rowbottom cardinals and jonsson cardinals are almost the same

6
Rowbottom Cardinals and Jonsson Cardinals are Almost the Same Author(s): E. M. Kleinberg Source: The Journal of Symbolic Logic, Vol. 38, No. 3 (Dec., 1973), pp. 423-427 Published by: Association for Symbolic Logic Stable URL: http://www.jstor.org/stable/2273038 . Accessed: 14/06/2014 10:26 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. . Association for Symbolic Logic is collaborating with JSTOR to digitize, preserve and extend access to The Journal of Symbolic Logic. http://www.jstor.org This content downloaded from 188.72.126.108 on Sat, 14 Jun 2014 10:26:12 AM All use subject to JSTOR Terms and Conditions

Upload: e-m-kleinberg

Post on 16-Jan-2017

213 views

Category:

Documents


1 download

TRANSCRIPT

Page 1: Rowbottom Cardinals and Jonsson Cardinals are Almost the Same

Rowbottom Cardinals and Jonsson Cardinals are Almost the SameAuthor(s): E. M. KleinbergSource: The Journal of Symbolic Logic, Vol. 38, No. 3 (Dec., 1973), pp. 423-427Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2273038 .

Accessed: 14/06/2014 10:26

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

Association for Symbolic Logic is collaborating with JSTOR to digitize, preserve and extend access to TheJournal of Symbolic Logic.

http://www.jstor.org

This content downloaded from 188.72.126.108 on Sat, 14 Jun 2014 10:26:12 AMAll use subject to JSTOR Terms and Conditions

Page 2: Rowbottom Cardinals and Jonsson Cardinals are Almost the Same

THE JOURNAL OF SYMBOLIC LOGIC Volume 38, Number 3, Sept. 1973

ROWBOTTOM CARDINALS AND JONSSON CARDINALS ARE ALMOST THE SAME1

E. M. KLEINBERG

Each of the various "large cardinal" axioms currently studied in set theory owes its inspiration to concrete phenomena in various fields. For example, the statement of the well-known compactness theorem for first-order logic can be generalized in various ways to infinitary languages to yield definitions of compact cardinals, and the reflection principles provable in ZF, when modified in the appropriate way, yields indescribable cardinals.2

In this paper we concern ourselves with two kinds of large cardinals which are probably the two best known of those whose origins lie in model theory. They are the Rowbottom cardinals and the Jonsson cardinals.

Let us be more specific. A cardinal K is said to be a Jonsson cardinal if every structure of cardinality K has a proper elementary substructure of cardinality K.3

(It is routine to see that only uncountable cardinals can be Jonsson. Erdds and Hajnal have shown [2] that for n < co no M,, is Jonsson. (In fact, they showed that if K is not Jonsson then neither is the successor cardinal of K and that, assuming GCH, no successor cardinal can be Jonsson.) Keisler and Rowbottom first showed that the existence of a Jonsson cardinal contradicts V = L.) The definition of a Rowbottom cardinal is only slightly more intricate. We assume for the moment that our similarity type has a designated one-place relation. A structure now is said to be of type <K, A> where K > A are cardinals, if its domain has cardinality K

and its designated one-place relation has cardinality A. If K is a cardinal and 8 is an uncountable cardinal less than K, then K is said to be a 3-Rowbottom cardinal if, for any cardinal A < K, every structure of type <K, A> has an elementary sub- structure of type <K, y> where y < 3. X1-Rowbottom cardinals are simply called Rowbottom. Obviously any 3-Rowbottom cardinal is Jonsson-indeed to be 3-Rowbottom we require very special proper elementary substructures of the same cardinality. (A neoclassical result of Rowbottom showed that the existence of 3-Rowbottom cardinals contradicted V = L. His initial theorem was that if a Rowbottom cardinal exists, then any ordinal definable in L (such as card (20TL)) is countable. On the other hand, he showed that Rowbottom cardinals do exist. In fact, he proved that any measurable cardinal is Rowbottom. It is simple to use this to see that if K is measurable and It is any normal measure on K, then almost every (with respect to It) cardinal less than K is Rowbottom.) A very good back-

Received July 10, 1972. 'This research was partially supported by NSF grant GP-29079. 2 The entire rationale behind formulating new axioms for set theory is extremely complex

and will not be discussed here. 3 Throughout, all similarity types are at most countable.

