- Email: [email protected]
A hierarchy of ramsey cardinals
Annals of Pure and Applied Logic 49 (1990) 257-277 North-Holland
A HIERARCHY
OF RAMSEY
257
CARDINALS*
Qi FENG** Institute of Sofiware, Academia Sinica, P. 0. Box 8718, Beijing, People’s Rep. of China Communicated by T. Jech Received 15 June 1989 Assuming the existence of a measurable cardinal, we define a hierarchy of Ramsey cardinals and a hierarchy of normal filters. We study some combinatorial properties of this hierarchy. We show that this hierarchy is absolute with respect to the Dodd-Jensen core model, extending a result of Mitchell which says that being Ramsey is absolute with respect to the core model.
1. Introduction Our set-theoretical usage is standard (see [S]). If x is a set, P(x) is the power set of x, and lx] is the cardinality of x. If x is a set and d is a cardinal, then [x]“={y~x:JyJ=J.}, and [x]‘“={ycx:Iyl<)L}. In particular, [xl’” is the family of finite subsets of x, and [xl” is the family of subsets of x with exactly II elements, n < w. If x, y are sets, then yx is the set of functions from y to x. Iffis a function on A and B E A, flB is the restriction of f to B, and f”B = {x : 3y E B x =f(y)} is the set of images of B under 5 If H is a subset of the range off, then f_‘(H) = {x :f(x) E H}. Let (Ybe an ordinal. V, is the set of all sets of rank smaller than (Y. That is, and if A is a limit ordinal then V, = V,= 0, the empty set; V,,, = P(V,); tJ {V, : 1y< A}. Then our universe of set theory ZFC, V, is the union of all Vol’s. Let X be a set of ordinals. A function f defined on X is regressive if for all a E X, (Y> 0 implies f(a) < CY.A function f defined on [X] 0 implies f(a) < min(a). Let X be a set. If f is a function on [X]“, l~n~ to denote that for each f :[I$” --*2 there is an H E [rclX so that H is homogeneous for J A cardinal K is Ramsey if K--* (K)>~. *This work is a part of the author’s thesis in Penn State in 1988, supervised by Professor T. Jech. ** Current address: Dept. of Math., Natl. Univ. Singapore, Lower Kent Ridge Rd., Singapore 0511. 0168-0072/90/$03.50 0 1990-
Elsevier Science Publishers B.V. (North-Holland)
258
Q. Feng
Later, we will see that if K is Ramsey then K is the Kth weakly compact cardinal. Also there might be lots of Ramsey cardinals. Let us consider a model A = (A, . . . ) of an infinitary language, with A 2 K, where K is an infinite cardinal. We say that a set H E K is a set of indiscernibfes for A if for every n < o, for every formula cp(vr , . . . , v,) of the language of A of n free variables, we have Akq(o~,...,
4
@
Ab &PI> . . . > Bn)
whenever q < * - - < an, PI < - * - < /3n are from H. The following theorem is folklore (see [8]). Theorem. Let K be an infinite cardinal. Then K is Ramsey if and only if every model A = (A, . . . ) of a language of size < K, with K G A, has a set H E [K] K of indiscernibles.
Jensen noticed that one need only consider a special kind of models [3,5]. The following definition is due to Jensen [5]. Let AI,. . . , A ,cKandletA=(L,[A, ,..., A,],E,A1 ,..., A,). FOrA
(f))s
Theorem
(Jensen [5]). ~isZ3amseyifandonlyifforeveryA,,...,A,~~, A=
(L,[AI,
. . . , A,],
E, AI,
. . . , A,)
has a good set of indiscernibles of order type
The following interesting Ramsey cardinals. Theorem
theorem
K.
of Mitchell stimulates our work to study
(Mitchell [12]). Zf K is Ramsey, then K is Ramsey in the Dodd-Jensen
core model
K.
For a limit ordinal a>O, a subset X E (Y is unbounded in (Y if Vy< (Y 3/!IEXY
A hierarchyof Ramsey cardinals
259
on K, we denote by I+ the collection of subsets of K which are not in Z, i.e., I+ = {x E K :x $ Z}. Then Z is nontrivial iff I+ is not empty. An ideal Z on K is A-complete (A < K) if Z is closed under unions of fewer than A. of its elements. Z is normal if for every S E I+ and every regressive f : S- K there is S’ E I+ so that S’ is a subset of S and f is constant on S’. F c P(K) is a fiber if X, Y E F j X n Y E F, and X E F implies that every superset of X is in F, and K - { cu} E F for each (Y< K. A filter F is nontrivial if 0 $ F. Notice that we do not consider those so-called principal filters, i.e., F={X~K:LYEX} for some (YCK. Let Z(F)={~EK:K-~EF}. Then F is a nontrivial filter iff Z(F) is a nontrivial ideal. We call Z(F) the dual ideal of F in that case. Conversely, if Z is a nontrivial ideal on K, let F(Z) = {x E K : K -x E Z}, then F(Z) is a nontrivial filter. We call this filter the dual filter of I. Now a nontrivial filter F is A-complete if its dual ideal is A-complete. From now on by filter we always mean nontrivial filter. Then a filter F is A-complete if and only if F is closed under intersections of fewer than Iz of its elements. F is normal if its dual ideal is normal. Then F is normal if and only if F is closed under diagonal intersections, i.e., if (Xol : a < K) is a sequence from F, then A ,cKX, = {/3< ~:tlc~
Theorem
that on S’ f intersection.
is constant.
Consequently,
the club filter is closed under diagonal
Let Z be an ideal on K. Let X be a subset of K. The notation X+ (I+): means that for every f : [Xl” + m there is YE I+ so that Y G X and Y is homogeneous for f. And X-t (Z’):@ means that for each f : [X] <@+ 3c there is YE I+ so that Y c_X and Y is homogeneous for f. The notation X% (Z’)z” means that for every regressive function f : [X] co+ K there is YE I+ so that YE X and Y is homogeneous for $ Let K be an infinite cardinal. A filter U on K is an ultrafilter if for any X E K either X E U or K -X E U. An uncountable cardinal K is measurable if there is a K-complete ultrafilter on K. The following two theorems are basic in the studying of measurable cardinals.
260
Q. Feng
Theorem (Scott [15]). Zf
K
is
measurable,
then there is a normal
K-complete
ultrafilter on K.
Theorem (Rowbottom
[14]). Let U be a u-complete normal ultrafilter on
x E u.
is
Zff :[xy ---,K
regressive,
K.
Let
then there is Y E U so that Y is homogeneous
for f-
(See [8] for proofs of both theorems.) Let Z7: @A, respectively) be the class of formulae of second-order language which in prenex form has at most n alternating blocks of quantifiers of second-order variables, starting with the universal (existential, respectively) quantifier. A cardinal K is ZZk-indescribable if and only if for every rrf, formula #(X) with one free second-order variable, for each R c V,, if (V,, E, R) t=e(R), then there is (Y< K so that (V,, E, R fl V,) b $(R tl V,). Define EA-indescribability analogously. Remark. It is useful to notice that K is ZIt-indescribable if an only if K is inaccessible and for any ZZf,formula #(X) with one free second-order variable, for each R c K, if (K, E, R)l= @(R), then for some cy< K we have ((u, E, ~II R) tr$r(R II a).
More generally, a subset A G K is Z7:-indescribable if and only if for each Z7: formula #(X) with one free second-order variable, for each R c K, if (K, E, R) k #(R), then &YEA ((u, E, Rfla)k$(anR). Theorem (Baumgartner
[l], Levy [ll]).
(1) Let K be a regular
uncountable normal ideal
cardinal. Then {x c K :x is not Z7:-indescribable} is a K-complete on rc. (2) Zf K is IZ!,-indescribable, then the ideal defined in (1) is nontrivial. (3) Zf K is inaccessible, and the ideal defbted in (1) is nontrivial, then K is ZZi-indescribable.
We will call the ideal defined in (1) the ZZA-indescribable ideal on K. If it is nontrivial, then we call its dual filter the IZA-filter. In [l, 21, Baumgartner generalized this, among other things, to Ramsey ideals. Let K be regular uncountable. An X E K is Ramsey if for every regressive f on WI- and any club C G K there is an H G C n X so that lZZ1= K and H is homogeneous for f. Z = {x G rc :x is not Ramsey} is defined to be the Ramsey ideal on K, which is K-complete and normal. Theorem (Baumgartner
[2]). Zf K is Ramsey, then the Ramsey ideal on K is nontrivial, and {a < K :a is weakly compact} is in the Ramsey filter, the dual of the Ramsey ideal.
A hierarchy of Ramsey cardinals
2. A Ramsey operator
261
on ideals
In this section, we define an operator 9? on ideals so that for any rc-complete ideal I on K, 3(Z) is a K-complete normal ideal and extends I. We will also give some equivalent definitions of the operator. This operator will be used in the next section to define a hierarchy of normal filters in the theory of large cardinals. Definition 2.1. Let K be a regular uncountable cardinal. Let I be an ideal on K For every X E K, X E 9+(Z) if and only if for every regressive such that Z z~[K]? function f : [X]
P(K) {xs
-
%+(I), K:K--XE
s(l)}.
We call the operator 5%the Ramsey operator. Immediately, W(Z) is an ideal extending both ideals Z and the nonstationary ideal ZVS,. Also, if Z E J, both are ideals, then 99(Z) E 9?(J). Theorem 2.1. a(Z) is a K-complete normal ideal. Proof. Without loss of generality, we assume that K $93(Z), that is, 9?(Z) is nontrivial. Notice that the K-completeness follows from the normality. Suppose that X, E %(I) for each (Y< y < K, X = Uncy X, and X is in 3+(Z). We may assume + K by f(B) = the least LYsuch that j3 E X, for each that Xrly=0. Definef:X /3 E X. Then f is regressive. By normality, for some Y s X, Y E 9+(Z), and for some a; El/3E Y f(/3) = (Y. We then have Y s X,. But X, E 9?(Z) and 3(Z) is an ideal. This is impossible. Hence X E R(Z) and 5?(Z) is K-complete. We need to show that 93(Z) normal. Given X c_ K, h :X+ K is a regressive function. Assume that for all (Y< K, h-‘({a}) E 3(Z). We prove that X E 3(Z). For each LY< K, fix fn, C, to witness that h-‘( { a}) E S(Z). Let Ed: K X K * K be a G6del pairing function. Define D to be the set {(Y< K :n"cx X LY G cu}, Let C = A,,& fl D. Then C is a club. Suppose that X E 9+(Z). Then X n C E 9+(Z). Define a regressive function f:[xnc]
f ({a>>= 4h(d
fwWN;
Let YE I+ be such that YE X n C and Y is homogeneous for J Let 77be such that V& E Y f ({a}) = q. Then for some j3, tla! E Y h(a) = j3 and /3 < (Y. But on [Ylcof =fs. Since Y c C, n h-‘({/I}), Y cannot be homogeneous for fs. We get a contradiction. This is the end of the proof. Cl
Q. Feng
262
Consequently, if 9(Z) is not trivial, then W*(Z) is a normal K-complete filter. We will be interested in such filters later on. In the rest of this section, we will be proving some results which tell us how to redefine the Ramsey operator in different ways. The following theorem is a straightforward generalization of a standard theorem. Theorem 2.2. Let
K
be a regular cardinal. Let NS, G Z and Z G P(K)
be an ideal.
