2),thismeansthat I , = and K must be a successor cardinal. As CT can be chosen arbitrarily large K , we may assume that K = a+;but then, p being as long as it is less than io= the least cardinal for which d ” a + = ~ GCH , implies that p = c f ( a ) (cf. e.g. Theorem 6.12). If a is regular, then p = c, and so K = p + = 2” by G C H ; we have 2,+(3): according to (19.17), which implies the desired result. Assume now that c i s singular. According to Theorem 20.2, we have a’+(a+, (3),,,,,)’; note that G C H is again used here. Observing that K=A,= a + and p=cf (a)in the present case, t h s establishes the desired result.
218
ORDINARY PARTITION RELATION WITH SUPERSCRIPT
2
CH.
VlII
Ad (vi). The result in (iv) with cf ( K ) replacing K says that cf ( ~ ) f , ( 3 ) : , ( ~ ~ ( ~ ) ) , and so we have a fortiori cf (K)+(cf (ti), (3)L3(c1(K)))2. Corollary 21.2 now implies K + ( K , (3)L3(cf(K))); since 1,=~and each 1, is 2 3 according to (2), this implies K + ( A ( ) ~ < ~ which , we wanted to show. Ad (vii). Replacing K with cf ( K ) in (v) and putting &=cf ( K ) there, we obtain cf ( ~ ) % ( c( Kf ) , ( w ) ~ ~ ~ , ~ , Corollary ( ~ ~ , ~ ) ) 21.2 ) ’ . now entails K % ( K , (W)L,,h,,cf ( A ) ) ) 2 , which implies that ( 1 ) indeed fails. Ad (vii’). We proceed similarly as before, but we use the stronger relation cf (K)f,(cf (K),(3)L,,A,(cf(h)))’, which holds under the assumption of GCH according to (v’) if we make the same substitutions there as just before. We obtain by Corollary 21.2, which we wanted to show. K-/+(K,(~)~,~,,+~ Ad (viii). In case L 3 ( t i )is regular we have 9 2 L , ( ~ )and , so (1) fails by (iv). Assume therefore that p = L 3 ( ~ is )singular. We have 2pf,(~z)5,cf(p) according to (19.15) in Corollary 19.8; as 2 ” 2 by ~ the definition of p and L e 2 ~ by : our assumptions, this relation implies ~ - / t ( & ) f <which ~, we wanted to show. Ad (ix). Write ~ = L , ( K ) Then . ~ P = K by our assumption. We have 2e+(p’,p)’ according to (21.11), which implies a). To see b), observe that 2e%(p, p, (3)L,(cr(p,,)Z by (21.10). c) follows from Corollary 21S(iii), which says that cf (p)f,(cf (p)): ensures 2P+(p, p)’. d ) is a consequence of (21.13), which says that 2ef,(p, p, (cf ( p ) ) ’ ) ’ holds. The proof of the lemma is complete. Our next result analyses the situation when the conditions of the preceding lemma fail in terms of the relation 4 given in Definition 35.1. This will help us to describe a condition which is as close to being necessary and sufficient for ( 1 ) to hold as is possible by our present knowledge.
LEMMA35.3. Assume that all the conditions (ik(ix) of Lemma 35.2 fail with any
rearrangement (A: : 5 < 9’) of
a , B < p and a < c f (p), and
k,
furthermore, T < L , ( K ) , and if ~ < m i n{ L 3 ( ~L,(cf ),
a=O provided either (i) k
for every v < z ;
A=K,
then
(5)
(K))).
p or (ii) A=K, p=cf(K), and p+(p):.
(6)
DISCUSSION OF ORDINARY PARTITION SYMBOL I N CASE
r=2
219
provided A>p, and 2 e = ~ , and at least one of the following conditions holds:
cr=O
(i)
A>P,
or
(ii) A=p
(7)
and cf (p)+(cf(p)):.
If 2e=K, A=p, and p>cf(p), then a i l ; if
a = 1 here, then we also have crscf ( p ) and
p, T < L3 (cf (p)).
(81
If (v’) and (vii’) (see immediately after (vii) of Lemma 35.2) aiso fail, then
(9)
7=O.
All these conditions are satisfied with
A = I rfor
some
5 < 9.
(10)
REMARK. (9) is important when GCH holds. PROOF. For the sake of brevity, throughout the proof we shall refer to the assumption that the conditions (i), . . ., (ix) of Lemma 35.2 fail with any rearrangement (A: : 5 < 9’) of (At : 5 < 9) replacing this latter sequence by using phrases such as “we obtain by 1(i),. . ., 1(ix)”. For any ordinaly put where Put
-
-
M . = ( E < 9 : 1. v : . is any of the relation symbols =, <, I,
>, and 1.
P0=L3(K).
By using 1(viii), it is easy to see that there exists a cardinal oo<po such that (i) Q=M5bouMLP0 and (ii) IM.{,ol
a,=sup {A,.: < < 9 & 1 , < p o j . . $
(12)
1(i) implies I M>Po1 I 1 ; otherwise we could choose ,lo XI >, po, and we would have A& A‘, >Lio(~)sinceL i b ( ~ ) i L 3 ( ~ ) =Itp follows 0. that thereisa maximal A, with 5 E MLPo,provided this latter set is not empty. Put A=& with this 5 (so that (10)will hold), or, if MZPo is empty, then put A = A< with an arbitrary (. Without loss of generality, we may assume 5 = 0 in both cases; we shall then have A = lo, which will make our notations below simpler. The choice of A implies by (1 l)(i) that:
220
ORDINARY PARTITION RELATION WITH SUPERSCRIPT
2 CH. VIll
Either (i) I=E.,
or
< with 1 5 < < 9, i.e.,
(13)
9 = (0;LJ M=,,"LJMS,(, . Note that the first relation in (4) is satisfied according to ( 2 ) , since ;.=E.,. Define p by the second relation in (4).To verify (5b-(9),we distinguish two cases: (a) p=po and (b) P < P o . Case (a). Putting ~1=1M=~,\(O}l,( T = ( T ~ , p=IM
<
i-E.,2p0=p.
We may also suppose that 2p=h.,
since this is given as an assumption in both (7) and (8). As for (7)(i), if 1= I o > > p = po, then M \ ( 0 )= 0 by 1 (ix)a), and so indeed a = 0. As for (7)(ii), the assumptions in this case are that E+,=p0 and cf (po)+(cf ( p o ) ) : ; thus 1(ix)c) implies that M = Y s\ {0)= O again, and so we have a=O in this case as well. To s1 by? (ix)d), settle (8),suppose that i = I o = p = p o > c f (po).ThenIM=,(,\ (0)1 andsousl.Assumecc=l. Noting thatwe put a,=sup{A::5
DISCUSSION OF ORDINARY PARTITION SYMBOL IN CASE r = 2
221
As A= A, 2 po > p, the definition of p in (4)together with 1(i) and 7(ii) (for a permutation (1;: < 9) of the ,It’s with 2, = A,) implies that M = ( 0 } ,and so g={O)uM,pu(M,onM
L,( K ) ’ LCf
( K ) 2 Lcfj h ) (Cf (K))= p
(cf. (4)),and the right-hand side here is regular again by Theorem 7.15, since cf ( K ) is regular. Thus the regularity of p is established. a + #? c p follows from 1(v) and 1(vii) (only the former is needed when 1= = 1, < K , but both are needed in case 1= A, = K); to verify (9) (when #? was defined differenlly), use 1(v’) and 1(vii’) instead ((v’) and (vii’) were mentioned immediately after (vii)). Now (5) easily follows: a < p by the definition of CF (cf. (ll)(i));/3
holds whenever (4k(8) are satisfied. Our next theorem, the main result of this section, collects the cases when we can really prove this. K , A,((<9) be curdinals with Consider the,following condition:
THEOREM 35.4. Let 9 2 2 be an ordinal, and let K>
w and 3 CE.,<
ti.
There are cardinals A, p, a, cc, /3, T, and k,, ( v < r ) satisfying (3)-(8).
a ) (14) is necessaryfor K
lo
hold.
--+
(J.:
1s<s
(14)
222
ORDINARY PARTITION RELATION WITH SUPERSCRIPT
2
CH.
VIII
b) (14)is suficient for (1 5 ) to hold provided T = 0 and at least one of the,following conditions hold: (i) K is reguiar; (ii) 3, < K ; (iii) K is strong limit, i.e., VK’< K[F< K ] ; (iv) p = w . The difficulty in applying this result is t o check whether (14) holds, especially because we are faced here with the problem of how to choose a 1for which (3b(8) are satisfied. We claim that (3)- (8) are satisfied with some I. exactly if they are satisfied with ;.=sup {&: ( < 8 ) , (16) and if (3b(8)are satisfied with T = O and some 2, then they are also satisfied with T = O and the above value of 1.. After having chosen this value for I., there should be no difficulty in checking (3k(8)provided the cardinal exponentation function is known. Before we go on to establish our claim, we point out that choosing 2 as given in (16)does not adversely affect the applicability of Theorem 35.4. What we mean by this is the following: assume (3)-(8) are satisfied with two sets of parameters: I., p , . . . and A’, p’, . . . where I. is as given in (1 6), and A‘ # 1.If our theorem is applicable with the second set of parameters, then it is also applicable with the first set of parameters. There are two reasons why this is not immediately clear: it may happen that CL)I,’ < K and I,= K or p ) p’ = w and p # w . In case tl), b(ii) of the theorem is applicable with the parameters i’,p’, . . . while it is not applicable with the parameters I., p , . . . . We have K = ( i = ) s u p [&: r<s; in this case; as we also have
(&: ( < 9 ) G ( i . ‘ ,
(p’),., ( C r ’ ) @ . )
according to (3) and r’=O holds, we must have max {I.’, p ’ , 0 ‘ ; = K ; as I.’
DISCUSSION OF O R D I N A R Y PARTITIONSYMROL I N CASE
22:
r=2
We now turn to establishing our claim that if (3b(8)are satisfiable, then they are satisfiable with the value of A given in (16).To this end assume that (3k(8)are satisfied by A‘, p’, . . . . We distinguish two cases: 1 ) i. < w and 2) A? o,where I is given by (16). In Case 1 ) we have p = L , ( t i ) = L 3 ( ~ ) 2 L , , ( ~ ) = Putting p’. a=O, a=>., f i = a ’ + / ? ’ ( < p ’ I p ) , T = T ’ , and k , = k : , (3)--(8) are satisfied ((4)-(8) are rather easy to check; the reason why (3) holds is that a=i.=sup {A:: <<9)). In Case 2) we distinguish two subcases: 2a) A‘>). and 2b) I‘
-=
here the supremum of the sequence on the right-hand side must be greater than the supremum of the sequence on the left-hand side, which is A. But >,’ I (note that A ‘ # K as A‘<E.
PROOF OF THEOREM 35.4. (14) is a necessary condition for (15) according to Lemmas 35.2 and 35.3. This settles a). We now turn to the verification of b). Assume therefore that r=O. We shall have to prove that
holds under the various conditions considered.
224
ORDINARY PARTITION RELATION WITH SUPERSCRIPT
2
CH.
VIII
Ad (i). (4)and the regularity of ti implies that p = L I ( ~holds ) in this case. If p < ti, then Corollary 17.2 implies that (1 8 ) holds. Assume therefore that p = K . In this case ti is inaccessible; in fact, it is regular and L l ( t i ) = K . If a=O, then (18)
follows from Corollary 17.3. If a>O, then we cannot have % < p according to (6)(i),i.e., we must have E.=p=h-=cf ( t i ) ; hence p - ( p ) : , i.e., ti-(ic);, holds by (6)(ii).Theequivalence of (iv) and (vi) in Theorem 29.6 then implies that ti+(^):. holds for any ti’< ti, and so (18) holds. Ad (ii).Assume that (18) is false. Then ti is singular according to (i),which was established just before. As I < ti in this case, we have p = L , ( K ) If . there is a regular cardinal ti, with ti such that 2 L L h ) < ~ (i.e., 0 L 3 ( t i O ) = L 3 ( ~and )) LI(tio)=Li(ic),then conditions ( 3 t ( 8 )hold with K , replacing K , which implies according to (i) that (18)holds with ti, in place of ti; but then (18)alsoholds with ti. This contradicts our assumption; no such ti, can therefore exist, and so p=Li(ic)=L3(ti) and
2P=ti
(19)
hold according to Theorem 7.16. We then have cf(p)=cf (ti)
(20)
by Theorem 6.10.d (since the alternative c(i) of this theorem fails with 2P=ti replacing ti?, as 2 P ‘ < holds ~ for any p ’ < p = L 3 ( t i ) by the definition of the logarithm operation). If a= 0, then choose a p , < p such that 21’0 2 i,and po 2 cr, B (this is possible in view of (19) and ( 5 ) ) ; we have (2p0)++((2p0)+,( p J ) , , , ) 2according to Corollary 17.5,and this implies (18)under the assumption of a =0, which is a contradiction. We must therefore have a>0. Thus (6)(i),(19), and (7) imply that
I=p
and
cf (p)-(cf ( p ) ) : ;
(21)
hence cf ( p )is inaccessible according to Theorem 29.1. Thus p cannot be regular; in fact, using (19), (20), and the singularity of K , the regularity of p would entail that cf ( p )= 29’= 2e= K > cf ( K ) = cf ( p ) , which is a contradiction. Thus we have cf ( p )< p ; therefore (8) implies with the aid of (19) and (21) that a s 1; as we saw a > O above, we must have a= 1; then (8) further implies that c r l c f ( p ) and p< (L3(cf ( p ) ) ~cf) ( p ) ; so we have K = P- ( ( P I 2 7
(a)p)2
by Corollary 27.4 plus (19), (20) and (21). As we obtained a= 1 just before, and I = p according to (21), t h s means that (18) holds. This is again a contradiction, settling the case of condition (ii).
DISCUSS~ONOF ORDINARY PARTITION SYMBOL I N CASE
r=2
22 5
Ad (iii). By (i) and (ii) we may assume cf ( K )< K and K = 1.As K is assumed to be strong limit in this case, we have L,(K)= K ; hence p = Lc,(,t(cf(K)) according to (4) in this case. We claim that
cf (K)--r(Cf
(K), (P)a,
(4pl2
(22)
holds. T o see this, it is enough to show according to (i) that (3b(8)are satisfied with ~’=E.’=cf( K ) replacing K and 1. This is obvious for (4) and ( 5 ) since p=Lcr,,,(cf (K)); (6)(i) is vacuous, and (6)(ii) is trivial since cf ( K ’ ) = =cf(cf ( ~ ) ) = c(K), f and so the assertion in (6)(ii) with the new parameters is identical to the one with the old parameters. To check (7) and (8), assume now that 2P=cf ( K ) holds. Then L3(cf(K))Ip= = Lcf(K) cf (K)), and so L , (cf (K))= p, as the reverse inequality also holds by the logarithm’s being a monotonically nonincreasing function of its base (cf. Theorem 7.1.b). p = Ld(K) (cf(K))is regular according to Theorem 7.15. We have cf(K)=cf(2P)=cf(p) in view of Theorem 6.10.d (as in the case of (20) above, the alternative c(i) in this theorem must fail since p = L3(2P) holds). Thus K’
=
z = Cf(K) =cf(p) = p ,
and so (7)(i)holds vacuously. As for (7)(ii),we have to show that in case a>O we have cf(p)-+(cf(p))i,which is the same as p - ( p ) : , as p is regular. This is indeed true, since in case p + ( p ) ; condition (6)(ii)involving the original parameters K, I, etc. implies a=O. Finally, (8) holds vacuously because p=cf(p). Thus (22) is established. Using the assumption that K is strong limit, (22) implies (18) via Corollary 27.5. Ad (iv). a, 8, and CT are finite in this case, hence the result follows from the Erdos-Dushnik-Miller theorem (Theorem 11.1; see (11.2) in particular). This completes the proof of the theorem. REMARKS. Lemma 35.2is very likely not the best possible, and (v)and (vii)there can probably be replaced with (v’) and (vii’)even if GCH is not assumed. (This leads to Problem 20.1 stated above.) If this is indeed the case, then we can take z =0 in Lemma 35.3; this would mean that z =0 is also necessary for (1)=(15 ) to hold. The other question left open by Theorem 35.4 is what happens if K = i > c f (K), K is not a strong limit cardinal, and p > w. We formulate the simplest instance of this question.
PROBLEM 35.5. Assume hold? 15 Combinn!orial
= K,
and Np
Kw2 (Km2K,IZ -+
9
226
ORDINARY PARTITION RELATION WITH SUPERSCRIPT
2
CH.
VIIl
Of course we have a positive answer here by Theorem 35.4.b(iii) unless
2K; >Nu,for some 5 <02. We shall state this problem in a more general form and
shall take a closer look at it in Section 37 (see Problem 37.1 below). If we assume GCH, then our discussion is complete. We have
COROLLARY 35.6. Assume GCH and ( 2 ) . Then (1) holds ifand only if there are cardinals satisfying conditions ( 3 t ( 8 ) of Lemma 35.3 with t = O . PROOF. Lemmas 35.2, 35.3 (especially (9) there), and GCH entail that (1) implies the existence of cardinals satisfying ( 3 t ( 8 ) with t = O ; this settles the “only if” part. T o see the “if’ part, one only has to observe that if GCH holds, then K is either regular or strong limit, and so the result follows from Theorem 35.4.b(i) and (iii). The proof is complete. There is a more direct way of formulating this corollary. This will be shown in the next section.
36. DISCUSSION OF THE ORDINARY PARTITION RELATION IN CASE r = 2 UNDER THE ASSUMPTION OF GCH The main result of the preceding section can be formulated in a relatively simple way if we assume GCH. To this end, for any cardinal K put cr (K)=cf (cf ( K ) - ) , which is called the critical number of
(1)
K.
LEMMA 36.1. If GCH holds, then
PROOF. We distinguish two cases: a ) K is a successor cardinal, and b) K is a limit cardinal. In case a), K - < K and cf ( K ) = K . Hence Cr ( K ) = C f (K-)=L,(K) ; the second equality here follows from GCH and Konig’s theorem (cf. Corollary 6.8 in particular). In case b), here
L , ( K ) = K ~ (K)1LC,(,,(Cf C~ (K)) ; L,,(,,(cf ( ~ ) ) = c( f~ ) = c (rK )
ORDINARY PARTInON RELATION IN CASE r = 2 UNDER GCH
227
ifcf ( K ) is a limit cardinal, and, by Konig’s theorem we obtain in the same way as above that f!.cf(K,(Cf
if cf
(K)
(Cf ( K ) - ) = C C
(K))’Cf
(K)
is a successor cardinal. The proof is complete.
We shall also need the following observation, easily checked by the reader: Cr(K)=Cf(K)
iff
We introduce two more operations: for any infinite cardinal SUC ( K ) =
(3)
Cf(K)-=Cf(K).
if 0 if 1
K
we put
K-
where ‘suc’ is the first syllable of the word ‘successor’, and com
(K) =
1 if
Cf (K)+(Cf
(K)):
Of
Cf ( K ) - < C f ( K )
0 if
Cf (K)+(Cf
(K)):
and
Cf ( K ) - = C f ( K f ;
(5 1
here the syllable ‘com’ is used on the pretext that the cardinals with K + ( K ) $ are exactly those which are inaccessible and weakly compact (seeTheorem 30.3).We can now state the main result of the present section (recall the meaning of the relation e, given in Definition 35.1):
THEOREM 36.2. (Erd6s-Hajnal-Rado [1965]) Let K, 9, and 1, (5<9) be cardinals such that K > o;9 2 2, and 3 5 At IK for any 5 < 9. Assume GCH. Then (6)
K+ (4)f<9
holds $and only if the sequence or (8):
(A? : 5 < 9) i s 6 one of the sequences satisfying (7)
( K , ( ~ ~ ( K ) ) , , ( B )where ~ ) , a , / ? < c r ( ~ ) and ( ( K - ) , , ( o ) @ ) , where
a,/?
and
Here a, /?, and o are cardinals, and
a < c r ( ~ ) . c o r n ( ~ ) + l , (7)
a
(8 1
+ denotes cardinal addition.
PROOF.We assume GCH throughout the proof without necessarily mentioning this explicitly. According t o Corollary 35.6, it is enough to show that a) any sequence satisfying (7) or (8) is dominated by (i.e., is < ) some-sequence (A, ( P ) ~(a)8) , satisfying (35.4)-(35.8), and b) any sequence (1,( P ) ~(o)#) , satisfying (35.4)-(35.8) is dominated by some sequence satisfying (7) or (8). Ad a). We deal separately with sequences satisfying (7) and (8). First we show:
228
ORDINARY PARTITION RELATION WITH SUPERSCRIPT
Assume ( K , (cr ( K ) ) , , (a)B) satisfies (7). Then it satisfies (35.4k(35.8) with % = K and p=cr
2
CH.
VIII
(K).
To see this, observe that (35.4) holds according to Lemma 36.1. cr ( K ) is regular, and so (35.5)holds by (7). (35.6)(i)is vacuous. (35.6)(ii)can be verified as follows: s u p p o s e c f ( ~ ) = p ( = c r and p+(p)i.Thencf ( ~ ) - = c f ( ~ ) a c c o r d i n g to (3); as cf (K)fi(cf( K ) ) : also holds by our assumptions, we have com ( K ) = O in view of (5). Hence a=O according to (7), which establishes (35.6)(ii). If P = K , then K=p(=Cr ( K ) ) by GCH, and so K=cf (K)=cf ( K ) - by (3). Thus (35.7)(ii) holds; in fact, if cf (p)f,fcf(p))i, then com ( K ) = O by ( S ) , and so a=O by (7). (35.7)(i) and (35.8) are vacuous since we saw that p = c f ( p ) = ~under the assumption 2e= K. This establishes (9). Next we prove: Any sequence satisfying (8) is dominated by the sequence (A, (p),, (a)a) satisfying (35.4)-(35.8) with p = where (i) R = K - if K - < K and (ii) A= 3 if K - = K .
K-,
(10)
We only have to show that (A, ( p ) , , (a)#)satisfies (35.4)-(35.8). We consider the cases (i) and (ii) separately. Ad (i). A= K - < K , and L 1 ( ~ ) K= - = p by GCH, and so (35.4) follows. (35.5) follows from (8). (35.6) is vacuous. 2e= p = K - < K holds by GCH, and so (35.7) and (35.8) are also vacuous. Ad ( i i ) . R = 3 a n d L 1 ( ~ ) = L 3 ( ~ ) = ~ = ~ - = p b y G C H , a n(35.4)istrue.B~ dso (8), we can see that (35.5) holds and a=O, as suc ( K ) = O in this case. Thus (35.6), (35.7),and (35.8)hold as well. (10)is established, and so part a ) of the proof of the theorem is complete. Ad b). Let (4(p),, (a)#) be a sequence satisfying (35.4b(35.8). We have to show that it is dominated by a sequence satisfying (7) or (8). We distinguish two cases: I ) ; I = K and 2) A < K . Ad 1). We prove that in this case
(A, (p),, (a)@) is a sequence satisfying (7). (11) In fact, as A= K in t h s case, we have p = min (L,(K),&,(,,(cf (K)): by (35.4), i.e., p=cr (K) according to Lemma 36.1. (35.5) implies a,B, a < c r (K). Assume
com ( K ) = O . Then cf (K)=cr ( K ) = P and p f r ( p ) i by (5) and (3); hence a=O in view of (356)(ii). Thus we always have a < c r ( K ) . com ( K ) + 1, which settles (11). Ad 2). We distinguish two cases: a ) 1< K - and 8) 1= K - . First we show that In case 2x1, (A, (PI=,(a)s) is dominated by ( ( K - ) ~ , ( ~ + c ) ~ + ~ ) , (12) and this latter sequence satisfies (8).
229
SIERPINSKIPARTITIONS
In fact, by (35.4) and GCH we have P = L ~ ( K ) = K as - ; E.
(2.9
= p i n this case,
( i + ~ ) p + I )
Now (35.5) implies I +u, P + 1 < K - , establishing (12). It remains to show that In case 2P), ( I , (pIa, (a)p) is dominated by and this latter sequence satisfies (8).
( ( K - ) ~ + , ,(a),&,
(13)
As I. < K in case 2) and I = K - in case 28), we have K - < K. Hence suc ( K ) = 1. ~ = L * ( K ) = Kholds by (35.4) and GCH. Thus (%, (p),, ( t ~ ) ~ ) 4 ( ( (a)p). ~'-)~+~, (35.5)now implies a, P < ~ = K - and a
37. SIERPIfiSKI PARTITIONS One of the main problems that remained open in the above discyssion of the ordinary partition relation was the following
PROBLEM 37.1. Let
K
be a singular cardinal and E.>w a cardinal such that Cf (K)+(Cf
holds and
(K),
(1)
pa
holds for any p K K. Does then K+ (K,
n)'
(3)
hold? (We stated this as Problem 35.5 above in the particular case when K = K , ~ , I = K,, 2 K o = K,,and Kyo < Kw, for every < w2.) One hasevery reasonto believe that the answer to this question is yes, though we have no idea how one could prove this. We can, however, show that one cannot prove for any choice of K and I that the answer is negative by using Sierpinski partitions only. We described Sierpinski partitions in Definition 19.5, but we give the definition here again, in a slightly modified form:
DEFINITION 37.2. Let
K,
p, and u be infinite cardinals. We say that
(in words: K S-arrows . . .) if any ordered set of cardinality K includes a set having
230
ORDINARY PARTITION RELATION WITH SUPERSCRIPT
2
CH.
VIII
either order type p or order type u*. The negation of this relation is written as S
K + ( ~ , o )in~ case ; this latter relation holds we may say that the relation ~ + ( p ,a)2is established by a Sierpinski partition (this agrees with the terminology
introduced in Definition 19.5).
S
Lemma 19.4 above says that the relation ~ f r ( pa)2 , implies ~ f r ( p , It is worth noting that, in each case above, when we proved a relation of the latter type, then we actually proved the stronger, i.e., the former relation instead. The main result of this section says that in Problem 37.1 at least K +S( K , A)2 holds; hence, even if (3) might fail in certain cases, an entirely new idea is needed to prove this.
THEOREM 37.3. Let
K
be a singular cardinal and Cf (K):((Cf
A)2
(K),
Azo a cardinal such that (4 1
and p’
for any p < K . Then S K+(K,
A)2
holds.
PROOF. Write
K
in the form K =
1
K,
a
where K, < K. Let (X, <) be an ordered set of cardinality K. We have to prove that X includes a set Y having either order type K or order type A*. Assuming that the second alternative fails, we shall prove the following CLAIM. There are pairwise disjoint sets X,C X , ci
Note that K,” < K according to ( 5 ) ; instead of (8) we might have required only IX,I > K,, but there is a technical advantage in requiring the former. It is easy to see that the above claim implies that X contains a subset of order type K, establishing (6). In fact, we have
231
SERPINSKI PARTITIONS
according to Corollary 17.5 (we have even more, but this is all we need); hence, a fortiori, ( K 3 + %K,,
holds by Lemma 19.4 (cf. the remark after Definition 37.2). Applying this to the set ( X u , < ), we can see by (8) that X u includes a set Y, having order type either K, or I*.The latter alternative fails according to our assumption; hence Y, must have order type K,. Then Y, has order type K by (7). This verifies (6).
-
u
u
We have yet to prove the above Claim. To this end, define the equivalence relation on X as follows: for any two x , y E X with x - l y put: x-yo({zE
x :X - l Z l Y ) l < K .
We distinguish two cases: a ) X consists of a single equivalence class, or b) X has at least two equivalence classes. Ad a). In this case, we have ) ( zE x : x - j z - j y ) ) < K
(9)
for any x, y E X with x < y . Take a set Y E X that is wellordered in > and cofinal in ( X , >) (i.e., Y is cofinal in X downward). 1 Y 1 < I by our assumption; and so I Y I
u
I{ZEX:Z
[{ZE
Y:X
xx
Using (9) again, we can see that l{z E
x :Z i X ) l < K
holds for any x E X . As I X = K , it is easy to see from this that there is a sequence ( x , : a
(a
has cardinality > ~f (note that K: < K by ( 5 ) ) . This establishes our Claim in case a). Ad b). If there is an equivalence class X’ of cardinality K of X , then replacing X with X’, our claim follows in exactly the same way as in case a); hence we may assume that each equivalence class of X has cardinality < K. As X is the union of its equivalence classes, it follows that there are at least cf (K) equivalence classes in X. Let z, E X , a
232
ORDINARY PARTITION RELAllON WITH SUPERSCRIPT
c f ( ~ )say , (x, : a
(K)).
2 CH. VIII
Then the sets
x,= ( z E x:X a j Z 4 X &
,}
satisfy the requirements of our Claim; in fact, as x, and x,il are not equivalent, we have I X,I = K > K,”. This proves the Claim also in case b), completing the proof of the theorem. S
The discussion of the Sierpinski partition relation ~ - + ( p a , )’ presents no new problems. Namely, Theorem 35.4 is valid for this relation, as rc-(p, a)’ implies S
~ - + ( p a, )’ by Lemma 19.4, as we pointed out this in the remark after Definition 37.2, and, as mentioned in the same paragraph, whenever we proved ~ f , ( p ,a)’, S
we actually always proved the stronger relation ~ + ( p a , )’. As far as Sierpinski partition relations are concerned, Theorem 37.3 fills the only gap in Theorem 35.4; so, putting these two results together, one can give a complete discussion of the Sierpiriski partition relation. We leave the details to the reader.
