Download The Hanf Number for Complete Lω1, ω-Sentences

The Hanf Number for Complete Lω1, ω-Sentences (Without GCH)
Author(s): James E. Baumgartner
Source: The Journal of Symbolic Logic, Vol. 39, No. 3 (Sep., 1974), pp. 575-578
Published by: Association for Symbolic Logic
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THE JOURNALOF SYmBoLc LoGIc
Volume 39, Number 3, Sept. 1974
THE HANF
NUMBER
FOR COMPLETE
(WITHOUT GCH)
L4,1.U,-SENTENCES
JAMESE. BAUMGARTNER1
The Hanf number for sentences of a language L is defined to be the least cardinal
with the property that for any sentence q' of L, if p has a model of power > K then
p has models of arbitrarily large cardinality. We shall be interested in the language
LUXU (see [3]), which is obtained by adding to the formation rules for first-order
logic the rule that the conjunction of countably many formulas is also a formula.
Lopez-Escobar proved [4] that the Hanf number for sentences of Lt,,J, is IC015
where the cardinals a,, are defined recursively by no = mOand m< = E{22 ,e
< cc}
for all cardinals a > 0. Here w, denotes the least uncountable ordinal.
A sentence of L
is complete if all its models satisfy the same L,,,U,1-sentences.
In [5], Malitz proved that the Hanf number for complete sentences of LG,1okis
also , ,, but his proof required the generalized continuum hypothesis (GCH).
The purpose of this paper is to give a proof that does not require GCH.
More precisely, we will prove the following:
THEOREM1. For any countable ordinal a, there is a complete Lc,,1,t,-sentence'a
K
which has models of power TVbut no models of higher cardinality.
Our basic approach is identical with Malitz's. We simply use a different combinatorial fact at the crucial point.
A subset A of the universe of a structure 21 is homogeneous for 21 provided that
every permutation of A can be extended to an automorphism of 21. It is shown
in [5] that Theorem 1 is a consequence of the following theorem:
THEOREM2. For any countable successor ordinal a, there is a complete L,1.,sentence a. involving the unarypredicate symbol A, which satisfies
(i) a. has a model 21in whichA' (the denotation of Ao in 21)has cardinality M.,
(ii) a. has no models of cardinality > n, and
(iii) if 21 is a countable model of ao, then Al is infinite and homogeneousfor 21.
Theorem 2 will be proved by induction on a. The cases a = I and a the successor
of a limit ordinal are dispensed with in [5], and the proofs do not require GCH.
Therefore we confine ourselves to the case when a = , + 1 and ,8 is a successor
ordinal;
To treat this case, Malitz used the following proposition, which is a consequence
of GCH:
(I) For any infinite cardinal K, there is a densely ordered set of power 21Cwhich
has a subset of power K dense in it.
Received September 19, 1973.
1 The preparation of this paper was partially supported by National Science Foundation
grant GP-38026.
575
?
1974, Association
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for Symbolic Logic
576
JAMESE. BAUMGARTNER
Unfortunately, if GCH is not assumed then (1) is not provable, as Mitchell has
shown. See [6] and [1].
We shall replace (1) with a combinatorial theorem due to Hausdorff.
A family F of subsets of a given set A is independent iff, for any natural numbers
m and n and any distinct members AO,Al, - * , Am,Bo, B1, * , B. of F,
(Ao n A1 (P.no Am) - (Bo U B1 U.. *u Bn) is nonempty. Note that if F is
infinite and independent, then in the above situation (AOn... rl Am)...
u Bn) is always infinite.
(Bo U
LEMMA1. If K is an infinite cardinal and A is a set of cardinality K, then there
exists an independentfamily F of subsets of A such that F has cardinality2'. Moreover, F can be chosen so that, for any pair s, t of disjoint finite subsets of A,
{B e F: s c B and B c) t = O}has power 2c.
