γ-ramsey and γ-ineffable cardinals

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  • ISRAEL JOURNAL OF MATHEMATICS, Vol. 30, Nos. 1-2, 1978

    y-RAMSEY AND y-INEFFABLE CARDINALS*

    BY

    JAMES M. HENLE

    ABSTRACT

    A number of related combinatorial properties of a cardinal K contradicting AC are examined. Chief results include: (1) For many ordinals % K --~ (K) ~ implies K --~ (K)

  • 86 J.M. HENLE Israel J. Math.

    Section 1 is concerned with 3,-Ramsey cardinals. The first and only previous

    result on these is due to E. M. Kleinberg [7] who proved:

    TrtEOREM. I f K, 3' are any infinite ordinals, then

    (a) K ---> (K)~+~ implies K --> (K)7* for all a < K, and

    (b) K ---> (K )'~o implies K ---> (K ) (K)~'o for all a < K implies that K is 3,-weakly

    ineffable. We will also prove that for all cardinals r, if K is 3,-Ramsey, then K is

    < 3,-weakly ineffable. We will show further that it is not often possible for K to

    be 3'- or < 3,-ineffable.

    00. Definitions

    In this paper, K will always denote an uncountable cardinal. All other Greek

    letters may represent arbitrary ordinals. For any K, 3', the set [K ]~ is the set of all

    subsets of K of order- type 3'. We will somet imes view a member of [K]~ as a

    subset and somet imes as an increasing function from A to K. It will always be

    clear f rom the context, however, which meaning is intended.

    An infinite ordinal 3' with the property that ct < 3, implies c~ + ct < 3' is said to

    be indecomposable (sometimes called a "power of to").

    Given any ordinal a and set A, a cardinal K satisfies K---> (K)~, if[ for all

    partit ions F: [K] ~ ---> A, there is a set X _C K, .~ = K such that F" [X] ~ = 1.

    In this notation, r ---> (m (~ (K)~, then we write K -/, (K)7,, and if F if a partit ion which fails to have a

    homogeneous set, we say it is a bad partition.

    Certain facts are immediate f rom the definitions:

  • Vol. 30, 1978 y-RAMSEY CARDINALS 87

    FACt 1. If r is a cardinal satisfying K ~ (K) a, and /3 < a, then K satisfies

    K --, (K)~.

    FACT 2. If a cardinal r satisfies K ---* (K)', then K satisfies K ~ (K)7, for all n.

    The proofs of these facts are elementary. For the first, note that a --> 2 implies

    K is regular by standard methods. Induction on n suffices for the second. Fact 1

    implies among other things that K is weakly compact, and hence regular.

    We say that K is y-Ramsey, or that K satisfies the relation K ~ (K)

  • 88 J.M. HENLE Israel J. Math.

    Our first goal will be to show that all indecomposable ordinals are obliging. To this end, we require the following lemma:

    LEMMA 1. Given that an indecomposable ordinal 3' is expressible as a 9 where a and/3 are limit ordinals less than % then 3" is obliging.

    PROOF. Suppose we are given a partition F: [K]

  • Vol. 30, 1978 y-RAMSEY CARDINALS 89

    where in general, the/3th equation for/3 < 3' is

    F(p(a~),p(a~ + 1) , . . . ,p (a# + 8),... ), 3'. Let a be the least ordinal such that for some 8 < % t~ 9 8 _-> 3',

    and let 8 be the least ordinal such that ct 9 8 _-> 3'- By the indecomposability of 3',

    B is a limit ordinal, and hence so is a. Since t~ 9 it < 3' for all A < 8, it then follows

    that a 9 8 = 3'. Lemma 1, then, shows that 3' is obliging.

    Now, case 2. We define a partition G: [ r ] ~- -2 by:

    for any p E [K] *, G(p)=0 iff all the equations in the list are true. Let

    X E [K ]~, .~ = r be homogeneous for G.

    CLAIM. G"[X] "r = {0}.

    PROOF OF CLAIM. Simply define a sequence p E [X] ~ such that G(p) - - 0 as

    follows:

    Let p(0), p(1) be the least elements of X such that F (p(0) )= F(p(1)) and

    continue in this way. To avoid any use of the Axiom of Choice, choose the

    members of p consecutively: suppose we have chosen enough elements of p to

    satisfy the first/3 equations for all/3 < a. To satisfy the a th equation, take the

    first a .3 members of X greater than all members of p chosen so far:

    80, 81,'" ", 87," "7

  • 90 J.M. HENLE Israel J. Math.

    and such that for s2, the ath equation in the list reads "F (q)= F(r)" . Since

    G(s 0 = G(s2) = 0, these equations are both true, and hence F(p) = F(q). []

    The situation where y = a +/3, a, /3 < y is more difficult. The following

    results cover a number of cases, but by no means all.

    THEOREM 2. I f y is obliging, then y + n is obliging for all n < to.

    PROOF. By induction on n. If y+n-1 is obliging, any partition

    F: [K]

  • VOI. 30, 1978 3,-RAMSEY CARDINALS 91

    that s~tos3C_ql, s2Us~C_q2. Then singe G(p Uql) = G(p Uqz)=0, we have F(p U sl) = F(p U s3) = F(p tO s2), proving the claim.

    As a consequence of this, for any p E [Y] ' , there is associated a subset 1(/7) of

    /3 defined by:

    8 E l (p) iff for all q ~ [Y - Up] ~, F(p U q) = 1.

    This defines a partition I : [Y]~ --~ 2 ~. Let Z C Y, 2 = K, be homogeneous for I.