CHAPTER I X
DISCUSSION OF THE ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3
In the next five sections we collect the results known to us concerning the ordinary partition relation for the case r 2 3. Our discussion is based on the discussion of the case r = 2 given in the preceding chapter, the Stepping-up Lemma, the special negative relations obtained in Section 25, the addition results for negative relations in Section 22, and on the Negative Stepping-up Lemma in Section 24. The main difficulty is that these two latter results are applicable only if certain inconvenient conditions are satisfied, and the main concern of Sections 38 and 39 belsw is to analyse these conditions so that we obtain results that can be efficiently used to derive a partition relation from an analogous one with a smaller superscript, and Sections 4 3 4 2 pick the fruits of these preparations. Even more than in the case of the first two sections of the preceding chapter, one might expect little enjoymeni from reading these chapters, and so the reader will perhaps do well by consulting only the main results and problems obtained, and skipping the proofs of the former. As in the case of the preceding chapter, we feel it necessary to point out here that the importance of this chapter is, apart from giving a practically usable test to decide whether an ordinary partition relation with r 2 3 holds or fails whenever this is known to us, that it locates the most important problems left open by our study. As interesting by-products of our discussion we obtain that there are no genuinely new problems for the case r 2 4 , i.e., all gaps in our knowledge in this case issue from gaps in the cases r = 2 or 3; and, moreover, that the only problem remaining open with GCH is of the type given in Problem 25.8, which concerns the case r = 3.
38. REDUCTION OF THE SUPERSCRIPT The aim of tlus section is to show that the question whether or not the relation K+
holds, where we shall usually assume that
<,9
(11
234
ORDINARY PARTITION RELATION WITH SUPERSCRIFT 2 3 CH.
Ix
can be reduced to an analogous question involving a smaller r under fairly general circumstances. To this end, define the iterated logarithm L : ( K ) by recursion on i < o as follows:
L ~ ( K )K= and
L';+'(K)=L~(L(;(K)).
(3)
(cf. Theorems 7.lb and 7.2). Moreover, one can easily check that L\(K)=min { p : exp, ( p ) > ~ j
(5)
holds for any integer i, where the iterated exponential function is defined by expo ( K ) = K
and
exp, +
( K ) = exp,
(2")
(6)
for any i <w. We shall also need the following two concepts.
DEFINITION 38.1. Let r 2 3 be an integer and consider the sequence X = ( l , : <<9), where l t 2 r + 1. a ) We say that ns(r, X) holds if for any cardinal p 2 0 , the relation implies (Note that Lemma 24.1 says that ns(r, 1)often holds. The name Negative Stepping-up Lemma given to this lemma explains the meaning of the letters 'ns'.) b) We say that ad (r, holds if for every sequence ( p u : a < q ) of infinite cardinals, the relations pu+(Ac&<s for all a < q
x)
and
im PlY
Itl I
c
a
ecn,>;<
Pu%
(&K
0
.
(Theorem 22.1 says that ad (r, 1)often holds; the letters 'ad' are justified by the fact that this theorem was obtained by addition of negative relations.) We now prove a simple lemma that gives a sufficient condition for (1) to hold. It is a direct consequence of the Stepping-up Lemma (Lemma 16.1),but it will be useful to state it for further reference. The assumptions in (2) are not yet used here; in particular, the lemma is valid for r = 2, and we shall exploit this fact below.
REDUCTION OF THE SUPERSCRIPT
235
LEMMA 38.2. Let i be an integer with O
(A: - i):;;
Then (1) holds.
PROOF. Writing po = p and pi= (expj- (2P))' for any integer ,j >0, we claim that (7) implies -+(Az+ + j - i r5+<4!. j - i Pj
for any integer .j. In fact, this directly follows from the Stepping-up Lemma (Lemma 16.1) by induction onj; for the induction step one only has to observe that p,+ =(23)+ holds for every j. As p i s IC by our assumptions, the above relation with j = i gives (1). The proof is complete. Here is a partial converse to the lemma just proved: LEMMA 38.3. Let i be an integer with 0 < i Ir - 2 and suppose that cf ( L ~ ( K<) )
+
+
furthermore, that ad (r - i 1, ( A , - i 1 : t < 9)) and ns ( j , hold for any .j with r - i < , j S r (see Definition 38.1 above). I f P%@:
-9&i
(A2 -r,+,j: 5
holds whenever expi- (2e) < K, then (1 ) fails.
PROOF. We distinguish two cases: a ) 2L&)
In case a).we clearly have expi - (2W))
(9)
holdsforanyjwithr-ii,jSr. In fact, f o r j = r - i thisisjust (8)with ~ = L ' ; ( K ) , andifrLi<jsrand(9)holdswithj- 1 replacingj,thenns 6, (A{-r+j: ( < 8 ) ) implies that (9) itself holds with j (note that we have 2L;-'+1(K)>L>-j(~) by the definition of the logarithm). This verifies (9). Now (9) withj=r says that K = G(K)%(A,K<,
holds; hence (1) indeed fails. In case b) we claim that holds. T o see this, observe first that
236
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3 CH.
Ix
holds for any ~ < L $ ( Kaccording ) to (8). In fact, we have exp, - (2e)I exp, ( p )< K for any suc hp by (5). ( 1 l ) a n d ns(r-i+l,<~~-i+l:~~<))implythat 2p+(AS-i+
1)::r'
(12)
holds for any p with w < p < L ' ; ( ~ ) We . can substitute p = min { L F ' ( K ) , cf (L';(K)))
here, because t h s p is < L $ ( K ) .In fact, assuming ~ > L ' ; ( K we ) , obtain in particular that L';t ' ( K ) = L ' ; ( Kholds; ) this, in conjunction with the equality 2%5)= L\-'(K), which also holds in the case considered, implies that
Li(K)
zL%)=2 L ; 3 )
i.e., L';'(K)=L;(K).Now we have cf (L\(K))
l)&F',
(13)
sinceclearly 2P2cf (L';(K)) holds with this p. Noting that L';(K)>the above value of p, which is z w , let now ( p a : a
As
holds in case b), which is the case now being considered, t h s means that (10)does indeed hold. Using ns (j, (At-r+j: (t9)) for each j with r - i + l S j < r , we obtain from (10) by induction on j that L;-j(K)+(J-t-r+j&s
holds for any such j (note that 2L;-'"(")2 L . - ~ ( K )Withj ). = r this just means that K = L ~ ( K ) J , ( A ; E ,which ~, we wanted to show. The proof of the lemma is complete.
237
RWUCTION OF THE SUPERSCRIPT
We next turn to thecase L';'(~)=cf (L\-'(K))=L ~ ( Kwhich ) , was left open by the lemma just proved. This case is relatively easy to handle, as L';(K) = Li- ( K ) is an inaccessible cardinal in this case.
'
LEMMA38.4. Let i be an integer with O < i ~ r - 2 , and assume that
cf (L';(K))= L\-'(K). Assume, further, that (2) is satisfied. Then the following condition is suficient for (1) to hold: either there is a p < L'; '( K ) such that (14)
p-+(At-i&;i, or we have
L'; (K ) +(L;- (K ) ) ; and
~ < L \ - ' ( K ) and A,sL';'(K)
for any
5<9.
(16)
This condition is also necessary for (1) provided ns (j,
PROOF. Sufjiciency. L';'(K) is inaccessible in view of our asiumptions.
Therefore we have expi-l ( 2 p ) < L \ - 1 ( ~for ) ~ any ~ ~
'
9 2 L'; (K ) or
35 < $[A, 2 L ~ ; ' ( K ) ]
holds. Infact,in thecontrarycasethereis a a < L ~ - ' ( ~ ) s u c h t h a91aand t for any <<9. Then
(17)
Acso
+ --* (Q;
holds. Using Lemma 38.2 with r - i and r -i - 1 replacing rand i, respectively,we obtain that (exp,-,
-
(a))'
--*(A,+ r - i - 1fi:;
holds. Putting p=(exp,-,-,(a))+, which is
L';'(K)+(A, - i+ l ) & y
(18)
238
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3 CH.
Ix
holds. In fact, if (16) fails, then this is obvious. Assume therefore that (16) holds and (15) fails. The failure of ( 1 5 ) entails that L i - ' ( K ) f , (Lip'( K ) , 4)3 by the implication (iv)*(v) in Theorem 29.6, since L\-'(K) is inaccessible (actually, inaccessibility is not needed for this implication, as remarked at the beginning of the proof of Theorem 29.6). This relation, however, implies (18). In fact, Az - i + 1 = 2: 2 L i p ( K ) for some t < 9 according to the second relation in (17) (this must hold, as the first one fails in view of (16));we have r - i + 1 2 3 by virtue of our assumptions on i; and, moreover, 9 2 2 and At, - i 1 2 ( r + 1) -i + 1 2 4 holds for any 5' < 9 according to (2). We claim that L;-j(K)%(&+.i-r&s (19)
'
+
holdsforany jwith r - i + l s j s r . I n f a c t , f o r j = r - i + l thisisjust (18),and if (19)bolds withj-1 replacing j , where r - i + 2 < , j s r , thenit also holds withlin W i t h j = r (19) just gives view of ns 0, ( A t + j - r : (<9)), as 2L5-J+'(K)2L;-i(~). K + ( A & < ~ , which we wanted to show. The proof is complete. The next theorem sums up the results obtained.
THEOREM 38.5 (Reduction Theorem). Let r I3, ~2 w, 9 2 2, and E.: > r for every 5 < 9. Let i be an integer with 0 < i s r - 2. Consider the relation K+
(A&
<&
and the following conditions : (i) Either a ) there is an infinite cardinal p with expi-l ( 2 e ) < ~and
) any t<9, and o r b ) we have ~ < L ' ; ' ( K ) , A e 5 L i - ' ( ~for
(ii) For every sequence ( p . : a < q) of infinite cardinals, the relations
(20)
239
REDUCTION OF THE SUPERSCRIPT
(iii) For every integer j with r - i <,I 5 r and every infinite cardinal p , the relation implies (i) is a sufficient conditionfor (20)to hold. Zf(ii) and (iii) are satisfied, then (i) is also a necessary condition for (20). A detailed discussion when (ii) and (iii) are satisfied is given in the next section. We note that the requirement in (i) on p being infinite is unimportqnt, and was included only in order to stress that we are considering partitions' on infinite cardinals. In fact, if (21) is satisfied with a finite p , then it is also satisfied with p = o ; the only trouble might be that expiPl( 2 9 2 ~ in ; this case, however, p; ' ( K ) = oby ( 5 ) , and so (22) holds in view of Ramsey's theorem (Theorem 10.2), which means that (i) is satisfied.
PROOF. Suflciency. If (i)a) holds, then Lemma 38.2 implies (20), and if (i)b) holds, then the 'if part of Lemma 38.4 implies (20); in fact, the Assumption ) inaccessible by (22) and cf ( L ' ( K ) ) =L ' - ' ( K )of that lemma holds since P - ' ( K is Theorem 29.1, and, moreover, (15) and (16) hold in view of the assumptions in (i)b). Necessity. Assume that (20) holds and (i)b)is false, and, moreover, that (ii) and (iii) are satisfied. If L\-*(K) is not inaccessible, then (i)a) holds according to Lemma 38.3, and if L'; ' ( K ) is inaccessible, then it holds according to Lemma 38.4. In fact, to see the latter assertion, the conjunction of (15) and (16) in this lemma fails as (i)b) was assumed to be false; hence (14) must hold with some p < L ~ - ' ( K )but ; then expi- (2P)< L';'(K) I K , and so (i)a) indeed holds. The proof is complete. It might be interesting to Formulate the above result in the case i = 1:
COROLLARY 38.6. Let r l 3 , 1 ~ 2 0922, , and A,>r for every (<S. Consider the relation (23)
K--,(&S
and the following conditions: (i) Either there is an infinite cardinal p with 2P<
OT
we have 9 < K and I t I K for every 5
-=9, and
K
and
240
0RI)INARY PARTITION RELATION WITH SUPERSCRIFT 2 3 CH.
Ix
(ii) For every sequence ( p a : u < q ) of infinite cardinals, the relations and imply
c
Pa%(&):<&.
a<4
(iii) For every infinite cardinal p, the relation p% (i< - 11;:
implies 2p%(I:):.
(i) is a mflcient conditionfor ( 2 3 )to hold. If&) and (iii) are satisfied, then (i) is also a necessary condition for (23). As we already mentioned in Definition 38.1, conditions (ii) and (iii) in this corollary as well as in the preceding theorem are often satisfied. In fact, it follows from Theorem 22.1 and Lemma 24.1 that it is enough to assumee.g. that r - i r 4 in Theorem 38.5, i.e., r r 5 in Corollary 38.6. For the remaining cases, some special assumptions are needed (e.g. cf (I,)= I, 2 o and I, 2 o;cf. the results of the next section). For these cases, Theorem 38.5 does not contain all our results, since some gaps can be filled by using the special results obtained in Section 25. The filling of these gaps will be one of the aims of the next few sections. There remain however some problems which we are unable to settle. In case GCH is assumed, it will turn out that only one typical cask cannot be covered by our results, and this case leads to Problem 25.8 stated above.
39. APPLICABILITY OF THE REDUCTION THEOREM
Our first aim now is to study the applicability of Theorem 38.5, i.e., to consider when conditions (ii)and (iii)of that theorem are satisfied.To this end we need the following simple lemma :
LEMMA39.1. For any cardinal
p 2 w and any integer r 2 1 we have
expr- 1 (d%(r+ 1& *
(1)
h m ~ . T h e c a s e r =l i s trivial,andifr=2, thentheresultisgivenin (19.17).Let r = 3. Then relation (24.5) in Lemma 24.1 with K = 2p, t = p and Ir= 3 implies
APPLICABILITY OF THE REDUCTION THEOREM
241
(here we used the fact that ( 1 ) holds with r=2); as p 2 0 , this is just ( 1 ) with r = 3. Let now r 2 4, and assume that ( 1 ) holds with r - 1 replacing r. Then, using (24.3) in Lemma 24.1 with K = exp, - (p), z =p , and 1, = 3, we obtain that expr - 1 ( ~ 1 %( ( I + 1 ) p , (r + 1 ) m y
9
where m is finite; as p 2 0 , ( 1 ) follows. The proof is complete. The next lemma achieves a partial fulfilment of our first aim mentioned above:
LEMMA39.2. Let r 2 3 , K ~ W9 2, 2 , L c 2 r for every (<9, and let i be an integer with 0 < i Ir - 2. Condition (ii) of Theorem 38.5 holds except if i = r -2 and 9<w, 1,2w for some p < 9 , and (21 2: .CIc" for any 5 < 9 diferent from p, PROOF. Assume that condition (ii) of Theorem 38.5 fails, i.e., that there is a sequence ( p a : u
(3 )
lqI+(&-i+1E:F1,
(4)
for all u < q, and and yet
1 p.-+(I,-i+
1)f;F'.
Q-=q
By t h s last relation and the preceding lemma we certainly have
here. We first show that, as claimed in (2), there is a p<9 with L,zw. By (6), either there is an o! < q with exp, - i (I 91) < p a or we have exp, -i(l 91) < I q 1. Hence, if 9 <w then by Ramsey's theorem (Theorem 10.2), and if 9 2 0 then by (17.32),we obtainthat eitherp,-+(w);-'+'or I ~ ( - + ( w & - ~ + ' holds.Sotheremust indeedbea p < 9 with L,,?w, since otherwise these relations would contradict ( 3 )or (4).We can therefore assume that e.g. l o20 . (7) Using this inequality, (22.5)in Theorem 22.1 implies that (3),(4), and (5)can hold simultaneously (i.e., (ii) of Theorem 38.5 can fail) only if 1,<0for any t with 1 1 ~ < 9 a n d r - i + + = 3 , i . e . , i = r - 2 . Wehave yet toprovethat 9<w.But (3) and ( 4 )with i=r-2 plus ( 7 ) imply via (22.6) of Theorem 22.1 that
16 Combinatoriel
242
ORDINARY PARTITION RELATION WITH SUPERSCRIPT t3 CH.
Ix
holds, and in case 9 2 0 this contradicts ( 5 ) , since we then clearly have
((A: -r
+ 3)f,s,
+
-=
4 ) 6 ( A z - r 3: 5 9),
The proof is complete. The following lemma gives the analogous result for condition (iii): LEMMA 39.3. Let r 2 3, I C w,~ 9 2 2 , As > r for every 5 < 9, and let i be an integer with 0 < i I r - 2. Condition (iii) of Theorem 38.5 holds except y either i = r - 3 and 9<w, o s c f (&)
or i=r-2 and we have (*)
((;is -r
for some p ~ 9 , 5<9 differentfrom
p,
+ 3),.,9, (As - r + 3): <s, 4 4 ) # ( A c -r + 3: 5 < 9);
(9) moreover, l c 2 w for some 5 <w; and either all l e 2 w are singular, or there is exactly one p < 9 with 1,2 w, and if this 1, is regular then 9 <w. PROOF. Let p 2 o, j > r - i, and assume
We shall prove that then
p+ ( A f - r + j
- 1%;;.
2P%(Af-r+j&s
holds provided that the exceptions in the lemma fail (the additional assumption
j l r in (iii) of Theorem 38.5 will not be used). Assume that, on the contrary, we
have
2p+(Is -r +j%<,.
(11)
By Lemma 39.1, t h s implies that expj-,(191)<2P, i.e., that
in fact, if As < w for all 5 < 9, then (10) fails in view of the relation P+
(Wg;;
9
which holds by Ramsey’s theorem (Theorem 10.2) in case 9 < o,and by (12) and (17.32) in case 920.
CONSEQUENCE5 OF THE REDUCTION THEOREM
243
I f j 2 5 , then (10) and (13) imply via (24.2) with assumption d ) in Lemma 24.1 that (11) is false. Hence we must have j r 4 . This means that r-i equals 2 or 3, since 2 5 r - i <j . We consider two cases: 1) r - i= 3 and 2) r -i = 2. Case 1). We shall prove that (8) holds in this case. Instead of assuming r - i = 3 we shall only make the weaker assumption j = 4. If A,, which is infinite according to (13), is regular, then (10) implies that (11) fails in view of (24.2) with assumption b) in Lemma 24.1;this is a contradiction, and so A, must be singular. If there was a with 1 s ( < 9 such that A,zw, then we would obtain a contradiction again, as (24.2) with assumption c) in Lemma 24.1 entails that (10) and (11) cannot simultaneously hold. We have yet to prove that 9*<w holds. Now (10) implies via (24.3) in Lemma 24.1 that
<
2’fr ((At- r +.ik <sr
O’+ 1)mY
holds, where rn is finite. If 9 z w then this contradicts ( l l ) , since clearly ((1,--I + j ) <
<sy
O’+ 1)m> 6 (1c-r
+j : t < 9)
holds in this case. This shows that (8)is valid in casej =4, and so a fortiori in case i=r-3. Case 2). We have to show that (9) holds in this case. As r -i = 2 now, we have j = 3 or 4. We saw above that (8) is valid in case j = 4. Since (8)clearly implies (9), we may assume that j = 3 holds. (10) implies
2’%((~<--+33)5<s, (+-r+3),<s,4,4)3 via (24.5) in Lemma 24.1; this contradicts (11)unless inequality (*) in (9) holds. If there are two distinct 5, t‘< 9 such that I,, Le 2 0 and A< is regular, then (10) and (11)again contradict each other in view of (24.2) with assumption a) in Lemma 24.1. We have yet to prove that 9 < w holds in case Lo is regular (cf. (13)). Now (10) implies 2 P f r ( ( 4- r + 31, <s, 413 in this case in view of (24.4) in Lemma 24.1. In case 9 2 w this contradicts (1l), since clearly ((1, - r 3),,,, 4) d (A, - r 3 : 5 < 9)
+
+
holds then. The proof of the lemma is complete.
40. CONSEQUENCES OF THE REDUCTION THEOREM The lemmas of the preceding section supply us with an explicit way ofchecking whether Theorem 38.5 is applicable, i.e., whether its conditions (ii) and (iii) are satisfied. With this tool in our possession, we can proceed to establish the
244
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 5 3 CH.
Ix
analogue of Lemma 35.2. Throughout our discussion we shall assume that KZW,
r 2 3 , 9 2 2 , and I r > r forevery < < 8 .
(1 1
First we establish a simple corollary of the results of the preceding sections that will be needed in the next two lemmas. COROLLARY 40.1. Assume (1). Let i be un integer with O s i ~ r - 2 and , suppose that L;(K)+(A<-i)&;. (2) Then (3 ) K f ,(1:): < 9 holds except if i = r - 3 and (39.8) holds or i = r - 2 and (39.9) holds.
PROOF. If i = O , then (2) and (3) are identical; so we may assume that i > O . Lemmas 39.2 and 39.3 imply that (ii) and (iii) in Theorem 38.5 hold (to see this, observe that (39.2) implies (39.9)). (i) a ) in Theorem 38.5 fails in view of (2), since expi- (2e) < K there implies p~ L’;(K) by (38.5).So does (i) b) in that theorem, as (38.22)implies L‘;(K)=Li-‘ (K) via Theorem 29.1,and so (38.22)contradicts (2). Hence this theoreni implies (3), which we wanted to prove. Using ths, we can get an orientation as to when we can expect a positive arrow relation : LEMMA 40.2. Assume (1). If K
+
(A&
3
then I , ~ L ; - ~ ( ~ ) j all i o <<9. r
PROOF.It is enough to show that
( ~ ) Using the preceding in view of Corollary 25.2, as 2 L ; - ’ ( K ) 2 L ; - 3 holds. corollary with i = r - 3 , we can see that (5) implies (4), since the sequence ( ( L > - ’ ( K ) ) +r,+ 1) does not satisfy (39.8). This completes the proof.
Next we give an analogue of Lemma 35.2, which will be derived from that lemma with the aid of Corollary 40.1. This will not cover all the cases in which we know that the relation rc-+(A,)~ holds; the remaining cases, which involve some special assumptions about the iterated logarithms of K, will be dealt with later, at the end of this section.
CONSEQUENCES OF THE REDUCTION THEOREM
LEMMA 40.3. Assume (1). Write ic0 = L;-’(K). The relation K%
(A,): < Y
As in Lemma 35.2, (vii) can be strengthened if GCH holds, though we cannot quite get here the analogue of (vii’)in that lemma, mentioned immediately after (vii). This strengthening will be given in Lemma 40.8 below. We cannot strengthen (v) even under the assumption of GCH. The closest we can get to (v’) (mentioned after (vii)) in Lemma 35.2 is (iv). (ix) c in Lemma 35.2 has no exact analogue; (ix) c* and c** here cover somewhat less ground. A typical problem that remains open is the following:
PROBLEM 40.4. Assume 2%> K,, and 22%1>22”“ for every a
246
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3 CH.
Ix
As S. Shelah pointed out, the answer to a similar problem with K, replacing K,, easily follows from his result Corollary 27.4.That is, we have the following: if 2% > K, and 2 2 5 > 22N”for every n
2 2 5 -+(K,, K,)3
(8)
holds. In fact, we have 2%+(K,, KJ2 according to the result of Shelah just mentioned, and so ( 2 2 ! ) ++(Ha,K 3 3 follows by Corollary 38.6. As we have
here by Konig’s theorem (cf. Corollary 6.2), the above relation implies (8). The essential difference between (7) and (8) is that cf(L3(2*5))=Ko and and while Ko-+(Ko),2 holds, we have K,+(K,): (cf. Ramsey’s cf (L:(22N_’)=K1, theorem and Theorem 29.1). Note that we have and by (ix) b in Lemma 40.3 and 2251
%R,~, K,~, K ~ ) ~
by (ix) c* in that lemma under the given conditions, while (ix) c** is not applicable here.
PROOF OF LEMMA 40.3. We usually proceed by combining Lemma 35.2 with Corollary 40.1, although occasionally we shall use other results as well. That is, where the Negative Stepping-up Lemma (or, rather, its consequence as formulated in Corollary 40.1) does not suffice, we shall use what might be called the Pseudo Negative Stepping-up Lemma, i.e., Theorem 25.6. Ad (i)-(v) and (vii). We have to prove the relations (i’) (ii’)
K%(&, (L,(KO)+)I if no>L,o(KO); K%(Ko7
(min { L K o ( K O ) 9
Ld(Ko)
(cf ( K o ) ) > > ’ r
(iii’)
K % ( K (~c, f ( ~ ~ ) ) + if) I cf(Ko)%(Cf(K0)):
(iv’)
K%(r
+
13(Ko);
CONSEQIJENCIS OF THE REDUCTION THEOREM
241
According to (i)-(v) and (vii),respectively, in Lemma 35.2 with K~ replacing K, we have the following analogous relations (these can be formally obtained from the above relations by replacing K by K~ and r by 2): (i”)
K ~ + ( I Z ~ , (Lie ( K ~ ) ) ’ ) ~ if & > L , , , ( K ~ ) ;
(ir’)
KO%(%,
(iii”)
K ~ + ( K ~(cf , ( K ~ ) ) + ) if ~
(iv’’)
KO%(~);&);
(v”)
KO
(Vii’‘)
K-o%(Ko, ( 4 . & , ( d ( K o ) )
% Go,
(min {Lko(Ko)r J ! t ~ ( k o ~ ( C f ( K o )1) ( )+ )2 ; cf (KO)+(cf( K ~ ) ) : ;
(&A,(ro))2;
12.
It is easy to check that each sequence on the right-hand side here violates (39.9). Hence relations (i”)-(v”) and (vii”) imply (i’)-(v’) and (vii‘), respectively, via Corollary 40.1 with i = r - 2. Ad (vi). It is sufficient to show that (vi’)
K%(KO,
(r + 1)L3(cf(no))Y
holds. The trouble here is that the sequence on the right-hand side here might satisfy (39.9)and so (vi) in Lemma 35.2 is not applicable. But noting that it violates (39.8), we can use Corollary 40.1 with i = r - 3, provided we can show that L ; - 3 ( ~ ) % ( (4)L,,d(k0)))3 ~0, (9) holds. w e have
2“.% ( K O (4)L,(d(no))l3 9
according to Corollary 25.7. As 2“. = 2 L ; - p ( K L ) 1; - 3 ( ~ the ) , relation in (9) follows from here. Hence (vi) is established. Ad (viii). If L 3 ( ~ ois) regular, then the result follows from (iv), which has already been established. Hence we may assume that is singular. In this case it means n o restriction of generality to assume that the K ~ ( Sare regular, and then we have to prove that holds. We have
K%
(K&
(KF)?
248
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3 CH.
Ix
according to (viii) in Lemma 35.2 with K~ replacing K. As the sequence on the right-hand side violates (39.9)in view of the regularity of the K < S , this relation implies the former one via Corollary 40.1 with i = r - 2. Ad (ix) a, c*, and d. It is sufficient to show that a’) ‘*’) and d’1
K % W 3 (K))+,
L3 ( K d r ;
wr
if
cf (L31K0))%(Cf(L3(K0))):;
K%(L3(KO)9
L3(KO)r
K%(L3(K0)?
L3(KO),cf (L3(KO)))r.
According to (ix)a,c, and d, respectively, in Lemma 35.2 with K~ replacing K, we have a‘‘)
KO%((L3(KO))+>
c’‘)
KO%(L3(KO),
L3(K0))2;
L3(K0))z
if cf (L3(KO))%(cf
(L3(K0))):;
The sequences on the right-hand sides of a”) and d”) violate (39.9).This is not necessarily true in the case of c“), but it is true for its followingweakened version: c*’’)
KO%(L3(KO),L3(KO), if cf (L3(K0))%(Cf(L3(K0))):.
a”), c*”), and d ) imply a’), c*’), and d’), respectively, via Corollary 40.1 with i=r-2. Ad (ix) c**. We have to show that
c**’) K % ( L 3 (KO),L3 fKO)F holds, provided L3(KO)+(L3(KO)): andeither L , ( ~ ~ ) i s r e g u lor a rI n the former case, the sequence on the right-hand side violates (39.9),and so the result followsfrom c”) above (in the proof of (ix)a, c*, and d) via Corollary 40.1 with i = r - 2. So we may suppose that cf ( K ~<) K~ = L , ( K ~ ) i.e., , that K~ is a singular strong limit cardinal. We are then going to prove the stronger version K
% ((cf ( K O 1) +,
KO)I
instead of c**’).The right-hand side here violates (39.9), and so it is sufficient to prove .O%((Cf(KO))+,
KOl2
according to Corollary 40.1 with i = r - 2 . This is, however, confirmed by Corollary 21.3. Ad (ix) b. We have to prove that b’)
~ % ( L 3 ( ~ 0~) (9 K o ()r +, 1)~.,~cr(~,(~~)))Y
c0"EQUENCES OF TIE REDUCTlON THEOREM
249
holds. As the sequence on the right-hand side violates (39.8), it is sufficient to show according to Corollary 40.1 with i = r - 3 that L;-3(K)%(L3
L3
( 4 ) L 3 ( C f ( L 3 ( K o ))3 ))
(10)
holds. We are going to use Corollary 25.9 with A= L 3 ( ~ replacing O) K . This says that 22A%(A I , (413 (11) holds provided (a) 2P2cf (A) and holds with
(fl) 2"<24 for any cardinal a
p = L , ( c f ( I ) ) = L , (cf(L3 ( K O ) ) ) ,
and (fi) holds in view of the definition of I , since 2 L & 0 ) = ~ 0holds by our assumptions in (ix)of the lemma to be proved. Hence ( 1 1)is valid with the above value of p ; and since we have 22&= 2ZL&Q) =2Ko =2&*(K)>Lr-3 -
3
(4,
the relation in (10)is also valid, and this is what we wanted to show. The proof of the lemma is complete. So far we made use only of Corollary 40.1 instead of the stronger Theorem 38.5. The following result strengthens Lemma 40.2 provided certain special assumptions on the iterated logarithms of K are satisfied. LEMMA40.5. Assume (1). Write L'= L ~ ( Kfor) any integeri. Assume, further, that 2Lz2 =L r - 3 and L'-~+;(L~-~);. (12) Suppose that Io-> Lr-2. Then K%
(I&
<9
(13)
holds provided at least one of the following conditions is satisfied:
6)
A12r+2;
(ii)
~ r - 2 , ~ r - 3 .
(iii)
cf(Lr-2)=Lr-2;
(iv)
r=3
(v)
923.