For X c K, the following are equivalent:
(1) (2) (3) (4)
x E 9?+(z). X% (z+),‘o. x+ (z+),(o. kfA< KX-@+);?
(5) x-+(1+),6".
(6) For any structure A= (A, <, . . . ) of a countable language, X c A, A has a set of indiscernibles B c_ X which is in I+. Let X c K. By a (w, X)-sequence we mean a sequence S of the form We (~~~.**~cr~,cr, ,..., LY,EX), where S, ,...n,~a,. S=(S,, ...=n:lsn 0, for all ,...am= alnS, ,...B.. Bl, * - - , #& from Y, (~~~/3i*S~ al, * * * , cu,, Theorem 2.3. Let containing
[K]?
K be a regular uncountable cardinal. Let Z be an ideal Then for any X E K the following are equivalent:
(1) x E 9+(z). (7) For any (w, X)-sequence
S there is a Y E I+ so that Y is a subset of X and Y
is homogeneous for S. (8) For any (0, X)-sequence S there exist a Y E X with Y E I+ and a sequence (A,,: 1
Sal...an= a-1n A,. Proof. (1) 3 (7)‘. Given a (0, X)-sequence S, we want Y. Assume that so that for aO> =
0 [ l+y
if S, = S, II aO, if y=min{b
Let YE I+, YE X, be homogeneous for g. Then, of course, S, = S, n min(a) for all a, b E [Y]” such that max(a) < min(b). But then given arbitrary a, b E [Y]“, min(a) < min(b), choose c E [Y]” so that max(a), max(b) < min(c). Then S, = S, n min(a) and S, = S, rl min(b), so S, n min(a) = S,. ’ The proof here is given by the referee.
Originally,
we had a rather complicated
argument.
A hierarchy of Ramsey cardinals
263
+ K be regressive. For u E [X]
2.4. Let
[K]‘“.
K be regular uncountable. Assume that Z is an ideal on Then for any X E K the following are equivalent:
Then
K
and
(1) x E 9’(Z). (9) For AI, . . . , A, G good set of indkcernibles
K,
A = (L,[X,
The proof follows from Theorem Corollary 2.5. On/J’ if K $
Let
K
AI, . . . , A,],
E, X, AI, . . . , A,)
has a
YE X which is in I+.
2.3 and standard arguments.
be a regular uncountable
cardinal. Then
K is
Ramsey if and
%([K]--).
is Ramsey, we call %([K]<~) the Ramsey ideal on K, and the dual of it, ~F([K]<~), the Ramsey jilter. By Theorem 2.1, it is K-complete normal filter. We call every X E W([K]<~) a Ramsey subset of K. If
K
3. The hierarchy
In this section, we iterate the Ramsey operator to define a hierarchy of Ramsey cardinals. Then in the next sections we show that the hierarchy defined here is really a proper hierarchy. Also we will define some terminologies which will be used later. Definition 3.1. Let K be a regular uncountable cardinal. We define I?* = [K]<“; I?1 = NS,; for n < o, Zi= 9?(ZE_2); for (Y?=w, if a=/3+1, thenZ”,=%(Z;);if Ly is a limit ordinal, then Z”,= Us<&.
For any regular uncountable K, there exists some LY< (2”)+ so that I:+1 = Zz, because the Z”,‘s form an increasing chain of ideals and there are only 2” many subsets of K. Let us denote by 0, the least (Ywith this property. Notice that for many K the ideals defined above are trivial. We will be only interested in those cases where some of them are nontrivial. This leads the following definition. Definition 3.2. LA 17,-Ramsey if K $ Zz; Remark.
K K
be a regular uncountable cardinal. For is completely Ramsey if for all a, K $ ZZ,.
(Y2 0,
K
is
For a limit ordinal (Y> 0, K is Z7,-Ramsey if and only if K is ZJ@-Ramsey for all /!I< CY.And K is completely Ramsey if and only if K $ Z:Kif and only if 3a Z”,+1= Z”,# P(K).
Q. Feng
264
We define the Ramsey rank of K, Rk(K), to be the least (Yso that K E Z”,if such exists; otherwise, Rk(K) = =J. Notice that if Rk(K) is an ordinal, then Rk(K) = f3, and it is either 0 or a succesor ordinal. In particular, Rk(rc) > 0 if and only if K is Ramsey if and only if K is ZZO-Ramsey. Therefore, the nontriviality of any Z”, for as0 is a large cardinal property. On the other hand, the following theorem tells us that the nontriviality of the Ramsey operator is worth to study. Theorem
3.1.
ZfK
is
measurable,
then
K is
completely Ramsey.
Proof. Let U be a normal K-complete nonprincipal ultrafilter on K. By Rowbottom’s theorem, for each X E U, for any regressive function f :[X]<“+- K, there is Y E U such that Y is homogeneous for f. Let Z = {x G K :K - x E U}. By a simple induction on CY,we have Zlc,G I. Therefore, none of them is trivial. K is then completely Ramsey. 0 3.3. Assume that K is Z&-Ramsey. We call Z”,the ZZ,-Ramsey ideal on K, and the dual of it the &-Ramsey filter on K. We call each subset X of K which is not in the Z7=-Ramsey ideal a &-Ramsey subset of K. If K is completely Ramsey, then we call ZGzthe completely Ramsey ideal and the dual of it the
Definition
completely Ramsey filter, each x G
K
not in the ideal a completely Ramsey subset
Of K.
Remark. Let K be fla-Ramsey. If (Y= /I + 1, by Theorem 2.1. If (Yis limit ordinal > 0, then may not be K-complete. If (Y< K+, cf (Y= K, (sometimes it happens, e.g., if K is completely is normal as the following fact shows.
then I^, is normal and K-complete Zlc,is min(cf a; K)-complete, and it then Z”, is not normal. If cf & > K, Ramsey, then ZE+c ZE++i), then Zz
Fact. Zf K is ZIa-Ramsey, (Y is a limit ordinal, and cf (Y> K, then Z”, is normal.
Let XE(Z~))+ and f:X+K be regressive. increasing sequence convergent to (Y.Then
Proof.
v~
)+ 3Ys
Let
((us:/3
be an
= y} is unbounded in cf cr. Let Y = { 9 E Hence 3y-C~ so that {/I
4. The hierarchy theorem
In this section, we finite stages. We will the hierarchy defined We start with a
for cardinals of finite Ramsey
rank
prove that the hierarchy defined in Section 3 is proper for also introduce a new type of Ramsey cardinals, related to before. lemma of Baumgartner, which allows us to use the
A hierarchy of Ramsey cardinals
265
indescribability argument in addition to pure combinatorial argument. Let K be a cardinal and let A c K. A (1, A)-sequence S is a sequence S=(S,:CKEA) withindexsetA andS,scuforeach (YEA. Lemma 4.1 (Baumgartner [l]). Let K be a cardinal. Let A E K. Assume that for every (1, A)-sequence S there is a B E Q such that B is homogeneous for S. Zf m 3 0 and every member of Q is ZIh-indescribable, then A is ZI!,,+,-indescribable. Proof. Let R be a relation on K. Let # be a flk+,-sentence. And assume that (K, E, R) k 9. Write 9 as V.r 3y q(x, y), where r+!~(x, y) is a n!,, formula. Suppose a, X,) LVY~W(xY, Y).
VCYEA~X&,E,RI
Let S, code X, as a subset of a. Let B E Q be homogeneous X=IJ{X,:CYEB}. Thenforsome YsKwehave
for S and let
(K E, R, X, Y) I=W(X, Y). Since B is Z7;-indescribable,
there is a E B such that
(a; E, R 1 a, X 1a, Y 14 b v(X But X 1 (Y=X,.
A contradiction.
1 a, Y 1~1.
•i
It is easily seen by induction that the definition of “K being D,,-Ramsey” is a n’ n+2 sentence. Our next theorem says that there is no simpler (even no ZA+2) definition. Theorem 4.2. Zf K is Z&-Ramsey, then K nn-Rumsey, then X is ZIi+,-indescribable.
is
ZZ!,+l-indescribable. And if X s K i.~
Consequently, if K is K&-Ramsey, then the 17:+,-filter over the n,-Ramsey filter over K.
K
is contained
in
Proof. By induction on n we prove that if X c K is I&-Ramsey then X is ni+ i-indescribable. For n = 0, K is Ramsey, hence K is weakly compact. It follows from a theorem of Baumgartner [l, 21 that every Ramsey subset of K is fl:-indescribable. For n = m + 1, K is nn-Ramsey implies that K is IT,-Ramsey. By induction hypothesis, every II,,,_,-Ramsey subset of K is II;-indescribable. If X c K is I7m+l Ramsey, by Theorem 2.3, then every (1, X) sequence S has a homogeneous n,_,-Ramsey subset Y cX. By the lemma of Baumgartner, X is D&+,indescribable. •i Applying this indescribability theorem, we can now prove our first hierarchy theorem for cardinals with finite Ramsey rank which says that if K is n,,+,Ramsey then there are stationary many nj-Ramsey cardinals below K.
Q. Feng
266
Theorem 4.3. If is I&,-,-Ramsey
K is
I&,-Ramsey, and A c
K is
Iln_l-Ramsey,
then {a! <
K
:A tl LY
in a} is in the l&-Ramsey filter on K. (Notice that II_,-Ramsey
subsets of 3care exactly the stationary subsets of A.)
Proof. Let A c K be a &,-Ramsey subset of K. Then “A is &,-Ramsey” is a l7rn+l sentence over (K, m, A). Therefore, Y = {d < K :A n CYis not II,,_,-Ramsey in a} is in the n!,+, -ideal on K. Since K is n--Ramsey, by the Indescribability Theorem 4.2, the II:+, -ideal on K is not trivial. Hence the complement of Y in K is in the flA+, -filter on K. Again by Theorem 4.2, this filter is contained in the L&-Ramsey filter on K. We are done. 0 Immediately,
we have the following hierarchy theorem. then {cx < K : (Y is I&-Ramsey} is in the K is n,,,, -Ramsey, filter. In particular, there are stationary many I&,-Ramsey cardinals
COrOkUy 4.4. If l&+,-Ramsey below
K.
The following fact says simply that the indiscernibles you could get for any structure of countable language over a &+,-Ramsey cardinal are from the smaller L&-Ramsey cardinals essentially. Fact. If K is &+, -Ramsey and X E K is a lI,,,1-Ramsey subset, then for any regressive function f : [X] cm+ K, there is an A G X such that A is l&+,-Ramsey, A is homogeneous for f, and Va EA cx is II,,-Ramsey. Proof. It simply follows that if X c_ K is D”+r-Ramsey n,-Ramsey} is fin+,-Ramsey. Cl We are now ready to prove our second hierarchy theorem. the following theorem. Theorem 4.5. Assume that
K is
l&-Ramsey.
If X G
K is
then
{(YE X: (Y is
First, let us prove
fin-Ramsey
then
{(Y E X : either a is not II,,-Ramsey or X n (Y is not I&-Ramsey} is l&-Ramsey.