423 D 1973, Association for Symbolic Logic

This content downloaded from 188.72.126.108 on Sat, 14 Jun 2014 10:26:12 AMAll use subject to JSTOR Terms and Conditions

Page 3: Rowbottom Cardinals and Jonsson Cardinals are Almost the Same

424 E. M. KLEINBERG

ground source for Jonsson and Rowbottom cardinals is Devlin's thesis [1]. In this paper we show that Jonsson cardinals and Rowbottom cardinals are

almost the same. Here are our theorems. THEOREM 1. Assume that K is the least Jonsson cardinal. Then there is a 8 less

than K such that K is 3-Rowbottom. THEOREM 2. Assume that K is any Jonsson cardinal. Then there is a 8 less than K

such that for infinitely many A between K and 8, every structure of a type <K, A> has an elementary substructure of type <K, 8>.

Somewhat surprisingly, the proofs of these two theorems are entirely different from one another.4

Our first step along the way to establishing these theorems is to characterize Jonsson and Rowbottom cardinals in pure-set theoretic terms: Let K, A and 8 be infinite cardinals. If x is any set, let [x] ' @ denote the collection of finite subsets of x. Then K -+ [K]< '0 (K -+ [K]

< d') ((K -* [K]< C)6d)) denotes the assertion for each function

Ffrom [K] <"' into A there exists a subset C of K of cardinality K such that F"[C]<'? is a proper subset of A (has cardinality < 8) ((has cardinality < 3)). The characteriza- tions, now, are simply as follows: K is a Jonsson cardinal iff K -* [K] <c--every model of type <K, A> has an elementary submodel of type <K, y> for y < 8 iff K -* [K]L<'o

(and SO K iS 3-Rowbottom (for NO < 8 < K) if, for any cardinal A < K, K -* [K]A,? 6).5

The proofs of these equivalences are quite well known and so we will only sketch them. Assume that K is a Jonsson cardinal and that F: [K] < ' K. Consider the structure

21 = <K; ; F1, F2, ... , Fw ... >n<co

where each function F, is given by F,(al,. , cza) = F({caj , cz j). Then if <C; ; Fl, F2 *.X Fan.. >n< co is any proper substructure of 21 of cardinality K,

C is our desired set, i.e., card (C) = K and F"[C]<w ' C i K. Now assume that every structure of type <K, A> has an elementary substructure of type <K, 8>, and suppose that F: [K] <' - A. Consider the structure

Q3= <K; A; F1, F2, ** ,Fw .. >n <co

where each function F2 is given by Fi(lg,... , a) = F({a1,l , aj). Then it is routine to check that if C is the domain of any type <K, 8> elementary substructure of 3, then card (C) = K and card (F'[C]<c0) < 8. To prove the converses of the characterizations one simply looks at Skolem functions. If

X = <A; P, Q, - *;f, g, -.->

is a given structure, let $F be the result of taking the collection of Skolem functions for C, closing it under composition, permutation of variables and identification of variables, and adding dummy variables so that each function takes a different number of arguments. Then if x c A and S"x denotes {f(x1*,. Xn x) If e I, xi E x},

- =df <K X; P ri X, Q rn9 7x, ;f [ K7x, g [ 9fx,. * >

I In [4] we use these results to prove that the axioms "there exists a Jonsson cardinal" and "there exists a Rowbottom cardinal" are equiconsistent over Zermelo-Fraenkel set theory plus the axiom of choice.