9
and cf(Lr-2)+(cf(Lr-2)): ;
The ideal result would be obtained if we could prove (13)without any of the assumptions (i)-(v); the relevant open problems will be mentioned after the proof. The rationale behind the assumption in (12)is that, as the reader can easily
250
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3 CH.
Ix
check, it guarantees that, in case i = r - 2, (2) in Corollary 40.1 is an essentially stronger requirement than the failure of (i) in Theorem 38.5; hence we can expect to prove more by applying this latter. We shall nonetheless get along by using only the former result and not the latter, plus some special tricks.
PROOF. It is sufficient to prove that, under the assumption of (12), we have (it)
K7qLr-2, r+2)' ;
(ii')
K+(L'-2,r+l)'
if L'-2=Lr-3.
(iii')
~ j r ( L ' - ~r +, 1)'
if cf ( L r - 2 ) = ~ r -; 2
(iv')
K + ( L ' , ~ ) ~if cf (L')+(cf (L')); (r=3) ;
(v')
K++(L'-2,r+l,T+1)I.
To this end, we first show that K jr (L' - 2 ) y 1
and K+(L'-~);-' provided either L'-2 = L'-3 or cf(Lr-2)=Lr-2 or cf (L')+ (cf (I,')); and r = 3. (15)
To see these, observe that we have
by (ix)d in Lemma 35.2 or 40.3 according as r = 3 or r > 3 . This proves (14) in case Lr-2 is singular. Assume therefore that L'-2 is regular. Then cf (IT2)+(cf(L'-2)); in view of (12), and so we even obtain (15) (which is clearly stronger than (14)) in this case by (ix) c in Lemma 35.2 or (ix) c** in Lemma 40.3 according as r = 3 or r > 3 . It remains to see that (15) holds if - ~ , this follows from (ix) c in either r = 3 and cf(L')+(cf(L')):or J ~ ' - ~ = L ' but Lemma 35.2 and (ix)c** in Lemma 40.3, respectively (note that in the latter case there is nothng to prove ifr = 3, since then we have L'-2 = L'-3 =Lo = K , and so (15)simplymeansthat L'-2fr(L'-2):,whichisassumedin (12)). So (14)and (15) are established. Using (25.3) in Theorem 25.1 with I = L'-2, T = 3 or 2, and with r - 1 replacing r, we obtain that ~ + ( L ' - ~ , r + 2 y if (14) holds, and K+(E-',r+l)'
if (15) holds.
CONSEQUENCES OF THE REDUCTION THEOREM
251
Thus (i')-(iv') are satisfied. (v') follows from (14) by (25.2) in Theorem 25.1 with I = L'-2, t = 3 , and with r - 1 replacing r. The proof is complete. We now formulate two typical problems that remain open: PROBLEM 40.6. (i) Assume that 2% > K O and 2 2 5> 22H"for all n < m 1s it true that (a) 2%+(K,, 4)3 or (b) 22%+(K,, 5)4 ?
(16)
(ii) Assume 251>K,, and 2 2 5 1 > 2 2 n mfor all a < q . Does then
,514
2251+(K,,
(17)
hold? For an appreciation of the difference between (i) and (ii) here see the remarks after Problem 40.4. The question whether (16)(a) holds was already asked in Problem 22.2. Note also that Problems 40.4 and 40.6 are closely related. In fact, (7) implies (17) in view of (25.3) in Theorem 25.1. By Lemma 40.5 we know only that the foliowing relations hold under the assumptions of ji) and (ii), respectively: 2S+(N,,
5)3;
25+(K,, 4,4)3; 25l+(N;,, 4)3
225f*(K,, 6)4; 22%+(K,, 5, 5)4 22k1f*(K,, ,6)4,225'f*(K,, , 5 , 5)4,
(17) remains an open problem despite the relation 2"&+(KW,, 4)3, as the sequence (K,,,5) satisfies (39.Q and so Lemma 40.1 with i = r - 3 is not applicable. The next result answers a slightly simpler question than that asked in Problem 40.4 LEMMA 40.7. Assume (1). Write L ' = L ; ( K )for any integer i. Assume, further,
that
and
2LZZ = L?- 3
and L'
-2
f ,(L' -2);,
2LZ1 = L ' - 2
If I,, I l >L'-' then Kjr
<9
(18)= (12) (19)
holds. This says, in particular, that under the assumptions of Problem 40.4 we have 2 2 9 1 + ( ~ , 1 ,K , , ) ~
252
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3 CH.
Ix
and under the assumptions of (8) we have
PROOF. As the sequence (Lr-', Lr-') violates (39.8), it is enough to prove Lr - 3 +(Lr - 1, Lr - 1)3
(20)
in view of Corollary 40.1 with i = r - 3. Let ( p a : a
c
u
22pQ
by (19) and the first relation in (18). We have 2p=+(p:)$ by (19.14), and so 22p'+(p:)23 follows by (24.2) with assumption a ) in Lemma 24.1; hence we have a fortiori 22Pa +(Lr - 1, L r - 1)3 (22) for any a
(23)
In fact, in the contrary case we must have cf (Lr-') = C - ' , and we would have a fortiori Lr-i+(Lr-l,Lr-l)2. A s L ~ - ~ ==Lr-' ~ ~ Lin~thiscaseby (19)andTheorem29.1, thiscontradictsthe second relation in (18). Thus (23) is established. Now (21), (22), and (23) imply (20) via (22.5) in Theorem 22.1. The proof is complete. We now turn to the strengthenings of (vii) in Lemma 40.3 in case GCH holds (cf. (vii'), mentioned immediately after (vii), in Lemma 35.2). LEMMA 40.8. Assume GCH. Assume (l), and write K ~ = L ; - ~ ( K Suppose, ). further, that lo= K~ and 9 2 Lcf(,Jcf ( K ~ ).) (24) Then K%(&s
holds provided either (i)
K~
is regular or (ii) 1' 2 0 .
PROOF. We have to prove the following relations: and
+
K%(KO,
(r l ) L c f ~ m o ~ ~ifc f ~c ~x (, K , ~o~) ~ =KO
Kfr(KO3
w, ( r + l ) & & f ( K o ) J .
THE MAIN RESULT FOR
r23
253
The sequences on the right-hand side here do not satisfy (39.9); hence it follows from Corollary 40.1 with i = r - 2 that it is sufficient to prove that we have KO%(KO,
and
(3)L,,,,)(cf(K,),)2if cf ( K O ) =
KO
.
K " f r ( K 0 7 0, (3)L,,,,,(cr(K,),)2
Both of these relations hold by (vii') (mentioned after (vii)) of Lemma 35.2. The proof is complete. 41. THE MAIN RESULT FOR
THE CASE r 2 3
The following rather lengthy theorem does not contain any new results. It is just a systematic recapitulation of results already proved.
THEOREM 41.1.Assume K ~ W ,9 2 2 , r 2 3 , and l t > r for (<9. Put L'=L:(K) for any integer i. Consider the following conditions: (0)
2L2
-2
or
Lr-2-+(~r-2)i,
(00)
2LL2=L'
-3
and
Lr-2+(Lr-2):,
(+)
2LL1 < L r - 2 ,
(+ + )
YL'= L . ' - Z .
Then exactly one of (0)and (oo),and exactly one of (+ ) and ( + + ) holds. Consider the relation (0) K--r(&9. W e have the following results:
(i)A,IL'-~ for all t < S is always necessaryfor (5). (ii)Each of the following conditions implies that (0)is false: (ii.1) ( A , L ~ ( L ~ - ~ ) +(<s> ) ~ < Afor ~ :some A>L,(L'-'); (ii.2) (Lr-2,Ld(L.-2)(cf( L r - ' ) ) + ) 4 ( l t (<9) : ; (ii.3) ( L r - 2 , L L , - 1 ( L ' - 2 ) + ) 4 ( A (<9) t: ; (ii.4) (L'-2, cf (L'-')) -g(A:: (< 9)
(ii.5) 92L'-', i.e., (ii.6)
and
cf (Lr-2)+(cf ( L r - 2 ) ) i;
( r + 1 ) L r - 1 g ( , ? t : ( < 9 ;)
(A, ( t ~ ) L , ( L r - z ) )4 (A,: 5 < 9) for some Iz with r
254
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3 CH.
Ix
(ii.7) (Lr-', (r+l)L,(~f(L.-z)))+(At: 5<9) ; (ii.8) (Lr-2,( o ) ~ 2 ~ , ( c f,~ ~~- z l l ) + ( ~ t :
5<W
;
(ii.9) There are pa
' + (At : 5 <9).
( p , : a
(
1
'
a < d (L'- )
p.=Lr-' and
+ + ) is assumed for (ii. 1O)-(ii.15):
(ii.10) ( ( L r - ' ) + , L r - ' ) 4 ( A5<9) t: ; (ii.11) ( ~ ~ - l , ~ ~ - ' , c f ( L ' - ' ) + ) + (5<9) A~: ; (ii.12) (Lr-l,,!,-',m)4(At: (ii.13) ( L r - l , L'-
5 ~ 9 ) and
cf(Lr-')fr(cf(Lr-')): ;
',(r + l)L,(d(L'- -4 (A:: 5 <9) ; 1)))
(ii.14)(Lr-l,Lr-l)+(At: 5 < 9 ) , L r - ' f , ( L r - ' ) : , u n d either cf(C-')=Lr-' or
Lr-1 = ~ r - 2
.
7
(ii.15) ( L r - ' , L r - ' ) + ( A t : 5 < 9 ) and
(00) holds.
GCH is assumed for (ii.16) and (ii.17): (ii.16) ( L r - ' , ~(r+l)L,,Lr , 2 , ( c f ( ~ - ~ ~ )5) ~ < (9~;)t : (ii.17) (C-', (r+ l)Lr.z(L,-z))<(At:
5<9)
and cf ( L r - 2 ) = L ' - 2 .
Put (Exception 1) (Lr-', Lr-')4(Ac: <<$)e(L.-', Lr-', (kv)v
and consider the following condition (a)
There is a sequence ( A , (p)a,(a)B,(kv)v
<
( A t : < 9) 4 (A, (PI., (a)&?,( U V < 7 ) .
<
(iii) (0) implies (*) with A= A, for some < 9 provided (Exception 1) is false. If GCH holds then ( 0 )implies that (*) holds with 1=Lr- and z =0 provided there is a p c 9 such that A,, =Lr- and either cf (A,,) = A, or there is a v < 9 with v # p and A, 2 0.
(iv)Assume (0)holds. Then (*) is a sufficientconditionfor (0)provided T =0 and at least one of the following conditions holds:
THE MAIN RESULT FOR
(iv.1) L'-'
255
is regular;
(iv.2) A < L'-' (iv.3) L'-'
r23
;
is a strong limit cardinal;
(iv.4) p = o . As a corollary of these results we have
(v) Zf
(0)holds
then ( 0 )holds provided nteorem 35.4 confirms I,-'+(A<
-r
+ 2)fCs
and if(Exception 1) is false then ( 0 ) f a i l s provided Lemma 35.2 confirms
L'-'+(A,
-r
+ 2)f<s.
Write (Exception 2 ) 9=2, {Lr-',r+l)={&,,Al), and either r > 3 or cf(Lr-2)+(cf(L'-2))$.
Lr-2
We have (vi) If (00)holds and (Exception 2) is false, then I , < LrP2for every (<$ is ). necessary for (0 (vii) Assume (00)and (+) hold. The following condition is necessary for (0) provided (Exception 2) is false:
(**I
(A<:t<$><
where A
T
V
Condition (**) is suflcient provided either T = O or LA(Lr-*)=Lr-'. (viii) Zf(00) and (+ + ) hold and (Exception 2) isfalse, then thefollowing condition is necessary and sqflcient for (0): (***)
(At: <<$)<(A,
(a)D),
where A
and a, B
REMARKS. (Exception 1) leads to Problem 40.4. Make the assumption now that (which assumption in a sense covers the nondegenerate cases), and suppose also that (Exception 1) is false. Then (v) above says that the general problem can be reduced to the case r = 2 whenever we know the answer in this latter case. That is why this result, though formally incomparable with that given by the Reduction Theorem (Theorem 38.5), seems to be stronger and more effective than the latter. (0)holds
256
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3 CH.
Ix
(Exception 2) leads to Problem 40.6. We should like to point out that if GCH is assumed then (vii)is vacuous; in fact ZLZ'=L'-3 holds according to (oo),and so L'-2=L'-3 is a strong limit by GCH, i.e., L'-'=L3(L'-2)=Lr-2, which contradicts ( + ). The problem left open in (vii) is not a genuinely new one, and it corresponds to the analogous problem for r=2, which leads to Problem 20.1. The reason that in case (00)and ( + + ) hold we have a simple result according to (viii) is that this is the case if GCH is assumed and LrP2is a strong limit cardinal with L'-2+(L'-2)$;cf. (36.7) in Theorem 36.2.
PROOF. To see that exactly one of
(oo), and exactly one of ( + ) and (+ +) holds, one has to observe that 2LZ2I L ' - ~ and 2Lc2I L ' - ~ , respectively, hold by the definition of the (iterated)logarithm, and the rest follows by a simple exercise in propositional calculus. (i) holds by Lemma 40.2; (ii)under conditions ( i l k (ii2) v (ii.3), (ii.4j(ii.9) holds in turn by (i), (ii), (iiij(viii) in Lemma 40.3; (ii) holds with (ii.10j(ii.14) by Lemma 40.9(ix)a,d, c*, b, and c**, respectively; (ii) holds with (ii.15) by Lemma 40.7, and with (ii.16)and (ii.17)by Lemma 40.8. Ad (iii).Assume (0) holds and yet (*)fails with I = I t for any 5 < 9. This means in view of Lemma 35.3 (cf. (35.10)especially)that at least one of conditions ( i j -(ix) of Lemma 35.2 is fulfilled with a suitable rearrangement (2;: 5 < 9') of the sequence (I: : t < 9) and with L'-2 replacing K. Note that these conditions are identical to conditions (ib(ix) in Lemma 40.3, except that (ix)c is replaced by (ix)c* and c** in the latter. Yet none of the conditions (ij(ix) in Lemma 40.3 holds for (4: <<9'), since the contrary would mean by this lemma that (0) fails. Hence (ix)cof Lemma 35.2 with L r W replacing 2 K must hold for ( I ; : 5 < 9'). That is, we have
(L'-',L'-')4(It:<<9),
( 0 ) and
(++)holds, and cf(L'-')+cf(L'-')i.
(1)
We may assume here that I,, I 1 > L r - ' . As ( 0 )holds, we also have A,, A I s L r - ' by (ii.10) in the theorem being proved; A e < o for any t with 2 s 5 < 9 by (ii.l2), 9
( A (r + 1)L$
* ( A , : 5 <9)
THE MAIN RESULT FOR
r23
257
for some 1with r + 1 II< Lr-', or
If GCH holds, then L,(L'-2)=L,(L'-2)=Lr-1 for any 1with 311
(Lr-',
0, (I+ l
) L ~ ~ , ~ . * , ( ~ ( L , - z ) ) )5~ <( lW ~:
or Lr-' is regular, which is impossible by (ii.16)or (ii.17)above, respektively.To see that we can take I = L'-2 in.(*) one needs only to consider the remark made immediately after (35.16). The proof of (iii) is complete. Ad (iv). (*) and T = 0 imply
Lr-2+(1;):<9 via Theorem 35.4, and so L.-+ ~ ( I z -r
+2); <s 2
holdsa fortiori. Observe that we haveeithere~p,-,(2~Z')
(Q: <9
holds, which we wanted to show. (v) immediately follows from (iii) and (iv). Ad (vi). Assume that (00)and (0) hold, and yet we have A , 2 L ' - 2 for some < 3 or cf (Lr-2)+(cf ( L r - 2 ) ) i ,respectively. Thus (Exception 2) holds; this is what we wanted to prove. Ad (vii). We begin by discussing some of the consequences of (00)and (+ ):
If (00) holds, then Lr-2 is a limit cardinal and cf (L'-2)=cf (L'-,).
(2)
This follows by Theorem 6.10 d. If (00)holds, then (Exception 1) is false.
Indeed, (Exception 1) requires that 17 Cornbinatorial
( 0 ) holds,
and this contradicts (00).
(31
258
ORDINARY PARTITION RELATION WITH SUPERSCRIF'T 2 3 CH.
Ix
If ( + ) holds, then the sequence (A, ( p ) , , (o)Br( k v ) v < r ) satisfies conditions (35.4)-(35.8)of Lemma 35.3 with A< Lr-2 and with L'-' replacing K if and only if P = L , ( L ' - ~ ) , a
(4)
To see this, one only has to observe that (35.6)(ii)is vacuous because A
(5)
This follows from Theorem 7.16, as Lr-2 is a limit cardinal by (2). We now embark upon the proof of the first part of (vii). Suppose to this end hold and (Exception2)fails. As (Exception 1)is then false that (oo),( + ),and (0) by (3),we can conclude by the first sentence of (iii) that (a) holds for a sequence (A, (pIu, ( o ) ~ (, l ~ ~ ) , , <with ~ ) A=Ac for some {<9. We have A=A,
(Ln(Lr-2))u3(o)BI(k,),
holds. Choose a regular cardinal K' satisfying (5). Then by Theorem 35.4.b(i)we have K ' 4 J - Y (LZ(K')),, (.)sl2 in any case, and K'+(A,
(L.I(K')),Y( o ) p( ~ v ) v < z ) ~
in case LI(~')=LA(Lr-')= Lr-' (in this case we can use that theorem with F = =fi T, T' =Oh since (35.41 (35.51 and (35.6xi) hold with K' and L,(K') replacing K and p in view of the equality L , ( L r - 2 ) = L A ( ~(cf. ' ) ( 5 ) ) and (**), and, finally, (35.6)(ii), (35.7), and (35.8) are vacuous with these substitutions since A
+
THE MAIN RESULT FOR r 2 3 WITH GCH
259
that there is a P < L ' - ~such that p+(4 ( 4 p ) 2
+
(6)
holds. We have 2 C 1 = Lr-2 according to (+ ). Hence there is an infinite po
Let K be a n infinite cardinal. For any integer i, define K i by recursion as follows: K-O=K, K - ( i + l ) -- ( K - i ) - , (1) where we recall that 1- is the immediate cardinal predecessor of the cardinal 1if thisexists, and 1- = A otherwise. Clearly, if K = K . then K - ' = & - , . It iseasy to see that, under the assumption of GCH, we have K - = ~ ti,( K)
for any integer i. Put
h(K)=min { i < w : K
- ~is
a limit cardinal].
(2) (3)
Note that it is clear by (2) that if GCH holds then h ( K ) = min
{ i < o:L y '( K ) = L\(K)).
(4)
Finally, we remind the reader that the critical number cr (K) of K was defined by formula (36.1), that is, we have cr (K)=cf (cf ( r c - ) .
(5)
We nowturn to the corollary of the theorem in the preceding section for the case when GCH holds: THFQREM 42.1 (ErdBs-Hajnal-Rado [19651). Assume GCH, and let K 2 2, 9 2 2 , 3 1 r r for any t < 9. Consider the following conditions: (0')
h ( ~ ) ~ r - or 2
(00')
h ( ~ ) < r - 2 and
Then exactly one of
(0 1 17'
K-(~-')-+(K-(~-~))$,
K-('-~)+(K-(~-'))$.
(0') and (00')
holds. Consider the relation K--r(Q;
260
ORDINARY PARTITION RELATION WITH SUPERSCRIPT-? 3 CH.
Ix
and the following conditions lfor the definition of the operations com and suc see (36.4) and (36.5)): there are cardinals a, B, and ~7 with (A<:5 <$)+ < K - ( - ~ ) ,(cr ( K - " - ~ ) ) ) = , ( o ) ~ ) (61 such that a < c r (K-(r-2)).com( ~ - ( ~ - ~and ) ) p, + lo < c r ( K - ( ~ - ~ ) ) (note that + denotes cardinal addition), and there are cardinals a, B, and u with
( I < :' % $ ) < ( ( K - ' r - l ) ) a ,
such that a
(O)B>
and
B, O < K - " - ' ) .
(7)
Put (Exception 3) K-(r-2)>cf (K-(r-2))>cf( K - ' ~ - ~ ) )>cr ( K - ( r - 2 ) ) , and ( - 0. - 2), (r 1lCr ( K - + z , l ) 6 (A, : 5 < 8 ) < ( K - ' ~-2), (KA< ,,> , where f
+
( i ct :< 9 ) 4 ( a ) ,
holds with some p, u < K - ( ~ - ~ ) .
REMARK. If (Exception 3) holds then we d o not know whether or not (0) holds even in case r = 3 . This leads to Problem 25.8 discussed earlier. PROOF. It is easy to see by elementary propositional calculus that exactly one of (0')and (00')holds; it is also important to note the fact u k d below often, without explicitly mentioning it, that under the assumption of GCH,(0') and (00') are equivalent to conditions ( 0 )and (oo),respectively, of the preceding section; this easily follows from (3) and from the equality 22= I,true for any infinite cardinal 2 if GCH is assumed. a) follows from Theorem 41.1(i). Ad b). Assume that ( 0 ' ) holds and either (6) or (7) is satisfied. Then (*) of Theorem 41.1 holds with T = O by (36.9) and (36.10). Taking into consideration that, under the assumption of GCH,every infinite cardinal is either regular or strong limit, (0) follows by Theorem 41.1(iv.l) or (iv.3). This proves the sufficiency of (6) or (7). We now turn to the proof of necessity. Assume to this end that (0') and (0) hold, and (Exception 3) fails. Noting that (Exception 1)always fails under GCH
THE MAIN RESULT FOR
r 2 3 WITH
261
GCH
(in fact, (Exception 1) requires in particular that (+ + ) hold, which simply becomes C-' = Lr-2 under the assumption of GCH, and it also requires the inequality I,'-' < L'-2, which contradicts the former equality), we can conclude by the first sentence of (iii) in Theorem 41.1 that (*) holds there with 2=AZ for some 5 <9 (we may assume A=A, without any loss of generality), i.e., that we have ( A c : 5<9> 4 ( 2 , @ ) a , (a)@,( k v ) v < z ) (8) 9
and the sequence on the right-hand side satisfies conditions (35.4b(35.8) of Lemma 35.3 with K - ( ' - ~ ) replacing K. Here we have * < ~ ~ ( ~ - ( ' - z ) ) = ~ - 1( )r -
and
for any v < z
kv
according to (35.5). We distinguish two cases: 1) 2) A = A - K - ( * - 2 )
A=A,
(91 and
0-
Case 1 . By (36.12) and (36.13) with
replacing
K-('-')
(A ( P I a , ( 0 ) D ) e < ( K - ( r - ' ) ) a , ,
K
we have
(0')B')
,
with some a',p', and u' satisfying Co
(K-('-')).
suc ( K - ( ' - ' ) ) +
1 and
p', O ' < K - ( ' - ' ) .
(10)
By (8) we then have
( A t : t < 9 ) ~ ( ( ~ - ( ~ - ' ) ) a *(0')@,, , (kv)v
K-(I-~)>o
(0'
(7) is confirmed by the
+ HlJp +,) .
Thus (7) holds in Case 1.
Case 2. We shall prove that (6) holds in this case; assume, on the contrary, that (6) fails. We have A= A, = K-(' - ') in the case considered and, by the second clause
of (iii) in Theorem 41.1, we may assume that
either cf ( K - ( ~ - ' ) ) < K - @ - ' ) and & < a for any {with 1 < { < 9 or z=O in (8).
By
(36.11) we obtain that
(11)
262
ORDINARY PARTITION RELATION WITH SUPERSCRIPT 2 3
CH.IX
p=cr ( K - ( ~ - ~ ) ) ,D , j < c r ( K - ( ~ - ~ ) ) ,and a < c r (K-(r-2)).com( K - ( ‘ - ~ ) ) + 1.
(12)
If ~ < c (rK - ( ~ - ’ ) ) , then we can conclude from (8) and (12) and the equality A= K - ( r - 2 ) , valid in the present case, that (6) holds with /3’=j+r replacing j (D should also be replaced by D’ = o+ sup { k,: v < r> or o’= D + No according as cr ( K - ( r - 2 ) ) equals or is greater than No). This contradicts our assumption that (6) fails; hence we must have r2cr that is, we cannot have r=O, and so cf ( K - ( r - 2 ) ) <
(13)
( K - ( ~ - ~ ) )
K-(r-2)
and
(14)
A,
for any 5 with l r ( < 9 by (11). As A=K-(‘-’) and (35.5) holds with replacing K according to what was said after (S), we have T
(K-(r-2)))=Cf
(K-‘‘-2)
I-.
]c-(*-~)
(15)
Comparing this with (13), we obtain that >cr
(..J(K-(r-2))-
(K-(r-2)).
(16)
Thls implies via (36.3) that cf ( K - ( r - 2 ) ) ,
Now we have by the equality A =A, the theorem being proved that < K - ( r - 2 ), ( I +
f.,
( K - (r - 2 ) ) -
= K - ( - ~ ) , (13), and the
l ~ ( p 2 b ) ) + ( A <5<9). :
(17) assumption 2: > r of (18)
On the other hand, we have 1 9 l l l + ~ + ~ + T < C f (K-‘r-2))-,
wherethe first inequality holds by (8), and the second one by (12), (15), and (16). Hence, by (14) and (8) we can conclude that k : < o for any v < T ’ . The first inequality in where t’-191-1
CHAPTER X
SOME APPLICATIONS OF COMBINATORIAL METHODS
The next few sections contain some applications of partition relations and combinatorial methods connected with them in topology and in the,theory of set mappings, etc., and the culmination of this chapter is the proof of a recent inequality on cardinal exponentiation due to F. Galvin and A. Hajnal. The chapter ends with a discussion of results of J. Ketonen, T. J. Jech and K. Prikry on the relationship between cardinal exponentiation and saturated ideals.
43. APPLICATIONS IN TOPOLOGY Here we give some applications in topology of the results and methods relating to the ordinary partition relation. We do not strive.at obtaining the most general results; our aim is, rather, to show the usefulness of partition relations in topology by picking characteristic examples. We first give a few basic concepts.A topological space is a pair (E, 0)such that E is a set (the underlying set) and 0s B(E)is such that 0 E 0, E E 0,and if H E 0 then H E 0 and, finally, if H E 8, H # 0 is finite then H E 0. The elements of E are called points,0 is said to be 'the topology on E, and the elements of 0 are called open sets. X c E is closed if E \ X is open. The closure cl (X) of a set X G E is the least closed set including X, i.e.,
n
u
cl ( X ) = (){YE E: X E Y & E\ Y E 0}. If x E E and x E X E 8, then X is called a neighborhood of the point x. More generally, if X c E and X E X' E 8, then X' is called a neighborhood of the set X. One often writes briefly E instead of ( E , 0).E is a Hausdorfspace if any two distinct points in E have disjoint neighborhoods. All spaces considered below will be assumed to be Hausdorff. Given a topological space (E, 0) and a set E' E E, the pair ( E ' , {XnE': X E 0}) is also a topological space. This space is called a subspace of E; the topology on E' is called its relatiue (or subspace) topology. Given a point x of a topological space (E, O), a set BE 0 is called a neighborhood base of (or at) x if for every X E 0 with
264
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
x E X there is a Y E B such that x E Y G X. B is called a base for the space (E, 0)if. it is a neighborhood base at each point x of E. This concludes our review of the main concepts of topology. What comes next is a series of characteristic applications of combinatorial methods in topology. When necessary, a short paragraph preceding a result will explain the topological concepts not explained so far. Our first result, due to A. Hajnal and I. Juhasz [1967], is obtained by an application of the partition relation
(2Ho)+4K):
(1)
(see Corollary 17.5). To explain the topological concepts needed, a topological space is first countable if each point has a countable neighborhood base; it has the Suslin property if it has no N, pairwise disjoint open sets. THEOREM 43.1. A $rst countable Hausdorff space with Suslin property has cardinality I2'o.
PROOF. Let (E, 0) be a topological space with the required properties and assume, on the contrary, that I E I > 2'0. (2) For each point x E E let { U:: n co)be a countable neighborhood base of x; we may assume without loss of generality that x E U:. Put
then, clearly, {V;: l s n < w )
is also a neighborhood base of x . As E is a Hausdorf€space, any two distinct points x and y have disjoint neighborhoods X and I: respectively. We have and
VZcX
VYGY
for any large enough integer n. Then, obviously,
v;n v;=o holds. Let f ( { x y j )be the least integer n for which this relation holds. Then
f: [E]2+w is a coloring, and according to, (1) and (2), there is a homogeneous set H c E of cardinality K1;let m be the color of this set (i.e., f"[HI2 = { m } ) . Then we have V;n
V;=O
APPLICATIONS IN T o P o m y
for any two distinct x, y
EH
265
by the definition of the coloring f. Hence
{ v;: X E H ]
N, of pairwise disjoint open sets, contradicting our assumption that the space (E, 0) has the Suslin property. Our next result, also due to Hajnal and Juhasz [1967], gives a more sophisticated application of the partition relation
is a set of cardinality
(225)f+(K):,
(3)
which holds for any infinite K (cf. (17.30) with I = 3 and p = K). It usesthe notion of discrete topological’spaee: (.E, 0)is discrete if 0 = g ( E ) ;for this, it is obviously enough to require that each one-element set be open. THEOREM 43.2. ff K is infinite, then a Hausdorffspace of cardinality >22‘has a discrete subspace of cardinality K . This result can be sharpened as follows: For any infinite cafdinal K , a 22’ has a discrete subspace of cardinality K; Hausdorff space of cardinality > A
i.e., one can “usually”lower the bound for the cardinality of the space in case K is not a successorcardinal. This can be proved by a minor modification of the proof of Theorem 43.4; see our remark right after that proof. By improving the general Canonization Lemma, Shelah [198 11 can prove this result even if K is singular and the cardinality of the space is > 22’ for all I < K (cf. Theorem 43.3 for the role of the General Canonization Lemma).
PROOF. Let (E, 0) be a Hausdorff space such that IEI > 22p,and let < be a wellordering of E. For any x, y E E with x < y let U!y and U i ybe neighborhoods of x and y, respectively, such that UzynU i y = O .