The first half of Lemma 1 is due to Hausdorff [2]. Given such an F,
PROOF.
we obtain a family F' satisfying all of Lemma 1 as follows: For each pair s, t
of disjoint finite subsets of A, let A&be a subset of F of power 2'. Since there are
only K such pairs s, t, we may assume that the sets Aa are all pairwise disjoint.
Let Ba = {(B u s) - t : B E As}. It is easily checked that the family F' =
U{Bt : s, t disjoint, finite} is as desired.
LEMMA 2. Let N = {O, 1, 2, * * *}, the set of natural numbers. Suppose that, for
each n e N, An is a collection of subsets of N satisfying
(i) each An is countable,
(ii) Amr AAn= 0 when m # n,
(iii) U{An: n E N} is independent,
(iv) for every pair s, t of disjoint finite subsets of N and every n e N,
{X e An: s c X and X r t = 0} is infinite.
Suppose that {B,,:n e N} is another collection satisfying (i)-{iv). Let A=
U{An : n e N} and B = U{B,,: n e N}. Then there is a permutationp of N and a
function j mappingA one-one onto B such thatj carries each An onto Bn and,for all
Xe A, j(X) = {p(n): n e X}.
The proof is a back-and-forth argument, rather like Cantor's proof of
PROOF.
the theorem that any two countable dense linear orderings without endpoints are
isomorphic. Let {Xn : n e N} and { Yn: n e N} be enumerations of A and B respectively. At step 2n, we determine p(n) and j(X,,); at step 2n + 1 we determine
p - l(n) and j- '( Yn).We describe step 2n; step 2n + 1 is just the same except for the
fact that the roles of A and B are interchanged.
Step 2n. Let XO,X1, - *, Xk?_1be an enumeration of the set
{X0,
X1,
--,
Xn,.1,_1'(Y0),ji-(YA)*
-.
1(Yn)}
Suppose p(n) has not yet been defined. We must determine p(n) so that, for all
i < k, n e Xi' iff p(n) Ej(X/'). For each i < k, let Zi = j(Xi) if n E Xi', and let
Zi = N - j(Xi') otherwise. Since B is infinite and independent, ZOr) Z1 nr..
r1 Zk- 1 is infinite. Choose m e ZOr Z1 r*..* n Zk_ 1 so that m 2 n and m # p(i)
for any i < n. Let p(n) = m.
Now assume j(Xn) has not yet been defined. Say Xn E A,. Let s =
{Q, *, * *, n, p-'(0), - - , p-1(n - 1)} r) Xn and let t = {O. 1, *.* -, n
(0), * *
p '(n - I)} - Xn. By condition (iv) of Lemma 2, there is Ye B, such that
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THE HANF NUMBER FOR COMPLETELCO1,,o-SENTENCES
577
YO {Y0, **.* Yn- i(X0) ...* j(Xnl)} and {p(i): i e s} c Y and {p(i): i e t} n
Y = 0. Let j(Xn) = Y. This concludes step 2n.
It is easy to check that p and j satisfy Lemma 2.
Now we turn to the proof of Theorem 2. Let a = , + 1, where P is a successor
ordinal, and assume that Theorem 2 holds for P.
It is not difficult to see that there is an L31,co-sentenceat, the models of which
are precisely those structures of the form
M = (BuB'UBe,B,B',BeBB,Ro,Ri,-Rn,
,-,EeFe)
which satisfy the following:
(i) If Z = (B, B8, R0, R1, *.,,
* *), then 2 is a model ofao
with Al = B8.
(ii) B, B' and Be are pairwise disjoint.
(iii) The relation Ee holds only between members of BOand members of B';
cEeb}forbeB',thenb # b'implies Xb # Xband{Xb: beB'}
if Xb = {ceB,:
is an independent set.
(iv) The relation Fe holds only between members of B' and members of Be; if
c e B. then, for every pair s, t of disjoint finite subsets of B8, {b e B': b Fe c and
s a Xb and Xb n t = O}is infinite; if c, c' e Be then
{b e B': b F c} C){b e B': b Fec'} = 0.