5 These characterizations are due to Rowbottom.

This content downloaded from 188.72.126.108 on Sat, 14 Jun 2014 10:26:12 AMAll use subject to JSTOR Terms and Conditions

Page 4: Rowbottom Cardinals and Jonsson Cardinals are Almost the Same

ROWBOTTOM AND JONSSON CARDINALS 425

is an elementary submodel of C. Now suppose that K -+ [K]K and that 21 is a given structure of power K. By isomorphic translation we may assume the domain of 21 to be K. Let F: [K]<" -+ K as follows:

F({a1,., a*}) = f(al,* *, a.) if fe Y, dmn (f) = Kn and a1 < ...< an,

= 0 otherwise.

Then if C c K is such that card (C) = K and F"[C]<" i K, 2iC is a proper ele- mentary submodel of 2I of power K. Suppose that K -+ [K]' d and that Q3 is a given structure of type <K, A>. We may, as before, assume that the domain of 23 is K

and that the extension of the designated relation is A. Now let F: [K] <C' -*A as follows:

F({f1,, a*n}) = f(t.l , an) iff E A, dmn (f) = K, Cal < ... < Can and f(a?ll * an) < A,

- 0 otherwise.

Then if C c K is such that card (C) = K and card (F"[C] < ) = y < 8, Q3cT is a desired elementary submodel of Z of type <K, Y>. We have thus established our set- theoretic characterizations of Jonsson and Rowbottom cardinals, and so may pro- ceed with the matters at hand.

LEMMA 1. Assume that K -+ [K]K' . Then, for some y < K, K -+K]Y K]

PROOF. Assume that, for each y < K, K + [K] "o, i.e., let G : [K]< -'c y be such that for any size K subset C of K, G'[C]<' = y. Now let F: [K]<'o -C- K be given by F({f1, , an}) = G.1({Ca2,. *, aCn) where a1 < *. < an. Then if C is a given size K subset of K, F"[C] <' = K. For if a < K is given, let a1 be a member of C greater than a and let a2 < . < an be members of C greater than a1 such that Gai({2 , * *, aXn}) = a. Then {a,, . * *, an} E [C]<w9 and F({a1, - - , aj) = a. We now see that this function F contradicts K -* [K]K c) and so we have proved Lemma 1.ut

LEMMA 2. Assume that K - [K]< and y + [y] <'?. Then K -* [K]y, <?

PROOF. Assume that K+ [K],<"7~ and that y + [y] <'. We shall show that K + [K]" C'. To begin with, let us assume that we have coded, with positive integers, all ways to extract subsets from finite ordered sets. For example, the number 5 might tell us that one should extract from any two-element ordered set {a, b}, a < b, the subsets {a} and {a, b}. Let us further assume that we have arranged things in such a way that, for any positive integer n, the cardinality of the finite set n deals with, g(n), is always at most n. This was the case with our example above where g(5) = 2 < 5. Now let G mapping [y]<', into y attest to y + [y]y < (i.e., C c y and card (C) = y imply G"[C]'@ = y) and let H mapping [K]<'o into y attest to K + [K]<,?y (i.e., C c K and C = K imply card (H"[C]<"?) = y). We can define a function Ffrom [K] <" into y by F({a1,... , an}) = G({H(A1), H(A2), H(m)}) where a, < ... < an and Al, A2, * - *, Am are the subsets of {.,, , Iag(n)} the code n tells us to extract. We now claim that F attests to K + [K]" c. For suppose that C is a size K subset of K. We wish to show that F"[C]<'' = y. So suppose a < y is given. Since card (H"[C]<'1) = y, G"[H"[C]<'1] < = y. So let {flj, * *, fm} e [H"[C]< ]i<' be such that G({f1, * * *, fm}) = a and let, for each i, A, be a member

This content downloaded from 188.72.126.108 on Sat, 14 Jun 2014 10:26:12 AMAll use subject to JSTOR Terms and Conditions

