Published on

24-Aug-2016View

212Download

0

Embed Size (px)

Transcript

<ul><li><p>ISRAEL JOURNAL OF MATHEMATICS, Vol. 30, Nos. 1-2, 1978 </p><p>y-RAMSEY AND y-INEFFABLE CARDINALS* </p><p>BY </p><p>JAMES M. HENLE </p><p>ABSTRACT </p><p>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) </p></li><li><p>86 J.M. HENLE Israel J. Math. </p><p>Section 1 is concerned with 3,-Ramsey cardinals. The first and only previous </p><p>result on these is due to E. M. Kleinberg [7] who proved: </p><p>TrtEOREM. I f K, 3' are any infinite ordinals, then </p><p>(a) K ---> (K)~+~ implies K --> (K)7* for all a < K, and </p><p>(b) K ---> (K )'~o implies K ---> (K ) (K)~'o for all a < K implies that K is 3,-weakly </p><p>ineffable. We will also prove that for all cardinals r, if K is 3,-Ramsey, then K is </p><p>< 3,-weakly ineffable. We will show further that it is not often possible for K to </p><p>be 3'- or < 3,-ineffable. </p><p>00. Definitions </p><p>In this paper, K will always denote an uncountable cardinal. All other Greek </p><p>letters may represent arbitrary ordinals. For any K, 3', the set [K ]~ is the set of all </p><p>subsets of K of order- type 3'. We will somet imes view a member of [K]~ as a </p><p>subset and somet imes as an increasing function from A to K. It will always be </p><p>clear f rom the context, however, which meaning is intended. </p><p>An infinite ordinal 3' with the property that ct < 3, implies c~ + ct < 3' is said to </p><p>be indecomposable (sometimes called a "power of to"). </p><p>Given any ordinal a and set A, a cardinal K satisfies K---> (K)~, if[ for all </p><p>partit ions F: [K] ~ ---> A, there is a set X _C K, .~ = K such that F" [X] ~ = 1. </p><p>In this notation, r ---> (m (~ (K)~, then we write K -/, (K)7,, and if F if a partit ion which fails to have a </p><p>homogeneous set, we say it is a bad partition. </p><p>Certain facts are immediate f rom the definitions: </p></li><li><p>Vol. 30, 1978 y-RAMSEY CARDINALS 87 </p><p>FACt 1. If r is a cardinal satisfying K ~ (K) a, and /3 < a, then K satisfies </p><p>K --, (K)~. </p><p>FACT 2. If a cardinal r satisfies K ---* (K)', then K satisfies K ~ (K)7, for all n. </p><p>The proofs of these facts are elementary. For the first, note that a --> 2 implies </p><p>K is regular by standard methods. Induction on n suffices for the second. Fact 1 </p><p>implies among other things that K is weakly compact, and hence regular. </p><p>We say that K is y-Ramsey, or that K satisfies the relation K ~ (K) </p></li><li><p>88 J.M. HENLE Israel J. Math. </p><p>Our first goal will be to show that all indecomposable ordinals are obliging. To this end, we require the following lemma: </p><p>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. </p><p>PROOF. Suppose we are given a partition F: [K] </p></li><li><p>Vol. 30, 1978 y-RAMSEY CARDINALS 89 </p><p>where in general, the/3th equation for/3 < 3' is </p><p>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', </p><p>and let 8 be the least ordinal such that ct 9 8 _-> 3'- By the indecomposability of 3', </p><p>B is a limit ordinal, and hence so is a. Since t~ 9 it < 3' for all A < 8, it then follows </p><p>that a 9 8 = 3'. Lemma 1, then, shows that 3' is obliging. </p><p>Now, case 2. We define a partition G: [ r ] ~- -2 by: </p><p>for any p E [K] *, G(p)=0 iff all the equations in the list are true. Let </p><p>X E [K ]~, .