    CLAIM. Z is homogeneous for F.

    PROOFOF CLAIM. Suppose sl, s2 E [Z] a, and p~ = s~ [ a, p2 = s21 a, let ql, q2 be such

    that p~ tO qm = s~ and p2 U q2 = s2, Up~ _-< nq~, Up~ _-< nq2 and let 8 = t71 = t~2.

    Then,

    F(s~) = 1 iff 6 e 1(1Ol ) iff 6 ~ I(p2) iff F(sz) = 1.

    This completes the proof of the lemma. []

    THEOREM 3. If y is indecomposable, then for all n >= 1, y 9 n is obliging.

    PROOF. The proof is by induction on n. The case for n = 1 is covered by

    Theorem 1.

    Suppose 7 " n is obliging, n < to. By the previous lemma, to prove 1' "(n + 1)

    obliging, it suffices to show that r ~ (r)2\". In the interests of generality, we will

    show K ~ (K)~s for all a < K in two stages. Our first step will be to show that

    K --+ (K)~" " for all o~ < K is a consequence of K --> (K) "("

    Suppose F: [K]""---> Ol is any partition. Let G: [K]" . . . . --~ 3 be the following

    partition: if p0, pl, " 9 " , p . E

    G(poU- ' " Up . ) ={

    [K]% and Up~ ~ rip,+1 for all i < n, then:

    0 iffF(poU'"Up.-1)=F(plU"'Up.) 1 i f f F (poU" .Up , -O>F(p lU ' "Up , ) 2 iff F(poU. . .Up , -~) FLo, u - - -up . )>. . . > F(p U . . . U 9 9 9

    an impossibility.

  • 92 J.M. HENLE Israel J. Math.

    Suppose O"[X] " '" {2}. Let {q,}a

  • VoL 30, 1978 y-RAMSEY CARDINALS 93

    H(p) = 1 iff/3 E F(p) , for all p ~ [X] ~". Let Y _C X, f" = K, be homogeneous

    for H. It is clear that for any p E [Y ]~" and q ~ [p]~",

    F(p) fq /3 = F (q) n by the homogenei ty of X,

    F (p ) 71 (/3 + 1) = F(q) N (/3 + 1) by the homogenei ty of Y,

    hence G(p)>/3 + 1, a contradiction, and the claim is proved.

    Suppose now that p, q ~ [X] ~'" are such that the consecutive supremums of

    their consecutive y-sequences are the same, i.e., that p = po tO. . . tO p,-1 and

    q = qo to 9 9 9 tA q,_~, p,, q~ E [X] ~, and that Up, = Uq, , for all i. It then follows

    that F (p ) = F(q) , because each p, tO q, is a member of [X] ~ (a consequence of

    y ' s indecomposabi l i ty) and so p tA q E [X] ~". Thus G(p U q) = 0 implying that

    F (p )= F (p tO q)= F(q) . Hence, we may make the following unambiguous

    definition: H : [K]" ---* 2 ~ is defined by: for all (/30,/3~,'" " , /3 . -0 ~ [K]",

    /-/((/30,/32,..-,/3.-~)) = { A if for a l l i

  • 94 j.M. HENLE Israel J. Math.

    Since proving this obliging will require proving that K--~ (K) . . . . . implies

    K --~ (K)~", it does not look easy.

    In closing this section, it should be noted that a simple application of the

    techniques used above will produce results of the following sort: If 3' is

    sufficiently large (a cardinal, for example, or a Ao-admissable, etc.) then

    K --* (K)

  • Vol. 30, 1978 y-RAMSEY CARDINALS 95

    PaooFov CLAIM. Let a be the least element of X. Let X~ be homogeneous for

    F~, and choose p E [X - a] ~, q E [X~ - a] ~ such that ,p = ~q. Then we have:

    F,,(q) = a,,,

    so A,oqn a = A~

    so A j N a = A~

    so F({a} Up) = 0,

    and hence the claim is proved.

    Next consider {A~ I oc E X}. We claim that these sets all cohere, that is, if

    a,/3 E X, a

  • 96 J .M . HENLE Israel J. Math.

    THEOREM 5. Given K, 3' ordinals, if 3" is greater than all regular cardinals below

    K, then K is not ),-ineffable.

    PROOF. For each p E[K]*, let

    I{0} if p contains a limit point of itself, Ap |

    l [ 1} otherwise.

    We will show that {At,}eEl~j~ cannot have a stationary cohering set, for if X is

    stationary, and A is such that for all p ~ [X] ~, At, = A N p(0), then A must he

    either {0} or {1}. Clearly, A cannot be {0}, since a p E [X] ~ can easily be found containing no

    limit points. On the other hand, consider (X) =- the set of limit points of X. (X) is

    a closed, unbounded set, hence X fq (X) ~ ~. Let a E X M (X). By hypothesis,

    there is a sequence of points in X of length less than 3' with sup equal to a, thus

    there is a p E [X] ~ such that At, = {0}. This proves the theorem. []

    Since < 3' ineffability implies 8 ineffability for all 8 < % a similar result holds

    for < 3' ineffability.

    Our last theorem concerns < 3'-weak ineffability:

    THEOREM 6. I f 3" >--tO is a cardinal, then K--~ (K)

  • Vol. 30, 1978 3,-RAMSEY CARDINALS 97

    Given any possible homogeneous set X ~ [K ]', X = {/31,/32," 9 }, let p = X r a,

    q = (X - p) [ a. If A~ and Aq do not cohere, then fa (17 U q)