Page 5: Rowbottom Cardinals and Jonsson Cardinals are Almost the Same

426 E. M. KLEINBERG

of [C] <t such that H(A4) = Pi. Finally let n be a number telling one to extract A1, * * *, Am from U{Ai I 1 < i < m} and let {a.,,- , An} E [C]<co be such that the least g(n) members of {a1,. , An} are U{A, I 1 < i < m}. Then F({a1,... , anj) = a. Since a was arbitrary, F"[C] < = y and so as C was arbitrary K + [K]<Y ". Thus our lemma is proved. D

PROOF OF THEOREM 1. Theorem 1 follows quite easily from Lemmas 1 and 2. For suppose that K is a Jonsson cardinal. Then K -* [K]< C) and so, by Lemma 1, K- [K]Y <' for some y less than K. Let 8 be the least such y. If, now, K is the least Jonsson cardinal, then A + [A]< co for every A < K. Thus, by Lemma 2, K -*[K]<

for every A such that 8 < A < K. This easily gives that K -+ [K]<" d5 for every A < K.

For if not, let A be the least counterexample, i.e., A < K, K + [K]L,5D, and A is the least such. Let F attest to K + [K] <06. As K -+ [K]", A. let C be a size K subset of K such that card (F"[C]'c) = y < A. Then F r [C]<o: [C]<w -t F"[C]<'0 and since card (F'[C] <) = y < A, we can use K -> [K]1",56 to find a size K subset D of C such that card (F'[D] < ) < 3. Since D is a size K subset of K and card (F'[D] < c) < 8 we contradict our choice of F. The theorem thus follows. D

LEMMA 3. Assume K -> [K]' '? Jor some y < K. Let 8 be the least such y. Then 8 is a regular uncountable cardinal and -* [K] CoD

PROOF. We use a sophisticated version of the proof of Lemma 1. It is easy to see that we need only show K -+ [K]C), < cf(6 (where cf(3) denotes the cofinality of 3). So suppose K e [K]C), <cf(" ) Let H: [K]<' -*cf(8) attest to this and, for each a less than 8, let Ga: [K] < a' -a attest to K + [K]< co. Finally let f map cf(8) in an order-preserving way onto an unbounded subset 8. We now define a function F from [K] <"' into 8 as follows:

F({(xl ...

( at2f3J}) = df Gf(H({aj,.. .aJ}))({?Ci + 1 O t t + , })

where a1 < < a2y3j. Then we claim that F attests to K + [K]" c'. For suppose C is a size K subset of K and a is an ordinal less than 8. Since card (H"[C] < ) = cf(8) let {a,, , aj} in [C] <w be such that f(H({aj,... , aj)) > a. Since

Gf (H({a , -a ))[C - (U{?i,, a* } + 1)]<'? = f(H({a1 . . ., aj))

let {a(i+, * , a(+,} E [C - (U{ai, ... , a2} + 1)] < be such that

Gf(H({al,.- ,ai))({aCi+1, ... * *ai + }) = aC

Then if ai+j+l <*.. < a2i3j are ordinals in C larger than any a,, for n < i + j, {aC1 ,... , a2*3j}e [C]<O and F({a1,.. , a2t3j}) = a. We have thus shown that K + [K]< co, a contradiction. Lemma 3 thus follows. D

LEMMA 4. Assume that A < K are cardinals. Then for every function F from [K] < C' into A + ,6 there exists a function GF from [K] < 0 into A such that for any subset Cof K card (F"[C] < w) < card (GF[C] <')+.