~ = r be homogeneous for G. </p><p>CLAIM. G"[X] "r = {0}. </p><p>PROOF OF CLAIM. Simply define a sequence p E [X] ~ such that G(p) - - 0 as </p><p>follows: </p><p>Let p(0), p(1) be the least elements of X such that F (p(0) )= F(p(1)) and </p><p>continue in this way. To avoid any use of the Axiom of Choice, choose the </p><p>members of p consecutively: suppose we have chosen enough elements of p to </p><p>satisfy the first/3 equations for all/3 < a. To satisfy the a th equation, take the </p><p>first a .3 members of X greater than all members of p chosen so far: </p><p>80, 81,'" ", 87," "7</p></li><li><p>90 J.M. HENLE Israel J. Math. </p><p>and such that for s2, the ath equation in the list reads "F (q)= F(r)" . Since </p><p>G(s 0 = G(s2) = 0, these equations are both true, and hence F(p) = F(q). [] </p><p>The situation where y = a +/3, a, /3 < y is more difficult. The following </p><p>results cover a number of cases, but by no means all. </p><p>THEOREM 2. I f y is obliging, then y + n is obliging for all n < to. </p><p>PROOF. By induction on n. If y+n-1 is obliging, any partition </p><p>F: [K]</p></li><li><p>VOI. 30, 1978 3,-RAMSEY CARDINALS 91 </p><p>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. </p><p>As a consequence of this, for any p E [Y] ' , there is associated a subset 1(/7) of </p><p>/3 defined by: </p><p>8 E l (p) iff for all q ~ [Y - Up] ~, F(p U q) = 1. </p><p>This defines a partition I : [Y]~ --~ 2 ~. Let Z C Y, 2 = K, be homogeneous for I. </p><p>CLAIM. Z is homogeneous for F. </p><p>PROOFOF CLAIM. Suppose sl, s2 E [Z] a, and p~ = s~ [ a, p2 = s21 a, let ql, q2 be such </p><p>that p~ tO qm = s~ and p2 U q2 = s2, Up~ _-< nq~, Up~ _-< nq2 and let 8 = t71 = t~2. </p><p>Then, </p><p>F(s~) = 1 iff 6 e 1(1Ol ) iff 6 ~ I(p2) iff F(sz) = 1. </p><p>This completes the proof of the lemma. [] </p><p>THEOREM 3. If y is indecomposable, then for all n >= 1, y 9 n is obliging. </p><p>PROOF. The proof is by induction on n. The case for n = 1 is covered by </p><p>Theorem 1. </p><p>Suppose 7 " n is obliging, n < to. By the previous lemma, to prove 1' "(n + 1) </p><p>obliging, it suffices to show that r ~ (r)2\". In the interests of generality, we will </p><p>show K ~ (K)~s for all a < K in two stages. Our first step will be to show that </p><p>K --+ (K)~" " for all o~ < K is a consequence of K --> (K) "(" </p><p>Suppose F: [K]""---> Ol is any partition. Let G: [K]" . . . . --~ 3 be the following </p><p>partition: if p0, pl, " 9 " , p . E </p><p>G(poU- ' " Up . ) ={ </p><p>[K]% and Up~ ~ rip,+1 for all i < n, then: </p><p>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 </p><p>an impossibility. </p></li><li><p>92 J.M. HENLE Israel J. Math. </p><p>Suppose O"[X] " '" {2}. Let {q,}a</p></li><li><p>VoL 30, 1978 y-RAMSEY CARDINALS 93 </p><p>H(p) = 1 iff/3 E F(p) , for all p ~ [X] ~". Let Y _C X, f" = K, be homogeneous </p><p>for H. It is clear that for any p E [Y ]~" and q ~ [p]~", </p><p>F(p) fq /3 = F (q) n by the homogenei ty of X, </p><p>F (p ) 71 (/3 + 1) = F(q) N (/3 + 1) by the homogenei ty of Y, </p><p>hence G(p)>/3 + 1, a