jorg brendle and saharon shelah- evasion and prediction iv: strong forms of constant prediction
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Evasion and prediction IV
Strong forms of constant prediction
Jorg Brendle
The Graduate School of Science and Technology
Kobe University
Rokkodai 11, NadakuKobe 6578501, Japan
email: [email protected]
Saharon Shelah
Institute of Mathematics
The Hebrew University of Jerusalem
91904 Jerusalem, Israel
and
Department of MathematicsRutgers University
New Brunswick, NJ 08903, USA
October 6, 2003
Abstract
Say that a function : n
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prediction number vconstn (k), that is, the size of the least family of
predictors needed to k-constantly predict all reals, for different valuesofn and k, and investigate their relationship.
Introduction
This work is about evasion and prediction, a combinatorial concept originallyintroduced by Blass when studying settheoretic aspects of the Specker phe-nomenon in abelian group theory [Bl1]. The motivation for our investigationcame from a (still open) question of Kamo, as well as from an argument ina proof by the first author. Let us explain this in some detail.
For our purposes, let n and call a function : n
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(see 1.4). Moreover, for k < , one may well have vconst2 () < vconst2 (k)
(Theorem 2.1). Any hope to use Theorem 1.4 as an intermediate step toanswer Kamos question is dashed, however, by Theorem 2.2 which says thatvconst2 may be strictly smaller than the minimum of all v
const2 (k)s.
In Section 3, we dualize Theorem 2.1 to a consistency result about evasionnumbers and establish a connection between those and Martins axiom for-k-linked partial orders (see Theorem 3.7).
We keep our notation fairly standard. For basics concerning the cardi-nal invariants considered here, as well as the forcing techniques, see [BJ]and [Bl2].
The results in this paper were obtained in September 2000 during and
shortly after the second authors visit to Kobe. The results in Sections 1 and2 are due to the second author. The remainder is the first authors work.
1 The ZFCresults
Temporarily say that : n
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Given = g,j ; (g, j) G k, a sequence of predictors for the space
2, and n
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predictor there is x F which is not k-constantly predicted by . Similarly,
define the constant evasion numbereconstn .Let vconstn (k) denote the size of the least family of predictors : n
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Theorem 2.1 Assume CH. Let2 k1 < ... < kn1. Also let i, i n,
be cardinals with i = i and n < . . . < 0. Then there is a genericextension satisfying vconst2 = min{v
const2 (k); k } = v
const2 (kn1 + 1) = n,
vconst2 (ki) = vconst2 (ki1 + 1) = i for 0 < i < n (where we put k0 = 1) and
c = 0.
Proof. We force with the countable support product P =
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such that
(a) if T 2mj , j D, and i \ i+1 (i < n), then |{ T 2mj+1; }| = 2ki,
(b) p f [T], and
(c) whenever q p where q = q; B with A B, T 2mj ,and j D are such that q f, then there are r q and T 2mj+1 with , such that r f where r = r; Bwith r = q for = .
Now let Gi0
be
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For 2, p S and s p, say s is a splitting node ofp if all immediate
successor nodes of s belong to p. Define recursively the nth splitting levelof p such that the 0th splitting level consists of the least splitting node andthe (n + 1)st splitting level consists of the least splitting nodes beyond thenth splitting level. For p, q S, say q n p if q p and the nth splittinglevels of p and q are the same.
Let f be a P2name for a function in 2. Notice given any p0 P2, we
can find p p0 and < 2 such that
p f V[G] \
k. The following is the main point.
Main Claim 2.3 There are q p and a predictor V such that
q -constantly predicts f .
Proof. We construct recursively
A countable (intended as the domain of the fusion q),
D; A, a partition of into countable sets (its purpose beingthat at step j of the construction we preserve one more splitting levelof the th coordinate of the condition where j D),
finite partial functions aj : A , j (keeping track of how oftenthe th coordinate has been worked through),
conditions pj P, j (intended as a fusion sequence),
a strictly increasing sequence mj , j ,
a tree T 2
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(b) a0 = ,
(c) if j D, then dom(aj+1) = dom(aj) {}; in case / dom(aj), wehave aj+1() = 0, otherwise aj+1() = aj() + 1; aj+1() = aj() for = ,
(d) p0 = p,
(e) pj+1 pj; furthermore for all dom(aj+1),pj+1 pj+1() aj+1() pj(),
(f)j dom(pj) =
j dom(aj) = A,
(g) if T 2mj , j D, then |{ T 2mj+1 ; }| = k,
(h) for each T 2mj , there is pj pj which forces f; furthermore
pj fmj T 2mj , and
(i) -constantly predicts all branches of T.