(4 )
For any x,, x,, x2 E E with x , < x , < x 2 and for i, j = O or 1 put and Writing fij=Zinfj,it is easy to see by (4) that f = (I,,, I,,, Zl0,Zl,) is apartition of [EJ3 into four (not necessarily disjoint) parts. According to (3), there is a homogeneous set X of cardinality IC; we may assume that X has brder type IC in the wellordering <. Let xa, a < y be the enumeration of the elements of X in
266
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
increasing order and let Y = { x u : a is an odd ordinal
here an odd ordinal is a limit ordinal (or 0) 4 an odd integer. We claim that Y is a discrete subspace of E. To this end we have to show that any xu E Y has a neighborhood U , such that U,n Y = { x , ) .
Let B=a-1 and y = a + l . We claim that satisfies (7). Note that U , is a neighborhood of xu since it is the intersection of two of its neighborhoods. To see that (7) holds, let x,: E Y be a point distinct from xu; we shall show that xt$
(91
u,.
To this end, we distinguish four cases: 1) < < a and [X13=Z,, 2) < < a and [X]3cZ,,3)<>aand [ X ] 3 ~ Z o , a n d 4 ) < > a a n d[ X l 3 ~ Z ' .(Notethat incases 1) and 2) < y holds since is an odd ordinal, as x,:E Y). In case l), we have x,:$ U:mx,by ( 5 ) , in case 2) x,: # U.&x,x. by (9,in case 3) x,: # U:sx, by (6),and in case 4) x,: 6 U:pxaby (6). So (9)follows from (8). Hence (7) holds; i.e., Y is a discrete subspace of cardinality K of E. This completes the proof.
<
Our next result exploits the same idea plus the Geieral Canonization Lemma
(Lemma 28.1). This is the first instance in this book that we canonize triples; so far, the Canonization Lemma was only applied in the case of pairs. THEOREM 43.3. Zf K is a singular strong limit cardinal, then a Hausdorffspace of cardinality 2 K has a discrete subspace of cardinality K.
<
PROOF. Let (K,:
u
x < y i f x ~ A ,Y, E A , ,and < < q for some <,q
(10)
As in the preceding proof, for any two points x , y ~ Ewith x < y pick neighborhoods V g and U i yof x and y , respectively, so that (4) hold, and define I,
267
APPLICATIONS 1N TOPOLOGY
and IJ by (5) and (6). We again obtain a partition I = (I,,, I , , , IIo,I , 1 ) of [El3 as above. Write
f ({xox,x,))=k-l
(11)
if the kth class of the above partition is the first one containing {xox1x2) (k= 1, 2, 3,4). By Lemma 28.1, there are sets B , E A ~ ,5
and a is odd}.
We claim that Y is a discrete subspace of E. To this end, we have to show that any Y has a neighborhood U: such that
X:E
U : n Y = {x;}
(12)
holds. We claim that the neighborhood U:= U $C L l xt nn U$,.il C t of x; satisfies (12). T o see (12), let
X;E
x;$i
(13)
Y be arbitrary; we shall show that
u;.
(14)
In case q = 5 the same argument works as in the preceding proof, since the set Y nB, is homogeneous with respect to the coloring f (and so with respect to the partition I defined after (6)) by the definition of canonicity. If e.g. q < { and holds for some k <4 (the first equality holds by canonicity; this is the point where canonicity is used), then in case k = 0 or 1 we have {x,"x;$+ E I, by (1l), and so x,"# U$,.+I by (5), and in case k = 2 or 3 we have {x,"x;~'x;} E I,, and so x,"$ U . & I ~by ~ (5), verifying (13} in case q e 5 The case q > 5 can be dealt with similarfy, :sing (6) instead of (5). The proof of the theorem is complete.
'>
Call a topologlcal space right(1eft))-separatedif it has a wellordering under which each initial (final) segment is open. The next result, due to J. de Groot [1965], applies a simple tree argument. As pointed out by Shelah, a proof using the relation (2%)' + ( K ) ~ (see (15.8)) and ideas similar to those used in the proof of Theorem 43.2 also works (see below for hints).
THEOREM 43.4. I f K is an infinite cardinal, then a Hausdor-space of cardinality >25 has a subspace of cardinality K that is right-separated.
268
SOhE APPLICATIONS OF COMBINATORIAL METHODS CH.
x
H m for Shelah' proof. Let 4 be a wellordering of the space (E, O), consider the partition 1 = ( I o , 1') of [El3 defined by (6), and assuming lEI>2&, let X = {xQ:a < K ) be end-homogeneous with respect to this partition. Given a < K odd, either x,i2 I U:+,or x,i2 $ U;mxmi,. Incase the former alternative happens throw away xa+l and otherwise throw away x,. The remaining set is rightseparated by end-homogeneity.
PROOF. Let (E, 0) be a Hausdorff space with 1~51>2~. We are going to construct a partition tree (E, S, R, T) by transfinite recursion as follows (cf. Section 14). We put "2; (15) dom ( S ) = aslSl+
for any function f~ dom (S) the set S(f) will be aclosed subset of E. Assume that S(f ) has already been defined and write dom (f ) = a. In case a < I S I we have to define S ( f u ( ( a 0 ) ) )and S ( f u ( ( a 1 ) ) ) . If IS(f)ls 1, define both sets as empty (so IR(f)l=l by (14.3));if IS(f)112, then let s(fu{(a0)))ands(fu{(al)])be two distinct elements of S ( f ) , and let U , and U , , respectively, be disjoint neighborhoods of them. Put +
and Note that in this case we have i.e., (14.3) implies that R(f)=O.
This completes the definition of S, and with it that of the whole partition tree (E, S, f,T).bbserve that S(f ) is always a closed set. In fact, S ( 0 )= E is closed. If g E dom (S) and S(g P <) is closed for all <dom (g), then we have
<
SW=
n
t < d o m (s)
SWS)
in case dom (g) is a limit ordinal, and so S ( g ) is also closed in this case. Finally, if dom ( g ) = a i l forsome a, thenghaseithertheformfu((a0)) or f ~ { ( a l ) ) , and so S ( g )is either empty (when IS(f)l< l ) ,or equals one of the sets in (16) and (17); in both cases, S ( g ) is closed provided that S(f)= S(g P a ) i s closed. Note that we always have
INf )Is 1;
269
APPLICATIONS IN MPOLOGY
in fact IR(f)l= 1 occurs in case IS(f)l= 1 and IR(f)l = O in all other cases (cf. (18)). Taking, as usual, T={f~dom(S):S ( f ) # O ) , we obtain by Theorem 14.3.b with (25))+and K replacing K and A that T has a path of length K + 1 (note for this that each element of T can have at most two immediate successors by (15)), i.e., there is an f e T with dom ( f ) = K. As S(f)#Oin view of fe7; the point s(f p a ) is defined for every successor ordinal a < K ; put x={s(f
p(ai 1)):a<~).
The subspace X of E is right-separated; in fact, for each a < K, the set E \ S(f p a ) is a neighborhood of (s(f /".(ti1)):< < a } that does not inters9ct the set ( s ( f p ( t i 1 ) ) : a s t < K ) . One can also see from here that the points s(f p ( a 4 1)) are pairwise distinct, and so X has cardinality K. The proof is complete. REMARK. If we want to prove the sharpening of Theorem 43.2 mentioned immediately after that theorem, then we have to construct almost the same tree as in the preceding proof, where now we have IEl> 22' instead of IEl>25.
1
d
I
The modification that we have to make is the following: having defined S(f),put where a = dom (f ), and replace S(f ) with S(f) in the definitions of s(fu { (aO)}) and s(fu {(awl}), and on the right-hand sides of (16) and (17) above. Noting that the closure of a set of cardinality I in a Hausdorff space has cardinality 12*', it is easy to conclude that there is an f e T with dom (f)=K. The sequence (s(f p ( a i1)): a < K)is left-separated in view of the definition of S', and it is rightseparated by the argument used in the preceding proof. Hence its elements form a discrete subspace of cardinality K. Our next aim is to establish A.V. Arhangel'skifs [19691 famous result saying that every first countable compact Hausdorff space has cardinality IPo, which settled a nearly fifty year old conjecture of P. S. Alexandrov. For this, we need the following concept: a sequence (x<: 5 < a ) of points of a topological space is free if for each q W be a regular cardinal, ( E , 0) a topological space of cardinality > 2!,' and assume that for every set X s E of cardinality < K we have and
ICl(X)l I 2 F
cl (X) is the intersection of at most 25 open sets. Then there is afree sequence of order type K in E.
(19)
(20)
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SDME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
PROOF. We shall define a partition tree ( E , S, R, T) such that dom (S)=
u "(2"; U l K
we shall take care that S(f)is a closed set for every f~ dom (S). For each g E T with dom (g) a successor ordinal we shall also define a point s(g) E S(g). Let u
be a representation of cl (Af) as an intersection of 25 open sets; such a representation exists according to (20). Write
S ( f u {(av))) = S ( f ) \A;
9
(22)
and if S(fu{(av))) is not empty, then select an element of this set as s ( f u { (av))) arbitrarily. This completes the definition of the partition tree ( E , s, R, Observe that the set S ( g ) is closed for every g E 7: Indeed, if dom (g) is a limit S k p 5 ) ,and so the closedness of S ( g ) follows from ordinal, then S k ) =
n.
0
t c d o m (8)
that of the S(g t)'s; and if g has the form fu{(av)), where a=dom(f) and v<2&,then S(B) is closed in view of (22), since S(f) is closed (by the induction hypothesis) and A; is open. Note also that we have cl ( A g p t ) n S ( g p ( t /1))=0
(231
for every g E T and every 5 <dom fg);indeed, this follows simply from (21) and (22) by taking f =g p 5 and v = g ( < ) there. This means that the sequence ( g ( 5 / 1): 5
/ 1<dom (g))
is a free sequence. Hence we only have to show that there is a g E T with dom (g)= K. Assume that this is not the case; then for any f E T there is an a < K such that dom Cf)=a; hence R(f)=S(f)\
by (21) and (22), and so
n
S(fu {
v-z2'?
IR(f)112F
n Aj=cl ( A / )
v<25
APPLICATIONS IN TOPOLOGY
271
according to (19). Observing that (25)p= 2F holds for any p < K in view of the regularity of K by Theorem 6.10.e, we can use Theorem 14.3.a with (29+ and K replacing K and 5 respectively. We obtain that T has a path of length K -k 1; this contradicts our assumption that dom (g)< K for every g E 7; completing the proof. Here the combinatorial part of this section is essentially over. As a final act, we show how to derive Arhangel'skii's result from the above lemma: COROLLARY 43.6. Everyfirst countable compact Hausdorffspace has cardinality I2HO. By an argument similar to that given in the proof below, one can ekablish the stronger results in which compact' is replaced by 'Lindelof (a space is Lindelol if every open cover has a countable subcover); for the proof see the remark at the end OF this subsection.
PROOF. Let (E, 0) be a first countable compact Hausdorff space. First observe
that (19)and (20)are satisfied with K = K ~by (E, 0). Infact,asfor (19),let X c E be an arbitrary set and let x E cl (X). Then, by first countability, it is edsy to find a sequence (x,: n
( X ) = U(c1 (Y):Y E X & I Y l I W } ,
(24)
and secondly, Icl ( X ) ( I I X I;~ ~ this latter holds because in a Hausdorff space a convergent sequence can have only one limit point, and the cardinal on the right-hand side is an upper bound of the number of convergent sequences that can be formed by elements of X. If we ) from (25). assume IXllK0, then (19) (with K = K ~ follows To see (20) with K = K,,we show that any closed set Y c E of cardinality I20' is the intersection of I open sets. To this end, for each y E Y let IVY:n w}be a countable neighborhood base of y (assume y E U ; for n
I:by the compactness of E there is a finite set
YE
(J Y E Y(X)
u y.
212
SOME APPLICATIONS OF COMBlNATORlAL METHODS CH.
x
As x does not belong to the set on the right-hand side, we have Y=n{
u
Z/;W~~E\Y~..
Y E Y(Xl
The sets on the right-hand side are open, and there are only I2uo different ones among them. This establishes (20) with K = K , . Hence IE112uo will follow from Lemma 43.5 with K = K , if we show that E cannot have a free sequence of order type K,.Assume, on the contrary, that ( x ~( : < K , ) is such a sequence. Writing X={x,: (
by (24). As
cl ({x:: ( < a j . ) n c l ( ( x : : u ~ 5 < K 1 ) ) = 0
holds for any u < K , since the the sequence (x:: ( < K , )is free, we can see from here that
n cl ( { X g : u i t < K 1 ; ) s c i(XI\
a 4 ,
u Cqx5:
c
a
no intersection of countably many sets on the left-hand side is empty, however. This contradicts the compactness of the space E (it would contradict even Lindelofness). Hence E cannot indeed have a free sequence of order type K,, which completes the proof.
44. FODOR'S AND HAJNAL'S SET-MAPPING THEOREMS Given a set E, a set mapping on E is a function f :E + P ( E ) such that x 6 f ( x ) holds for every x E E. (Set mappings ouer E will be considered in the next section.) A subset H of E is called free with respect to f if H nf ( x )= 0 holds for every x E H. Note that set mappings were briefly considered earlier (cf. Lemma 20.3). In this section we shall consider set mappings where a cardinality restriction is imposed upon the images, and we shall consider the question whether large free sets exist. Iff is a set mapping on K, then we might first ask how large a free set must exist if we assume only If(u)l< K for every u E K. A brief reflection shows, however, that this assumption does not even ensure the existence of a two element free set (clearly, a one element set is always free); indeed, take f ( u ) = = u( = (/3: /3 < a } ) for every u E K. One might expect more if we assume that I f ( u ) l < l holds with some cardinal A < K for all U E K . Ruziewicz [1936] conjectured in 1936 that this assumption ensures the existence of a free set of
FODOR’S AND HAJNAL’S SET MAPPING THEOREMS
213
cardinality K; this conjecture was settled affirmatively by Hajnal in 1960 (see Theorem 44.3 below). The following theorem, which answers Ruziewicz’s conjecture for regular K (seeCorollary44.2 below),isduetoFodor [1952].ThecaseA=oofthistheorem was already known t o de Bruijn and Erdos [1951]. THEOREM 44.1. Let K and A be infinite cardinals, and let f be a set mapping on K such that If ( x ) ]< I holdsfor every x E K . Then K can be represented as the sum of 1 sets free with respect t o f . PROOF. We may assume A < K. We define a matrix ( x a s: a < p & p<1) of elements of K such that the sets { x , ~a: < p > ,p < I , are pairwise disjoibt free sets and K = { x a s a:
,
COROLLARY 44.2. Let K and 1 be cardinals, K regular and I < K , and let f be a set mapping on K such that If (x)l< I holds for every x E K. Then there exists a set of cardinality K that i s f f e e with respect to f .
PROOF. If 1 2 w , then the assertion follows immediately from the preceding theorem. If K > w, then we may assume ,420;so the only case that remains to be 18 Combinatorial
274
SDME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
settled is when K = W and I < o.In this case Theorem 4411is not applicable, since A ~ isw assumed there (though an extension of this theorem to fintie A’s, due to de Bruijn and Erdos [19Sl], which says that, under the same assumptions, K splits up into 2A - 1 free sets, would be applicable). But a simple argument invoking Ramsey’s theorem settles this case as well. To see this, define a coloring g : [wI2-3 as follows: for each m, n
THEOREM 44.3. Let K and I be cardinals, K infinite and I < K , and let f be a set mapping on K such that I f ( x ) l < A holds for all x E K. Then there is a set of cardinality K that isfLee with respect t o f.
PROOF. By the preceding result we may assume that K is singular; hence we may also assume that I is regular and I > c f ( K ) (indeed, if this is not the case already, then replace A with a larger X < K that satisfies these requirements). Let
( K ~ t:
‘u u
act
it is clear that I B , ( = K , holds, as the set subtracted from A, has smaller cardinality. We shall define a matrix (C,, : 5 < cf (K) & p < A) and a sequence ( q t : t c c f (K)) by transfinite recursion such that, for any <
hold for all p, p‘ < I with p # p‘, and
SET MAPPINGS OF TYPE
>1
215
holds for all x E () C,, and for all a < <.Assume that for some 5 <.I the matrix P
.'
< 5 & p
Bz={xEB,:f(x)n
u
giv
U C,,=O) a<:
for each r]
C u v , v < I , are U
pairwise disjoint, the assumption I f ( x ) l < A implies that each x E B, must belong to some Bz. Hence, as lB,l = K , is regular and K , > I,it follows that [B;l= K , for some r]
It is easy now to finish the proof of the theorem: As I > c f (K) and 4. is regular, C,, is a there must be an r] < I such that 11,Ir] for all 5
u
C
free set by (1) and (4) and because each C,, is free; this latter is the case since K , = K by (2).The proof C,,C B,c A,, and A, itself is free. C has cardinality
1
is complete.
C
45. SET MAPPINGS OF TYPE > 1
In the preceding section we discussed questions involving set mappings on a set. In this section we turn to the usually more intricate questions involving set mappings over a set: given a set E, a set mapping ouer E is a function f from a subset of 9 ( E )into 9 ( E )such that X nf ( X )= 0 for every X in the domain off. A subset H of E is called free with respect to f if H n f ( X ) = O holds for every X E 9 ( H )ndom cf ). Given cardinals K, I,p , and a, we shall study the following question: for an arbitrary set mapping f l [ K ] ' + [ K ] < ~ , does there exist a free set of cardinality a? The string of symbols (K,
3, P ) +a
(11
will mean that the answer to this question is affirmative for all such f, and ( K , 3, p)+a will mean that this is not the case. For such set mappings f, I is called the rype and p , the order, of f. The case I=1 was covered in the preceding section: in fact, a set mapping f :K - @ ( K ) on K can be identified with the set mapping f': [K]'+SCK) over K
276
SOME APPLICATIONS OF COMBlNATORlAL METHODS CH.
x
by putting f ' ( { x ) ) = f ( x ) for all x E K. (For example, Theorem 44.3 can be expressed in symbols by saying that (K,1, A)+K holds if K >and I K ~ hold.) W The case il20 is entirely settled by the following theorem: THEOREM 45.1 (ErdBs-Hajnal [1958]).
If
K
and I are infinite cardinals, then
(K, A92)f'A (2) holds. Note that every set of cardinality <1 here is free by definition, and so ( K , I, 2)+a trivially holds for any a
PROOF. First we prove the relation
be a 1-1 mapping such that h ( X ) c X for every To this end, let h: [A]'+[I]' XE[I]' (c means proper inclusion). It is easy to construct such an h by transfinite recursion: in fact, let ( A ? : 5 ~ 2 ' ) be a wellordering of [A]', and choose h ( A J as the first A , c A: ( q <2i) that has not been chosen as h(A,) for any Or<<. For any X E [I]', let f ' ( h ( X ) )be an arbitrary element of X\h(X), and for any Y E [I]'\ra (h) define f ' ( Y ) arbitrarily. Put f ( Y ) = { f ' ( Y ) ) for every Y E [A]'. It is easy to see that no subset of A of cardinality is free with respect to the set mapping f ;this verifies relation (3). We now turn to the proof of (2). To this end, let S G [K]' be a maximal system of almost disjoint sets; i.e., let S be such that and
IXn Y I < A for any two distinct X, Y E S ,
if XE[K]',
then
IXnYI=I
for some
YES.
(4) (5 )
For each X E S, let fx: [X]'+[X]' be a set mapping such that no subset of X of cardinality I is free with respect to fx ;there is such a set mapping according to (3). Given a set Z E [K]', put f(Z)=f,(Z)
(6)
if Z G X for some X E S (note that if there is such an X, then it is uniquely determined by Z in view of (4)); if there is no such X E S, then put e.g. f(Z)=O. We claim that f verifies relation (2). In fact, assume on the contrary that the set X~L~]'isfreewithrespectto$Wehave ( X n Y(=Iforsome Y ~ S b y ( 5 ) , a n d i t is easy to see by (6) that the' set X n Y must be free with respect to fu; this, however, contradicts the definition of fu. The proof is complete.
SET MAPPINGS OF TYPE
>1
277
An interesting intermediate case between A
p)+a
+
(7)
if in case there is a free set of cardinality c for all such f , and write instead of i this is not the case. Baumgartner, and later Devlin [1973a], proved that V = L implies (K,
(8)
i f i < k is the least integer such that x iE f (X \ {xi]) holds; if no such i exists, then we put g,(X)=k. If dom (f )= [ K ] < ~then , we shall define g, on the whole of
[ ~ ] < ~ , a n d i f d o r(f n for someintegern, then weshalldefineg/on [K]"". It is obvious in both cases that a set H c K is free with respect to f if and only if g,(X) = IX I holds for every set X c H , X E dom (g,); hence in particular, a set free with respect to f is homogeneous with respect to g,.. Note that if X E dom (g,)and I X I I1 then g,(X) = 1 X 1 holds trivially; hence this equality has to be checked only in case 1x122.
PROOFOF THEOREM 45.2. Given a set mapping f : [K]'"-+[K]'",
define the coloring g,: [JC]<"-+W according to (8). We assumed that K-+(A);O holds; by Lemma 34.13 the relation ~ - + ( 2 ) ;also ~ holds; hence there is a set H of order type 1that is homogeneous with respect to g,.; i.e., there are integers in, n < w , such thatgf(X)=i, hold whenever X G H and IXI=n. Weclaim that theset His free. Indeed, assume, on the contrary, that in< n, holds for some n with 2 I n < w.
2 78
9 0 M E APPLICATIONS OF COMBlNATORlAL METHODS CH.
x
Denote by h: the 5th element of H for any 5
<
For each ordinal 5 with inI < 0, the set Y u { h , ) is an n-element subset of H the inth element of which is h,. So we have gr( Y u { h J ) = i n , i.e.,
<
holds by the definition of g r . As 5 is an arbitrary ordinal here with in I
holds.
PROOF. Let f :[o]'-+[w]'" be a set mapping; we have to show that there is an infinite free set. Define the coloring gr:[ o ] ' + ' - + r + 2 as describqd in (a). According to Ramsey's theorem, there exists an infinite homogeneous set H with respect to this coloring. Let i be the color of H; this means that g ( X )= i holds for all X E [H]"'. We claim that H is a free set with respect to f.i.e., that i = r 1 holds. In fact, assume i s r , and put
+
Y = ( h j : j< i ) u { h , :i +risk < n + r - 1) , where hj denotes the jth element of H . As g ( Y u { h , ) j = i holds for each 1 with i II t i n, we have h, Ef ( Y ) for these I according to the definition of g . This, however, contradicts the assumption that If ( Y ) l
+
The tricks used above exploited the existence of a large homogeneous set. The existence of a large canonical set (see definition 27.1) can also be exploited in a similar way. This is illustrated by the following result; it should be noted, however, that this result will be strengthened below (cf. Theorem 45.6). THEOREM 45.4. Assume
K
is a singular strong limit cardinal and A < K . Then (4
A)+K
holds for every integer r.
PROOF. Given a set mapping f :[ K ] ' - + [ K ] < ~ , define the coloring g r : [K]'+l+Y+2 as given by (8). According to Corollary 28.2, there are pairwise disjoint
SET MAPPINGS OF TYPE
sets Be, 5
(K),
of cardinality < K such that
>1
279
u
B, has cardinality K and the
C
sequence ( B : : 5
< <
u
on the contrary, that g f ( X )= i holds with some i Ir for an (r + 1)-element subset c
X of B. Let xo, . . .,x, be the elements of X in increaging order, and define trby xkE B,, ( k l r ) ; clearly, to<. .. St,. Define 'i as the largest integer I r for
which ti.=&. Denote by b; the ath element of €3;, which x i = b & . Put
let q be the brdinal for
Y = (X \ {xj: isjs i'j)u { b aii.4.j. c, . i<,jIit;
Note that all the br,'s are defined because the order type of Bt, is an infinite cardinal > 1 according to our assumptions. By canonicity, we have i =gr(X) =gf(Y v { b f o r e a c h a < l ( / i / 1); henceb~,+"Ef(Y)holdsforalla
K
and A be cardinals such that K > A and K 2 w, and let r 2 2 be
((ex~,-,(2~))+, r, A)+K
PROOF.Write K~ = K and K ~ =+ (25%)' ~ for each integer i2 1. Then =exp,-,(2!)+ holds for each integer r z 2 . It is enough to show that (Kr,
r, A)-*K
K,=
(91
holds for each r 2 1; we shall use induction on r. The case r = 1 follows from Hajnal's Theorem 44.3. Assume now that r 2 2 and (9) holds with r - 1 replacing r. Let f : [ ~ , l ' + [ ~ ,be ] ' ~a set mapping; we want to define a coloring analogously to the definition of g, under (8), but now we need a sligthly more complicated definition. We shall specify a mappingg;: [rcrlr++,+'2 as follows: given a set X={x,, . . .,x,)GK,, where x o < x l < . . . <x,, put whereg;(X)(i)= 1 ifandonlyifxiE f(X\(~~)),anditisOotherwise.(O~i~r).As
280
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
(2% )+, it follows from the relation ( 2 ” e ) + + ( K , - , i (cf. (15.18) in Corollary 15.3) that there is a set H G K , of order type K , _ , i1 that is endhomogeneous with respect to g;. Let [ be the maximal element of H, and write H’=H\{().Let XcH’beanarbitrarysetofcardinalityr- 1,andlet ( E H‘bean ordinal exceeding max X. We claim that K,=
f (X u { 5 ) 1nH‘E .f (X u { i1.)
(11)
holds. To see this, we first show that if a < ( and a E H‘ \ X, then a E .fWu
{t;)
(12)
Indeed, assume (12) and let i < r + l be the integer such that a is the ( i + l ) s t element ofthe set X u { 5 ) u { a ) .Theng;(Xu{().u{a))(i)= 1 follows from (12) according to (13); hence we have g;(Xu {[)u { a ) ) ( i ) =1 by end-homogeneity, and so (13) holds again by (10). So it is enough to show in order to complete the proof of (1 1) that (12) fails for the remaining values of a, i.e., that a 6 f ( X u ((1.)
(14)
whenever either a- E H‘ and a > 5 or a E X u {c;. In the latter case this holds by the stipulation that the image under a set mapping of a set is disjoint from this set, made in the definition of set mappings; hence we only have to deal with the case when a E H’ and a> (. In this case a is the last (i.e., the (r 1)-st) element of the set Xu{i;).u{a), and so, assuming that (14) fails, we have g;(Xu (g;u { a ) ) ( r ) =1. By end-homogeneity, g;(Xu { ( ) u{a’))(r)=1 holds then for any a ’ H‘ ~ with a’> 5 ; i.e., a’ E f ( X u { ( I ) holds for any such a’. This, however, contradicts our assumption that I f ( X u { < ) ) ~ < A < K , - ~ , verifying (14). Thus (11) is established. Define now the set mapping f’: [H’]r-l+[H’]<‘ by putting
+
f ’ ( X ) = f ( X u {i).)nH‘ for every X E [H’]‘-’. As JH‘I= K,- the induction hypothesis implies that there exists a set F G H’ of cardinality K that is free with respect to f’.It follows from (11) that F is free also with respect to f . The proof of the theorem is complete. Later on we shall comment on the strength of the above theorem, but a t present we give another, more sophisticated, application of the method used here in order to give a generalization of ‘Theorem 45.4. Instead of obtaining a large end-homogeneous set now, we shall obtain a set that might be called ‘endcanonical’.
SET MAPPINGS OF TYPE
>I
28 I
THEOREM 45.6. Let r 2 1 be an integer and T 2 o a cardinal, and let (K : : 5 < T ) be a sequence of cardinals such that T < K , , and exPr-1 (K:)<exPr-I holds whenever
( ~ q )
(15)
4 < q < T , and let 2 be a cardinal < K,,. Then we have
PROOF. Instead of (16), we shall prove the apparently (but not in fact) stronger statemen t expr-1 ( ~ : ) , r , l + ) + C xf. (16')
(C
1< I
CCI
We shall use induction on r. (16) holds in case r = 1 by Hajnal's Theorem 44.3. Let now r > 1 and assume that (16) holds with r - 1 replacing r and with any A' < K,, replacing l (note that (1 5) obviously remains valid if we replace r with r - 1 there). Writing
1 expi-l ( K ~ )
K ( ~ '=
€
for any integer i2 1, assume that we are given a set mapping f : [ K ' ~ ) ] ~ + [ K ' ~ ) ] ' ' . . We havetoprovethat thereisaset ofcardinalitydl'thatisfreewithrespect t o f . To this end, choose sets A,EK('), a < ? , such that I A , l = ( e ~ p , - ~(K,))' and such that each element of A, precedes any element of A, whenever a < j?< T . Define the coloringg;: [ ~ ( ' ) ] ' + l + ~ + l 2 as was done in (10) (replace K, there with K ' ~ ) ) For . each a
whenever X E [
u
it)) = & ( X u { l a ) )
B J , X E 5, and
t E B,
(17)
(the sequence (B, : v < T ) might be
V i r
called endcanonical with respect to the coloring g f ;the arguments used here do in fact form a part of the proof of the General Canonization Lemma [Lemma 28.1]).Fix a
g;(zux)=g;(zuY ) holds for every Z E [
u ,
BJ'l-";
it is easy to see that, for each n, there are at
most 2(r+1).('+28<.1841)<2(r+').exp,-2(k.m)=expr1 ( ~ , ) equivalence classes. Let gl;: [ A , ] ~ r + l - + e x p r - l ( ~be a )a coloring such that two sets have the same color
282
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
exactly if they have the same cardinality and they are equivalent in the sense describedjust before. It followsfrom the relation ( 2 " ) + + ( ~ i + l ) T (cf. (15.17) in Corollary 15.3)with ~ = e x p ~ - ~ ( ~ , )thereisaset that ofordertypeexp,-,(~,); 1 that isend-homogeneous with respect tog:. Choose this set as B,.. It is easi to see that (17) holds for this B,. Writing B =
u B , , we claim that we have
1' < I
f ( y u { 5 I 1nB c f ( y LJ { 5, ). ) nB
(18)
'
whenever Y E [El'- and Y c 5 E B, holds (a < t).In fact, if y E B\ Y and y < <, then (17) with X = Y u { y > implies that y ~ f ( Y u { < l ) holds exactly if y E f(Y u { holds. If y E Y or y = c, then y $ f(Y u { { I ) by the definition of set mappings. Finally, if ~ E B with , a s / ? < t and y > < , then y $ f ( Y u { < ] ) also holds. In fact, (17) with y replacing and with X = Yu{r). implies that y E f ( Y u ( 5 ) ) holds if and only if [, E f (Yu So, if the former relation holds with one y, then it must hold with all y satisfying y E B, and y (;this is, however, impossible,as B, hasordertypeexp,-,(rc&/ Land I f ( Y u {<))111accordingto our assumption. Therefore y q! f(Y u { <>)indeed holds in this case. Thus (18) is com pletely verified. Consider now the set mapping f': [BTJ'-' +[B]5'.'Tdefined as follows: given any Y E [ B ] ' - ' let a Y < ? be the least ordinal such that Y G [ , ~ , and put
[,I)
<
As the cardinality of B equals
{
exp,-,(K:) and 1.T
=-
< K ~ ,there
is a set F free
<
with respect to f ' by the induction hypothesis. It followsfrom (18) that F is also free with respect to f . The proof is complete. Some remarks about the strength of the above theorems, especially of Theorem 45.5, seem appropriate here. We first mention the following simple result of Kuratowski and Sierpihski (see Kuratowski [ 19511): THEOREM 45.7. For any integer k and any ordinal a w e have
(N,+ k , k + 1, N,)%k + 2 .
