Moti Gitik and Saharon Shelah- On Densities of Box Products
Transcript of Moti Gitik and Saharon Shelah- On Densities of Box Products
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ON DENSITIES OF BOX PRODUCTS
by
Moti Gitik Saharon Shelah
School of Mathematical Sciences Hebrew University of JerusalemSackler Faculty of Exact Sciences Department of Mathematics
Tel Aviv University Givat Ram, Jerusalem
Ramat Aviv 69978 Israel Israel
Abstract.
We construct two universes V1, V2 satisfying the following GCH below ,
2 = +2 and the topological density of the space 2 with 0 box product
topology d
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W. Comfort asked the following question: Assume is a strong limit singular, >
cf. Is d + when 2 > +?
d
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(c) for every < d
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Lemma 1.4. Let D be a normal ultrafilter over , PD the Prikry forcing with D, a
PD-name of a partial function of cardinality< ( < ) from to 2. Then there areA and
f satisfying the conditions (a), (c) of 1.1 so that , A
=
n
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and D D1 we are done.
Now the plan will be as follows: Well blow up the power of to some cardinal of
cofinality + using < -support iteration of forcings of the type QD. Using hugeness, a
sequenceD0 D 1 D 2 D ( <
+)
will be generated and QD s will be used cofinally. The final step will be to use the Prikry
forcing with
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with p, for < . At stage just pick < , A such that p is incompatible with
p and every p,( < ) if there is such an . Otherwise we stop. Let , | < be
such a sequence. Then, by the claim, < (2||+||)+ < . Hence, if we take a regressive
function g() = , then whenever g(1) = g(2) p1 , p2 will be compatible. So, we obtaina stationary +-c.c.
Proof of the Claim. Let i | < (2||+||)+ be a sequence from A. Set B0 =
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then forcings QD will preserve its supercompactness and hence also the measurability. So
new ultrafilters will appear over forever.
Let us work now over . Let G P be generic. We consider in M j(P) = Pj()
and Pj()G in M[G]. Let us split Pj()G into Q0Q1 and P>. The generic objectfor Q0 is just any ordinal F() < j() = . By standard arguments on backwards Easton
forcing (see, for example, A. Kamamori [Ka]), for every F() the length ofQ1 will be
F(). For a while set F() = , i.e. we like to deal with iteration Q1 of the length . We
consider an enumeration A | < of Q1-names of subsets of in M[G][{}], such
that 1 < 2 < implies that A 1 . depends on the part of Q1 of the length then those
of A 2 . Since is measurable and Q1 has < -support there will be C U consisting
of inaccessibles such that for every CA | < enumerates the names of all subsets
of appearing before the stage , i.e. Q1
-names. Equivalently, all the subsets for the
Q1 with F() = . Now let be in C. For every D
appearing in Q1
let rD
Qj(D) be
defined as follows. rD
= f
, , A
where
A = AD ({j
(B ) | B D }) ,
f
= fD and above we take f() = {j(f)() | f appear in a condition in GD }.Let q Q1j() consists of this rD
s sitting in the right place.
Clearly, that if > is also in C, then q
= q.
Let be in C. Pick a master condition p P> (j() Q1j()) deciding all the
statements j(A) for < and stronger than q, i.e. p satisfies the following: for
every < there is s G {} Q1
so that s, p j(A ).
Shrink the set C to a set C U so that for any two 1 2 C decisions are the
same, i.e. for every s, < 1 as above
s, p1 j(A )
iff
s, p2 j(A )
and
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s, p1 j(A )
iff
s, p2 j(A ) .
For every C we define in V[G {} G(Q1)] a normal ultrafilter D() over
, where G(Q1) Q1 generic and Q1 has length . Let us set A D() iff for some
s G {} G(Q1) s, p j(A) where the interpretation of A is A for < ,
Suppose now that < 1 are two elements of C. Work in V[G {1} G(Q1)].
Then, clearly, G(Q1)
will V[G {}] generic for Q1 (or Q1
in the sense of the
iteration to 1). So D() V[G {1} G(Q1)].
Claim. D() D(1).
Proof: Let A D(). Pick < , A and s to be as in the definition of D(). By the
choice ofC, then s, p1 j(A ). So, iG(Q1)(A ) = iG(Q1 )(A ) = A is in D(1)
as well as in D(), where iG is the function interpreting names.
Now we are about to complete the construction. Thus, let be a limit of an increasing
sequence i | i < + of elements of C. We consider V[G {}], i.e. the iteration Q1
will be of the length . By the claim,
D(0) D(1) D(i) (i < +) .
For every i < +, D(i) is a normal ultrafilter over in V[G {} G(Q1)
i]. Hence,
the forcing QD(i) was used at the stage i + 1. Finally set D =
i +.
Proof: 2 = since at each stage of the iteration Q1 a new subset of is produced and
is a limit of inaccessibles of cofinality +.
Notice that D is a normal ultrafilter over since it is an increasing union of+ normal
ultrafilters D(i) (D(i) is such in V[G {} Q1
i]) and Q1 satisfies +-c.c. It is
+-generated since for every i < + a set AD(i) generating D(i) is added at stage i + 1.