Proof. Let that
K
In particular,
{ LY< K :(Y is not I&,-Ramsey)
be the least counterexample.
is K&,-Ramsey.
Let X G K be a n”-Ramsey
subset so
A = { LYE X : either (Yis not K&-Ramsey or X rl (Yis not n,,-Ramsey}
is not nn-Ramsey. Then X - A = {a E X : a is L&-Ramsey & (Yn X is K&-Ramsey} is L&,-Ramsey. For CYE X -A, by the minimality of K, X, = (Yn A is fl”-Ramsey in LY.Set q Y = lJ {X, : (1:E X - A}. Y is then I&-Ramsey. Contradiction.
A hierarchy of Ramsey cardinals
By Corollary 4.4 and this theorem, hierarchy theorem. Theorem
4.6.
If K is
Z&+,-Ramsey,
we have immediately
then Vi Sn
267
the following second
the Z&Ramsey filter on
K
ih
properly contained in the ZJ+l-Ramsey filter on K. Proof. For i =Sn, {(Y < K : a is Z&-Ramsey} is in the Z&+,-Ramsey filter on K but
not in the Z&-Ramsey filter on
K.
0
The following definition is essentially due to Baumgartner [2], where he considered the cases when n = 0 or 1. Also the following two theorems for n = 0 or 1 are due to him [2], where he uses the names “pre-Ramsey,” “pre-ineffably Ramsey” and “ineffably Ramsey,“. Definition 4.1. For a Z&-z-Ramsey
cardinal K, for X c K, (1) X E %E,+(Z,“_,)if and only if for every regressive f : [X]<“-* K, for each club C c K, there is LYE C n X and there is A c C fl cx rl X so that (Yis I&,_,-Ramsey and A is ZIn_z-Ramsey in a and A is homogeneous for f and V/3 E A f ({p}) =
f ({a>)* (2) x E%K-2) ex
$ %x-*).
Notice that by “IT-i-Ramsey cardinal (i = 1, 2)” we mean regular uncountable, and by “E,-Ramsey subset” we mean unbounded subset, as well as by “I7_,-Ramsey subset” we mean stationary subset. Theorem 4.7. Zf K is Z&-Ramsey, then 9&(Z,“_,) is a nontrivial xc-complete normal ideal, and it is contained in the Z&-Ramsey ideal. Proof. The proof of the normality
of ‘&,(Zk,) is the same as the proof of the normality of 9(Z). We omit the argument. We need to show that every Z7,-Ramsey subset of K is in $%e,‘(ZE_,). First, assume n 2 1. Fix a ZZn-Ramsey subset X of K. Let f be a regressive function from [X]
is in the Z&-,-Ramsey filter on K, hence it is in the Z7,-Ramsey filter on K. Therefore, B fl C fl Y #0. Let a be in this intersection; then A fl LY is Z&_,Ramsey in (t: We are done for n 9 1.
268
Q. Feng
For n = 0, one gets the A as above in the same way. Then B={cx<~:Ancrisunbounded} is a club in
K.
Hence B n C n Y # 8. Then any (Yin this intersection
works.
Cl
Theorem 4.8. (1)
K is Z&-Ramsey if and only if the ZZ~+I-ideal and the ideal %&(I:_,) generate a nontrivial ideal.
(2) Zf $?$(I:_,)
K is
I?,-Ramsey,
then the Z&-Ramsey ideal is the ideal generated
by
and the II:+,-ideal.
Proof. (+) It follows from Theorem 4.2 and Theorem 4.7 that both the ITf+,-ideal and 9&(Z,“_,) are contained in the &-Ramsey ideal. (e) Let Z be th e generated ideal. It suffices to show that for any X G K if X $ Z then X is Z&,-Ramsey, which implies that Z,”E I. Assume that X $ I, f : [X]4w+ K is regressive, and C E K is a club. Then
cnx4z. Suppose that VA E C n X if A is ZZ,,_,?-Ramsey then A is not homogeneous f. Then this is a ZZA,, sentence @ over the structure
for
A=(w,C,X,f).
Thus Y={cx
I[a]‘“)k$} is in the ZZA+i-filter on This is because
K.
xnc=(XncnY)U(XnCn(K-Y)),
and the second part of the right-hand side of the equality is in the Z7:+,-ideal on K. Let cueXnCflY, and let AsXnCrlY be Z7k_z-Ramsey in LYand be homogeneous for f. But (YE Y. This is a contradiction. Hence X is Z&-Ramsey. 0
5. The hierarchy theorem for cardinals of infinite Ramsey rank In this section we prove a hierarchy theorem for cardinals of infinite Ramsey rank. How we are going to do it is the reason why we make the distinction between the case of finite Ramsey rank and the case of infinite Ramsey rank. Notice that in the proof of the first hierarchy theorem for cardinals of finite Ramsey rank we essentially use the indescribability argument. To prove a hierarchy theorem for cardinals of infinite Ramsey rank, we use canonical sequences, which were also used by Baumgartner in [4] to establish his ineffability hierarchy. Let K be a regular uncountable cardinal. For functions f, g E Icy, we define f
A hierarchy of Ramsey cardinals
everywhere)
269
if
{a < K :f(4 contains a club in
< g(a)>
K,
and f C g if
{a
contains a club in K. A sequence & :a' < K+) of elements of "K iS a canonical sequence on K if the following two conditions are satisfied. (1) For all cu, B < K, LY < /3 implies f= Cfs. (2) For any other sequence (g,: CY < K+) of elements of K~ with the property thatVa
: a < K+)
K so
is
that h,(n)
Fact 4. For all ~3<
K+
a canonical sequence on
K,
then for all cx < K+ there is a
< Inl+ for all n E C,.
there is a club Ce c K such that for all 2 E Ce if A > o and ),
is regular then {Y < A :f&u(y)
=f 2(Y))
contains a club in 3L, where f”, fK are canonical sequences on A and K respectively.
Now we are going to prove the hierarchy Ramsey rank.
theorem
for cardinals
Theorem 5.1. Assume that (& : CY< p) is a sequence of elements of a
"K
of infinite
such that if
then {Y < K:f,(Y)
is in the ZIP-Ramsey filter on K, and fa( y) S a for all y < all X c xc, if { y < K :X fl y is IT&)-Ramsey
in y}
is Z&-Ramsey, then X is IZ*-Ramsey.
Proof. We prove this theorem by induction on (Y. If Q is finite, we are done. So assume (Yis infinite.
K.
Then for all a < u and
270
Q. Feng
Suppose LY= b + 1. Then Y = {y C K :X II y is 17__e(,,-Ramsey in y & fs(y)
~(~e’“~~=(~~eleastcEXI_x~, ifxEE.xqJ otherwise. and g({y,,
- - * 7 Y~+~})
=
the value off
on [X,,]“.
Now for some Q-Ramsey set A c C’ n Y, A is homogeneous for g. Then for all&q~A ~ 0 is a limit ordinal. Fix any /3 < (Y.Then {?I < ~:f~(rl)
& X n v is fl&,-Ramsey
in rl]
is &-Ramsey. By induction hypothesis, X is flP-Ramsey. Thus for all /3 < (YX is flP-Ramsey. Hence X is &-Ramsey. This completes the proof. 0 Theorem
5.2. Let ($n : rj C K+) be the canonical sequence
on
K.
Assume that K is
I&-Ramsey with p < K+. Then for all a S p X, = {y < K : y is not l7,-(,,)-Ramsey} is I&-Ramsey. Proof. We prove this by induction on K. If A = { y < K : y is n_*(,) -Ramsey} is not n-Ramsey, assume that A is &-Ramsey.
then we are done.
So
Let C, c_ K be a club such that for all regular uncountable A E C, we have contains a club D in A. Then C, n A is &-Ramsey. If {Y < A:f30dY) =fa(r)l AEC,rlA, Then X fl A = { y < A: y is not 17f,(vj-Ramsey}.
271
A hierarchy of Ramsey cardinals
Since A is IT&,- Ramsey, L(Y) %(4,
and DA is a club in A, and for all y E II*, f&&y)
=
we have
{y E DA: is not KIfo(,)-Ramsey} is II__a(i,,-Ramseyin A.by induction hypothesis. Thus X, fl I. is II&-Ramsey By the previous theorem, X, is &-Ramsey. This finishes the proof. Cl
in 3L.
More generally, the same argument shows the following. Theorem 5.2’. If &-Ramsey
K is L$-Ramsey, implies that
@ < K+, then for all 0: S /I, for all X c K, X i.s
{ y E X : either y is not II&~-Ramsey
or X fl y is not 17f,Cvj-Ramsey}
ti l&-Ramsey.
Next we want to show that higher rank Ramsey filters always concentrate on the set of smaller Ramsey cardinals, which is then our first hierarchy theorem for cardinals of infinite Ramsey rank. Theorem 5.3. Let (f, : ?j < K+) be the canonical sequence on Q-Ramsey,
K.
Assume that
with (Y < K+. Then for all p S (Y, for all subsets X of
II c-r+1 -Ramsey
K is
K,
if X is
XGK
be a
then
{y <
K :X
tl y is 17fp(Yj-Ramsey}
is in the I&+, -Ramsey jilter. Proof. We prove this by induction on #I.
For /3 = 0, it is true. Suppose /?=q+l. Then fs(y)=fv(y)+l Q-Ramsey set. By induction hypothesis, {y <
K :X
for each Y
Let
fl y is I&(,,)-Ramsey}
is in the IIq+, -Ramsey filter on K. Let T = {y < K : either y or X fI y is not Eff,~,,~+l-Ramsey}. Suppose that T were fl,+,- Ramsey. We want to get a contradiction. Then A = {y
< K :X rl y is II&,,)-Ramsey
but not I$-,,,+,-Ramsey}
is &+, -Ramsey. For each y E A let g, : [X fl y]‘” + y be regressive and let 0 4 C, be a club in y such that no II f,(,,j-Ramsey subset of C, rl X is homogeneous for g,. Let D = {/3 < K : f3 is closed under the Gijdel pairing function}. Then D rl A is
Q. Feng
272
&+,-Ramsey.
Define g : [D n Alcm+
K
by
where s,={(n,B1,...,~~,gy({B1,...,~~8n))>:1~n
K :E
rl y is nfq(,,-Ramsey}
is in the Q-Ramsey filter. Let y E B be such that E rl y is nfc,,-Ramsey. Then cm = g, and C rl y 2 C,, is a club. But E n y is homogeneous for g, and fI[Xnrl Since C, fl E rl y is still n&,)-Ramsey, we get a it is nf,(vj-Ramsey. contradiction. Therefore, T can not be &+,-Ramsey. Now assume that /3 is a limit ordinal, and X z K is I$-Ramsey. If rl< /3, then X is II,-Ramsey and D,, = {y C K: X fl y is ZIf,(,,)-Ramsey} is in the n,+,Ramsey filter on K by the induction hypothesis. Hence D,, is in the flfl-Ramsey filter, and therefore in the n,+,- Ramsey filter on K, which is K-complete and normal. Let D c K be a club so that Vy E D f@(y) is a limit ordinal. Also let (& : q < cf /I) be a normal sequence approaching to /I. Case 1: cf /3 c K. Since f&y) = sup{fs, (y) : q < cf p} on a club C, and C fl D fl (n {Ds,,: q < cf /!I}) is in the II,+,- Ramsey filter on K, then we have that { y < K :X fl y is II&,)-Ramsey}
is in the I&+,-Ramsey filter on K. Case 2: cf 6 = K. Similarly for some club C c VY E Cfs(Y) = sup{&(r): Also { y <
K : Vq<
K
II < Y)*
y y E Do,} is in the &+,-Ramsey
filter. Thus
{ y < K :X n y is flf6s(vj-Ramsey}
is in the Q+r-Ramsey filter on This is the end of the proof.