(19)
PROOF. We use induction on k. The case k = Ois simple and has been mentioned in the introduction of the preceding section (the following set mapping fo: [N,]'+[NJ"= has no twoelement free set: for each ( E N , , let f({{))= = <[= { y : y < <]I).Let now k > O and assume that the assertion is valid with k - 1 replacing k; let X - : [ V , i ( k - I ) ] ' - ' +[N,+(k-I]sNabe a set mapping which has
SET MAPPINGS OF TYPE
283
>1
no free set of k + 1 elements. For each ( < H a + , , let h,: (-+Ka+,k-l, be a 1-1 function, and for any set X & K a i k of k- 1 elements and any ordinal with X S r < K , i , put fk(xu { t i ) = { q < t : h:(q) E fk- 1(hYX)).
r
Then fk is a set mapping from [K,+r]k into [K,+k]
K
be an infinite cardinal and assume that 2 K = ~ +Then . (K',
2,2)frK+.
PROOF. Let A,, u < K', be an enumeration of the elements of [K']'', and for each ( < K + define the set mapping gp r-.[(I1 so that no set A,E with a < be free with respect to g,. This can easily be done as follows. Rearrange the set {&: a < < & A , c t ) into order type V ~ S K : q < v , ) . Using transfinite recursion, pick two distinct elements yp 6, E Bc,\ {yr, 65:(
<
{q:
GCH + (Ka+z, 3,2)%Ko,+ 1
r
284
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
holds in a cardinal preserving Cohen extension (Hajnal-Mate [1 9 7 5 1 ) . Recently, J. Burgess [ 1 9 7 9 ] and K. J. Devlin showed that this relation also holds at least in casea=Oif I/=L.Therearenogoodresultsincaser24;onewouldexpect that GCH + (K3,4,2)j)Ki is consistent, but we cannot prove this. We shall further study set mappings in the next section; to conclude this one we should like to indicate an alternative approach to Theorem 45.5. It works only in case K is regular, since it makes use of the partition relation ( e ~ p , _ , ( 2 ~ ) ) + + (which ~ h , was established only.for I < c f ( K ) (see ( 1 7 . 3 0 ) ) . For the sake of simplicity, we shall confine ourselves only t o the instructive particular case of Theorem 45.5 when r=2, K = K ~and 1 = K , ; that is we shall give an alternative proof of the relation ((2'0)+,
2, w ) + K , .
(20)
PROOFOF (20). Write K = ( ~ ' o ) + . Given a set mapping f : [ K ] * + [ K ] ' " , we have to prove that there exists a free set of cardinality K,. For E < K let 9.: K-+[K]<~ be the set mapping defined by ( B < K and P f a ; for B=a put g,(fi)=O). Theorem 4 4 . 1 of Fodor with I = w implies that K can be represented as the sum of pairwise disjoint sets E,,, n <w, each of which is free with respect to g,. Define a coloring h: [ K ] ~x ~w as w follows: for any u, fl with U < ~ < K put h({aP))=(mn) if U E E,, and PEE,, (m, n < 0). As K = (2'0)+, it follows from the partition relation (2'0)' +(Hi ho(cf. Corollary 1 7 . 5 ) that there is a set H C K of cardinality K , that is homogeneous with respect to h ; i.e., there are integers m, n such that h({a/?))=( m , n) holds for any two distinct a, j?E H.We claim that the set H is free with respect to f. In fact, and let a, B, y E H be ordinals such that a < fl< y. Then a $ f ( { b y ) ) ,as a, B E Eymr we have a $g,(p)=j ( { P y ) ) since E,, is free with respect tog,. We can establish /3 $ f((ay}) and y $ f({a/?)) similarly. The proof is complete. This proof of course works for any regular K . As for the case r > 2, given a set mapping f :[ K ] ~ + [ K ] < ~ we , have to define the set mappings gx: K + [ K ] < ' for any X E [K ] , - by putting g, (a)= f ( X u {a))for any a E K \ X (sx(a) = 0 if a E X), and then the coloring h: [ K ] ' + ~ L can be defined analogously as we did above, with appropriate changes. The details are left to the reader (see also Erdos-Hajnal [19583).
FINITE FREE SETS WITH RESPECT TO SET MAPPINGS OF TYPE
>1
28 5
46. FINITE FREE SETS WITH RESPECT TO SET MAPPINGS OF TYPE > 1 We start out with the counterpart of Theorem 45.7, due to Kuratowski and Sierpiliski (see Kuratowski [1951]). THEOREM 46.1. For any integer k and any ordinal
ci
we
have
(Ka+k, k, K a ) - + k+ 1 .
(1)
PROOF. We use induction on k. In case k = O relation (1) holds vacuously (in fact, the case k = 1 is also easy, and is settled by Theorem 44.3). Let f be a .set mapping, J [Ka+k]k+[Ka+k]
u
t
f(Xu{<',).
Then Ig(X)l < N u +1, and so the relation (Ka+k,k-l,Ka+i)+k, which holds in view of the induction hypothesis, implies that there exists a set Y C Ka+k\Ka of k elements that is free with respect to g . Let q be an arbitrary element of the set K a \ f ( Y ) (this set is nonempty since f ( Y ) has cardinality < H a ) . Itiseasytoseethat Yu {qlisaset ofk+ 1elementsthatisfreewithrespect to f. The proof is complete. The above theorem can be improved in the case k = 2 or 3 (and, of course, in case k = l), but nothing is known in case k 2 4 ; for instance, it is not known whether or not the relation
.
(K4,4,Ko)+6
can be proved in ZFC. If GCH holds, then of course we have (K4, 4, Ko)+Kl in view of Theorem 45.5. The result that improves Theorem 46.1 in case k = 2 and 3 is the following:
THEOREM 46.2. For any integer n and any ordinal
a: we have
PROOF OF (3). Let f: [Ka+2]2+[Ka+2]
286
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
s e q u e n c e H o 2 H 1 2 .. . ? H , ? . properties:
x
..(m
(i )
IH,,,l=Ka+, and {x,,,~: ~ < K , ] G H , , ,holds for every m<w,
(ii)
u $ f ( { x m c v )holds ) for every m<w, < < H a , and u, U E H , , , + ~ .
The construction proceeds by finite recursion as follows. Take H , =Ka+,,and, having defined H, for m Ir and x,; for m < r and 5 < K,, where r < o,choose pairwise distinct elements of H , as x,+ <
=Ka+,and Ig,(u)l
u
i< j <j'sn - 1
f ({xjxj,)).
(5)
Itiseasy toseethat theset { x i : i < n ]isfreewithrespecttoJInfact,ifi<j
u Ei,
where
i
Ei is a set of cardinality > A for every i < k . Let f : E + B ( E ) be a set mapping such that If (x)l A holds for every i < k.
PROOF. First we prove the weaker assertion that there is a free transversal with respect to f; more precisely, we prove that the following Claim holds: Under the assumptions of the above lemma, there is afree set Y such that CLAIM. Y nEiis nonempty for any i < k.
FINITE FREE SETS WITH RESPECT TO SET MAPPINGS OF TYPE
287
>1
PROOF OF CLAIM. Define the sets F i c Ei, i < k, by recursion as follows. Given i < k , if Fj has already been defined for allJ
Now definethe elements yi E Fi, i < k , by recursion as follows. Given i < k, ifyj has already been defined for allj with i <,j < k, then choose yi as an arbitrary element of the set Fi\ f(yj).
u
i<j
It is easy to see that Y = (yi: i < k ) is a free set, completing the proof of the c!aim. We now return to the proof of the lemma. Define the pairwise disjoint finite sets Yt, <
u
s<<
Define now the set mapping g: A'-+P(A') g(t)={rl<J+:y,,n
as follows: for each
u
XE
e
put
f(x)#O)..
Y;
Clearly, ( $ g ( < ) (since & is free) and lg(5)(< A for all (<.A+. According to e.g. Theorem 44.3, there is a &t H E A+ of cardinality A' that is free with respect tog. It is easy to see that the set x = { Y,: 5 E H )
u
satisfies the requirements of tfie lemma, completing the proof.
PROOF OF (4). k t f: [K,+3]3-[K,+3]"* be a set mapping. We start out in a way analogousto the proofof (3)andconstructamatrix ( ~ , , , , : m < w & < < K , + ~ ) of pairwise distinct elements of Nu+,and a sequence H , z HI 2 . . . 3 H , 2 . . . ( m < o ) of subsets of Ha+, with the following properties: (i)
( H m ( = K a + 3and {xmc:< < K U i l ) ~ Hfor m every m < w .
(ii).
~ ~ ~ ( { x ~ ~ x , ,holds , , , u )whenever ) I<m<w, <,q < K , + , and u, U E H , , , . ~ .
The construction proceeds by finite recursion as follows. Take H,=K,+,, xo,=5 (<
,,
gr(u)=
u.If
( { x m t x r q u } ): m
288
SOME APPLICATIONS OF COMBlNATORlAL METHODS CH.
x
for any u E H i . As IHit=K,+, and Ig,(u)l by Theorem 44.3 there exists a set H ; E Hi of cardinality Km+3that is free with respect to 9,. Put H,+ = H;. This finishes the construction of the matrix (xme:m
u
by putting
j l m
Clearly Ih,(u)l
for,jlm. Pick an arbitrary element of E: as x,. This finishes the construction of the set { x i : i < n f . It is easy to see that this set is free with respect to f. In fact, let i,,
After Cohen's result on the independence of the continuum hypothesis, it was soon proved by W.B. Ea'ston [1970] that powers of regular cardinals can be simultaneously given 'arbitrary reasonable' values (that is, Konig's theorem and the monotonicity of cardinal exponentiation has to be obeyed, and the values have to be prescribed by an 'absolute' operation); for about a decade, however, nothing was known about how powers of singular cardinals depend on powers of smaller cardinals except for the relatively simple Bukovsky-Hechler result (see
289
INEQUALITIES FOR POWERS OF SINGULAR CARDINALS
Corollary 6.1 l), and Solovay’s [I9741 result that the continuum hypothesis holds for every singular strong limit cardinal exceeding a strongly compact cardinal (see also Drake’s book [1974]). Silver’s theorem that the continuum hypothesis cannot first. fail at a singular cardinal of uncountable cofinality (see below) was the first real breakthrough. Silver’s results have been further improved by the results of Galvin and Hajnal. Important contributions were made by Baumgartner, Benda, Jech, Magidor, Prikry, Shelah, and Solovay. In another direction, Jensen [1974b] showed that the failure of the GCH at a singular strong limit cardinal implies that O X exists (see e.g. Devlin-Jensen [1974] and Jech [1978], p. 356).Later Dodd and Jensen [1981] also broved that the same implies the existence of a measurable cardinal in an inner model. Silver’s original prooflused model-theoretic methods; Baumgartnkr, Prikry, and Jensen observed that this proof can also be formulated in terms of elementary combinatorial set theory, .using almost disjoint transversals. The results of Galvin and Hajnal are obtained in this latter way. DEFINITION 47.1. Let K be an uncountable cardinal, and let ( A , : u,
ci
< K ) if
,-=K
I{ci < K :
f ( c i ) = g ( a ) )I < K
holds for any two distinct f, g E F . The usefulness of a. d. t.’s is illustrated by the following simple
LEMMA 47.2.. Let K be a cardinal, ( K ~ a: < K ) a sequence of cardinals, and let A = ( A , : a < ~ ) ,where A,= )( K ~ Then . there i s an a. d . t . F ,for A with IFI=
n
B
Ka.
a
PROOF. For each h E )(
K,
and a < K put
U
f,,(ci)= h fi a .
ThenF={f,:he
x ~,)isana.d.t.forAandIF1=1)( K ~ I = n K,.Theproof U
U
U < K
is complete. From now on in this section K will denote a fixed regular cardinal > w. For any sequence A = (A,: U < K ) put T(A)=sup { I F ( :F is an a. d. t. for A ) . 19 Combinatorial
(1)
290
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
We make a trivial but important remark : if A = ( A , : a < K) and B = ( B , : a < are such that ~ A , ~ for ~ all ~ B a < Km, then ~
K)
(2)
T(A)I T ( B ) .
In particular, T ( ( A , :a <
K))=
T((I A,1: c(<
K)).
Note that if cp is a function in “On,then the notation T(cp)makes good sense, since cp is the same as the sequence (cp(a): a < K ) . If cp z.d for some ordinal 6, i.e. if cp(a)= 6 for all a < K, then put
(3 1
T(40)=T(K,6).
In order to obtain good estimates for T(cp),where cp is an arbitrary function in ‘On,we are going to define a partial ordering < I , x on “On,depending on a normal ideal I on K (see Definition 34.1) and on a set X E ~ ( K ) \ I as follows: given q, II/ E ‘On, we put cp < t,h if and only if { a E X : cp(a)2II/(a))E I .
Note that <,,, is well-founded, i.e., there is no infinite ,,-descending sequence (cf. the proofs of Theorem 34.9 and Lemma 31.2). For cp E “On, we define the I,X-rank ~~cp~~,,,of by generalized transfinite recursion as follows: ~ ~ c p ~ is ~ I the , , least ordinal exceeding Ilt,hllr,, for all II/ with t,h ,cp, i.e., ’
II cp I I/,
= SUP {II t,hII/,X
41 I//
(4)
If X 2 Y E P(K)\I , then clearly
II40 (I,., holds since the partial order by
5 II cp 1L.Y
is finer than < I , x . Define the I-rank
(5 1
IlcpII, of cp
II cp Il/= II cp IIIJ . Also, write
INEQUALITIES FOR POWERS OF SINGULAR CARDINALS
291
LEMMA 41.3. Let cp E "On. (a) l f p < ~ + then , there is a function ( P , E ~ Knot depending on 1 such that llcppllI=p and, for all $ E "On, I ( $ ( J r > pifand only if$>/cp,,. If@, is a function satisfying the conditions imposed on cp,, then { a < K : cp,(a) # cpL(a))E 1. (b) I f p < ~ ,then we can put cp,,=p; i.e., ( ( c p l ( , > p if and only if (a
(ncl
The assertion in ( f ) can be stated more generally for arbitrary uncountable K as follows: (f) If1 is the ideal ofnonstationary sets in K, then we have llcpllr< K + for all cp E 'K ifand only iffor every mapping h: [K+]<"'+K there is Q [< K such that for all a < K there is a set A c K + with tp A 2 u and SUP
{ h ( X ) :X E [ A ] ' " ) =(.
This generalization is due to Galvin. To see that the 'if part here includes (f) of the above lemma, one only has to note that the well-known fact that Chang's conjecture is equivalent to the partition relation NZ+[NI]H<~No; the proof of the 'if part of (T)is essentially the same as the proof of (f), given below, in the above lemma. The 'only if part of ( f ) sheds additional light on the situation.
PROOF OF LEMMA 47.3. Ad (a). Put cp,(a)= 0 for all a < K. Let 0 < p < K + , and assume that cpv has already been defined for all v < p. As cf (p)5 K, let ( p c : ( < K) a sequence of ordinals such that p=sup {p:+ 1: < K ) and put
This defines the functions cp,, p < K + . One can easily prove by induction on p that cpy
292
SOME APPLICATIONS OF COMBINATORIAL METHODS CH
x
(see (6) for p?). As p , < p , we have ll$’llI>p2, and so $’>Icpt by the induction hypothesis; hence X , E I. As X = ( X t \ (5 i1)) by the definition of cp,, in (6),
u
:
we have X E I by the normality of I, which contradicts (7). This contradiction proves the ‘only‘if’part of the assertion in the first sentence of this paragraph. Toshow that IIcp,,l(,=pit remains to be verified that llcp,,llI~p.But [[cp,,[lI>p would imply (P,,>~(P,, by what we have just proved, and this latter is absurd. Finally, we are going to establish the assertion in the second sentence of (a); writing cp; + 1 for the function defined by ( c p l i 1) ( a ) = q$(a) i1 for a < K, we have (1 cpk 4 1 (II > )IcppI 1 , = p, and so cp; i1 >]cp,,. Similarly, cp,, 4 1 > I cpl. Hence { a < K : cp,,(a) # cp;(a)) E I, which we wanted to show. Ad (b). Given p < K, put p c = 5 if 5 < p and p: = 0 if p I5 < K in the definition of cp,, in (6). Assuming that cp,,.-p’ holds for all p ’ < ~ we , get cp,,(a)=p for all a with p s a < K according to (6). Changing the values of (p,, at fewer than K places does not affect the claims made about it in (a); hence we may assume cp,(a)=p for all a < p . The assertion in the second clause follows from (a). Ad (c). Let p = K, and put p c = 5 for all 5 < K in (6). Then (6) with cpr, = p’ for p‘ < K (cf. (b)) gives cpK = id,. The assertion in the second clause again follows from (a). Ad (d). For any $, $’ E “On put
$=,$’
iff
{a
E
I.
Clearly, if $ =,$‘, then Il$ll,= Il$’llI. Note that if $<,cp, then there is a $‘ with $‘ s I $such that @’(a)< cp(a) or $’(a) = cp(a)= 0 holds for every a < IC,and such that the number of these ordinals is 5 7 , where we put < I d a )I :a < K & cp@) #@, hence there are c r I-ranks below IIcpll,. Since all ordinals
J=
{ y s K : X n y fi~*llcpll,.xnu>v).
INEQUAUTIES FOR POWERS OF SINGULAR CARDINALS
293
We claim that J is a nontrivial K-complete ideal on K ; in fact, it is also normal, but we shall not show this since we shall not need it. J is clearly nontrivial, as. IG J and X $ J ; the latter holds as =v. It is also easy to see by ( 5 ) that if ZG Y and YE J then also Z E J. For K-completeness, we only have to show that
whenever y < K and Y, E J for all a c y; here we may assume without loss of generality that K G X and Y,n Ys = O whenever a, < y are distinct. We may also assume that Ya4 I for all a < y, since those Y,’s which belong to I will not make any difference (namely, if I: Y‘cK, Y $1, and Y ’ E I then clearly IIcpllr,y = = v‘). Now Y, E J under these assumptions means that lJcpllr,ya,> v, and we V. will be easy; in shall have to show that Y= U Y, E J , i.e., that J J ( P J J ~ , ~ >This a
fact, we are going to show by induction on p that, for any ordinal p and any function $ E “On, if II$ 2 p holds for all a
u
a
Next we establish the following CLAIM. There is a set Y E P(X)\ J and a set F c “1of cardinality < A such that F is cofinal to cp in the partial ordering < I , y , i.e.,for every $ E ”On such that $ < I , y cp there is an f E F such that $ < I , y f v as Y $ I. So Ilfllr,r < v for the f ’ s considered on the right-hand side, and then 11f Il r , u < A as v 2 rZ was chosen minimal. As 1 F I <1and 1is regular, this means that the supremum on the right-hand side is less than 1.Thus the above equality cannot hold; this is the contradiction we wanted to derive. Next we turn to the proof of the above Claim. To this end, consider the function h on K defined by h(t)=cf (cp(t))( C < K ) . We distinguish two cases: 1) there is an Y E B ( X ) \ J such that .his constant on I:and 2) there is XIQsuch K Ad 1). Let the constant value of h on Y be p ( < l ) . Clearly, p is a regular cardinal. For each t E Y let
294
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
( ~ ( 5 (outside ) Y define ha arbitrarily). If
P > K , then put F = {ha:a < p ) . Then If ~ I K then , let F be the set (FI=p
As {< E X: h(<)=a}E J for all ordinals a in this case, it is clear by the K-completeness of J that r] must be the supremum of an increasing sequence of type K of values assumed by h ; hence cf ( q ) = K, and since h only assumes cardinals as values (cofinalities are cardinals), r] must be a cardinal. Also, r] K is regularandqs1,Notingthat h(t)=cf(cp(r)),foreach<E Y let FSscp(5)beaset of cardinality h(c)
3"
F= on 1.This completes the proof of the Claim, and hence of part (e) of the lemma. Ad (f). Assume Chang's conjecture (see Subsection 8.5). It is easy to see from the original, model-theoretic formulation of Chang's conjecture that the following partition relation holds then : K2+"11;:t40
(8 1
7
which means that given any coloring h: [K2]
<,
h(X)=min n { C , , : p , v E X & p < v ) . Then h is a mapping from'[K,] into K,,so by (8) (or, rather, by the clause after (8)) there is a Set A S K , of cardinality H, such that
< = s u p {h(X):x E [ A ] < " ] < o l .
(9)
It is easy to see that 5 E C,, whenever p, v E A and p < v. In fact, assuming this is not the case for some p, v as above, we have 5'= sup (C,,n (5 41)) < 5. Namely, we cannot have <'=5, as 5' E C,, by the closedness of the latter. By (9), there is X E [ A ] < ~such that h(X)><'. Then h ( X u ( p v J ) >
<
INEQUALITIES FOR POWERS OF SINGULAR CARDINALS
295
Given a cardinal p and an ordinal y, we recall that p + ? denotes the 7th successor of p ; i.e. p+O=p, p + ( Y + ' ) = ( p + Y ) + ,and, for limit 7, p + y = s u p p < y p + p . We are now in a position to formulate the Galvin-Hajnal [1975] result that, together with the last lemma above, gives most of the results included in this book concerning powers of singular cardinals. In the applications I will be the ideal of nonstationary subsets of K. THEOREM 47.4. Let K be an uncountable regular cardinal, and let I be a normal ideal on K . Let (oa: a < K ) be a nondecreasing continuous sequence of infinite cardinals (i.e.,om= os+ ,for a > 0), and let cp E "On. Put $(a) =:u ?(')for a < K
u
B
and let (see (3)). Then
PROOF. We use induction on IJcpllI. Assume first that )Icp)lI=O (the assertion with this assumption is a theorem of Erdos, Hajnal, and Milner [1$68] in case the ob)s are strictly increasing). According to Lemma 47.3(b), this assumption implies that { a < K : cp(a)=O) 4 I, and therefore we also have X,
= {a < K:
cp(a)=O& a is a limit ordinal). # I,
as 1 is a normal ideal, which implies that the set of successor ordinals < K belongs to I. Let F be an a. d. t. for $. For f~ F and a E X, we have f ( a ) < om;by the continuity of the sequence om,there is an h,(a)
, A. Since there are only 2" pairs ( X , fl), it It is easy to see that lFx,fllI7 ( ~O@)I follows that IF1 5 2 " .A = A. Assume now JJcpIJI = v >O, and suppose that the assertion holds for any cp' with II cp'll, < v replacing cp. Let F be an a. d. t. for $. For each f~ F, define cp, $fE "On by putting cpf(a)=min {fl: If(a)IIo:p) and $f(a)=o:vf(a) for all a < K . Note that we have either cp,(a)
u
P
-
IF)1A+~~~1~~=A+'itwillsufficetoshowthat IF,,lsA+'foreachp
296
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
To this end, let p < v be fixed. We define a set mapping H on F , as follows: we Put H(f)={gEF,:va
+
for every f~ F , . Then H ( f )is an a. d. t. for f + 1. Since lf(a) 1 I < aTvf(a) = $,(a) for a
PROOF. Define $ E "On by putting $(a)= aTV(lr)for any a < K. According to Lemma 47.5 it is enough to show that T($)I A+11V11' (cf. also ( 2 )above). This is, however, confirmed by Theorem 47.4. The proof is complete. We are going to derive various corollaries of this result. Corollaries 47.W7.13 are improvements by Shelah of the original Galvin-Hajnal results, brought about by his improvement of Lemma 47.3(e). COROLLARY 47.6 (Galvin-Hajnal-Shelah). Let K and 1 be uncountable regular cardinals such that 1> K and T" < 1 holds for all cardinals 7 <1 with cf ( 7 )= K . Let, further, u be an infinite cardinal such that r~' < u", and ( K ~ :a < K) a sequence of
291
INEQUALITIES FOR POWERS OF SINGULAR CARDINALS
cardinals such that Kg
holds for every a < K . Then
PROOF.Choose aa=a for all a < u in the preceding theorem, and define
cp E "On by putting
cp(a)=min for all a < K (i.e., either q(a)= 0 or
n
(r:
~ K , I C T + ~ >
B
K~
= a++"')). Then we have 11 cp [Ip<
1for any
B
normal ideal I on K (the simplest is to choose I as the ideal of nonstationary sets in K ) according to Lemma 47.3(e). It is easy to see that with the A in (10) we have A I 2" * Q" < a+', where the second inequality holds by our assumption a" < a". So, using the preceding theorem, we get
n
K , I A + ~ ~ ' P ~ ~-~ < A + ~ < Q + ~
U
(as A 2 Q obviously, we actually have equality instead of the last inequality), which completes the proof.
COROLLARY 47.7 (Galvin-Hajnal-Shelah). Let K , 1,and Q be as in Cofollary 47.6. T < K , then p"
I f p r < a + ' for all
PROOF. Put ~ * = pin the preceding result. COROLLARY 47.8 (Galvin-Hajnal-Shelah). Let K , A,and Q be as in Corollary 47.6, and let p b e a cardinal such that cf(p)=u. I f 2 r < ~ +for ' all r e p , then 2 p < ~ + ' .
PROOF. Write p =
n B
1 pa, a<"
I C ~ = ~ ~ < Q where + ~ ,
Corollary 47.6.
z=
pu
1 pa
~,=2Pa.
Hence 2p=
Then, for any
a
fl K,
COROLLARY 47.9 (Galvin-Hajnal-Shelah). Let K be an uncountable regular cardinal p and z cardinals 222, and 5 and t,~ ordinals, V < K . If p'<&i(rK)++l for all cardinals Q < K , and k $ < K < i ( r W ) + V i l then p ' < & i ( r ~ ) + V ~ l . PROOF.Put o=K, and A=(z")+V+' in Corollary 47.7. Using the same substitutions, Corollary 47.8 gives the following:
298
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
COROLLARY 47.10 (Calvin-Hajnal-Shelah). Let K be an uncountable regular cardinal, p and T cardinals, assume T 2 2 and cf ( p ) = K , and let 4 and q be for all cardinals 9 < p and K;
PROOF. Replace 5, q, T, p, and preceding corollary.
K
with 0, q, 151, K,,and cf (0 respectively, , in the
COROLLARY 47.12. Let ( and q be ordinals with cf(t)>q, w. Assume K~
PROOF. The result follows from Corollary 47.9 with 0, q, 151, cf (5), and K: replacing 5, q, T, K , and p, since we have K: I 1 K,"
1
a<,
for any cardinal a
C~ROLLARY 47.13. Let q be an ordinal with q<w,. for all a < q , 2*a<8(Znl)+li~ (i) (ii) r f K,No
then 2 H w 1 < K ( ~ w l ) + + ~ . then K ~ : < K ( 2 w l , + 7 + 1 .
The last result in this series (there are more to come later) involves Chang's conjecture. Its original proof involved more complicated ideas than used here; the Baumgartnei-Benda result (Lemma 47.3(f)) made it possible to treat this result in the present framework.
COROLLARY 47.14 (M. Magidor [19771). Assume Chang's conjecture (see Subsection 8.5), and assume that 2N-< KmIholds for all a < wl. Then 2',1< KW1.
PROOF. We are going to apply Theorem 47.5. To this end, write K = K,,6,=No, K ~ = K (, a < o l ) and define cp~'lOn by stipulating that
n
hCp(@ = K,( I2'.
holds for a < w,. Then cp(a)
INEQUALITIES FOR POWERS OF SINGULAR CARDINALS
299
that A = 2 K ~
n K, < (Nu!
)+K2 =Nu*
a
holds. Here the left-hand side equals
n K,= n 2K=2=.
a
a
which completes the proof. These results leave the following problems open: 47.15. (i) If 2K"
COROLLARY 47.16. Let K and 1 be infinite cardinals with K = C ~@)>a such that p"<1 for all p < l . Let
300
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
ideal on
K.
for every
lf we have
U
n 1,s
x (16)
il,ya)
8<0
then
A K
5 ~ f l l ‘ P l l r.
(17)
What we really need instead of the assumption that (A,: cc<~)‘is strictly increasing is that 2,
PROOF. Take o a = ~ = = Aina Theorem 47.5. It is then easy to see that for the cardinal A in (10) we have AS
cilI5il,
a
and so (13) in Theorem 47.5 implies (17), completing the proof.
