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Saharon Shelah and Jouko Vaananen- Stationary Sets and Infinitary Logic
Transcript of Saharon Shelah and Jouko Vaananen- Stationary Sets and Infinitary Logic
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Stationary Sets and Infinitary Logic
Saharon Shelah
Institute of Mathematics
Hebrew University
Jerusalem, Israel
Jouko Vaananen
Department of Mathematics
University of Helsinki
Helsinki, Finland
October 6, 2003
Abstract
Let K0 be the class of structures ,
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If =
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where g is the canonical name for g. It is easy to use =
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Theorem 1 gives a new proof of the result, referred to above, that if
=
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Theorem 6 Con(ZF) implies Con(ZFC + CH + there is no P C(L21)
such that for all A 1: 1, < , A |= A is stationary).
Proof. We start with a model of GCH and add 2 Cohen subsets to 1. Inthe extension GCH continues to hold. Suppose there is in the extension a P C(L21) such that for all A 1:
1, < , A |= A is stationary.
Since the forcing to add 2 Cohen subsets of1 satisfies the 2-c.c., belongsto the extension of the universe by 1 of the subsets. By first adding all butone of the subsets we can work in V[A] where A is a Cohen subset of 1 and
is in V. Note that A is a bi-stationary subset of 1. Let P be in V theforcing for adding a Cohen generic subset of 1 and let A be the P-name forA. Let p force 1,
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open set B N1 N1 such that f(f A g((f, g) B). A set is 11
if its complement is 11.Let CUB be the set of characteristic functions of closed unbounded sub-
sets of1, and NON-STAT the set of characteristic functions of non-stationarysubsets of 1. Clearly, CUB and NON-STAT are disjoint
11. It was proved
in [4] that, assuming CH, CUB and NON-STAT are 11 if and only if there isa Canary tree. Another result on [4] says that the sets CUB and NON-STATcannot be separated by any 03 or
03 set.
Theorem 7 Assuming CH, the sets CUB and NON-STAT cannot be sepa-rated by a Borel set.
Proof. Let {s : < 1} enumerate all s
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We shall construct two P-generic sets, G0 and G1, over M. For this end,
list open dense D P with D M as D : < 1. Define Gl = {pl : < 1} so that pl0 = p, p
l+1 p
l with p
l+1 D M, p
l+1(pl
) = l, and
pl =
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Proof. There is a set {Bi : i < L1 } of almost disjoint subsets of in L.
Since L1 = 1, this set is really of cardinality 1. Let (x, y) be a 1-formulaof set theory such that for all and x, y L, x
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Let
B(z, y) xuv(wf(x, u, v) cor(x, u, v, z)cor(x, u, v, y) 0(z, y, x, u, v)).
The point is that if and , then B if and only if ,
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Lemma 12 f/D is independent of the choice of the sequence ai : i < .
Theorem 13 Suppose
(i) = +, where = 0.
(ii) For every club C there is some X such that
C {i < : (i, f(i)) X} contains a club
C {i < : (i, f(i)) X} contains a club.
Then there is a sentence L such that for all A :
,
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Proof. Let a club C be given. For < < , let C D so that
f|C < f|C. Let P consist of conditions
p = (Bp, fp, cp, gp),
where
(i) Bp . |Bp| < .
(ii) fp is a partial mapping with Dom(fp) , |Dom(fp)| < , andRng(fp) {0, 1}.
(iii) If Bp, then {i < : (i, f(i)) Dom(fp)} is an ordinal jp.
(iv) cp = cp : Bp, where cp is a closed subset of j
p. We denote
max(cp) by p.
(v) If Bp C and i Cp, then fp(i, f(i)) = 1. If Bp \ C and
i Cp, then fp(i, f(i)) = 0.
(vi) gp is a partial mapping with Dom(gp) [Bp]2 and Rng(gp) .
(vii) If < Dom(gp), then = cP \ g(, ) C.
The partial ordering q extends p is defined as follows:
p q Bp Bq, fp fq, gp gq,
Bp(cp is an initial segment of cq),
and if p < q, then Dom(gq) [Bp]2.
We show now that P satisfies conditions (GMA1) and (GMA2).
Lemma 15 P satisfies (GMA1).
Proof. Let po . . . pi . . . (i < ) in P with < . We may assumep0 < p1 < . . .. Let = sup{pi : i < }. Let B = i
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and i < i. Since pi < pi , g(, ) is defined and cpi \ g(, ) C. Hence
C, whence f() < f().Let c = c : B where c =
i c
pi {}. Let j =
ij
pi {}. Nowthe condition p = (B,f, c, g) is the needed l.u.b. of (pi)i
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Then B = and each c is a club of . Let X = {(, ) : f(, ) =
1}. Suppose C and i c. Then f(i, f(i)) = 1 whence (i, f(i)) X.Suppose C and i c. Then f(i, f(i)) = 0 whence (i, f(i)) X.
Corollary 17 Suppose = +, where = 0, and GMA. Thenthere is a sentence L such that for all A :
,