K.
0
A special case of the theorem is the following corollary, hierarchy theorem for cardinals of infinite Ramsey rank.
which is our first
A hierarchy of Ramsey cardinals
Corollary 5.4. Zf K
{y <
is H~+l-Ramsey,
K : y is
273
(Y< K+, then
ITfn(,)-Ramsey}
in in the ZIw+,-Ramsey filter on K. Our second hierarchy following.
theorem
for cardinals of infinite Ramsey rank is the
Theorem 5.5. Zf K is 17,+, -Ramsey, a < K, then the Z&-Ramsey filter is properly contained in the Z7a+l-Ramsey filter on K. Proof. It follows from Theorem 5.2 and the previous corollary that { y < K : y is II fm(,,-Ramsey} is in the &+, -Ramsey filter but not in the &-Ramsey filter. Coroby 5.6. not normal.
If (Y< K+, cf a =
K, K is
&-Ramsey,
then the Z&-Ramsey filter is
Proof. Let (No : /3 < K) be an increasing sequence converging to (y. For each #I
6. Inner model for completely Ramsey cardinals In this section, we follow Mitchell and Jensen to show that the core model of Dodd-Jensen is a canonical inner model of completely Ramsey cardinals. Let K be the Dodd-Jensen core model (see [3,4,5,12] for more information). Indiscemibility Lemma (Jensen [5]). Let A c K be such that L,[A] E K,. Let A = (K,, E, D CIK, A). Zf A has a good set X E K of indiscernibles such that cf(]X]) > o, then there is an X’ E K such that X c X’ G K and X’ is good for A. Where D is the coding of mice, and K = L[D], K, = L,[D rl K]. (See [4]). Lemma 6.1 (Absoluteness). Let K be a regular uncountable cardinal. Let Z G P(K) be an ideal on K. Assume that Z is definable in ZFC, and ZFCF [K]<~ s 1. Zf vx E K (x E Z+ n Kjx
E (z+)K),
then tlx E K (x E w+(Z) fl KJx
E (%+(z))K).
274
Q. Feng
Proof. Fix K, Z as above. Notice that by our assumption, (Z)K, as the ideal defined in K, and (C%+(Z))“, as defined in K, make perfect sense. Let X E W(Z) n K, let A E P(K) fl K. By Theorem 2.5,
has a good set of indiscernibles Y E X such that Y E I+. Applying the Indiscernibility Lemma, we get Y’ E K, Y s Y’ E K, such that Y’ is good for A. Then X fl Y’ E I+ tl K. Therefore, X tl Y’ E (Z+)K. Hence X E P+(Z))“. This completes the proof.
0
Theorem 6.2. Zf K is &-Ramsey, Proof.
then
K
is
&-Ramsey
in K.
By induction on a; we prove the following. Vx (x E (I:)+ rI K+x
E ((1:)‘)“).
(1)
For Q:= 0, Z$ = %(ZQ. Since “being unbounded” is absolute, by the Absoluteness Lemma, Vx E (la+ fl Kx E ((13’)“. For cr = 1, Zr = 9(12r). Since “being stationary” is downward absolute, by the Absoluteness Lemma, we have our conclusion. Forcu=j3+1, bal, let/3’=/3-1if/3Co, andB’=pif/3>o. Byinduction hypothesis, Vx E (Zi,) n Kx E ((Zi,)‘)“. By the Absoluteness Lemma, since I”,= W(Z;.), we have our conclusion. When (Yis a limit ordinal, (Ii))+ = n {(Z3+ : p < a}. LA X E (I”,)+ fl K. Then V/3 < CUXE (I;)+ n K. Now by induction hypothesis, we conclude that X E ((13’)“. This finishes the proof of (1). Now the theorem follows by induction, Corollary 6.3. Zf K
is completely
applying (1) to x =
K.
0
Ramsey, then
K k K is completely Ramsey. Definition 6.1. Let K be a regular uncountable cardinal. Let f = & : a < K+) be the canonical sequence on K. (1) K is 0-hyper completely Ramsey if K is completely Ramsey. K is ((Y+ 1)-hyper completely Ramsey if K is a-hyper (2) For a’<~+, completely Ramsey and there is a completely Ramsey subset X so that for all il E X, h is f,(l)-hyper completely Ramsey. (3) For (YG K+, a! is a limit ordinal, K is cy-hyper completely Ramsey if for all 6 < cu, K is /3-hyper completely Ramsey. (4) K is super completely Ramsey if K is K+-hyper completely Ramsey.
275
A hierarchy of Ramsey cardincils
Theorem
6.4. {A <
If K is measurable, K : k is
then K is super completely Ramsey and
super completely Ramsey)
is stationary.
Proof. Let u be a normal K-complete nonprincipal ultrafilter on K. Let j: V-* M = Vk/U be the canonical elementary embedding, where Vk/U is the ultrapower of the universe V by 17, and M is the transitive collapse of the ultrapower. Let f be the canonical sequence on K. Then for all cy< K+, [fa] = a in M, and if g(cu) = (Y+for all LY< K, then [g] = K+ in M. In particular, if D < K+ is a limit ordinal, then the cofinality of & in V = the cofinality of a in M. Also, since M is closed under K-sequences, the definition off is absolute with respect to M. With the remarks above in mind, let, for LY< K+, X, = {A <
K :A
is f,(A)-hyper
completely
Ramsey}.
We now prove VcY
(*)
By induction on a < K+, we show that X, E U. For LY= 0, it is clear. Let (Y= /3 + 1. By induction hypothesis, X, E U. Then M k K is P-hyper completely
since /3 = [fs] in M and and for il E X,,
K =
Ramsey
[id] in M with id(y) = y for y <
M k I. is fs(3L)-hyper completely
K.
Now X, is in M
Ramsey,
also M LX, is a completely
Ramsey subset of
K.
Therefore, Ml= K is (/3 + 1)-hyper completely
Ramsey.
Consequently, since /3 + 1 = [fs+J in M, X, E U. Let (Ybe a limit ordinal. cf (YG K. Let (a: #I E cf (u) be an increasing sequence converging to (Y. Then this sequence is in M, and a6 = [fn,] in M for each p < cf a: By induction hypothesis, for each p < cf (Y, X,, E U. Therefore, M k K is a;6-hyper completely
Ramsey.
Thus, M k K is Lu-hyper completely
Ramsey.
Since (Y= [fJ in M, we have that X, E U. This finishes the proof of (*).
Q. Feng
276
Now by (*), M k K is K+-hyper completely M E K is super completely
Ramsey, i.e.,
Ramsey.
Hence, {A < K :A is super completely Ramsey} E U. Since each X, is completely Ramsey, K is super completely Theorem 6.5. K kK
Proof.
ZfK is
is
Ramsey in V.
0
super completely Ramsey, then
super completely Ramsey.
To prove this corollary,
we first prove one fact on the canonical sequence
f First, let us observe that if a is a limit ordinal, then cf (cfc(cu)) = cf (Y. Fact. Let K be regular uncountable. Let a< (K+)~. Zf fa is the cuth canonical function on K, and if f b, is the cuth canonical function on K in K, then {A < K : f,(n) = f h(n)} COnthIS a club in K.
It is clear for the cases when LY= 0 or (Yis a successor assuming the Fact for smaller ordinals. Let (Y< (K+)~ be a limit ordinal. Case 1: &a < K. Let (Es : /3 < cfK&) E K be a normal sequence converging to fx. Let ( oq : q < cf a) be a normal sequence converging to ~?‘a. Then (5, : @< cf o) is normal cofinal in (Y. So there is a club C E K so that for each A E C, fu(A) = sup{ fs,(A) : p < cf a}. By induction hypothesis, V/l < cf (Y, there is a club C, so that VA E C, we have fE,(A) =f&,(A). Let D = C n fl {C,:/3
completely
Ramsey
assuming that K is super completely Ramsey. When (Y= 0 or a limit ordinal, it follows from Corollary hypothesis. Let Q!= /3 + 1. By induction hypothesis, K I=K is /3-hyper completely
Ramsey.
6.3 or the induction
A hierarchy of Ramsey cardinals
277
Let A = {A < K : K k 3cis f;(A)-hyper completely Ramsey}. Then A E K. Let C G K be a club so that VA E Cfs(A) =fb(A). Let B = {A < K :il is &(A)-hyper completely Ramsey},
Since K is super completely Ramsey, B fl C is then completely Ramsey. Therefore A is completely Ramsey. By (1) of the proof of Theorem 6.2, A is completely Ramsey in K. Hence, K k K is rr-hyper completely Ramsey. Therefore, K k K is super completely Ramsey.
Acknowledgement The author is grateful to Professor T. Jech, Professor W. Mitchell and Professor S. Simpson for their teaching and guidance.
References [l] J. Baumgartner, Ineffability properties of cardinals I, in: A. Hajnal et al., eds., Colloq. Math. Sot. Janos Bolyai 10, Infinite and Finite Sets, Vol. III (North-Holland, Amsterdam, 1973) 109-130. [2] J. Baumgartner, Ineffability properties of cardinals II, in: Butts and Hintikka, eds., Logic, Foundations of Mathematics and Computation Theory (Reidel, Dordrecht, 1977) 87-106. [3] T. Dodd, The Core Model, London Math. Sot. Lecture Note Series 61 (Cambridge Univ. Press, Cambridge, 1982). [4] T. Dodd and R. Jensen, The core model, Ann. Math. Logic 20 (1981) 43-75. [5] D. Donder, R. Jensen and B. Koppelberg, Some applications of the core model, in: R. Jensen and A. Prestel, eds., Set Theory and Model Theory, Lecture Notes in Math. 872 (Springer, Berlin, 1981) 55-97. [6] P. Erdijs and R. Rado, A partition calculus in set theory, Bull. Amer. Math. Sot. 62 (1956) 427-489. [7] G. Fodor, Eine Bemerkung xur T’heorie der Regressiven Funktionen, Acta Sci. Math. (Szeged) 17 (1956) 139-142. [8] T. Jech, Set Theory (Academic Press, New York, 1978). [9] T. Jech, Stationary subsets of inaccessible cardinals, in: J. Baumgartner, ed., Axiomatic Set Theory, Contemporary Math. 31 (Amer. Math. Sot., Providence, RI, 1984) 115-142. (lo] T. Jech and K. Prikry, Ideals over uncountable sets: application of almost disjoint functions and generic ultrapowers, Mem. Amer. Math. Sot. 214 (1979). (111 A. Levy, The sizes of indescribable cardinals, in: Axiomatic Set Theory, Proc. Symp. Pure Math. 13, Part 1 (Amer. Math. Sot., Providence, RI, 1971) 205-218. [12] W. Mitchell, Ramsey cardinals and constructibility, J. Symbolic Logic 44 (1979) 260-266. [13] F. Ramsey, On a problem of formal logic, Proc. London Math. Sot. (2) 30 (1930) 264-286. [14] F. Rowbottom, Large cardinals and constructible sets, Doctoral Dissertation, University of Wisconsin, Madison, WI, 1964. [15] D. Scott, Measurable cardinals and constructible sets, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 7 (l%l) 145-149.