COROLLARY 47.17 (Silver [1975]). Let K and il be infinite cardinals with K = ~ ( I ) > osuch that p K < l for ail p < l . Let (La: a < K ) be a strictly increasing continuous sequence ofinfinite cardinals converging to 1,and suppose that the set
{U < K :
n 1,s I:”)
P<m
is stationary in
K
for some ordinal p <
K.
Then
I“sA+’. If also 2‘51“ for all T < A then 2111”. PROOF. The first assertion follows from the preceding corollary with 1 chosen
astheideal of nonstationary sets in view of Lemma 47.3(b). As for theassertion in the last sentence, this follows from the first assertion by noting that under the additional assumption VT
The proof is complete.
n a
2i* 5
n 1“=1“. U
COROLLARY 47.18 (Silver [1975]). Let A be a singular strong limit cardinal with cf (A)= K > w, let (1,: a < K ) be a strictly increasing sequence of infinite cardinals converging to I , and let p < u be an ordinal. If ( u < K : &I ,I:’is), stationary in
n
K,
then PIA+’.
PROOF. The assertion follows from the last sentence of the preceding corollary. COROLLARY 47.19 (Silver [1975]). Let I be a singular cardinal with cf(l)>w, and let p
PROOF. Note that 2 is a strong limit cardinal. Let (Aa: a < K ) be a strictly increasing continuous sequence ofcardinals tending to 1.Then {u < K : 2 ’ - I A:”]
301
CARDINAL EXPONENTIATION AND SATURATED IDEALS
is stationary in
K
(see Theorem 5.2). Since
n 1,
I%== 2i=, the result follows
,
from the preceding corollary.
COROLLARY 47.20 (Silver [1975]). Let p < w , . I f is stationary in K,,then K::
p l < K u I
and {o!<w,: K ~ O I
PROOF. By Corollary 47.17, since we have EE;I = 2 K .~K> < Kw, for any o! < m i , as is easily seen. In fact, the equality above is trivial in case a s 1; if a> 1, then, using induction on a < o1and applying Corollary 6.4, we obtain
(2KI.Ky3.KC+])5
= lIC
fl
(2KI.K>.K,)=2h’,.~OI~I,
1sCia
which we wanted to show. The proof is complete.
COROLLARY 47.21 (Silver [1975]). Let p t o , . If rC*
PROOF. By Corollary 47.18. COROLLARY 47.22 (Silver [1975]). Let p < w , . I f { a < o l : p * < K a + p ] .i s stationary in o,, then 2K-1
PROOF. By Corollary 47.19. I The next result is meant, only as an illustration of further consequences of Corollary 47.!6. COROLLARY 47.23 (Silver [19751).If 2Nn
IN,
zl
PROOF.Corollary 47.16 with K = K , , A=K,,,, and I being the ideal of nonstationary sets on K, g i v e ~ 2 ~ ~ ~ 1 K .,in , , view of Lemma 47.3(c).Equality here is impossible in view of Konig’s theorem (Theorem 6.9), completing the proof.
48. CARDINAL EXPONENTIATION AND SATURATED IDEALS The moral of the results of the preceding section is that cardinal powers of singular cardinals (of cofinality >w, at least) are usually affected by cardinal powers of smaller cardinals. Something similar happens at regular cardinals carrying certain saturated ideals (for saturated ideals, see Section 34). The first
302
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
result in this direction is Scott's result (see below) saying that the GCH cannot first fail at a measurable cardinal; this result was recently generalized by Ketonen. That something similar can happen for small cardinals was observed by Baumgartner, who found a connection between the values of 2uo and 2u~,and the maximal number of almost disjoint subsets of K , (i.e., sets such that any two of them have countable intersection). This result has been generalized by Jech and Prikry to the case in which the ideal of countable sets on K, is replaced by an arbitrary K,-complete ideal. Before we can discuss the result outlined above, we need some preliminaries about saturated ideals. To this end, let K be a,yfnfinite cardinal and let I be a (3-complete) ideal on K . The saturation sat (I)of I is a cardinal defined as follows: sat (I)= min {A: I is A-saturated) .
(1)
We need the following simple lemma:
LEMMA48.1 (Tarski). Let K K.
~ be Wa cardinal, and let I be a (3-complete)ideal on Then s a t (I) is either finite or regular.
PROOF. Assume that A=sat (I) is singular, and let (I,: C
303
CARDINAL EXPONENTIATION A N D SATURATED IDEALS
Again, as in the preceding section, the results here below can be approached either by elementary methods or by model-theoretic methods using generic ultrapowers. Unlike the situation in the preceding section where the elementary method gave more general results than can at present be obtained by modeltheoretic methods, here the model-theoretic approach does not seem to be inferior to the elementary one, and all the results discussed below could just as well be obtained by using the former. In fact, the former approach here seems to have some distinct advantages. True to the spirit of this book, we shall nonetheless apply combinatorial methods to derive the results below. The framework here slightly differs from the one used in the precedilbg section; instead of full transversals, we shall have to use partial transversals here, as described in the following definition: DEFINITION 48.2. Let K be an uncountable regular cardinal, I a K-complete nontrivial ideal on K (see Definition 34.1(i)), X E ~ ( K )a \ set, ~ and A = = (A,: a E X) a sequence of sets. A set F is called a sad pat (set of almost disjoint partial transversals) for A and I if the following hold (i) each f~ F is,a function with dom (f)E 9 ( X ) \ l such that if a E dom (f)thenf(a) E a, and (ii) if f, g E F and f # g then dom (fng) E 1 (this latter is the same as saying that {aEdom ( f ) n d o m ( g ) : f ( a ) = g ( a ) )€ 1 ) . From now on to the end of this section, K wili denote an uncountable regular cardinal, and I a Kcomplete nontrivial ideal on K. The relevance of sad pats for cardinal exponentiation is illustrated by the following lemma, which is an analogue of Lemma 47.2. LEMMA 48.3. Let ( K ~ :a < u ) be a sequence of cardinals and write A= =(Aa: a < K ) , where A a = F u p . Then there is a sad pat F for A and 1 with
n
IFI=Q < K Ma. PROOF. The result follows directly from Lemma 47.2, as an a. d. t. for A is also a
Q
sad pat for A and 1 no matter what the ideal 1 is. (The result is also true in case K = O , but this case is of no interest to us here.) -For any sequence A = ( A a :a E X), where X $ I, put PT(A,I)=sup {IF'I': F is a sad pat f0T-A and I}.
(2)
It is worth comparing this to (47.1). We make similar remarks as we made after (47.1). If A=(AQ: a E X ) and it=(€?,: a E X ' ) are such that X E X ' and IA,IIIB,I for all a E X , then PT(A, l ) I P T ( B , I).
(3 1
304
SOME APPLICATIONS OF COMBlNATORlAL METHODS CH.
x
In particular P T ( ( A , : a E X ) , I ) = P T ( ( 1A.I: a E X ) , I ) .
Note that if cp is a function in ‘On, then the notation PT(cp, I) makes good sense, since cp is the same as the sequence (cp(a): a
(4 )
The following result, due to Jech and Prikry, will enable us to derive our main result concerning the relationship between cardinal exponentiation and saturated ideals on K in case K is a successor cardinal. THEOREM 48.4 (Jech-Prikry [1976]). Let K and p be infinite cardinals such that is a successor cardinal and K < p < K + ‘. Then
K
P T ( K ,K-, I)<sat (I) and
P T ( ~ , p , I ) < ( p . s a(I))+ t (sat (I)was defined in (1)).
PROOF. Write I=sat (I).Then A is regular by Lemma 48.1 and A>K
(7 )
in view of Theorem 34.4. For an ordinal p write cp= ( p : a < K ) ;
(8)
i.e., let c , denote the constant function on K with value p. We prove ( 5 ) first. To this end, let F be a sad pat for c,- ,and assume, contrary to what we want to prove, that IF122. For ~ E and F ( < K - put
Xf = { a E dom (f):f ( a ) = () . Clearly,
u Xf c
.
= dom ( f ) $ I; so, by the Kcompleteness of I we obtain that for
each f there is a tf< K - such that X{, q! I. By (7) and the regularity of I,there is a ( < K - such that the set F,={fEF:(/=(} hascardinality 2 1 . Then the set { X i : f~ F,} refutes the I-saturatedness of I; in fact if f # f’and f’E F,, then Xf, X.f’ $ Z;and X f n X C E I (this latter holds because f and f ’coincide on XfnXf’).This is a contradiction, establishing ( 5 ) . We are now going to prove (6). We distinguish three cases: 1) p = K , 2) p > K is regular, and 3) p > K is singular.
30 5
CARDINAL EXPONENTIATION AND SATURATED IDEALS
Ad 1).Assume, on the contrary, that (6)fails with p = K , and let F be a sad pat of cardinality ' 1 =(sat (I))' = ( p * sat (I))' (the latter equality here holds by (7)) for the constant function c, on K. Define the set mapping H on F as follows: For any f E F put H ( f ) = { g E F : { a E d o m ( f ) n d o m ( s ) : g ( a ) < f ( a ) ;$ 1 1 .
For a fixed 1; we want to estimate the cardinality of H ( f ) . To this end, write g = g p { a ~ d o m( f ) n d o m ( g ) : g ( a ) < f ( a ) ] . for any g E H(f ). Note that if g , , g, E H ( f )are distinct, then so are g; and g,, as g, and g, are almost disjoint partial transversals; and, moreover, the set of g's (s E H ( f ) )is a sad pat for f : Hence, using some earlier observations (see below), we obtain that H ( f )Ithe number of g's < PT(1; I ) ( < P T ( c , - pdom (f),I ) I P T ( K ,K - , I ) < 5 sat (I)=I;
here the third and fourth inequalities follow from (3), and the fifth holds in view of (5). Thus H ( f )< 1 for any ~ E Fas; IF I =A+, it follows from Hajnal's Theorem 44.3 that there is a set F'c F of cardinality 1 ' that is free with respect to H. We claim that the elements of the set
X = {dom ( f ) : f ~ F') are almost disjoint with respect to I, i.e., that dom (f,)ndom (f,) distinct f,,f, E F'. In fact,
E Ifor anytwo
dom (fi)ndom ( f z ) = = { a < K : f i(a)< f,( a)) u { a < K : f l( a )= f2 (a)>u { a < K : fl( a )> f2 ( a ) ),
and the first set on the right-hand side here belongs to I because f,$ H ( f , ) (this latter holds since F' is a free set), the third, because f2 $ H ( f , ) , and the second, because f, and f2 are almost disjoint partial transversals. The set X above has the same cardinality as F', i.e., IXI=A+. This, however contradicts the assumption that I is I-saturated. In the remaining two cases we use induction on p, that is, we assume that (6) holds if we replace p with any p' satisfying K I p'
F , = ' ( ~ F: E VaEdom ( f ) [ f ( a ) < < ] )
20 Cornbinatorid
306
SOME APPLICATIONS OF COMBINATORIAL METHODS C H .
for any ordinal t < p. We have
F=
x
u F,,
:
as p > K is regular in the case considered. So, using the induction hypothesis, we obtain that IFII
c IF,I<
S
1Itl.sat ( I ) s p - s a t( I ) < ( p . s a t ( I ) ) + ,
S
showing that (6) holds in the present case. Ad 3). As before, let F be a sad pat for cp. Let ( p S :t
F= We claim that In fact, for f e F , put then
{f:f e F , )
U
F:.
C
IF,I
f = f b{ctedom ( f ) : f ( a ) < p J ;
is a sad pat for cp,, and so
I F <= I ~ {f~ f :F;)I < P T ( K ,p S ,I ) ,
as was claimed. Hence, using the induction hypothesis, we obtain that
showing that (6) holds. The proof is complete. The above theorem and Lemma 48.3 give the following result due to Jech and Prikry [1976]. In case I is the set of all subsets of cardinality < K of K, this was obtained by Baumgartner [19701 earlier.
THEOREM 48.5 (Jech and Prikry [1976]). Let 7 be an infinite cardinal, K = T + , and assume that 2'< K + " . Let I be a K-complete nontrivial ideal on K . Then 2"12'.sat (I).
PROOF. Using Lemma 48.3 with ~ , = 2 'for all a< PT(K, 2',1)>2".
(9) K, we
can see that
CARDINAL EXPONENTIATION A N D SATURATED IDEALS
307
(6) in Theorem 48.4 implies that
PT(K, 2', 1)s(2'. Sat (1))' as we assumed that 2'< K + " . These two inequalities imply (9), completing the proof. Our next result sharpens Theorem 47.4 for certain specific cp's provided the ideal I has good saturation properties. This result will give us estimates of powers of certain regylar limit cardinals. In order to state our result, we define where 1 is a cardinal and a 2 1 an ordinal, as $ a < o and as A"ifa2o. In particular, for an ordinal p we have i - ' + t p + ' ) = l + p if p < o and A-'+(p+l)= = A + ( p ' + ' ) i f p 2 w . We shall simply write A-l+p+' for 1 - ' + ( p + ' ) We have 48.6. Let. K.be an uncountable regular cardinal, I a normal ideal on K , THEOREM and (nu: u < K ) a nondecreasing continuous sequence of infinite cardinals (i.e., nu=
u
ap+l for u>O). Let p < ~ +be an ordinal, and let cp,,~"Onbe the
@
function described by Lemma 47.3(a). Define $,, E "On by putting $,,(g)=a:s*(s) for all u < K and let A,=sup{( a c K P u ) + : P u < P T ~ K , % , l ) ] . (10)
c
Then (11)
PT($J)9l,?+~+'.
PROOF. The proof bears some similarity to that of Theorem 47.4. We use induction on p. Assume first that p=O. According to Lemma 47.3(b), {aeic: cp,(a) # 01 E I, so we may just as well assume that cp, = 0, i.e., that Il/,(u)= nufor all u < K. Let F be a sad pat for $o and I; we have to prove that 1 FI < A , . To this ~ be arbitrary. Putting A = { ~ < K a: is a limit ordinal), K \ A is end let f r F nonstationary, and so IC \ A E I since, as is easily seen, a normal ideal contains every nonstationary set. As dom ( f ) $ I, we therefore have dom ( f ) nA $ I as well. If a E dom ( f ) nA, then f ( a ) < $t0(a)= au,and, as a is limit, we have f(a).< for someh,(a)
1
20.
308
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
Assume now that O < ~ < K +and , that the assertion to be proved holds with any p' < p replacing p. Let F be a sad pat for (P,, . For each f~ F define (P, E "Onas follows: cpf(a)=min {B: If(a)lra,'p] provided a E dom ( f )and p,(a)= 0 if a $ dom (f).Put j d f
= IIp,ll,;
then we have (12)
P,
by Lemma 47.3(a), as we have either cp,(a) < cp,(a) or cpf(a)= 0 for every a < K. Moreover, Lemma 47.3(a) also implies that X,=(a
(13)
For v < p put F , = { fF ~ :p,=~); we have F =
u F , according to (12). To see
v
(ll),we have to show that IF1 <
< A;'+P++'.Asp
I F,I I (14) holds for every v < p. To see (14), let v < p be fixed. Define a set mapping H on F , by putting H ( f ) = { g EF,: {aEdom ( f ) n d o m ( g ) : g (a )< f ( a ) j $ 1 ) for every f~ F , . We want to have a good estimate for the cardinality of H ( f ) .To this end, fix f~ F, and put
g=g
/-
(dam ( f ) n d o m k ) \ X , )
for any g E H ( f ) (see (13) for X,). Clearly, if g#g', then g i g ' (cf. (13) and the definition of H(f)), and so IH(f)l= the cardinality of the set B(f)= = (9: 9 E H ( f ) } . Ths latter set is a sad pat for f i" (dom ( f ) \ X,) and I, and so
IH(f)I = I R(f)l< PT(f
(dam
To estimate the right-hand side here, define $, all a < K, and observe that
E
(f)\ XfL 1 ) .
(15)
"Onby putting $,,(a) = a,+VJa)for
I f ( a ) l s $,(a)
for all tz E dom ( f ) \ X , by (13), as p,=v in the present case. Hence, in virtue of (3) above, the right-hand side of (15) is sPT($,
/-
(dom ( f ) \ X , , I ) s P T ( $ v ,I ) ~ L I ; ~ " " ~ ,
CARDINAL EXPONENTIATION AND SATURATED IDEAIS
309
where the last inequality follows from (11) with v replacing p, which holds according to the induction hypothesis. Comparing this to (15) we obtain that IH(f)I
for every f~ F,. Assume now that (14) fails, i.e., that lF,l > A;14v41. Then using the above estimate for I H ( f ) l , it follows by Hajnal's Theorem 44.3 that there is a set F'G F, of cardinality A;'4vi2 that is free with respect to H. Then the set dom ( f ) n d o m (g)={ a < K : f ( a ) < g ( a ) ) u { a < ~ : f ( a ) = =g (a)>u{~<~:g@)
belongs to I for any g E F'; in fact, the first set on the right-hand side,belongs to I because f$ H ( g ) , the second one, because f and g are almost disjoint partial transversals, and the third one, because g $ H ( f ) . So the elements of the set {dom (f):f~ F') are almost disjoint with respect to I; in other words, defining f by,stipulating that dom (f) =dom ( f ) and f(a)= O for any a E dom (f), the set {f: f~ F']is a sad pat for the indentically 1 function on K ; i.e. IF'[ =
I{fl f E F')I < P T ( K ,1 , I ) I A , ,
which is a contradiction, since (F'(=d;'4'+2.The proof is complete. We are now going to list some corollaries of the above result. The first one is an analogue of Theorem 47.5; in fact, the only change here is that we use A, instead of A, but then we have to assume that I I ( P I ~ , < K + . THEOREM 48.7. Let K be an uncountable regular cardinal, I a normal ideal on K,
and assume that p
holds for every a < K. Then
x $1)
9
(16)
(u; a
310
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
x
PROOF. Define the functions $ and $,, on
K by putting $(a)=a,+q(u’ and where cp, is the function described in Lemma 47.3(a). asp<^+, (16)implies via Lemma47.3(a)that { a < K : cp(a)>cp,(a)j € 1 . Hence we also have { a < K: $ ( a ) > $,(a)] E I, and so, as is easily seen,
$p(a)=ca+(Pr(a) for every
a
W $ I,) s P T ( $ , , 1 ) holds. The right-hand side here is sd”+”++’by ( l l ) , and the left-hand side is > K, by Lemma 48.3 in view of (17).Thus (18) is established, completing the
n
01
proof. COROLLARY 48.8. Let K > O be a regular cardinal, I a normal ideal on K , and assume A> K is a cardinal such that I is A-saturated. Let cp E “On be afunction with II(PI~,
2 “ s1-1i l l r p l l l .
PROOF.We are going to use Theorem 48.7 with K , , c, ( a < K ) defined as follows: for each a < K put K, = I a I. The definition of the 0,’s is slightly more complicated,
as we have to make sure that the sequence (c,:a < ~ )is continuous. To this end, given a < K, put 0, = a if a i s a cardinal, put c,= I a I if a is not a cardinal and Ial+ < K, and, finally, put c,= I a1 if la1 = K (note that we have 6,= la1 for a closed unbounded set of a’s because either K is a successor cardinal, or the cardinals < K form a closed unbounded set). Observe that (19) implies (17) with the above choices of K, and 6,. Writing p= llpll,, (16) is, however, not necessarily satisfied, and to make it valid we shall have to replace the ideal I with another one. To this end let (P, be the function described in Lemma 47.3(a) (note that p < K + ) . We have +
+
A = { a < K : cp(a)s cp,(a))$ I
according to that lemma. Put
I’={YEK: A n Y E I ) . Then I’ is a I-saturated normal ideal on
K;
as
we have IIpIlr,xsIlcp,,llr,x for any X E ~ ’ ( K ) \ ~Note ’ . here that the function cp,
311
CARDINAL EXPONENTIATION A N D SATURATED IDEALS
in Lemma 47.3(a) is independent of the ideal I; hence, by that lemma, I I ( P ~ ~ ~ ~ , ~ = I I ( P ~ I I , . , =Y~ E, ~K~ :~ X ~ ~Y~I~"I=' ]{. T h a t i s ,(16)issatisfied with I' replacing I. Hence we have 2"= lal
n
a<"
according to (18) with I' replacing I, as K , = ( a (in the present case. Thus, to see (20)we shall only have to show that A,. 5 A. In doing so, we may choose A the least possible cardinal compatible with the assumptions above, i.e., we may assume that A=max { K + , sat ( I ) ; . Lemma 48.1 then implies that l i s regular. Noting that 6,< K holds for all u < K in the present case, A,, 51will therefore folios/ if we show that P T ( K , '5, 1')51 (21) holds for each cardinal 7 < K (cf. the definition of A, in (10) above). To this end let F be a sad pat for the constant function having 7 as value on K and for 1'; we have to show that (FI
and write
9
F , = { f F~: Xf,< $1;.
We have IF,] < A in view of the I-saturatedness of I', as the sets X,,,, f~ F,, are almost disjoint with respect to 1';and, mbreover, we have F = F, by the
u
<
x-completeness of I'.So IFI < A by the regularity of A, which we wanted to snow. The proof is complete. We mention some corollaries of the above result. The first of these was obtained by J. Ketonen [1974] in case p = 1. COROLLARY 48.9. Let
K,
A be cardinals, p an ordinal, and assume K >w is regular,
A > K , and p < ~ .Assume, further, that I is a I-saturated normal ideal on If {a
K.
PROOF. The result follows from the preceding corollary and Lemma 47.3(b). COROLLARY 48.10 (D. S. Scott). If K is measurable (see Definition 34.1qi)) and 2p= pt for any p < K , then 2" = K+. In other words, the GCH cannotfirstfail at a measurable cardinal.
PROOF. This follows from the preceding result with I = .'K (Note that saying that K is measurable means it carries a 2-saturated Kcomplete nontrivial ideal; the existence of a normal 2-saturated ideal on K follows from Theorem 34.9). The next result is now only of historical interest, because Theorem 48.5 gives the same conclusion under weaker assumptions.
312
SOME APPLICATIONS OF COMBINATORIAL METHODS CH.
COROLLARY 48.11 (J. Ketonen). lj 2'o=K1 and saturated K,-complete ideal on K , .
x
P I= K 3 ,then there is no K 2 -
PROOF. Use Corollary 48.9 with K = K ~ A, = K 2 , and p= 1; then it fdlows that there is no K,-saturated normal ideal on K,, and so the-result follows via Theorem 34.9. Theorem 48.5 gives the same conclusion under the assumption 2';IK2 instead of 2 K o =K,.In general, it may be observed that Theorem 48.5 is stronger than Corollary 48.9 in case K is a successor cardinal.
CHAPTER X1
A BRIEF SURVEY OF THE SQUARE BRACKET RELATION
Negative square bracket relations can be used to express stronger combinatorial assertions than the corresponding negative ordinary partition relations. We start with a discussion of the classical ErdBs-Hajnal-Rado results on such relations under the assumption of GCH, and will then go on Fo consider a result of Jensen and Shore about negative relations which follow from the existence of a Suslin tree. The theorems of Galvin and Shelah give important nontrivial negative relations without the assumption of GCH. Then we take a look at what positive relations can be obtained with the aid of the General Canonization Lemma; this will be the only application of the case r 2 3 of this lemma to partition relations in this book. The chapter is concluhed with a discussion of infinitary Jonsson algebras. 49. NEGATIVE SQUARE BRACKET RELATIONS AND THE GCH A theorem of ErdBs, Hajnal, and Rado [1965] says that if cardinal such that 2"= K + , then K++[(K:
K
is an infinite
K+)},'+
holds; this relation is much stronger than the Kuratowski-Sierpinski relation (cf. (19.14)). A further improvement on (1) is the following theorem, which we shall give with proof. This result cannot be expressed in terms of the partition relations introduced above:
2"74(~',)'K
THEOREM 49.1 (Erdbs-Hajnal-Milner).
If K is an infinite cardinal with 2" = K',
then there is a coloring f : [ K + ] ' - + K + such that for every X , Y S K + with ( X I = K , 1 Y 1 = K + there is a 5 E X such that { f ( ( t a } ) :a E Y',= u+ (i.e.,such thatfor every color there is an edge of that color from ( into Y ) . PROOF. Using the assumption 2" = K + , let ( h a : a < K ' ) be an enumeration of all functions of cardinality K (i.e., [ h a = [ ldom (h,)l = K ) the domains and ranges of which are subsets of K + . Given a< K', we are going to specify the colors f ( { ( a ) ) ,
314
A BRIEF SURVEY OF THE SQUARE BRACKET RELATION
(
I
by transfinite recursion, and specify f ({ tga,")a]) according to (2);for any 5
(3)
holds for any singular cardinal K (cf. (19.12)in Corollary 19.7).Here we are going to show that under the assumption2"= K + the following strengthening of this relation holds: K + %[Cf
(K)+, (K+
: K+)K+]2
(4)
(this result holds of course also in case K is regular but then it says less than (1)). Similarly as in case of (1) and Theorem 49.1, we shall prove a result stronger than (4).
THEOREM 49.2. Let K be a singular cardinal with 2"=K+. Then there is a coloring f :[ K + ] ' + K + such that (i)for any X G K + of cardinality cf ( K ) + we have 0 E f "[XI' (i.e., there is an edge of color 0 in X ) , and (ii) for any sets X,, X , c K + of cardinality K + , there is an i=O or 1 and a { E X isuch that K + \ (0)c { f ({ ( a ) ) : a E X , -i] (i.e., such that for every nonzero color there is an edge of that color from ( into X , -i).
PROOF. We have I & ' ( ~ ) = K + under the assumption 2 " = ~ +Let . <' be a vellordering of type K + of c f ( K ) ~ , and let <' be its lexicographic ordering (see Definition 19.2). Defining the Sierpinski partition Z'= ( I ; , Z;) by 1; = { (Xy ) 1 X, y
Ed ( K ) K & X
< ' y & X <' y)
(5)
and
NEGATIVE SQUARE BRACKET RELATIONS AND THE GCH
315
1; = { {Xy) : X, V, E cr‘K)K& X < ’ y &.y<’X). ,
(6)
we know from Corollary 19.7 that this partition establishes relation (3). Let the 1-1 onto mapping that takes the ordering ‘less than’ of ordinals over into the wellordering <‘ (ie., 5 < q o g ( < ) < ’ g ( q ) for any 5 , q c ~ + ) , and define the ordering < by
~ : K + + ‘ ~ ( ~ be ) K
5~V1-g(O<’gh) ;
(7)
then the partition I = ( I o , I,) defined by and clearly also establishes relation (3). By further subdividing the class I, in this partition we shall be able to establish the stronger relation in (4), and, in fact, the assertion of the theorem to be proved. In order to simplify our notation, if X C K + and < < K + then write X < < for V q ~ X [ q < t ] ;the expressions < < X , X < < , < < X , X
<
<
f({ta))=h,(<) ; ’
(10)
this can easily be done in a way similar to what we did in order to make (2) valid in the proof of the preceding theorem. This completes the definition of the coloring f. We are now going to show that this f satisfies assertions (i) and (ii) of the theorem to be proved. As for (i) this is obvious from the fact that, as remarked above, the partition I defined in (8) and (9) establishes relation (3). To see (ii), note first that if X E K + and ( X ( = K + then , there is a < < K + such that ){qEX:q<<)1=I{qEX: < < q ) l = K + . (11) In fact, putting
x’=( 5 EX:l { q EX:q i < ) l ( K ) ,
316
A BRIEF SURVEY OF THE SQUARE BRACKET RELATION
there is, by the Hausdorff Cofinality Lemma, a set Y c X ’ wellordered in the ordering < that is cofinal to X ’ in this ordering. We may also assume here that the order type of Y in < is a cardinal; this assumption immediately implies that I Y 15K, since < is isomorphic to 4’(cf. (7)), and there is no increasing sequence of type K + in the lexicographic order <’; we proved this latter fact when in Corollary 19.7 we showed that the partition in ( 5 ) and (6)establishes relation (3). As X’= (qEX:q
u
5E y
and each of the sets on the right-hand side has cardinality IK by the definition of X ’ , and I Y 11 K as remarked above, it follows that IX’I i K . Similarly, putting
x ” = { (E x : 1{q E x : ( < q ] l < K ’ , , we can conclude that I X”I IK . Any element 5 of the set X \ ( X ’ uX ” ) , which has now been shown to have cardinality K ’ , will satisfy (11). We are now ready to verify (ii). To this end, let X , , X , C K + be sets of cardinality K + , and let toE X , and t, E X , be elements for which (11 ) holds with these sets replacing X ; assume e.g. that to<(,.We are going to show that then (ii) holds with i = 1.Assume the contrary, i.e., that for every 5 E X, there is a color h(5) E K + \ (0) such that f(((a)) # h ( 5 ) holds with any a E X , . Put and then IY,I=IY,I=K+ by ( l l ) , and, moreover Yo
holds for a singular cardinal under the assumption of GCH. Here denotes the set of all graphs
[.“‘I
NEGATIVE SQUARE BRACKET RELATIONS AND THE GCH
317
where A, B E K + are sets of cardinality K + , and the relation in (12) is interpreted according to the convention set down in Subsection 8.8. In Section 25 we proved that for any infinite cardinal K the relation 2"+(ri+, 4)3
(13)
holds (cf. Corollary 25.2). Under the assumption 2" = K + this can be strengthened to K'%[4, ( K + ) , + I 3 . (14) In fact, similarly as before, we can prove an even stronger result: THEOREM. 49.3. Let K be an infinite cardinal with 2 " = ~ + .Then there exists a coloring f : [ K + ] ~ - + K + such that (i) jor any four-element'set X G K ' there are at least two triples u, us X such that f (u) = f (v)=O, and (ii)for any sets X , Y ~ K with + IXI=K, I Y I = K + there is an edge { t q ) ~ Xsuch that K + \ (0) E { f ( { t q a ] ) a : E Y ). PROOF. Let ( h u : a < K + ) be an enumeration of all functions of cardinality K with domainE[K+l2 and range E K + \ (0).In defining the colors f ( { < q a ) )with 5, q < a for an a < K + , for each B < a with dom ( h , ) s [.I2 pick ordinals ear q, (depending also on a )with { t,qs) E dom (h,) in such a way that all these ordinals are pairwise distinct and put S({ta'laQ).)=h,({Spfla)) ;
(15)
otherwise put f ( { t q a ) ) = O (5 < q < a ) . This completes the definition o f f : If is easy to see (i). To this end let to< t1 < < a < K + be four ordinals. Out of the three triples {To a ) , {to a). and {tL t2a). only one can have a nonzero color; in fact, it is clear from the above definition that if the triples {tq a ) and { t' q' a ) (5, q, t', q'< a ) have nonzero colors, then the edges ( 5 q).and {t'q') must be disjoint (i.e., we must have ( 5 q ) n {t'q') =O). So at least two of the above triples must have color zero, which is what we claimed. To see (ii), let X , Y E K + be sets with J X I = K and I Y I = K + . Assume that, contrary to what we claimed, for every distinct 5, q E X there is a color h ( { t q ) ) ~ ~ + \ (such O ) that f({Srp})#h({(q))foranyaE Y . Wehaveh=h@for some / ? < K + ; let CIE Y be such that dom ( h , ) E [ a I 2 (i.e., X r a ) and j?
r2
rz
fcr some distinct t, q E X ; this contradicts the choice of h, establishing (ii). The proof is complete. I t is clear from the above proof that (ii) in our last theorem can be strengthened as follows: (ii') for any set E c [ K ' ] ~ of cardinality K of pairwise disjoint
318
A BRIEF SURVEY OF THE SQUARE BRACKET RELATION
edges and any set Y c K + of cardinality K + , there is an edge e E E such that K + \ ( 0 ) s { f ( e u { a ) ) :a E Y',. Using exactly the same method as above, the reader can easily establish the following generalization of (14): if K is an infinite cardinal with 2"= K + , then K+
%C(r + 11, ( K + ) , + I *
116)
holds for any integer r L 3. There is also an analogous generalization of Theorem 49.3. 50. THE EFFECT OF A SUSLIN TREE ON
NEGATIVE RELATIONS Trees were defined in Section 13. Given a tree (7; < ), a set X c T is called an antichain it if consists of elements parirwise incomparable in the partial ordering <. For an infinite cardinal K , the tree (17: <) is called a Suslin K-tree if it has length K and has no chain or antichain of cardinality K. We prove THEOREM 50.1 (R.B. Jensen [1972], R.A. Shore [1974]). I ~ isK a regular cardinal holds. such that there is a Suslin K-tree, then K+[K]? According to a well-known theorem of Jensen, under the assumption of V = L there is a Suslin K-tree for regular K exactly if K + ( K ) $ holds (see e.g. Jech [I19783, Theorem 49 on p. 226, and the remarks. on p. 336; cf. also our Theorem 29.5).The proof relies on the following lemma:
50.2. Let K be a regular cardinal and assume that there is a Suslin K-tree. LEMMA Then there is a Suslin u-tree T such that, for every CK < K , each element in the ath level of T has at least la1 immediate successors.