contradiction, and the claim is proved. </p><p>Suppose now that p, q ~ [X] ~'" are such that the consecutive supremums of </p><p>their consecutive y-sequences are the same, i.e., that p = po tO. . . tO p,-1 and </p><p>q = qo to 9 9 9 tA q,_~, p,, q~ E [X] ~, and that Up, = Uq, , for all i. It then follows </p><p>that F (p ) = F(q) , because each p, tO q, is a member of [X] ~ (a consequence of </p><p>y ' s indecomposabi l i ty) and so p tA q E [X] ~". Thus G(p U q) = 0 implying that </p><p>F (p )= F (p tO q)= F(q) . Hence, we may make the following unambiguous </p><p>definition: H : [K]" ---* 2 ~ is defined by: for all (/30,/3~,'" " , /3 . -0 ~ [K]", </p><p>/-/((/30,/32,..-,/3.-~)) = { A if for a l l i</p></li><li><p>94 j.M. HENLE Israel J. Math. </p><p>Since proving this obliging will require proving that K--~ (K) . . . . . implies </p><p>K --~ (K)~", it does not look easy. </p><p>In closing this section, it should be noted that a simple application of the </p><p>techniques used above will produce results of the following sort: If 3' is </p><p>sufficiently large (a cardinal, for example, or a Ao-admissable, etc.) then </p><p>K --* (K)</p></li><li><p>Vol. 30, 1978 y-RAMSEY CARDINALS 95 </p><p>PaooFov CLAIM. Let a be the least element of X. Let X~ be homogeneous for </p><p>F~, and choose p E [X - a] ~, q E [X~ - a] ~ such that ,p = ~q. Then we have: </p><p>F,,(q) = a,,, </p><p>so A,oqn a = A~ </p><p>so A j N a = A~ </p><p>so F({a} Up) = 0, </p><p>and hence the claim is proved. </p><p>Next consider {A~ I oc E X}. We claim that these sets all cohere, that is, if </p><p>a,/3 E X, a </p></li><li><p>96 J .M . HENLE Israel J. Math. </p><p>THEOREM 5. Given K, 3' ordinals, if 3" is greater than all regular cardinals below </p><p>K, then K is not ),-ineffable. </p><p>PROOF. For each p E[K]*, let </p><p>I{0} if p contains a limit point of itself, Ap | </p><p>l [ 1} otherwise. </p><p>We will show that {At,}eEl~j~ cannot have a stationary cohering set, for if X is </p><p>stationary, and A is such that for all p ~ [X] ~, At, = A N p(0), then A must he </p><p>either {0} or {1}. Clearly, A cannot be {0}, since a p E [X] ~ can easily be found containing no </p><p>limit points. On the other hand, consider (X) =- the set of limit points of X. (X) is </p><p>a closed, unbounded set, hence X fq (X) ~ ~. Let a E X M (X). By hypothesis, </p><p>there is a sequence of points in X of length less than 3' with sup equal to a, thus </p><p>there is a p E [X] ~ such that At, = {0}. This proves the theorem. [] </p><p>Since < 3' ineffability implies 8 ineffability for all 8 < % a similar result holds </p><p>for < 3' ineffability. </p><p>Our last theorem concerns < 3'-weak ineffability: </p><p>THEOREM 6. I f 3" >--tO is a cardinal, then K--~ (K) </p></li><li><p>Vol. 30, 1978 3,-RAMSEY CARDINALS 97 </p><p>Given any possible homogeneous set X ~ [K ]', X = {/31,/32," 9 }, let p = X r a, </p><p>q = (X - p) [ a. If A~ and Aq do not cohere, then fa (17 U q) </p></li><li><p>98 a.M. HENLE Israel J. Math. </p><p>This shows that X is a cohering set for the Ap, and completes the proof of the </p><p>theorem. [] </p><p>The hypothesis that 3' is a cardinal may be weakened to: "there exists a </p><p>one-one function f:[3,] </p></li></ul>