A HIERARCHY
OF RAMSEY
257
CARDINALS*
Qi FENG** Institute of Sofiware, Academia Sinica, P. 0. Box 8718, Beijing, People’s Rep. of China Communicated by T. Jech Received 15 June 1989 Assuming the existence of a measurable cardinal, we define a hierarchy of Ramsey cardinals and a hierarchy of normal filters. We study some combinatorial properties of this hierarchy. We show that this hierarchy is absolute with respect to the Dodd-Jensen core model, extending a result of Mitchell which says that being Ramsey is absolute with respect to the core model.
1. Introduction Our set-theoretical usage is standard (see [S]). If x is a set, P(x) is the power set of x, and lx] is the cardinality of x. If x is a set and d is a cardinal, then [x]“={y~x:JyJ=J.}, and [x]‘“={ycx:Iyl<)L}. In particular, [xl’” is the family of finite subsets of x, and [xl” is the family of subsets of x with exactly II elements, n < w. If x, y are sets, then yx is the set of functions from y to x. Iffis a function on A and B E A, flB is the restriction of f to B, and f”B = {x : 3y E B x =f(y)} is the set of images of B under 5 If H is a subset of the range off, then f_‘(H) = {x :f(x) E H}. Let (Ybe an ordinal. V, is the set of all sets of rank smaller than (Y. That is, and if A is a limit ordinal then V, = V,= 0, the empty set; V,,, = P(V,); tJ {V, : 1y< A}. Then our universe of set theory ZFC, V, is the union of all Vol’s. Let X be a set of ordinals. A function f defined on X is regressive if for all a E X, (Y> 0 implies f(a) < CY.A function f defined on [X]
Elsevier Science Publishers B.V. (North-Holland)
258
Q. Feng
Later, we will see that if K is Ramsey then K is the Kth weakly compact cardinal. Also there might be lots of Ramsey cardinals. Let us consider a model A = (A, . . . ) of an infinitary language, with A 2 K, where K is an infinite cardinal. We say that a set H E K is a set of indiscernibfes for A if for every n < o, for every formula cp(vr , . . . , v,) of the language of A of n free variables, we have Akq(o~,...,
4
@
Ab &PI> . . . > Bn)
whenever q < * - - < an, PI < - * - < /3n are from H. The following theorem is folklore (see [8]). Theorem. Let K be an infinite cardinal. Then K is Ramsey if and only if every model A = (A, . . . ) of a language of size < K, with K G A, has a set H E [K] K of indiscernibles.
Jensen noticed that one need only consider a special kind of models [3,5]. The following definition is due to Jensen [5]. Let AI,. . . , A ,cKandletA=(L,[A, ,..., A,],E,A1 ,..., A,). FOrA
(f))s
Theorem
(Jensen [5]). ~isZ3amseyifandonlyifforeveryA,,...,A,~~, A=
(L,[AI,
. . . , A,],
E, AI,
. . . , A,)
has a good set of indiscernibles of order type
The following interesting Ramsey cardinals. Theorem
theorem
K.
of Mitchell stimulates our work to study
(Mitchell [12]). Zf K is Ramsey, then K is Ramsey in the Dodd-Jensen
core model
K.
For a limit ordinal a>O, a subset X E (Y is unbounded in (Y if Vy< (Y 3/!IEXY
A hierarchyof Ramsey cardinals
259
on K, we denote by I+ the collection of subsets of K which are not in Z, i.e., I+ = {x E K :x $ Z}. Then Z is nontrivial iff I+ is not empty. An ideal Z on K is A-complete (A < K) if Z is closed under unions of fewer than A. of its elements. Z is normal if for every S E I+ and every regressive f : S- K there is S’ E I+ so that S’ is a subset of S and f is constant on S’. F c P(K) is a fiber if X, Y E F j X n Y E F, and X E F implies that every superset of X is in F, and K - { cu} E F for each (Y< K. A filter F is nontrivial if 0 $ F. Notice that we do not consider those so-called principal filters, i.e., F={X~K:LYEX} for some (YCK. Let Z(F)={~EK:K-~EF}. Then F is a nontrivial filter iff Z(F) is a nontrivial ideal. We call Z(F) the dual ideal of F in that case. Conversely, if Z is a nontrivial ideal on K, let F(Z) = {x E K : K -x E Z}, then F(Z) is a nontrivial filter. We call this filter the dual filter of I. Now a nontrivial filter F is A-complete if its dual ideal is A-complete. From now on by filter we always mean nontrivial filter. Then a filter F is A-complete if and only if F is closed under intersections of fewer than Iz of its elements. F is normal if its dual ideal is normal. Then F is normal if and only if F is closed under diagonal intersections, i.e., if (Xol : a < K) is a sequence from F, then A ,cKX, = {/3< ~:tlc~
Theorem
that on S’ f intersection.
is constant.
Consequently,
the club filter is closed under diagonal
Let Z be an ideal on K. Let X be a subset of K. The notation X+ (I+): means that for every f : [Xl” + m there is YE I+ so that Y G X and Y is homogeneous for f. And X-t (Z’):@ means that for each f : [X] <@+ 3c there is YE I+ so that Y c_X and Y is homogeneous for f. The notation X% (Z’)z” means that for every regressive function f : [X] co+ K there is YE I+ so that YE X and Y is homogeneous for $ Let K be an infinite cardinal. A filter U on K is an ultrafilter if for any X E K either X E U or K -X E U. An uncountable cardinal K is measurable if there is a K-complete ultrafilter on K. The following two theorems are basic in the studying of measurable cardinals.
260
Q. Feng
Theorem (Scott [15]). Zf
K
is
measurable,
then there is a normal
K-complete
ultrafilter on K.
Theorem (Rowbottom
[14]). Let U be a u-complete normal ultrafilter on
x E u.
is
Zff :[xy ---,K
regressive,
K.
Let
then there is Y E U so that Y is homogeneous
for f-
(See [8] for proofs of both theorems.) Let Z7: @A, respectively) be the class of formulae of second-order language which in prenex form has at most n alternating blocks of quantifiers of second-order variables, starting with the universal (existential, respectively) quantifier. A cardinal K is ZZk-indescribable if and only if for every rrf, formula #(X) with one free second-order variable, for each R c V,, if (V,, E, R) t=e(R), then there is (Y< K so that (V,, E, R fl V,) b $(R tl V,). Define EA-indescribability analogously. Remark. It is useful to notice that K is ZIt-indescribable if an only if K is inaccessible and for any ZZf,formula #(X) with one free second-order variable, for each R c K, if (K, E, R)l= @(R), then for some cy< K we have ((u, E, ~II R) tr$r(R II a).
More generally, a subset A G K is Z7:-indescribable if and only if for each Z7: formula #(X) with one free second-order variable, for each R c K, if (K, E, R) k #(R), then &YEA ((u, E, Rfla)k$(anR). Theorem (Baumgartner
[l], Levy [ll]).
(1) Let K be a regular
uncountable normal ideal
cardinal. Then {x c K :x is not Z7:-indescribable} is a K-complete on rc. (2) Zf K is IZ!,-indescribable, then the ideal defined in (1) is nontrivial. (3) Zf K is inaccessible, and the ideal defbted in (1) is nontrivial, then K is ZZi-indescribable.
We will call the ideal defined in (1) the ZZA-indescribable ideal on K. If it is nontrivial, then we call its dual filter the IZA-filter. In [l, 21, Baumgartner generalized this, among other things, to Ramsey ideals. Let K be regular uncountable. An X E K is Ramsey if for every regressive f on WI- and any club C G K there is an H G C n X so that lZZ1= K and H is homogeneous for f. Z = {x G rc :x is not Ramsey} is defined to be the Ramsey ideal on K, which is K-complete and normal. Theorem (Baumgartner
[2]). Zf K is Ramsey, then the Ramsey ideal on K is nontrivial, and {a < K :a is weakly compact} is in the Ramsey filter, the dual of the Ramsey ideal.
A hierarchy of Ramsey cardinals
2. A Ramsey operator
261
on ideals
In this section, we define an operator 9? on ideals so that for any rc-complete ideal I on K, 3(Z) is a K-complete normal ideal and extends I. We will also give some equivalent definitions of the operator. This operator will be used in the next section to define a hierarchy of normal filters in the theory of large cardinals. Definition 2.1. Let K be a regular uncountable cardinal. Let I be an ideal on K For every X E K, X E 9+(Z) if and only if for every regressive such that Z z~[K]? function f : [X]
P(K) {xs
-
%+(I), K:K--XE
s(l)}.
We call the operator 5%the Ramsey operator. Immediately, W(Z) is an ideal extending both ideals Z and the nonstationary ideal ZVS,. Also, if Z E J, both are ideals, then 99(Z) E 9?(J). Theorem 2.1. a(Z) is a K-complete normal ideal. Proof. Without loss of generality, we assume that K $93(Z), that is, 9?(Z) is nontrivial. Notice that the K-completeness follows from the normality. Suppose that X, E %(I) for each (Y< y < K, X = Uncy X, and X is in 3+(Z). We may assume + K by f(B) = the least LYsuch that j3 E X, for each that Xrly=0. Definef:X /3 E X. Then f is regressive. By normality, for some Y s X, Y E 9+(Z), and for some a; El/3E Y f(/3) = (Y. We then have Y s X,. But X, E 9?(Z) and 3(Z) is an ideal. This is impossible. Hence X E R(Z) and 5?(Z) is K-complete. We need to show that 93(Z) normal. Given X c_ K, h :X+ K is a regressive function. Assume that for all (Y< K, h-‘({a}) E 3(Z). We prove that X E 3(Z). For each LY< K, fix fn, C, to witness that h-‘( { a}) E S(Z). Let Ed: K X K * K be a G6del pairing function. Define D to be the set {(Y< K :n"cx X LY G cu}, Let C = A,,& fl D. Then C is a club. Suppose that X E 9+(Z). Then X n C E 9+(Z). Define a regressive function f:[xnc]
f ({a>>= 4h(d
fwWN;
Let YE I+ be such that YE X n C and Y is homogeneous for J Let 77be such that V& E Y f ({a}) = q. Then for some j3, tla! E Y h(a) = j3 and /3 < (Y. But on [Ylcof =fs. Since Y c C, n h-‘({/I}), Y cannot be homogeneous for fs. We get a contradiction. This is the end of the proof. Cl
Q. Feng
262
Consequently, if 9(Z) is not trivial, then W*(Z) is a normal K-complete filter. We will be interested in such filters later on. In the rest of this section, we will be proving some results which tell us how to redefine the Ramsey operator in different ways. The following theorem is a straightforward generalization of a standard theorem. Theorem 2.2. Let
K
be a regular cardinal. Let NS, G Z and Z G P(K)
be an ideal.