PROOF. Let (S, < ) bea Suslin K-tree. We are going to construct T as a subtree of S. To this end, let So be the set of elements of S that have K successors, i.e., put so={X
E S : I { y E S :X < y ) l = K )
.
(1)
We claim that IS\SoI
In fact, for each x E S\So, there is a minimal y ~ with x y E S\So; that is, putting A = { y E S : I { u E S : y
we have
VZ < y l {U E S : z < u ) ~= K ) ,
THE EFFECT OF A SUSLIN TREE ON NEGATIVE RELATIONS
319
Each summand on the right-hand side here has cardinality < K , so by the regularity of K, (2) will follow if we show that ( A \ < K . But A is obviously an antichain, so this inequality must hold, as S is a Suslin tree. By (1)and (2) it is clear that (So, < ) is a Suslin tree where each element has K successors, that is ({yESo:X
(3)
for every x E So. To get closer to our goal of constructing T we want to find a 'large' subtree S , of So such that each element of S, splits, i.e., such'that lims (x, S,)l 2 2 for each x E S,, where ims (x, S,) denotes the set of.,immediate successors of x in S , (cf. Section 13).To this end, define the equivalence relation E on So as follows: for each x , y E S o , x = y if either (i) x I y and no element z E So with x Iz < y splits, or (ii) y_c x and no z E So with y s z < x splits. Clearly, each equivalence class C is,a chain, and so I C 1 < K. Define S, as a set of the least elements of the equivalence classes. Since, as was remarked just before, each equivalence class has cardinality < K , it is easy to see from (3) that I { y E S , :X < y ) / = K
(4 )
for each x E S1. Moreover, it is also easy to see that each element of S, splits, i.e., that lims (x, S , ) l 1 2 (5 ) for each x E S, .In fact, let x E S, and define aas the least ordinal < K such that the set So[cc] n { y E So: x < y j (6) has at least two elements, where So[.] denotes the ath level of the tree ( S o , <). There must be such an a since otherwise the set on the left-hand side of (3)would be a chain of cardinality K. Noting that each element of S, is the least in its equivalence class, it is easy to see then that the equivalence class under = of the x in question is {YE So: x s y & o ( y , So)
Si;.a = { y E S ,: x
320
A BRIEF SURVEY OF THE SQUARE BRACKET RELATION
ao=o(x,S1)/l/E.. Then
cx,i= { y E s, :x
is a chain of cardinality 1.Using ( 5 ) , each y E Cx,Ahas an (immediate) successor that is not comparable with u ; then
sg E S,
A,,i={sy:yECx,l)GSI
(7)
is an antichain of cardinality I consisting of successors of x. Thus we proved that, for any I < K, any x E S , has a set of I pairwise incomparable successors in S, . It is now easy to define a Suslin K-tree TE S, as required by the lemma to be proved. T will be defined as a union
u T,.
Put To= S, [O]; let a be an ordinal
a < K
with O
T,=
u
{A,,,?,:x E T,).
9
(8)
where the sets Ax.Awere defined in (7). If a is a limit ordinal, then let ya=sup{o(x,s,)/l: and put
X E
u T,;,
B
T,=S,[yal.
(9)
T, of T so defined satisfies the
It is easy t o see that the subtree T= M < K
requirements of the lemma. In fact, (9) makes sure that (?;< ) has length K . Moreover, it is clear that if B r a < K and x E To,y E T,, then y < x cannot hold. Hence if x E T, ( ~ c K ) , then o(x, T ) c a ; all elements of Ax,lalare immediate successors of x (there may be others, added at a limit step later, cf. (9)), and so x does indeed have at least lo(x, T)I immediate successors. The proof of the lemma is complete.
PROOF OF THEOREM 50.1. Let T be a Suslin K-tree such as described by the preceding lemma, and for each x E T let (x;: (
THE KUREPA HYPOTHESIS AND NEGATIVE STEPPING UP TO SUPERSCRIPT
3
321
immediate successor x t . If x , y~ X and x # y , then x e < y t would mean that xt-
We start with the definition of a Kurepa family: DEFINITION 51.1. Let K be an infinite cardinal. A Kurepa family op K is a set F c 9 ( such ~ ) that (i) IF I = K', and (ii)for every ordinal a with w 5 a .cK, we have J F n 9 ( a ) l I J a l . T h eK-Kurepa Hypothesis, abbreviated asKH(K), says that there is a K-Kurepa family. According to a theorem of Solovay, if V = L, then the KH(K)holds with every successor cardinal K > O . Jensen extended this result by proving thgt if V = L , then KH(K)holds if and only if K > O is not ineffable (K is called ineffable if for every coloring f:[~]'-+2 there is a homogeneous set stationary in K). The following theorem connects the Kurepa Hypothesis with negative squarebracket relations.
THEOREM 51.2.(R. A. Shore [1974]). Let
K
be an infinite cardinal such that the
relation (1)
K + f + [ ( K : K')]:+
and the
K
+
:Kurepa Hypothesis hold. Then K + ++[K+],3+
.
(2)
Therelationin(1)will holde.g.if2"=rcf (cf.(49.l));if V = L t h e n K H ( ~ + ) a l s o holds, as remarked above, and so (2) always holds if V = L.
PROOF. Let F be a K+-Kurepa family; for any two distinct x, y E F, write d ( { x y ) ) =rnin { a < K + : x n(a i1) # y n ( a 41)) .
For any three distinct x, y , z E: F put
The three elements on the right-hand side here are not all distinct; in fact, it is clear that d ( { x y z ) )contains exactly two distinct elements. Let nowg: [ K + ] * + K + be acoloring verifyingrelation (1).Then weclaim that thecoloring f :[ F - j 3 - + ~ + , 21 Combnatoriai
322
A BRIEF SURVEY OF THE SQUARE BRACKET RELATION
for any u E [FI3, verifies relation (2) (note that IFI=K++). To see this, let Ec F be an arbitrary set of cardinality K+,and consider the set
x = {d({xy)): x, y E E & x # y ) . Note that XEK'; moreover, X has cardinality K + .In fact, assume, on the contrary, that X c a for some a< K + ;then the mapping x b x n (a41) (x E E) is 1-1; that is K + = l E l = I E n q ( a + 1 ) l I I F n B ( a i1)1, which contradicts the assumption that F is a K+-Kurepa family. Next put
H = { ( x n a, a ) : x E E & 3y E E [ x # y & d((xy))= a]}. Clearly, IHI = K + , since for each a E X there is an x E E such that (x na, a ) E H. Define the partial ordering 4 on H as follows: if (u, a), (u, /3) E H, then put (u,a)<(v,/3> iffa3and u=uncr.
It is easy to see that (H,<) is a tree of length IK+. We claim that for each t
u
m
1
9
note that q < K + by our assumption on 5. Observing that for each a < K + we have { u : (u, a) E H[t]3r{ x n a : x E E ] E F n P ( a ) ,
and the set on the right-hand side has cardinality IK,the assumption IH[tJ1= K + ifiplies that the set
D = { (u, a ) : (u, a) E H[5]& a> q + 1). has cardinality K + .Note, however, that if (u, a), ( u , /3) E D are distinct, then u n ( q i l ) # v n ( q g l ) since otherwise (u, a ) or ( v , 8 ) would be above the 5th level in H ;this, however, means that the set { u o ( q + ~ )(: u , a ) ~ ~ }
also has cardinality K + .As t h s set is a subset of F n B ( q / l ) , this contradicts
the assumption that F is a K+-Kurepa family. This establishes our claim that ( H [ ~ ] I < Kfor each ((K.
THE KUREPA HYPOTHESIS AND NEGATIVE STEPPING UP TO SUPERSCRIPT
Observing that the set
u
3
323
{tEH:s
saHCh-1
contains all but I K elements of H,and so it has cardinality K', there is an s E H [ K ] for which { t E H: s
(i = 0, 1).
d( { x f x r ) )= d ( { x x , ) ) A={d({xx,)):r<s[
(4 )
(={?: 3 r < s 3 u [ r = ( u , y ) ] ) )
and
B = { d ( {xp x: >): s< t > ( = { 7: 3t>s3u[t = (u, y ) ]
1).
We have (A1= K and IBJ= K + by the choice of s (and, clearly, A, B c K'). Wecannoweasilyshow that foreveryy
which we wanted to show. This establishes our claim that f verifies relation (2). The proof of the theorem is complete. It is an open problem whether similar relations hold with higher superscripts under certain assumptions. The simplest instance of this problem is the following: assuming V = L, does the relation (5 1
hold.
B. TodorEevii: [1982] proved the following negative stepping up implication. Assume K 2 w and 0,. Let v 2 2 , 2 Ir < w and wlA,=cf ( Q S Kfor all O < < < v . Then K ~ , [ A & < ~ implies rc'+[Ac+ 11;:. Added in proof: Recently S.
The simplest instance of Theorem 51.2 is that K2 f,
1
I:,
(6)
provided V = L. The following result throws some light on the strength of this relation :
324
A BRlEF SURVEY OF THE SQUARE BRACKET RELATION
THEOREM 51.3 (Rowbottom). If (6) holds, then
As can be seen from Subsection 8.5, relation (7) means the following: there is an $[H2]*+K1 such that for any set X s H , of cardinality Hiwe have If"[X;IZI = =K1.Relation (7) is well known to be equivalent to the negation of Chang's conjecture, also discussed in Subsection 8.5. According to a theorem of Silver, Chang's conjecture is consistent with ZFC + GCH provided the existence of a cardinal K with K - ( K ) ; " is consistent with ZFC + GCH (as we saw in Section 34, ameasurable K satisfiesthis partition relation). So the above theorem shows that (6) cannot be expected to be provable in ZFC + GCH (unless, of course, large cardinal hypotheses generally considered 'sound' will prove to be inconsistent).
PROOFOF THEOREM 51.3. Let g: [Kzl3+K1 be a coloring that establishes (6), and fcr each a
(8)).
First note that the condition f.(y) < f.(B) allows countably many y's only, and so f ( { B a ) )< K l . We claim that the coloring f: [K2]2+K1 verifies relation (7). In fact, if X c K 2 has cardinality K, then g"[XI3 =K,. On the other hand, every triple in [XI3 can be obtained in the form {yfla), where fl, y
K1+[K1]:
and
2H0%[2~~]&.
The proof of the first relation relies on results derived in this section; that of the second does not need any of these results. The definition of Aronszajn K-trees was given in Definition 29.2; to recapitulate,an Aronszajn K-tree is a tree of length K having no branch of length K such that each of its levels has cardinality < K. The main result that we need here is the following
ARONSZAJN TREES AND SPECKER TYPES (PREPARATIONS FOR RESULTS WITHOUT GCH)
325
THEOREM 52.1 (N. Aronszajn-E. Specker [1951]; for Aronszajn’s proof see Kurepa [1935]). There is an Aronszujn EC,-tree.
PROOF. The elements of the Aronszajn K,-tree (‘I;<)will be countable sets of rationals wellordered by the ‘less than’ relation < on reals, and an element of T will precede another if it is an initial segment of the latter in this ordering. We shall construct T level by level by transfinite recursion. The bottom level T[O] of T will be defined as {O>, and T[l] as { { r ): r E Q),where Q is the set of rational’ numbers. The elements in the ath level ( a < o l ) will have order type a in the ordering ‘less than’. Assume that a <w1 and TI a= T[y] has been defined in
u
Y
such a way that for any y
(11
contains all rationals >sup x. We are now going to construct the level T[a]; for this, we distinguish two cases: a) a= q 4 1 for some q, and b) a is a limit ordinal. In case a ) p u t T [ a ] = { x u f r ) : x ~T [ q ] & r ~ Q & s u p x < r ) , and in case b), for each x E T I a and for each rational r >sup x choose a path p,.x of length a in Tla such that x E P , , ~and sup pr,x = r (this is possible according to what we said immediately after (l)), and put
u
T[a]
= { ( u p,,,)u
{ r ] :x E Tla & r E Q & sup x < r > .
This finishes the construction of the tree (?; <), provided we can show that the assertion about the set in (1) containing all rationals >supx is valid. Using transfinite induction, suppose that this is the case if we replace a by a’
u
n
establish our claim; all we need is to construct the sequence x, as described. To this end put x - = x, and if x,, - has already been constructed for some n <w in such a way that sup x,,-
326
A BRIEF SURVEY O F THE SQUARE BRACKET RELATION
Then x,,= u p , , E T[a,,]is such that sup x , , = r n < r n + , , and so we can finish the construction of the sequence (x,,: n <w). Thus our claim about the set in ( 1 ) is established, and so the definition of the tree (?;4)is complete. It is clear by this claim that, for each a<wl, T[a] is not empty; it is also clear from the construction of T that each T [ a ]is countable. Finally, T has no branch of type w1;in fact, if p were such a branch then p would be a set of rational numbers that has order type o1in the natural ordering of the rationals, which is absurd. The proof of the theorem is complete. Next we define Specker types:
u
DEFINITION 52.2. The order type of a totally ordered set (A, <) is called a Specker type if (i) IAI=N,, (ii) w,, w:$tp (A, <), and (iii) there is no uncountable subset B of A such that (B,<) is order isomorphic to a set of reals (in other words, there is no uncountable order type II/ with II/<tp(A, <) and $ 5 the order type of the reals).
52.3 (Specker). There is a Specker type; that. is, there is a totally THEOREM ordered set ( A , <) satisfying (ik(iii) in the above definition.
PROOF. Let (A, <) be an Aronszajn K,-tree. For each <
<
arbitrary (total) ordering of A[(]. Furthermore, for each x E A with o ( x ) 2 ( denote by X I ( the unique y E A [ ( ] with y < x . We are now going to define the ordering 4 on A. For any x , y E A write x 4 y if either x < y or x and y are incomparable in the ordering < and for the least ( for which X I # yl< we have XI<<^ yI<. This is the same ordering as described in the verification of implication (ii)*(iii) in the proof of Theorem 29.6 with I C = N ~(except , that there in case x
<
+
{b](a+l):bEB’),
being a subset of A[a 4 13, is countable, so there are two elements b,, bl of B’ with
327
NEGATIVE SQUARE BRACKET RELATIONS WITHOUT GCH
b,
53. NEGATIVE SQUARE BRACKET RELATIONS WITHOUT GCH After the preparations made in the preceding section, we can now prove THEOREM 53.1 (Galvin-Shelah [19733). W e have For the proof, we need the following lemma: LEMMA 53.2. Let (A, <) be an ordered set with tp ( A , <) a Specker type (see Definition 52.2), and let f : A+[O, 13 be a 1-1 mapping, where [0, 1) denotes the interval O
(2)
as is easily seen, the infimum here is assumed, i.e., g ( x )belongs to the set on the right-hand side here. g is a nonincreasing real-valued function on A ; therefore g can only have countably many different values. In fact, otherwise there would be an uncountable order type $ with $5tp ( A , <) and $ 5 the order type of the reals, which is not the case, as tp ( A , 4)was assumed to be a Specker type. Hence g is constant on an uncountable set A, E A. We can choose an x E A, so that ( y E A,: y > x ) is uncountable; otherwise A , would have a subset of order which is not the case. Now we can choose an x, E A, so that x,>x and type of, f ( x o )
PROOFOF THEOREM 53.1. Let IAI=K,. Choose orderings<,, < I , < 2 so that tP(A,
328
A BRIEF SURVEY OF THE SQUARE BRACKET RELATION
reals. Analogously to proofs of negative ordinary partition relations with the aid of Sierpinski partitions (cf. Section 19), we define a disjoint partition I= ( I i : i < 4 ) of CAI2 by putting
I,= I,
{{XYI
:x
C O Y , x < l Y 9 X<,Y’,
={{vI: X
I
x < IY1 Y <2xI
1 2 = { { X Y I : x < OY, y < 1x9 x < 2Y)
I , = { C X Y :x
4
9
9
1
where x and y range over elements of A. Consider any uncountable A ’ s A. Since tp (A’, < ) is still a Specker type and tp (A’, < 2 ) Ithe type of reals, by Lemma 53.3 there is an x E A’ such that {y E A’: x < , y & x < , y ]
is uncountable. Since tp (A, < o) = ol, we can choose any from this set such that x<,y; then { x y >E Zo,, showing that [A’]2n10#0. Since tp (A, > is also a Specker type and tp (A, >2)s the type of reals, it follows by symmetry that [A’I2nZi#O for i= 1, 2, 3 as well. This completes the proof of Theorem 53.1. The proof of the next theorem does not rely on the material discussed in the preceding section.
THEOREM 53.4 (Galvin-Shelah [1973)). We have
yo % C2uol:0
(3 )
Actually (3) can easily be derived from (4), but we shall not have to make the effort to dmo, as (3) will be obtained directly. The proof of the theorem proceeds through several lemmas, all due to Galvin and Shelah.
LEMMA 53.5. Let n be an integer, let fb, . . ., f,:F o + R be 1-1 mappings, where R. is the set of reals, and suppose that for every B E P ~cofinal in there exist p, V E B such thatfi(p)
PROOF. We write f ( p ) < f ( v ) to mean f;:(p)
(Pop=Po,we can write [PolK0= {A,,: p
329
NEGATIVE SQUARE BRACKET RELATIONS WITHOUT GCH
(iii) g(v)#g,*(p)for all p < v ; (iv) if p < v and A,Ev, then g ( v ) is neither the supremum nor the infimum of the set ( g ( a ) :a E A , & f ( a ) < . f ( v ) ) . So g : Fo-+R is defined in a 1-1 way. Consider any Bc Po cofinal in 2uo, and suppose that (i) fails; this means that, for everyp, V E B, f(p)
f ( a )< f ( v ) ) *
Indeed, = cannot hold here by (iv),and < cannot hold either as we assumed that (i) fails with B (cf. the second part of the first sentence in this paragraph). Recall that cf ( 2 N ~S)KO according to Konig's theorem. Consequently, if we partition a cofinal subset of 20' into countably many parts, then at least one of the parts will be cofinal; we are going to use this fact repeatedly. Choose E >0 and c B' so that B" is cofinal in 2uo and B
gtv)--E =-SUPf g ( a ) :a A, c!k f ( a )< f ( v ) )
j
(5)
holds for any v E B". Choose a set B " ' GB" cofinal in Yo so that (g(v,)-g(v,)( < < ~ / for 2 all y o , v 1 E B"'. Choose ao, v E B"' so that f ( a o ) < f ( v ) ;there are such a. and v according to the assumptions of the lemma. Since P A , is dense in k"B, we can choose an a E A , so that Ig(a)-g(ao)l < 4 2 and I J(a) - fi(ao)l < f i ( v )- J ( a o ) for all i 5 n. Then f ( a )
but this contradicts (5) since v E B ' . This proves that (i) holds; (ii) follows by symmetry. The proof of the lemma is complete.
LEMMA 53.6. There are 1-1 mappings f.: 2'O+R ( n < o ) such that, for any Bs2'" cofinal in 2O ' and for any n< w, there exist p, v E B such that f b ) c f x v ) for all i
so:Fo-+R, and apply the preceding lemma
~ M M A53.7. There are puirwise disjoint sets Z , , C [ F ~( ]n~ < o ) such that [B]2nI, #O for every n -cw and for every B S2'" that is cofinal in 2'O.
3 30
A BRIEF SURVEY OF T H E SQUARE BRACKET RELATION
PROOF. Using the mappings f,: Po+ R described in the preceding lemma, put
The sets I, fulfill all our requirements according to the preceding lemma.
PROOFOF THEOREM 53.4. The result in (3) immediately follows from the preceding lemma. As for (4), take a cofinal subset A of Po of cardinality cf (2'0); then the sets Inn[A]2, n <w, with I, as in the preceding lemma, form a disjoint partition of a subset of CAI2 that verifies (4). The proof is complete. The results in this section should be compared with (49.1), and also to a result of Baumgartner according to which 2 H o > > K 1 is anything reasonable and Po%C(Ko: K d X o
(6)
is consistent with ZFC. Moreover, a result of Solovay says that if 20' is realvalued measurable, then holds for every n <w; hence (3) above seems to be best possible. As for possible strengthenings of (1) it is an open problem whether any of the relations (8 1
K%"11,2
is provable in ZFC for 5 _ ( a s K , . As for higher superscripts, Shelah, using a result of Galvin, proved that
K%"13;,
holds in ZFC.
(9)
54. POSITIVE RELATIONS FOR SINGULAR STRONG LIMIT CARDINALS
The following result is an easy consequence of the General Canonization Lemma: THEOREM 54.1. Let
K
be a singular strong limit cardinal with
cf(K)+(Cf(K)): Let r > 1 be an integer and let
T
be a cardinal with 7-'
.
(1)
*
Then
(2)
331
POSITIVE RELATIONS FOR SINGULAR STRONG LIMIT CARDINALS
The definition of this symbol is given in Section 8.5, but the first sentence of the proof will also clarify its meaning.
PROOF. Let f : [K]'-+Z be a coloring; we have to prove that there is an X C K of cardinality K such that If"[Xl'I 5 2 ' - l . Using Corollary 28.2 of the General Canonization Lemma, we can see that there are pairwise disjoint sets B,E K, B, has cardinality K and ( B , : 5
u
C
canonical with respect to f: We may assume here that 1 B, I increases with 5. For any x E [B]' define the type typ x of x as follows: let &I, c . . . < 6 be the elements of the set {t
(3)
Here fyp,'x 5 [cf ( K ) ] ~ 'and typ, xis a finite sequence of positive integers whose sum equals r. It is easy to show by induction on r that the number N , of possibilitiesfor typa x is 2'- l , since we clearly have N o = 1 (the empty sequence is the only possibility), N , = 1, and N , =
r
1 Nr-i, where the
ith jerm here
i= 1
represents the case when the first integer in typz x is i. The canonicity of (Be: t < c f ( ~ ) )means that, for x E [ B r , f(x) depends only on typ x; hence it is sound to define a function g: {typx: x E [B]'}+t by puttingdtyp x)=f(x) for every x E [B]'. According to the implication (vi)*(iv) in Theorem 29.6 (cf. also Theorem 29.1), we have cf(+(cf(K)): in view of (1). Using this relation (e.g. repeatedly, once for each value of typ2 x), we can see that there is a set H s c f ( ~of ) cardinality d ( ~such ) that whenever typ, x c H , then d t y p x) is completely determined by typz x. Putting X = B,, this means, in particular,
u
that
~f"[X]'I=~g"{typx: typ, x c H 3 1 1 l t h e number of possibilities for typ, x = '2'-' ,
where the last equality here was established above. As X has cardinality K, this completes the proof of the lemma. Note that (2) is best possible here in a sense, since we have K%CKI$-1
(4 )
for every singular K and every integer r > 1. To see this, represent K as a B, of pairwise disjoint sets of cardinality less than K, and define the sum
u
t
function f on [K]' by putting f ( x ) = typ, x for any x E [K]'. Then, as remarked above, the cardinality of the range off is 7 - l .If X E Khas cardinality K, then there are at least r different ordinals 5
332
A BRIEF SURVEY OF THE SQUARE BRACKET RELATION
55. INFINITARY.J6NSSON ALGEBRAS The following problem is due to B. Jonsson: Given an infinite cardinal K, is there an algebra of cardinality K with countably many operations that has no proper subalgebra of cardinality K? An algebra with this property is called a Jonsson algebra. As far as finitary algebras are concerned, i.e., when the operations are assumed t o be finitary, this question has not yet been answered completely. H. J. Keisler and F. Rowbottom [1965] showed that the answer is yes for all K if V = L is assumed, P. ErdBs and A. Hajnal [1966] proved this for successor K under the assumption of GCH; they also proved this in ZFC without any additional assumption if K=K,, n <w. The question of the existence of a Jonsson algebra of cardinality K, (with or without the assumption of GCH) seems to be a very difficult open problem. If w a r y algebras are considered, i.e., If w-place operations are also allowed, then the question becomes much simpler, and we shall prove in ZFC that then the answer is yes for all K ~ W we ; shall actually prove the (apparently) stronger result that we have ~j*[~]:for all infinite K.A more comprehensive treatment of the topics dicussed in this chapter is to be found in GalGn-Prikry [1976]. Our first result confirms the existence of an w-ary Jonsson algebra for all K 2 w with finitely many operations.
LEMMA55.1 (ErdBs-HajnalC19661). Let K ~ beO a cardinal. Then there are an integer n, a set A of cardinalit) K, and functions fo, . . ., f, - : [A]".+ A such that whenever X E A and 1x1= K we have f [X]'o= A.
u
i
PROOF. According to Theorem 45.1, we have (K,KO,2)+K0, i.e., there is a set
mapping g: [ I C ] ~ ~ + [ K ] ~ such that there is no infinite free set with respect to g . Define f : [K]'~+K by writing g(x)= if(.); for all x E [K]'o. Assuming that the assertion of the lemma to be proved fails, we are going to define a sequence (tk: k < o ) of pairwise distinct ordinals and a sequence of sets K = A A,~ 2.. ~ . Z A k 2 . . . ( k t o ) as follows. Assuming that ti. i < k and A , , i l k , havealreadybeendefinedin suchawaythat {t,:i < k ) n A,=O,foreach b s {l,:i < k ) and for each x E [~\b]'a put f a ( x ) = f (bux). According to our assumption, the assertion of the lemma fails in particular with A = A, and with the functions f b , b~ it,:i
# fb(x)'
f(bux)
(1)
for every b c it,:i
333
INFINITMY J6NSSON ALGEBRAS
(1) means in particular that t k # f ( y ) , i.e., that t k & g ( y ) ,holds for every k < o and every infinite set y ~ : { t ~ : i < with w i C k # y . Thus { t i : i < w )is a free set with respect to g, which is a contradiction, proving the lemma. THEOREM 55.2 (ErdBs-Hajnal [1966)). W e have for every infinite cardinal
K.
PROOF. (2) says that there is a function f :[ K ] " o + K such that f ' : ~ X ] " o = ~ holds for any set X c K o f cardinality K ; in other words, it says that the preceding lemma is true already with n = 1 (we may obviously assume A = K there). Le., we have to show that the functions fo, . . .,f . - l in the preceding lemma can be 'coded' as a single function f. We shall have to do this only in case K >No as we have N O + [ K o ] according to Theorem 12.1, which implies (2) in case K = O . So assume K >No,and let fo, . . .,f,- : [K]"o-+ K be functions as described in the preceding lemma (i.e., we assume A = K there). Let g : [ ~ ] ~ o +ben adunction o , holds according to Theorem 12.1, in fact, verifying the relation K + [ K ~ ] ~which even the stronger relation K + [ N ~ ] ~holds & according to that theorem. I.e., g is such that, for any infinite set x E K , g"[x]"o = n holds. If x s K is a countable set of order type 2 o 4o,then denote by x' the set of its first o elements and putf(x)= =fd,.,(x\x'); if X E [ K ] " ~ has order type
so
f"[X]"o 2 ( f ( 2u x ) :z E
[zl"o
&x
E
[ X \ 2]"03. =
As g"[ZI"o=n by the choice of g, the last set here equals {1;:(x): i < n & x
E[X
\ Z - p j =K ,
where this latter equality holds since I X \ 21= K and in view of the fact that the functions 1;: satisfy the requirements of the preceding lemma. This means that f"[X]"o= K , completing the proof of the theorem.