(v) Be is infinite and Al
= Ba.
First we check that a. has a model 21in which Be has cardinality .e By inductive
hypothesis, acr has a model Z3= (B, B1, R0, * * , R,* *-) in which Bq has cardinality =8. Let F be a family of independent subsets of B8 satisfying Lemma 1.
Then F has cardinality24 = D. Let B' = F. For c E B8 and b e B', let c Ee b if
c e b. It is easy to see that F may be decomposed into a family G of disjoint subsets
such that G has power ne and, for every H e G and every pair s, t of disjoint finite
subsets of B8, {X e H: s c X and X n t = 0} is infinite. Let Be = G and if b e B'
and c e Be, let b Fe c iff b e c. Then clearly aL = (B U B' U B,, B, B', Bal,B8, R0,
R11,* I Rns * ,3Ecr,Fe) is a model of a,,cand Be has power ,.e
On the other hand, if 21 is a model of ea then, by inductive hypothesis, the
maximum cardinality that B (and hence B8) can have is zB. Since, for b, b' e B',
b # b' implies Xb : Xb', the maximum cardinality that B' can have is 2 =ns
Of course, Be cannot have greater cardinality than B', so ae has no models of
cardinality > >.
Now suppose 2t is a countable model of a.. We assert that Al = B. is homogeneous for 2t. Let h be any permutation of B.. By Lemma 2, h may be extended to
an automorphism h* of B' and B5 which preservesE. and F.. By inductive hypothesis the mapping h* restricted to B8 may be extended to an automorphism of
93 = (B, Be, R0,** *, Ran,.*). Therefore h may be extended to an automorphism of
Z.
Finally, we check that a. is complete. As remarked in [5], it follows from a
theorem of Scott [7] that a. is complete iff all countable models of a. are isomorphic. Let Zt = (B u B' U Ba, B, B', Ba, B8, R0, *., Ea, F.) be a countable
model of a.. Let 21' be another countable model. Since ao is complete by hypothesis, we may assume that 2t' = (B u B' u Bs, B, B", Ba, BR,R0, ... PEa,Fa).
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JAMESE. BAUMGARTNER
578
Let H be any function mapping Ba one-one onto By. By Lemma 2, h may be extended to a one-one function h* on B' and B8 which carries E. onto E.' and Fa
onto Fa. But now since Bo is homogeneousfor Z = (B, BO,RO, *. ), the permutation of BO determined by h* can be extended to an automorphism of Z. But this
means that h* can be extended to an isomorphism of 2t onto W'.
REFERENCES
[1] J. BAUMGARTNER,
Almost-disjoint sets, the dense-set problem, and the partition calculus
(to appear).
[2] F. HAUSDORFF,OJberzwei Satze von G. Fichtenholz and L. Kantorovich, Studia Mathematica, vol. 6 (1936), pp. 18-19.
[3] C. KARP, Languages with expressions of infinite length, North-Holland, Amsterdam, 1964.
[4] E. G. K. LOPEZ-ESCOBAR,
On defining well-orderings, Fwadamenta Mathematicae, vol. 59
(1966), pp. 13-21.
[5] J. MALITZ, The Hanf number for complete L.,,. sentences, The syntax and semantics of
infinitary languages (J. Barwise, Editor), Lecture Notes in Mathematics, vol. 72, Springer-Verlag,
Berlin and New York, 1968, pp. 166-181.
[6] W. MITCHELL, Aronszajn trees and the independence of the transfer property, Annals of
Mathematical Logic, vol. 5 (1972-73), pp. 21-46.
[7] D. Scorr, Logic with denumerably long formulas and finite strings of quantifiers, The
theory of models (J. Addison, L. Henkin and A. Tarski, Editors), North-Holland, Amsterdam,
1965, pp. 329-341.
DARTMOUTH
HANOVER,
COLLEGE
NEW HAMPSHIRE
03755
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