For X c K, the following are equivalent:
(1) (2) (3) (4)
x E 9?+(z). X% (z+),‘o. x+ (z+),(o. kfA< KX-@+);?
(5) x-+(1+),6".
(6) For any structure A= (A, <, . . . ) of a countable language, X c A, A has a set of indiscernibles B c_ X which is in I+. Let X c K. By a (w, X)-sequence we mean a sequence S of the form We (~~~.**~cr~,cr, ,..., LY,EX), where S, ,...n,~a,. S=(S,, ...=n:lsn
[K]?
K be a regular uncountable cardinal. Let Z be an ideal Then for any X E K the following are equivalent:
(1) x E 9+(z). (7) For any (w, X)-sequence
S there is a Y E I+ so that Y is a subset of X and Y
is homogeneous for S. (8) For any (0, X)-sequence S there exist a Y E X with Y E I+ and a sequence (A,,: 1
Sal...an= a-1n A,. Proof. (1) 3 (7)‘. Given a (0, X)-sequence S, we want Y. Assume that so that for aO
0 [ l+y
if S, = S, II aO, if y=min{b
Let YE I+, YE X, be homogeneous for g. Then, of course, S, = S, n min(a) for all a, b E [Y]” such that max(a) < min(b). But then given arbitrary a, b E [Y]“, min(a) < min(b), choose c E [Y]” so that max(a), max(b) < min(c). Then S, = S, n min(a) and S, = S, rl min(b), so S, n min(a) = S,. ’ The proof here is given by the referee.
Originally,
we had a rather complicated
argument.
A hierarchy of Ramsey cardinals
263
+ K be regressive. For u E [X]
2.4. Let
[K]‘“.
K be regular uncountable. Assume that Z is an ideal on Then for any X E K the following are equivalent:
Then
K
and
(1) x E 9’(Z). (9) For AI, . . . , A, G good set of indkcernibles
K,
A = (L,[X,
The proof follows from Theorem Corollary 2.5. On/J’ if K $
Let
K
AI, . . . , A,],
E, X, AI, . . . , A,)
has a
YE X which is in I+.
2.3 and standard arguments.
be a regular uncountable
cardinal. Then
K is
Ramsey if and
%([K]--).
is Ramsey, we call %([K]<~) the Ramsey ideal on K, and the dual of it, ~F([K]<~), the Ramsey jilter. By Theorem 2.1, it is K-complete normal filter. We call every X E W([K]<~) a Ramsey subset of K. If
K
3. The hierarchy
In this section, we iterate the Ramsey operator to define a hierarchy of Ramsey cardinals. Then in the next sections we show that the hierarchy defined here is really a proper hierarchy. Also we will define some terminologies which will be used later. Definition 3.1. Let K be a regular uncountable cardinal. We define I?* = [K]<“; I?1 = NS,; for n < o, Zi= 9?(ZE_2); for (Y?=w, if a=/3+1, thenZ”,=%(Z;);if Ly is a limit ordinal, then Z”,= Us<&.
For any regular uncountable K, there exists some LY< (2”)+ so that I:+1 = Zz, because the Z”,‘s form an increasing chain of ideals and there are only 2” many subsets of K. Let us denote by 0, the least (Ywith this property. Notice that for many K the ideals defined above are trivial. We will be only interested in those cases where some of them are nontrivial. This leads the following definition. Definition 3.2. LA 17,-Ramsey if K $ Zz; Remark.
K K
be a regular uncountable cardinal. For is completely Ramsey if for all a, K $ ZZ,.
(Y2 0,
K
is
For a limit ordinal (Y> 0, K is Z7,-Ramsey if and only if K is ZJ@-Ramsey for all /!I< CY.And K is completely Ramsey if and only if K $ Z:Kif and only if 3a Z”,+1= Z”,# P(K).
Q. Feng
264
We define the Ramsey rank of K, Rk(K), to be the least (Yso that K E Z”,if such exists; otherwise, Rk(K) = =J. Notice that if Rk(K) is an ordinal, then Rk(K) = f3, and it is either 0 or a succesor ordinal. In particular, Rk(rc) > 0 if and only if K is Ramsey if and only if K is ZZO-Ramsey. Therefore, the nontriviality of any Z”, for as0 is a large cardinal property. On the other hand, the following theorem tells us that the nontriviality of the Ramsey operator is worth to study. Theorem
3.1.
ZfK
is
measurable,
then
K is
completely Ramsey.
Proof. Let U be a normal K-complete nonprincipal ultrafilter on K. By Rowbottom’s theorem, for each X E U, for any regressive function f :[X]<“+- K, there is Y E U such that Y is homogeneous for f. Let Z = {x G K :K - x E U}. By a simple induction on CY,we have Zlc,G I. Therefore, none of them is trivial. K is then completely Ramsey. 0 3.3. Assume that K is Z&-Ramsey. We call Z”,the ZZ,-Ramsey ideal on K, and the dual of it the &-Ramsey filter on K. We call each subset X of K which is not in the Z7=-Ramsey ideal a &-Ramsey subset of K. If K is completely Ramsey, then we call ZGzthe completely Ramsey ideal and the dual of it the
Definition
completely Ramsey filter, each x G
K
not in the ideal a completely Ramsey subset
Of K.
Remark. Let K be fla-Ramsey. If (Y= /I + 1, by Theorem 2.1. If (Yis limit ordinal > 0, then may not be K-complete. If (Y< K+, cf (Y= K, (sometimes it happens, e.g., if K is completely is normal as the following fact shows.
then I^, is normal and K-complete Zlc,is min(cf a; K)-complete, and it then Z”, is not normal. If cf & > K, Ramsey, then ZE+c ZE++i), then Zz
Fact. Zf K is ZIa-Ramsey, (Y is a limit ordinal, and cf (Y> K, then Z”, is normal.
Let XE(Z~))+ and f:X+K be regressive. increasing sequence convergent to (Y.Then
Proof.
v~
)+ 3Ys
Let
((us:/3
be an
= y} is unbounded in cf cr. Let Y = { 9 E Hence 3y-C~ so that {/I
4. The hierarchy theorem
In this section, we finite stages. We will the hierarchy defined We start with a
for cardinals of finite Ramsey
rank
prove that the hierarchy defined in Section 3 is proper for also introduce a new type of Ramsey cardinals, related to before. lemma of Baumgartner, which allows us to use the
A hierarchy of Ramsey cardinals
265
indescribability argument in addition to pure combinatorial argument. Let K be a cardinal and let A c K. A (1, A)-sequence S is a sequence S=(S,:CKEA) withindexsetA andS,scuforeach (YEA. Lemma 4.1 (Baumgartner [l]). Let K be a cardinal. Let A E K. Assume that for every (1, A)-sequence S there is a B E Q such that B is homogeneous for S. Zf m 3 0 and every member of Q is ZIh-indescribable, then A is ZI!,,+,-indescribable. Proof. Let R be a relation on K. Let # be a flk+,-sentence. And assume that (K, E, R) k 9. Write 9 as V.r 3y q(x, y), where r+!~(x, y) is a n!,, formula. Suppose a, X,) LVY~W(xY, Y).
VCYEA~X&,E,RI
Let S, code X, as a subset of a. Let B E Q be homogeneous X=IJ{X,:CYEB}. Thenforsome YsKwehave
for S and let
(K E, R, X, Y) I=W(X, Y). Since B is Z7;-indescribable,
there is a E B such that
(a; E, R 1 a, X 1a, Y 14 b v(X But X 1 (Y=X,.
A contradiction.
1 a, Y 1~1.
•i
It is easily seen by induction that the definition of “K being D,,-Ramsey” is a n’ n+2 sentence. Our next theorem says that there is no simpler (even no ZA+2) definition. Theorem 4.2. Zf K is Z&-Ramsey, then K nn-Rumsey, then X is ZIi+,-indescribable.
is
ZZ!,+l-indescribable. And if X s K i.~
Consequently, if K is K&-Ramsey, then the 17:+,-filter over the n,-Ramsey filter over K.
K
is contained
in
Proof. By induction on n we prove that if X c K is I&-Ramsey then X is ni+ i-indescribable. For n = 0, K is Ramsey, hence K is weakly compact. It follows from a theorem of Baumgartner [l, 21 that every Ramsey subset of K is fl:-indescribable. For n = m + 1, K is nn-Ramsey implies that K is IT,-Ramsey. By induction hypothesis, every II,,,_,-Ramsey subset of K is II;-indescribable. If X c K is I7m+l Ramsey, by Theorem 2.3, then every (1, X) sequence S has a homogeneous n,_,-Ramsey subset Y cX. By the lemma of Baumgartner, X is D&+,indescribable. •i Applying this indescribability theorem, we can now prove our first hierarchy theorem for cardinals with finite Ramsey rank which says that if K is n,,+,Ramsey then there are stationary many nj-Ramsey cardinals below K.
Q. Feng
266
Theorem 4.3. If is I&,-,-Ramsey
K is
I&,-Ramsey, and A c
K is
Iln_l-Ramsey,
then {a! <
K
:A tl LY
in a} is in the l&-Ramsey filter on K. (Notice that II_,-Ramsey
subsets of 3care exactly the stationary subsets of A.)
Proof. Let A c K be a &,-Ramsey subset of K. Then “A is &,-Ramsey” is a l7rn+l sentence over (K, m, A). Therefore, Y = {d < K :A n CYis not II,,_,-Ramsey in a} is in the n!,+, -ideal on K. Since K is n--Ramsey, by the Indescribability Theorem 4.2, the II:+, -ideal on K is not trivial. Hence the complement of Y in K is in the flA+, -filter on K. Again by Theorem 4.2, this filter is contained in the L&-Ramsey filter on K. We are done. 0 Immediately,
we have the following hierarchy theorem. then {cx < K : (Y is I&-Ramsey} is in the K is n,,,, -Ramsey, filter. In particular, there are stationary many I&,-Ramsey cardinals
COrOkUy 4.4. If l&+,-Ramsey below
K.
The following fact says simply that the indiscernibles you could get for any structure of countable language over a &+,-Ramsey cardinal are from the smaller L&-Ramsey cardinals essentially. Fact. If K is &+, -Ramsey and X E K is a lI,,,1-Ramsey subset, then for any regressive function f : [X] cm+ K, there is an A G X such that A is l&+,-Ramsey, A is homogeneous for f, and Va EA cx is II,,-Ramsey. Proof. It simply follows that if X c_ K is D”+r-Ramsey n,-Ramsey} is fin+,-Ramsey. Cl We are now ready to prove our second hierarchy theorem. the following theorem. Theorem 4.5. Assume that
K is
l&-Ramsey.
If X G
K is
then
{(YE X: (Y is
First, let us prove
fin-Ramsey
then
{(Y E X : either a is not II,,-Ramsey or X n (Y is not I&-Ramsey} is l&-Ramsey.
Proof. Let that
K
In particular,
{ LY< K :(Y is not I&,-Ramsey)
be the least counterexample.
is K&,-Ramsey.
Let X G K be a n”-Ramsey
subset so
A = { LYE X : either (Yis not K&-Ramsey or X rl (Yis not n,,-Ramsey}
is not nn-Ramsey. Then X - A = {a E X : a is L&-Ramsey & (Yn X is K&-Ramsey} is L&,-Ramsey. For CYE X -A, by the minimality of K, X, = (Yn A is fl”-Ramsey in LY.Set q Y = lJ {X, : (1:E X - A}. Y is then I&-Ramsey. Contradiction.