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Let fD(i) | i < + be the generic functions added by QD(i)s. Use the Prikry
forcing with D. Let n | n < be the Prikry sequence. Then by Lemma 1.5 we obtain
the following:
Theorem 1.10. The following holds in the model V[G {} G(Q1) n | n < ]
(a) is a strong limit cardinal of cofinality
(b) 2 = > +
(c) the functions fD(i) | i < + are witnessing d
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(2) 2 = ++
(3) cf = 0
(4) d
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Proof: Let a sequence A | < + be given by Lemma 2.3. Assume also that it
is increasing. For every < + there is A N, such that k(A) = A and N |=
|A| = ++. Working in N and using GCH, we pick a dense subset F, with |F
| =
+
of the topological spaceA
with the topology generated by countable products. Thenlet F = k(F
) and F =
+. Clearly, F is as required.
The family F of Lemma 2.4 will be used to generate a dense set in the space 2
with countable product topology once the cofinality of is changed to and its power is
blown up to ++. Thus, if n | n < is the Prikry sequence for the normal measure
of the extender, i.e. for U0 and f = k(f) F, then let f be a function such that
(i)(f
)() = f
. The dense set will consist of functions n
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The following lemma is routine.
Lemma 2.5. For every < j0n() = n there are f : []n and < ++ such that
= j0n(f)(, j1(), j2(), . . . , jn1()).
Now fix n, 1 < n < . We like to describe one more way of constructing Mn. Thus,
we consider En1 and M1. En2 and even E is not in M1 but we still can from outside
measure subsets of of M1. So we can form Ult(M1, En1). Since V+2 M1 and
M1 M1, it is routine to check that Ult(M1, En1) Mn. Let be the corresponding
embedding. Then () = n1, (1) = n.
Lemma 2.6. For every < n there are g : []n1 1 and < ++ such that
= (g)(, j1(), . . . , jn2()).
Proof: Let g be a function representing in the ultrapower by En1, i.e. for some
< ++ jn1(g)(, j1(), . . . , jn2()) = . Then g : []n1 1, since < 1 and
jn1(1) = n. But then also (g)(, j1(), . . . , jn2()) = , sinceM = V.
Further let us add to such the subscript n.
Let F be the family given by Lemma 2.4. We define Fn = n(F) for every n, 0 < n < .
Let
Fnk = {f[k1, k) | f Fn} for every k, 0 < k n. For n, 0 < n < and
t nk=1 Fnk rngt is a partial function from n to 2 and it belongs to Mn as a finiteunion of its elements. Set Fn = {t| for some m, 0 < m n t
mk=1
Fmk}.Lemma 2.7. For every n, 1 < n < , Fn is dense in the topological space
n2 with
countable product topology.
Proof: Let m | m < be an -sequence of ordinals below n, for some n, 1 < n < .
Let {m|m 1. Since nothing happens between k and n, we can assume that
k = n. For every m < , by Lemma 2.6 there are gm : []n1 1 and m <
++ such
that m = n(gm)(m, j1(m), . . . , jn2(m)). Since (++)0 = ++ and E is -closed. So
we can code the sequence m | m < into one < ++. Hence, for every m <
m = n(gm)(, j1(), . . . , jn2()) .
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Since ms are all different, there will be A Un1 such that for every m = < and
s [A]n1 gm(s) = g(s). Let us also show that the ranges of gms can be made disjoint.
Let us do this for two g0 and g1. Using the completeness of U it is easy then to get the
full result.
Claim 1.8.1. There is B A in Un1 such that rng g0
B rng g1
B = .
Remark. It may not be true iff either 0, 1 are in different intervals [k, k+1) or if a
same measure appears in the extender several times.
Proof: In order to simplify the notation, let us assume that n = 3. So 2 0 < 1 .
Then the following holds in a generic extension V[G]:
(1) for every < or 2 = +.
(2) 2 = +
(3) cf = 0
(4) d
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For Theorem 3.3 we need also to replace 0-box products by 1-base products. Notice
that all the considerations of Section 2 are going smoothly if we replace 0-box product
by -box product for any < . Also instead of the space 2 we can work with for any
fixed < . So the following holds:
Theorem 3.4. Suppose that o() = ++ + 1, , < . Then the following holds in a
generic cardinal preserving extension:
(1) for every < or > 2 = +
(2) cf = 0
(3) 2 = ++
(4) the density of the topological space with the topology generated by -products is
+.
The analogs of 3.2 and 3.3 hold as well.
4. Reaching the Maximal Density and Wider Gaps
In previous sections, we constructed models with density less than the maximal pos-
sible value 2. Let us show now how to construct a model with the density 2 assuming
singularity of and 2 > +.
Theorem 4.1. Suppose o() = +3 + 1, then there is a generic extension V[G] satisfying
the following
(1) for every < or > 2 = +
(2) cf = 0
(3) 2 = ++
(4) d
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of cardinality + are added to sets ofV1. So there is no dense set in2 of cardinality +.
D is dense since there is no new basic clopen sets.
As in Section 3 it is possible to push this result down to and 1 .
Suppose now that one likes to have 2
big but still keep the density +
. A slightmodification of the construction of Section 2 will give the following:
Theorem 4.2. Suppose that > is a regular cardinal o() = + 1. Then there is a
generic cardinal preserving extension satisfying the following:
(1) is a strong limit
(2) cf = 0
(3) 2 =
(4) d
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