22 Combinatonal
BIBLIOGRAPHY
A. V. Arhangel'skii [1969] The power of bicompacta with first axiom of countability. Dokl. Akad. Nauk SSSR 187 (1969), %7-968. (Soviet Math. Dokl. 10 (1969). 951-955.) J. E. Baumgartner [ 19701 Results and independence proofs in combinational set theory. Doctoral Dissertation, University of California, Berkeley, Calif. ( 1970). [I9761 A new class of order types. Annals of Math. Logic 9 (1976), 187-222. J. E. Baumgartner-A. Hajnal-A. Mate (19751 Weak saturation propertiesof ideals. In: Infiniteand finitesets. (Colloq., Kkzthely 1973; dedicated to P. ErdBs on his 60th birthday), Part I, 137-158. Colloq. Math. Soc.Janos Bolyai, Vol. 10, North Holland, Amsterdam, 1975.
N. G. de Bruijn-P. ErdBs [I9511 A color problem for infinite graphs and a problem in the theory of relations. Akademia Amsterdam 13 (1951), 371-373.
L. Bukovsky [1965] The continuum problem and powers of alephs. Comrnentationes Math. Univ. Carolinae, 6 (1965), 181-197. J. P. Burgess [1979] On a set-mapping problem of Hajnal and Mate, Acta Sci. Math. 41 (1979), 283-288. P. J. Cohen [I9631 The independence of the continuum hypothesis. Proc. Nat. Acad. Sci. 50 (1963), 1143-1 148. [I9641 The independence ofthecontinuum hypothesis. Proc.Nat. Acad. Sci. 51 (l964), 105-1 10. [1966] Set theory and the continuum hypothesis. W. A. Benjamin, New York (1966).
K. J. Devlin (19731 Aspect of constructibility. Lecture Notes in Math. 354, Springer-Verlag, Berlin-Heidelberg-New York 1973. [1973a] Some weak versions of large cardinal axioms. Annals of Math. Logic 5 (l973), 291-325. K. J. Devlin-R. Jensen [I9751 Marginalia to a theorem of Silver, in: Proceedings of ISILC, Kiel 1974, Lecture Notes Vol. 499 (Springer Verlag, New York-Heidelberg-Berlin, 1975). I 15-142.
K. J. Devlin-J. B. Paris
[I9731 More on the free subset problem. Annals of Math. Logic 5 (1973), 327-336. 22'
336
BIBLIOGRAPHY
A. Dodd-R. Jensen [I9811 The core model, Annals of Math. Logic 20 (1981). 43-75. F. R. Drake [I9741 Set theory. North Holland, Amsterdam, 1974. B. Dushnik-E. W. Miller [I9411 Partially ordered sets. Amer. J. of Math. 63 (1941), 605. W. B. Easton (19701 Powers of regular cardinals. Annals of Math. Logic 1 (1970). 139-178. P. Erdds [I9471 Some remarks on the theory of graphs. Bull. Amer. Math. Soc. 53 (1947), 292-294. P. Erdos-A. Hajnal [I9581 On the structure of set mappings. Acta Math. Acad. Sci. Hung. 9 (1958), 1 1 1-131. [1966) On a problem of B. Jonsson. Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 14 (1966), 19-23. [I9701 Unsolved problems in set theory. In: Proc. Sympos. Pure Math. Vol. 13, Part 1, Amer. Math, Soc. Providence, R. 1. 1971, pp. 1748. [I9741 Unsolved and solved problems in set theory. In: Proc. Sympos. Pure Math. 25, Amer. Math. Soc. Providence, R. I. 1974, pp. 269-287. P. Erdos-A. Hajnal-E. C. Milner [I9681 On almost disjoint subsets of a set. Acta Math. Acad. Sci. Hung. 19 (1968), 209-218. P. ErdBs-A. Hajnal-R. Rado 119651 Partition relations for cardinal numbers. Acta Math. Acad. Sci. Hung. 16 (1965), 93-196. P. Erdos-R. Rado 119521 Combinatorial theorems on classifications of subsets of a given set. Proc. London Math. Soc. 3 (1952), 41 7439.. [I3561 A partition calculus in set theory. Bull. Amer. Math. Soc. 62 (1956), 427489. P. Erd6s-G. Szekeres (19351 A combinatorial problem in geometry. Comp. Math. 2 (1935), 463470. P. ErdBs-A. Tarski f1943j On families of mutually exclusive sets. Annals of Math. 44 (1943). 315-329. [I9611 On some problems involving inaccessible cardinals. In: Essays on the foundations of mathematics, Jerusalem, 1961, 50-82.
G. Fodor
119521 Proof of a conjecture of P. Erdos. Acta Sci. Math. 14 (1952), 219-227. [1956) Eine Bemerkung zur Theorie der regressiven Funktionen. Acta Sci. Math. 17 (1956), 139-142.
F. Galvin-A. Hajnal 119751 Inequalities for cardinal powers. Annals of Math. 101 (1957), 491498. F. Galvin-K. Prikry [I9761 Infinitary Jonsson algebras and partition relations. Algebra Universalis 6 (1976), 367-376.
BIBLIOGRAPHY
337
F. Calvin-S. Shelah [I9731 Some counterexamples in the partition calculu~.J. of Combinational Theory I5 (1973), 167-1 74.
K. Giidel [I9401 The consistency of the axiom of choice and of the generalized continuum hypothesis. Princeton, 1940. [I9471 What is Cantor's continuum problem. Amer. Math. Monthly 54 (1947). 515-525. R. E. Greenwood-A. M. Gleason [1955] Combinatorial relations and chromatic graphs. Canadian J. of Math. 7 (1955), 1-7. J. de Groot [I9651 Discrete subspaces of Hausdorff spaaes. Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 13 (1965), 537-544. A. Hajnal
[1960] [I9611 [1Y64] [1969]
Someresultsandproblemsinsettheory.ActaMath.Acad. Sci. Hung. 11 (1960),277-298. Proof of a conjecture of S. Ruziewicz. Fund. Math. 50 (1961), 123-128. Remarks on a theorem of W. P. Hanf. Fund. Math. 54 (1964), 109-113. Ulam matrices for inaccesible cardinals. Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 17 (1969), 683-688.
A. Hajnal-I. Juhasz [I967 Discrete subspaces of topological spaces. Indag. Math. 29 (l967), 343-356. A. Hajnal-A. Mati [1975] Set mappings, partitions, and chromatic numbers. In: Logic Colloq. '73, Proc. Bristol 1973. North Holland, Amsterdam, 1975. 347-379.
W.P. Hanf 119641 Incompactness in languages with infinitely long expressions. Fund. Math. 53 (1964), 309-324. [1964a] On a problem of ErdBs and Tarski. Fund. Math. 53 (1964). 325-334.
W.P. Hanf-D. S. Scott
[I9651 Classifying inaccessible cardinals. Amer. Math. SOC.Notices 12 (1963, 723.
F. HausdorfT
119041 Der Potenzbegriff in der Mengenlehre. Jahrber. d. Deutsch. Math. Ver. 13 (IW), 56!&57 1.
S. H. Hechler [I9731 Powers of singular cardinals and a strong form of negation of the generalized continuum hypothesis. Z. Math. Logik Grundlagen Math. 19 (1973), 83-84.
T.J. Jech
,
[I9711 Lectures in set theory with parficular emphasis on the method of forcing. M u r Notes in Math. 217, Springer-Verlag, Berlin-Heidelberg-New York, 1971. [ 19731 Some combinatorial problems concerning uncountable cardinals. Annals of Math. Logic 5 (1973), 165-198. [I9781 Set theory. Academic Press, New York-San Francisco-London, 1978.
338
BlBLlOGRAPHY
T. J. Jech-K. Prikry [I9761 On ideals of sets and the power set operation. Bull. Amer. Math. Soc. 82 (1976), 593-595.
R. B. Jensen
[I9721 The fine structure of the constructible hierarchy, Ann. Math. Logic 4 (1972), 229-308. Erratum: Ann. Math. Logic. 4 (1972). 443. 119741 Marginalia to a theorem of Silver. Dittoed notes. [1974a] More Marginalia. Dittoed notes. [1974b] Marginalia 111. Dittoed notes.
1. Juhasz
[I9711 Cardinal functions in topology. Math. Centre Tracts 34 (1971). Math. Centrum, Amsterdam.
H. J. Keisler-F. Rowbottom [I9651 Constructible sets and weakly compact cardinals. Amer. Math. Soc. Notices 12 (1965), 373. H. J. Keisler-A. Tarski [I9641 From accessible to inaccessible cardinals. Fund. Math. 53 (1964). 225-308. J. L. Kelley [I9551 General topology. Van Nostrand, New York, 1955 J. Ketonen [I9741 Some combinatorial principles. Trans. Amer. Math. Soc. 188 (1974). 387-394. J. Konig [I9051 Zum Kontinuum-Problem. Math. Annalen 60 (1905), 177-180. D. Konig [I9271 uber eine Schlussweise aus dem Endlichen ins Unendliche. Acta Sci. Math. 3 (1927), 121- 1 30.
K. Kunen [1970] Some applications of iterated ultrapowers in set theory. Annals of Math. Logic 1 (1970), 179-227. [I9711 A‘partition theorem. Notices of Amer. Math. Soc. 19 (1971), 425. 119781 Saturated ideals. J. Symbolic Logic 43 (1978), 65-76. K. Kuratowski [I9221 Une methode d’klimination des nombres transfinis des raisonnements mathtmafiques. Fund. Math. 3 (1922), 76108. [I9511 Sur une caracterisation de‘s alephs. Fund. Math. 38 (1951), 14-17. D. Kurepa [1935] Ensembles ordonnCs et ramifies. Publ. Math. Belgrade 4 (1935), 1-138. R. Layer [I9753 Partition relations for uncountable cardinals 5 25,. In: Infinite and finite sets. (Colloq., Keszthely 1973; dedicated to P. ErdBs on his 60th birthday), Part 11, 1029-1042. Co~loq. Math. Soc. Janos Bolyai Vol. 10, North Holland, Amsterdam, 1975.
D. LkAr [I9361 On a problem in the theory of aggregates. Compositio Math. 3 (1936), 304.
BIBLIOGRAPHY
339
M. Magidor [1977] Chang’s conjecture and powers of singular cardinals. J. Symbolic Logic 42 (1977), 272-276. [1977a] On the singular cardinal problem I. Israel J. Math. 28 (1977), 1-31. [1977b] On the singular cardinal problem. Ann. of Math. 106 (1977). 517-547.
P. Mahlo (I91 I ] uber lineare transfinite Mengen. Berichte uber die Verhandlungen der KBnigliche Sachischen Gesellschaft der Wissenschaften zu Leipzig, Mathematische-Physishe Klasse. 63 (1 9 I l), 187-225.
D. A. Martin-R. M. Solovay
(19721 Internal Cohen extensions. Annals of Math. Logic, 4 (1972), 121-130.
A. Mostowski [I9491 An undecidable arithmetical statement. Fund. Math. 36 (1949), 1-164 W. Neumer [1951] Verallgemeinerung eines Satzes von Alexandrov und Urysohn. Math. Z., 54 (1951), 2 5 6 2 6 1.
F. P. Ramsey [I9301 On a problem of formal logic. Proc. London Math. Soc. (2) 38 (1930), 264-286. W. N. Reinhardt-R. M. Solovay-A. Kanamori [I9781 Strong axioms of infinity and elementary embedding. Ann. Math. Logic. 13 (1978), 73-1 16.
F. Rowbottom [1971] Some strong axioms of infinity incompatible with the axiom of constructibihty. Ann. of Math. Logic 13 (1971), 1-44. S. Ruziewicz 119361 Une genkalisation d’un theoreme de M. Sierpinski. Publications Mathtmatiques de I’UniversitC de Belgrade 5 (1936), 23-27.
S. Shelah
[I9751 Notes on partition calculus. In: Infinite and finite sets. (Colloq., Keszthely 1973; dedicated to P. Erdos on his 60th birthday), Part 111, 1257-1276. Colloq. Math. Soc. Jinos Bolyai Vol. 10, North Holland, Amsterdam, 1975. (19791 Weakly compact cardinals: A combinatorial proof. J. Symbolic Logic 44, (1979), 559-562 [I9801 A note on cardinal exponentiation. J. Symbolic Logic 45 (1980), 56-66. [1981] Canonization theorems and applications. J. Symbolic Logic 46 (1981). 345-353.. [I9821 Proper forcing, M u r e Notes in Mathematics 940 (Springer, New York-HeidelbergBerlin, 1982).
J. R. Shoenfield 119711 UnramiRed forcing. In: D.Scott, ed., Axiomatic Set Theory. Proc. Sympos. Pun Math. XIII, Part 1 ( h e r . Math. Soc., Providence, R. I., 1971), 357-382. R. A. Shore [I9741 square bracket relations in L. Fund. Math. 84 (1974), 101-106.
340
BIBLIOGRAPHY
W.Sierpinski
[I9331 Sur un probleme de la theorie des relations. Ann. Scuolo Norm. Sup. Pisa 2 (1933), 285-287.
R. Sikorski [I 9601 Boolean algebras. Springer-Verlag, Berlin-Gottingen-Heidelberg, 1960. J. H. Silver [I9661 Some applications of model theory in set theory. Doctoral dissertation, University of California, Berkeley, 1966. 119711 Someapplicationsofmodel theoryinset theory. AnnalsofMath. Logic3(1971),45-110. [ 19751 On the singular cardinal problem. Proc. Internat. Congr. Mathematicians (Vancouver 1974) Vol. 1, 256-268, 1975.
R. M. Solovay
[I9671 A nonconstructible Ai-set of integers. Trans. h e r . Math. Soc. 127 (1967). S 7 5 . [I9711 Real-valued measurable cardinals. In: Axiomatic set theory. Proc. of Symposia in Pure Math., Vol. XIII, Part I, (Amer. Math. SOC.,Providence, R. I., 1971), 397428. [I9741 Strongly compact cardinals and the GCH. In: Proceedings of the Tarski Symposium (Proc. Sympos. Pur Math. Vol. XXV, Univ. of California, Berkeley, Calif. 1971), (Amer. Math. Soc. Providence, R. I., 1974), 365-372.
R. M. Solovay-S. Tennenbaum [I9711 Iterated Cohen extensions and Souslin’s problem. Annals of Math., 94 (1971), 201-245. E. Specker [I9511 Sur un probleme de Sikorski, Collouq. Math. 2 (1951), 9-12
J. Spencer [I9751 Ramsey’s theorem - a new lower bound. J. of Combinatonal Theory Ser. A 18 (1975), 108-1 15. G. Takeuti-W. M. Zaring [I9711 Introduction to axiomatic set theory. Springer, New York (1971). A. Tarski [I9251 Quelques theoremes sur les alephs. Fund. Math. 7 (1925), 1-14.
S. B. TodorEevic [I9791 Some results in set theory. 11. Abstract 79T-E39, Notices Amer. Math. Soc. M(1979). A 4 . [a)] Trees, subtrees and order types, to appear. [I9821 Some results in Set Theory, 111. Abtract. Notices Amer. Math. Soc.(1982), 185-186. S. Ulam [I9301 Zur Masstheorie in der allgemeinen Mengenlehre. Fund. Math. 16 (1930), 14&150.
N. H. Williams 119771 Combinatonal set theory. North Holland, Amsterdam-New York-Oxford, 1977. M. Zorn (19351 A remark on method in transfinite algebra. Bull. Amer. Math. Soc. 41 (1935), 667470.
Arhangel’skii, A. V. 6, 269, 271 Aronszajn, N. 169, 325 Baumgartner, J. E. 5,39,77,104,186,187,192, 277,289, 290,298, 302,306, 330 Benda, M. 289,290,298 Bemstein, F. 41 de Bruijn, N. G. 273, 274 Bukovsky, L. 44,288 Burgess, J. P. 284 Cantor, G. 40 Chang, G. C. 5 Cohen, P. J. 5, 15, 18 Devlin, K. J. 15, 16, 18, 277, 284, 289 Dodd, A. 289,299 Drake, F. R. 169, 289 Dushnik, B. 70 Easton, W.B. 288 ErdBs, P. 5,60,61,70,77,78,19,87,89,98,110, 149, 155, 159, 164, 168, 172, 175, 177, 211, 227, 259, 273, 274, 276, 283,284, 295, 313, 332, 333 Fodor, 0.36,37,202,273,284 Galvin, F.5,39,74,79,289,290,291,295,296, 297,298,313,324, 327,328, 330,332 Gleason, A. M. 155 Giidel, K. 11, 15, 18, 169 Greenwood, R. E. 155 de Groot, J. 267 Hajnal, A. 5,39,60,61,62,74,79,110.111,149, 159, 164, 168, 172, 175, 177, 205, 206, 211, 227, 259, 264, 265, 273, 274, 276, 283, 284, 285,289,290,295,296,297,298,313,332,333
Hanf, W.P. 168, 172, 174, 179, 183,200,211 Hausdorff, F. 30.41 Hechler, S. H.44,288 Jech,T. J. 15,16,18,79,212,289,299,304,306, 318 Jensen, R. B. 204,289,299, 313, 318,321 Jbnsson, B. 332 Juhbz, I. 264,265 Kanamori, A. 79 Keisler, H. J. 169, 177, 183, 332 Kelley, J. L. 15 Ketonen, J. 302, 311, 312 Konig, D. 68 KBnig, J. 40,42 Kunen, K. 77, 79, 86, 207 Kuratowski, K. 30, 31, 282,285 Kurepa, D. 109,325 Larson, J. 5 Laver, R. A. 5 , 74, 77 L M r , D. 273 Lovhsz. L. 148 Magidor, M.289,298,299 Mahlo, P. 183 Martin, D. A. 16 Mhte, A. 39, 111, 283,284, 285 Mathias, A. R. D. 212 Miller, E.W.70 Milner, E.C. 5, 295, 313 Mostowski, A. 26 von Neumann, 1.9, 16 Newer, W.35 Pans, J. B. 277 Prikry, K. 5,210,289,299, 302,304,306,332
342
AUTHOR INDEX
Rado, R. 5, 60,61, 77, 78.87, 90,98, 110, 149, 159, 164, 168, 227,259, 313 Ramsey, F. P. 66 Reinhardt, W. N. 79 Rowbottom, F. 21 1, 324, 332 Ruziewicz, S. 272 Scott, D. S.202, 207, 21 1, 302, 31 1 Shelah, S. 5,62,117,159,16l, 246,265,267,268, 289, 290, 296, 297, 298, 299, 313, 324, 327, 328, 330 Shoenfield, J. R. 18 Shore, R. A. 313, 318, 321 Sierpinski, W, 5, 107, 109, 282, 285 Sikorski, R. 86 Silver, J. H. 39,169,177,211,213,289,299,300, 301,324
Solovay, R. M. 16,37,77,79,177,204,210,277, 289, 321, 330 Specker, E. 5, 169, 176, 325 Spencer, J. 148 Szekeres, G. 155 Takeuti, G. IS, 16 Tarski, A. 41, 51, 168, 169, 172, 175, 177, 183, 302 Tennenbaum, S. 16 TodorkviC, S. B. 323 Ulam, S. 203
Zaring, W. M. 15, 16 Zermelo, E.28 Zorn, M.30, 31
SUBJECT INDEX
AC, 9, 15 accessible cardinal, 20 addition of negative partition relations for I =2, 113117 addition of negative partition relations for r 2 3, 117-122 a.d.t., 289 almost disjoint transversal system, 289 antichain, 318 Aronszajn-tree, 169,326326 atomic formulas of L,,(%‘,3), 177 axiom of choice, 9, 15 - of constructibility, 15 - ofdependent choice, 212 - of empty set, 13 - of extensionality, 13 - of infinity, 15 Martin’s, 15 - of pairing, 14 - of the power set, 14 - of regularity, I5 - of union, 14 axiom scheme, 14 -- of comprehension, 14 -- of replacement, 14
-.
base, 264 - of logarithm, 18 Baumgartner’s principle, 186, 188-194 Bernstein-HausdorE-Tarski theorem, 41 bounded quantifier, 10 branch of a tree, 68,80 canonical partition tree, 88, 100-104 canonical sequence, 158 c a n o h t i o n lemmas, 158-167 cardinal, 9, 17 -, accessible, 20
cardinal, inaccessible, 20 -, measurable, 210, 31 1 -, real-valued measurable 210, 330 -, regular, 20 -, singular, 20 -, strong limit, 20 -, strongly inaccessible, 20 -, weakly compact, 178 -, weakly inaccessible, 20 cardinal exponentiation, 18 -- and saturated ideals, 301-312 cardinal product, 17 cardinal subtraction, 17 cardinal sum, 17 Cartesian product, 12 case of exponent one of the ordinary partition relation, 65 cf, 20 CH, 18 chain, 28 - of a tree, 68,80 chain condition, 16 -- in Boolean algebra, 86 Chang’s conjecture, 57,291,294,298,324 choice function, 12 cl(X), 263 clash of variables, 10 class, 10 -, transitive, 22 -, universal, 1 1 classifier, 10 closed set, 34,263 closed unbounded set, 34 closure, 263 club, 34 cofinal, 20 cofinality. 20 co-finite filter, 66
344 collapsing lemma, 26 coloring, 52 - number, 62 common transfinite recursion, 22 complete bipartite graph, 61 complete field of sets, 170 complete ideal, 171,202 completely inhomogeneous set, 56 consistent, 178 constructibility, axiom of, 15 continuous sequence, 35 continuum hypothesis, 18 countable chain condition, 16 cr, 226 Critical number. 226
Dc,212 dense open sets, 16 ditTerence, 10 disjoint partition, 52 disjoint sets, 1 1 disjunction in partition symbols, 55 disjunction in polarized partition, 59 distributivity of a Boolean algebra, 86 divergent function, 35 domain, 11 dominating sequence, 216 double sequence, 13 empty set, 10,13 endeanonical set, 280 end-homogeneousset, 86-90 epsilon induction, 27 epsilon recursion, 27 equipollent sets, 17 ErdBs-Dushnik-Miller theorem, 70-77 ErdBs-Rado theorem, 98-100 essentially bounded function, 38 existential formula, 178 expm(4,99, 164 exponentiation, iterated, 18 field of sets, 170 filter, 66 -, co-finite, 66 -, generic, 16 -, maximal, 66
SUBJECT INDEX
final segment, 20 finite free sets, 28S288 finite negative stepping-up lemma, 136-138 finite Ramsey theorem, 69 first countable space, 264 first discrepancy, 105, 123 Fodor's set-mapping theorem, 273 Fodor's theorem, 36-37 free sequence, 269 free subset, 272,275 function, I1 GCH, 18 general canonization lemma, 163167 generalized continuum hypothesis, 18 generalized transfinite recursion theorem, 22 generated field, 170 generic filter, 16 &raph,60 Hajnal's lemma on triangular Ulam matrices, 204-205 Hajnal's set-mapping theorem, 274 Hanfs iteration scheme, 2W201 Hanf-Tarski theorem, 177-188, 194-199 Hausdorff's cofinality lemma, 31 HausdorK space, 263 homogeneous hypergraph, 61 - set, 53, 59 hypergraph, 60 -, in partition relation, 60 ideal, 171,201 -, complete, 171,202 -, nontrivial, 201 -, normal, 188,202 -, prime, 171,202 -, proper, 171 -, saturated, 202,302 immediate successor in a tree, 68,80 inaccessible cardinal, 20 inclusion, 10 index set, 12 indexed set, 12 einduction, 27 inequalities for cardinal exponentiation, 39
SUBJECT INDEX
inequalities for powers of singular cardinals, 288-301 infrnitary conjunction, 178 infinitary disjunction, 177 infnitary Jbnsson algebras, 332-333 infinite superscript, 78-79 infinity, axiom of. I5 initial ordinal, 17 initial segment, 20 integer, 17 intersection, 10 invariance of the ordinary partition relation under permutations of arguments, 63 isomorphic hypcrgraphs, 60 iterated exponentiation, 18 iterated logarithm, 18 (I,*rank, 290 Jksson algebra, 332
KH(rc), 321 kind, 53, 60 Kiinig’s lahma, 68. 169 Kiinig’s theorem,40 Kurepa family, 321 Kurepa hypothesis, 321 L i ( 4 45 language L,J%f,a),177 language of ZF,10 large cardinals, 168-214 <-least element, 19 least successor of a path, 80 left-separated space, 267 length of a me, 68,80 level ofa t r a , 68,80 lexicographic ordering, 105 lexicographic product, 105 limit ordinal, 17 Lindeliif space, 271 linear order, 19 linear order types, 19 logarithm, 18,45 -, iterated, 18 h s ’ s theorem, 181
MA,, 15
Mahlo cardinal, 77
345
Mahlo operation, 183, 195 Martin’s axiom, 15 matrix, 13 maximal chain. 28 -- theorem, 29 maximal element, 31 maximal filter, 66 measurable cardinal,210,31 1 metatheorem, 21 minimal element, 23 minimal function, 191 monotonicity in the superscript of the ordinary partition relation, 64 monotonicity properties of the ordinary partition relation, 64 Mostowski’s collapsing lemma, 26 multiplication of negative partition relations for r=2,105-110 multiplication of negative partitiop relations for r23, 122-132 negative partition relations and the GCH, 110-113 negative stepping-up lemma, 132-138 neighborhood, 263 - b w , 263 Neumer’s theorem, 35 nonstandard models of set theory, 27 nonstationary set, 37 nontrivial ideal, 201 nontrivial measure, 210 nontrivial ultrafiitcr, 66 normal ideal, 188,202 NRL, 206 nst. 37 omission rule, 63 omitting a class of a partition, 56 On, 16 one-to-one operation, 11 open set, 263 operation, 11 order, 19 , -, linear, 19 -, partial, 19 -, total, 19 order of an element in a tree, 68,80 order of a set mapping, 275
346
SUBJECT INDEX
order type, 19 ordered pair, 11 ordering, 19 ordinal, 9, 16 -, initial, 17 -, limit, 17 -, succesor, 17 ordinal product, 16 ordinal sum, 16 ordinary partition relation, finite case, 145-157 ordinary partition relation for inaccessible cardinals, 168-177 ordinary partition symbo!. 53
regular limit cardinal, 20 regularity, axiom of, 15 relation, 11 -, a + ( B ) p , 57 -, well-founded, 22 relative topology, 263 relatively homogeneous, 57 restriction, 11 - of a relation, 11 right-separated space, 267 R-induction principle, 25 rk, 27 Ruziewicz's conjecture, 273
path of a tree, 80 partial order, 19 partially ordered hypergraph, 60 partially ordered set, 19 partition, 52 partition tree, 83 polarized partition. 58 power, 1s prime ideal, 171,202 _ - property, 171 primitive language of ZF,10 product, cardinal, 17 -, Cartesian, 12 proper ideal, 171 PT(A,0,303
sad pat, 303 S-arrow, 229 sat(r), 302 saturated ideal, 202, 302 saturation of an ideal, 302 segment, final, 20 -, initial, 20 semantically consistent, 178 sentence, 178 sequence, 12 -, double, 13 set of almost disjoint partial transversals, 303 set mapping, 212 set mapping over E, 275 sets, disjoint, 11 Shelah's canonization, 158-163 Sierpinski partition, 107, 123,22!3-232 sieve method, 148 sim. arrow, 91 similar partial orders, 19 simultaneous ordinary partition symbol, 91 simultaneous stepping-up lemma, 92 singleton, 11 singular cardinal, 20 Skolem function, 179 Solovay's decomposition theorem, 37 Specker type, 326 square bracket symbol, 56 stationary set, 34 stepping-up lemma, 90-93 strong limit Fardinal, 20 strongly inaccessible cardinal, 20 subformula, 178 subgraph, 60
quantifier, 10 ramification system, 83 Ramsey's theorem, 6S70, 172 range, 11 range of parameters of the ordinary partition relation, 63 rank, 27 real class, 10 real-valued measurable cardinal, 210, 330 real-valued probability measure, 210 erecursion, 27 recursion on well-founded relations, 22 reduction rule, 63 reduction of the superscript, 23S240 reduction theorem, 238 regressive function, 35 regular cardinal, 20
SUBJECT INDEX
subspace, 263 - topology, 263 successor ordinal, 17 sum, cardinal, 17 Suslin property, 264 Suslin tree, 318
VA),289
Tarski’s function, 50 topological space, 263 - _ ,first countable, 264 - _ ,HausdortT, 263 - _ ,left-separated, 267 - _ ,Lindekf, 271 - _ ,right-separated, 267 topology, 263 total order, 19 transfinite induction, 25 transfinite recursion t h e o m , 21 transitive class, 22 transivite closure, 23 -- lemma, 22 Transitive sets, 16 transitivity rule for the ordinary partition r e lation, 65 fransversal, 13,40 tree, 67, 80 -, Aronszajn, 169,324-326 -, Suslin, 318 tree argument, 68-69,72-74,82-86 tree property, 169 triangular Ulam matrix, 204
trivial ultrafilter, 66 type, 57 type of a set mapping, 275
Ulam matrix, 204 ultrfiter, 66 -, nontrivial, 66 -, trivial, 66 unbounded set, 34 union, 10 universal class, 11 upper bound, 3 1 !I 27 V=L, 15
weak compactness, 177-188 weak power, 18 weak square bracket symbol, 57 weakly canonical sequence, 158 weakly compact cardinal, 178 weakly inaccessiblecardinal, 20 weakly Mahlo cardinal, 77 well-founded relation, 22 wellorder, 19 wellordering theorem, 28 Zcrmelo-Fraenkel set theory, 9, 13-15 ZF, 9, 15 ZFC, 9, 15 ZFC,, 182 Zorn’s lemma, 31
347