A hierarchy of Ramsey cardinals
By Corollary 4.4 and this theorem, hierarchy theorem. Theorem
4.6.
If K is
Z&+,-Ramsey,
we have immediately
then Vi Sn
267
the following second
the Z&Ramsey filter on
K
ih
properly contained in the ZJ+l-Ramsey filter on K. Proof. For i =Sn, {(Y < K : a is Z&-Ramsey} is in the Z&+,-Ramsey filter on K but
not in the Z&-Ramsey filter on
K.
0
The following definition is essentially due to Baumgartner [2], where he considered the cases when n = 0 or 1. Also the following two theorems for n = 0 or 1 are due to him [2], where he uses the names “pre-Ramsey,” “pre-ineffably Ramsey” and “ineffably Ramsey,“. Definition 4.1. For a Z&-z-Ramsey
cardinal K, for X c K, (1) X E %E,+(Z,“_,)if and only if for every regressive f : [X]<“-* K, for each club C c K, there is LYE C n X and there is A c C fl cx rl X so that (Yis I&,_,-Ramsey and A is ZIn_z-Ramsey in a and A is homogeneous for f and V/3 E A f ({p}) =
f ({a>)* (2) x E%K-2) ex
$ %x-*).
Notice that by “IT-i-Ramsey cardinal (i = 1, 2)” we mean regular uncountable, and by “E,-Ramsey subset” we mean unbounded subset, as well as by “I7_,-Ramsey subset” we mean stationary subset. Theorem 4.7. Zf K is Z&-Ramsey, then 9&(Z,“_,) is a nontrivial xc-complete normal ideal, and it is contained in the Z&-Ramsey ideal. Proof. The proof of the normality
of ‘&,(Zk,) is the same as the proof of the normality of 9(Z). We omit the argument. We need to show that every Z7,-Ramsey subset of K is in $%e,‘(ZE_,). First, assume n 2 1. Fix a ZZn-Ramsey subset X of K. Let f be a regressive function from [X]
is in the Z&-,-Ramsey filter on K, hence it is in the Z7,-Ramsey filter on K. Therefore, B fl C fl Y #0. Let a be in this intersection; then A fl LY is Z&_,Ramsey in (t: We are done for n 9 1.
268
Q. Feng
For n = 0, one gets the A as above in the same way. Then B={cx<~:Ancrisunbounded} is a club in
K.
Hence B n C n Y # 8. Then any (Yin this intersection
works.
Cl
Theorem 4.8. (1)
K is Z&-Ramsey if and only if the ZZ~+I-ideal and the ideal %&(I:_,) generate a nontrivial ideal.
(2) Zf $?$(I:_,)
K is
I?,-Ramsey,
then the Z&-Ramsey ideal is the ideal generated
by
and the II:+,-ideal.
Proof. (+) It follows from Theorem 4.2 and Theorem 4.7 that both the ITf+,-ideal and 9&(Z,“_,) are contained in the &-Ramsey ideal. (e) Let Z be th e generated ideal. It suffices to show that for any X G K if X $ Z then X is Z&,-Ramsey, which implies that Z,”E I. Assume that X $ I, f : [X]4w+ K is regressive, and C E K is a club. Then
cnx4z. Suppose that VA E C n X if A is ZZ,,_,?-Ramsey then A is not homogeneous f. Then this is a ZZA,, sentence @ over the structure
for
A=(w,C,X,f).
Thus Y={cx
I[a]‘“)k$} is in the ZZA+i-filter on This is because
K.
xnc=(XncnY)U(XnCn(K-Y)),
and the second part of the right-hand side of the equality is in the Z7:+,-ideal on K. Let cueXnCflY, and let AsXnCrlY be Z7k_z-Ramsey in LYand be homogeneous for f. But (YE Y. This is a contradiction. Hence X is Z&-Ramsey. 0
5. The hierarchy theorem for cardinals of infinite Ramsey rank In this section we prove a hierarchy theorem for cardinals of infinite Ramsey rank. How we are going to do it is the reason why we make the distinction between the case of finite Ramsey rank and the case of infinite Ramsey rank. Notice that in the proof of the first hierarchy theorem for cardinals of finite Ramsey rank we essentially use the indescribability argument. To prove a hierarchy theorem for cardinals of infinite Ramsey rank, we use canonical sequences, which were also used by Baumgartner in [4] to establish his ineffability hierarchy. Let K be a regular uncountable cardinal. For functions f, g E Icy, we define f
A hierarchy of Ramsey cardinals
everywhere)
269
if
{a < K :f(4 contains a club in
< g(a)>
K,
and f C g if
{a
contains a club in K. A sequence & :a' < K+) of elements of "K iS a canonical sequence on K if the following two conditions are satisfied. (1) For all cu, B < K, LY < /3 implies f= Cfs. (2) For any other sequence (g,: CY < K+) of elements of K~ with the property thatVa
: a < K+)
K so
is
that h,(n)
Fact 4. For all ~3<
K+
a canonical sequence on
K,
then for all cx < K+ there is a
< Inl+ for all n E C,.
there is a club Ce c K such that for all 2 E Ce if A > o and ),
is regular then {Y < A :f&u(y)
=f 2(Y))
contains a club in 3L, where f”, fK are canonical sequences on A and K respectively.
Now we are going to prove the hierarchy Ramsey rank.
theorem
for cardinals
Theorem 5.1. Assume that (& : CY< p) is a sequence of elements of a
"K
of infinite
such that if
then {Y < K:f,(Y)
is in the ZIP-Ramsey filter on K, and fa( y) S a for all y < all X c xc, if { y < K :X fl y is IT&)-Ramsey
in y}
is Z&-Ramsey, then X is IZ*-Ramsey.
Proof. We prove this theorem by induction on (Y. If Q is finite, we are done. So assume (Yis infinite.
K.
Then for all a < u and
270
Q. Feng
Suppose LY= b + 1. Then Y = {y C K :X II y is 17__e(,,-Ramsey in y & fs(y)
~(~e’“~~=(~~eleastcEXI_x~, ifxEE.xqJ otherwise. and g({y,,
- - * 7 Y~+~})
=
the value off
on [X,,]“.
Now for some Q-Ramsey set A c C’ n Y, A is homogeneous for g. Then for all&q~A ~
& X n v is fl&,-Ramsey
in rl]
is &-Ramsey. By induction hypothesis, X is flP-Ramsey. Thus for all /3 < (YX is flP-Ramsey. Hence X is &-Ramsey. This completes the proof. 0 Theorem
5.2. Let ($n : rj C K+) be the canonical sequence
on
K.
Assume that K is
I&-Ramsey with p < K+. Then for all a S p X, = {y < K : y is not l7,-(,,)-Ramsey} is I&-Ramsey. Proof. We prove this by induction on K. If A = { y < K : y is n_*(,) -Ramsey} is not n-Ramsey, assume that A is &-Ramsey.
then we are done.
So
Let C, c_ K be a club such that for all regular uncountable A E C, we have contains a club D in A. Then C, n A is &-Ramsey. If {Y < A:f30dY) =fa(r)l AEC,rlA, Then X fl A = { y < A: y is not 17f,(vj-Ramsey}.
271
A hierarchy of Ramsey cardinals
Since A is IT&,- Ramsey, L(Y) %(4,
and DA is a club in A, and for all y E II*, f&&y)
=
we have
{y E DA: is not KIfo(,)-Ramsey} is II__a(i,,-Ramseyin A.by induction hypothesis. Thus X, fl I. is II&-Ramsey By the previous theorem, X, is &-Ramsey. This finishes the proof. Cl
in 3L.
More generally, the same argument shows the following. Theorem 5.2’. If &-Ramsey
K is L$-Ramsey, implies that
@ < K+, then for all 0: S /I, for all X c K, X i.s
{ y E X : either y is not II&~-Ramsey
or X fl y is not 17f,Cvj-Ramsey}
ti l&-Ramsey.
Next we want to show that higher rank Ramsey filters always concentrate on the set of smaller Ramsey cardinals, which is then our first hierarchy theorem for cardinals of infinite Ramsey rank. Theorem 5.3. Let (f, : ?j < K+) be the canonical sequence on Q-Ramsey,
K.
Assume that
with (Y < K+. Then for all p S (Y, for all subsets X of
II c-r+1 -Ramsey
K is
K,
if X is
XGK
be a
then
{y <
K :X
tl y is 17fp(Yj-Ramsey}
is in the I&+, -Ramsey jilter. Proof. We prove this by induction on #I.
For /3 = 0, it is true. Suppose /?=q+l. Then fs(y)=fv(y)+l Q-Ramsey set. By induction hypothesis, {y <
K :X
for each Y
Let
fl y is I&(,,)-Ramsey}
is in the IIq+, -Ramsey filter on K. Let T = {y < K : either y or X fI y is not Eff,~,,~+l-Ramsey}. Suppose that T were fl,+,- Ramsey. We want to get a contradiction. Then A = {y
< K :X rl y is II&,,)-Ramsey
but not I$-,,,+,-Ramsey}
is &+, -Ramsey. For each y E A let g, : [X fl y]‘” + y be regressive and let 0 4 C, be a club in y such that no II f,(,,j-Ramsey subset of C, rl X is homogeneous for g,. Let D = {/3 < K : f3 is closed under the Gijdel pairing function}. Then D rl A is
Q. Feng
272
&+,-Ramsey.
Define g : [D n Alcm+
K
by
where s,={(n,B1,...,~~,gy({B1,...,~~8n))>:1~n
K :E
rl y is nfq(,,-Ramsey}
is in the Q-Ramsey filter. Let y E B be such that E rl y is nfc,,-Ramsey. Then cm = g, and C rl y 2 C,, is a club. But E n y is homogeneous for g, and fI[Xnrl Since C, fl E rl y is still n&,)-Ramsey, we get a it is nf,(vj-Ramsey. contradiction. Therefore, T can not be &+,-Ramsey. Now assume that /3 is a limit ordinal, and X z K is I$-Ramsey. If rl< /3, then X is II,-Ramsey and D,, = {y C K: X fl y is ZIf,(,,)-Ramsey} is in the n,+,Ramsey filter on K by the induction hypothesis. Hence D,, is in the flfl-Ramsey filter, and therefore in the n,+,- Ramsey filter on K, which is K-complete and normal. Let D c K be a club so that Vy E D f@(y) is a limit ordinal. Also let (& : q < cf /I) be a normal sequence approaching to /I. Case 1: cf /3 c K. Since f&y) = sup{fs, (y) : q < cf p} on a club C, and C fl D fl (n {Ds,,: q < cf /!I}) is in the II,+,- Ramsey filter on K, then we have that { y < K :X fl y is II&,)-Ramsey}
is in the I&+,-Ramsey filter on K. Case 2: cf 6 = K. Similarly for some club C c VY E Cfs(Y) = sup{&(r): Also { y <
K : Vq<
K
II < Y)*
y y E Do,} is in the &+,-Ramsey
filter. Thus
{ y < K :X n y is flf6s(vj-Ramsey}
is in the Q+r-Ramsey filter on This is the end of the proof.