PROOF. To begin, let us assume that we have a coding as in the proof of Lemma 2 but here our coding tells us how to extract precisely two subsets from each finite (nonempty) ordered set. In fact, let our notation and conventions concerning the coding be exactly as in the proof of that lemma. Also, for each ordinal a less than A+, let ga be a function mapping a + 1 1-1 into A. Now suppose that F: [K]<'o A+

6 For any y, V+ denotes the least cardinal greater than y.

This content downloaded from 188.72.126.108 on Sat, 14 Jun 2014 10:26:12 AMAll use subject to JSTOR Terms and Conditions

Page 6: Rowbottom Cardinals and Jonsson Cardinals are Almost the Same

ROWBOTTOM AND JONSSON CARDINALS 427

is given. Define GF: [K] <C -?A by GF({f1,.. , = gF(A1)(F(A2)) where a1 < ... < an, A1 and A2 are the subsets of {a1, , a,(n)} n tells one to extract, and F(A1) > F(A2). If now C is any subset of K it is easy to see how to map any initial segment of F"[C]<w 1-1 into GF[C]<O. For if A1 and A2 are any two mem- bers of [C] <w, F(A1) > F(A2), let tAl12 be a member of [C]'< such that GF(t,12) =

gF(Aj)(F(A2)). Then F(A2) ~- GF(tA1A2) is a 1-1 mapping of the initial segment of F"[C]< determined by F(A1) into GF[C] <'O. This now easily implies that card (F"[C] < ') < card (GF[C] < w)+. DG

PROOF OF THEOREM 2. Assume that K is a Jonsson cardinal. Let 8 be the least y such that K-* [K] <'?. By Lemma 1, 8 < K. By Lemma 3, K-* [K]< 6 C. We define cardinals Yn for n < w inductively by yo = 3 = y. It is now easy to check that K -- [K]]<6 for each n < w. We do this by induction, the case n = 0 being Lemma 3. So suppose that F: [K] <' ' Yn + 1 = y +. Let GF be as in Lemma 4 and, by the induction hypothesis, let C be a size K subset of K such that card (G'[C]<') < 3. Then by Lemma 4, card (F"[C]<w) < 3. We can now use K [K]

< ? 6 find a size K subset D of C (and hence of K) such that card (F"[D] < 0) < S.

Theorem 2 is thus proved. C: - REMARK. One can actually prove more for an arbitrary Jonsson cardinal than indicated in Theorem 2. For example, Lemma 2 still applies and so if K is Jonsson and 8 is the least y < K such that K -- [K]< 'I, K -- [K] <C) A for every non-Jonsson A which is at least 3. Indeed, contained in the proof of Theorem 1 is the following:

THEOREM. If K-- [K]< 'D and A > 8 ? y are cardinals such that no Jonsson cardinal appears between 7 A and 8, then K -- [K] A,

7iq is between a and 9 if a > X > P.

REFERENCES

[1] K. DEVLIN, Some weak versions of large cardinal axioms, Doctoral Dissertation, Univer-

sity of Bristol, 1971. [2] P. ERDOS and A. HAJNAL, On a problem of B. Jonsson, Bulletin de l'Acadimie Polonaise

des Sciences. Sirie des Sciences Mathe'matiques, Astronomiques et Physiques, vol. 14 (1966),

pp. 19-23. [3] E. M. KLEINBERG, Rowbottom cardinals and Jonsson cardinals are almost the same,

Notices of the American Mathematical Society, vol. 18 (1971), p. 827. Abstract 71T-E85. [4] , The equiconsistency of two large cardinal axioms (to appear). [5] , The abstract compactness idea, M.I.T. Logic Seminar, 1971-72 (to appear). [6] F. ROWBOTTOM, Some strong axioms of infinity incompatible with the axiom of con-

structibility, Annals of Mathematical Logic, vol. 3 (1971), pp. 1-44. [7] J. SILVER, Some applications of model theory in set theory, Annals of Mathematical Logic,

vol. 3 (1971), pp. 45-110.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY

CAMBRIDGE, MASSACHUSETTS 02139

This content downloaded from 188.72.126.108 on Sat, 14 Jun 2014 10:26:12 AMAll use subject to JSTOR Terms and Conditions