K.
0
A special case of the theorem is the following corollary, hierarchy theorem for cardinals of infinite Ramsey rank.
which is our first
A hierarchy of Ramsey cardinals
Corollary 5.4. Zf K
{y <
is H~+l-Ramsey,
K : y is
273
(Y< K+, then
ITfn(,)-Ramsey}
in in the ZIw+,-Ramsey filter on K. Our second hierarchy following.
theorem
for cardinals of infinite Ramsey rank is the
Theorem 5.5. Zf K is 17,+, -Ramsey, a < K, then the Z&-Ramsey filter is properly contained in the Z7a+l-Ramsey filter on K. Proof. It follows from Theorem 5.2 and the previous corollary that { y < K : y is II fm(,,-Ramsey} is in the &+, -Ramsey filter but not in the &-Ramsey filter. Coroby 5.6. not normal.
If (Y< K+, cf a =
K, K is
&-Ramsey,
then the Z&-Ramsey filter is
Proof. Let (No : /3 < K) be an increasing sequence converging to (y. For each #I
6. Inner model for completely Ramsey cardinals In this section, we follow Mitchell and Jensen to show that the core model of Dodd-Jensen is a canonical inner model of completely Ramsey cardinals. Let K be the Dodd-Jensen core model (see [3,4,5,12] for more information). Indiscemibility Lemma (Jensen [5]). Let A c K be such that L,[A] E K,. Let A = (K,, E, D CIK, A). Zf A has a good set X E K of indiscernibles such that cf(]X]) > o, then there is an X’ E K such that X c X’ G K and X’ is good for A. Where D is the coding of mice, and K = L[D], K, = L,[D rl K]. (See [4]). Lemma 6.1 (Absoluteness). Let K be a regular uncountable cardinal. Let Z G P(K) be an ideal on K. Assume that Z is definable in ZFC, and ZFCF [K]<~ s 1. Zf vx E K (x E Z+ n Kjx
E (z+)K),
then tlx E K (x E w+(Z) fl KJx
E (%+(z))K).
274
Q. Feng
Proof. Fix K, Z as above. Notice that by our assumption, (Z)K, as the ideal defined in K, and (C%+(Z))“, as defined in K, make perfect sense. Let X E W(Z) n K, let A E P(K) fl K. By Theorem 2.5,
has a good set of indiscernibles Y E X such that Y E I+. Applying the Indiscernibility Lemma, we get Y’ E K, Y s Y’ E K, such that Y’ is good for A. Then X fl Y’ E I+ tl K. Therefore, X tl Y’ E (Z+)K. Hence X E P+(Z))“. This completes the proof.
0
Theorem 6.2. Zf K is &-Ramsey, Proof.
then
K
is
&-Ramsey
in K.
By induction on a; we prove the following. Vx (x E (I:)+ rI K+x
E ((1:)‘)“).
(1)
For Q:= 0, Z$ = %(ZQ. Since “being unbounded” is absolute, by the Absoluteness Lemma, Vx E (la+ fl Kx E ((13’)“. For cr = 1, Zr = 9(12r). Since “being stationary” is downward absolute, by the Absoluteness Lemma, we have our conclusion. Forcu=j3+1, bal, let/3’=/3-1if/3Co, andB’=pif/3>o. Byinduction hypothesis, Vx E (Zi,) n Kx E ((Zi,)‘)“. By the Absoluteness Lemma, since I”,= W(Z;.), we have our conclusion. When (Yis a limit ordinal, (Ii))+ = n {(Z3+ : p < a}. LA X E (I”,)+ fl K. Then V/3 < CUXE (I;)+ n K. Now by induction hypothesis, we conclude that X E ((13’)“. This finishes the proof of (1). Now the theorem follows by induction, Corollary 6.3. Zf K
is completely
applying (1) to x =
K.
0
Ramsey, then
K k K is completely Ramsey. Definition 6.1. Let K be a regular uncountable cardinal. Let f = & : a < K+) be the canonical sequence on K. (1) K is 0-hyper completely Ramsey if K is completely Ramsey. K is ((Y+ 1)-hyper completely Ramsey if K is a-hyper (2) For a’<~+, completely Ramsey and there is a completely Ramsey subset X so that for all il E X, h is f,(l)-hyper completely Ramsey. (3) For (YG K+, a! is a limit ordinal, K is cy-hyper completely Ramsey if for all 6 < cu, K is /3-hyper completely Ramsey. (4) K is super completely Ramsey if K is K+-hyper completely Ramsey.
275
A hierarchy of Ramsey cardincils
Theorem
6.4. {A <
If K is measurable, K : k is
then K is super completely Ramsey and
super completely Ramsey)
is stationary.
Proof. Let u be a normal K-complete nonprincipal ultrafilter on K. Let j: V-* M = Vk/U be the canonical elementary embedding, where Vk/U is the ultrapower of the universe V by 17, and M is the transitive collapse of the ultrapower. Let f be the canonical sequence on K. Then for all cy< K+, [fa] = a in M, and if g(cu) = (Y+for all LY< K, then [g] = K+ in M. In particular, if D < K+ is a limit ordinal, then the cofinality of & in V = the cofinality of a in M. Also, since M is closed under K-sequences, the definition off is absolute with respect to M. With the remarks above in mind, let, for LY< K+, X, = {A <
K :A
is f,(A)-hyper
completely
Ramsey}.
We now prove VcY
(*)
By induction on a < K+, we show that X, E U. For LY= 0, it is clear. Let (Y= /3 + 1. By induction hypothesis, X, E U. Then M k K is P-hyper completely
since /3 = [fs] in M and and for il E X,,
K =
Ramsey
[id] in M with id(y) = y for y <
M k I. is fs(3L)-hyper completely
K.
Now X, is in M
Ramsey,
also M LX, is a completely
Ramsey subset of
K.
Therefore, Ml= K is (/3 + 1)-hyper completely
Ramsey.
Consequently, since /3 + 1 = [fs+J in M, X, E U. Let (Ybe a limit ordinal. cf (YG K. Let (a: #I E cf (u) be an increasing sequence converging to (Y. Then this sequence is in M, and a6 = [fn,] in M for each p < cf a: By induction hypothesis, for each p < cf (Y, X,, E U. Therefore, M k K is a;6-hyper completely
Ramsey.
Thus, M k K is Lu-hyper completely
Ramsey.
Since (Y= [fJ in M, we have that X, E U. This finishes the proof of (*).
Q. Feng
276
Now by (*), M k K is K+-hyper completely M E K is super completely
Ramsey, i.e.,
Ramsey.
Hence, {A < K :A is super completely Ramsey} E U. Since each X, is completely Ramsey, K is super completely Theorem 6.5. K kK
Proof.
ZfK is
is
Ramsey in V.
0
super completely Ramsey, then
super completely Ramsey.
To prove this corollary,
we first prove one fact on the canonical sequence
f First, let us observe that if a is a limit ordinal, then cf (cfc(cu)) = cf (Y. Fact. Let K be regular uncountable. Let a< (K+)~. Zf fa is the cuth canonical function on K, and if f b, is the cuth canonical function on K in K, then {A < K : f,(n) = f h(n)} COnthIS a club in K.
It is clear for the cases when LY= 0 or (Yis a successor assuming the Fact for smaller ordinals. Let (Y< (K+)~ be a limit ordinal. Case 1: &a < K. Let (Es : /3 < cfK&) E K be a normal sequence converging to fx. Let ( oq : q < cf a) be a normal sequence converging to ~?‘a. Then (5, : @< cf o) is normal cofinal in (Y. So there is a club C E K so that for each A E C, fu(A) = sup{ fs,(A) : p < cf a}. By induction hypothesis, V/l < cf (Y, there is a club C, so that VA E C, we have fE,(A) =f&,(A). Let D = C n fl {C,:/3
completely
Ramsey
assuming that K is super completely Ramsey. When (Y= 0 or a limit ordinal, it follows from Corollary hypothesis. Let Q!= /3 + 1. By induction hypothesis, K I=K is /3-hyper completely
Ramsey.
6.3 or the induction
A hierarchy of Ramsey cardinals
277
Let A = {A < K : K k 3cis f;(A)-hyper completely Ramsey}. Then A E K. Let C G K be a club so that VA E Cfs(A) =fb(A). Let B = {A < K :il is &(A)-hyper completely Ramsey},
Since K is super completely Ramsey, B fl C is then completely Ramsey. Therefore A is completely Ramsey. By (1) of the proof of Theorem 6.2, A is completely Ramsey in K. Hence, K k K is rr-hyper completely Ramsey. Therefore, K k K is super completely Ramsey.
Acknowledgement The author is grateful to Professor T. Jech, Professor W. Mitchell and Professor S. Simpson for their teaching and guidance.
References [l] J. Baumgartner, Ineffability properties of cardinals I, in: A. Hajnal et al., eds., Colloq. Math. Sot. Janos Bolyai 10, Infinite and Finite Sets, Vol. III (North-Holland, Amsterdam, 1973) 109-130. [2] J. Baumgartner, Ineffability properties of cardinals II, in: Butts and Hintikka, eds., Logic, Foundations of Mathematics and Computation Theory (Reidel, Dordrecht, 1977) 87-106. [3] T. Dodd, The Core Model, London Math. Sot. Lecture Note Series 61 (Cambridge Univ. Press, Cambridge, 1982). [4] T. Dodd and R. Jensen, The core model, Ann. Math. Logic 20 (1981) 43-75. [5] D. Donder, R. Jensen and B. Koppelberg, Some applications of the core model, in: R. Jensen and A. Prestel, eds., Set Theory and Model Theory, Lecture Notes in Math. 872 (Springer, Berlin, 1981) 55-97. [6] P. Erdijs and R. Rado, A partition calculus in set theory, Bull. Amer. Math. Sot. 62 (1956) 427-489. [7] G. Fodor, Eine Bemerkung xur T’heorie der Regressiven Funktionen, Acta Sci. Math. (Szeged) 17 (1956) 139-142. [8] T. Jech, Set Theory (Academic Press, New York, 1978). [9] T. Jech, Stationary subsets of inaccessible cardinals, in: J. Baumgartner, ed., Axiomatic Set Theory, Contemporary Math. 31 (Amer. Math. Sot., Providence, RI, 1984) 115-142. (lo] T. Jech and K. Prikry, Ideals over uncountable sets: application of almost disjoint functions and generic ultrapowers, Mem. Amer. Math. Sot. 214 (1979). (111 A. Levy, The sizes of indescribable cardinals, in: Axiomatic Set Theory, Proc. Symp. Pure Math. 13, Part 1 (Amer. Math. Sot., Providence, RI, 1971) 205-218. [12] W. Mitchell, Ramsey cardinals and constructibility, J. Symbolic Logic 44 (1979) 260-266. [13] F. Ramsey, On a problem of formal logic, Proc. London Math. Sot. (2) 30 (1930) 264-286. [14] F. Rowbottom, Large cardinals and constructible sets, Doctoral Dissertation, University of Wisconsin, Madison, WI, 1964. [15] D. Scott, Measurable cardinals and constructible sets, Bull. Acad. Polon. Sci., Ser. Sci. Math. Astronom. Phys. 7 (l%l) 145-149.