Most of this is standard. There is, however, one trick involved, and wedescribe the construction. For j = 0, there is nothing to do. So assume wearrived at stage j, and we are supposed to produce the required objects forj + 1. This proceeds by recursion on T 2mj . Since the recursion is
straightforward, we confine ourselves to describing a single step.Fix T 2mj . Let be such that j D. Without loss < (the
case = being easier). Consider pj . Step momentarily into V[G] with
pj G. Then pj () Q f. Since f is forced not to be in V[G], we
can find m , pairwise incompatible ri pj (), and distinct
i 2
m
extending where i < k such that ri Q
i f. As Q is kary Sacks
forcing, we may do this in such a way that the predictor can be extendedto -constantly predict all i .
Back in V, by extending the condition pj if necessary, we may withoutloss assume that it decides m and the i . We therefore have the extension of
which -constantly predicts all i already in the ground model V. We mayalso suppose that pj decides the stem ofp
j (), say p
j stem(p
j ()) =
t. For i < k define pij+1 such that
pi
j+1 = pj , p
i
j+1[+ 1, ) = pj [+ 1, ),
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pi
j+1 pi
j+1() = (pj ())t i,
pij+1 p
ij+1() = r
i .
Doing this (in a recursive construction) for all T2mj and increasing m
if necessary, we may assume there is mj+1 with mj+1 = m for all . Finally
pj+1 is the least upper bound of all the pij+1.
This completes the construction. By (c), (e), and (f), the sequence of pjshas a lower bound q P. By (d), q p. By (h), q f [T] whichmeans that (i) entails q f is -constantly predicted by , as required.
Now let be a limit ordinal. Using a similar argument and the fact that
below ,Q is cofinally often Sacks forcing, we see
Claim 2.4 There are q p and a predictor V such that
q 2-constantly predicts f .
This completes the proof of the theorem.
3 Evasion and fragments ofMA(-linked)
Let k 2. Recall that a partial order P is said to be -k-linked if it can be
written as a countable union of sets Pn such that each Pn is k-linked, thatis, any k many elements from Pn have a common extension. Clearly every-centered forcing is -k-linked for all k, and a -k-linked partial order isalso -(k 1)-linked. Random forcing is an example of a partial order whichis -k-linked for all k, yet not -centered. A partial order with the formerproperty shall be called --linked henceforth. We shall deal with partialorders which arise naturally in connection with constant prediction and whichare -(k 1)-linked but not -k-linked for some k. Let m(-k-linked) denotethe least cardinal such that for some -k-linked partial order P, Martinsaxiom MA fails for P.
Lemma 3.1 LetP be -2k-linked, and assume is aPname for a functioni 2
ik 2k. Then there is a countable set of functions
i 2ik 2k such
that whenever g 2 is such that for all there are infinitely many iwith (gik) = g[ik, (i + 1)k), then
there are infinitely many i with (gik) = g[ik, (i + 1)k).
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Proof. Assume P =
n Pn where each Pn is 2k-linked. Define n :
i 2
ik
2k such that, for each 2ik, n() is a such that no p Pn forces() = . (Such a clearly exists. For otherwise, for each 2k we couldfind p Pn forcing () = . Since Pn is 2k-linked, the p would havea common extension which would force () / 2k, a contradiction.) Let = {n; n }.
Now choose g 2 such that for all there are infinitely many iwith (gik) = g[ik, (i + 1)k). Fix i0 and p P. There is n such thatp Pn. We can find i i0 such that n(gik) = g[ik, (i + 1)k). Bydefinition of n, there is q p such that q (gik) = n(gik). Thusq (gik) = g[ik, (i + 1)k), as required.
Lemma 3.2 Let Pn, Qn; n be a finite support iteration, and assume is a Pname for a function
i 2
ik 2k. Also assume for each n andeach Pnname n for a function
i 2
ik 2k, there is a countable set nof functions
i 2
ik 2k such that g 2, if n i ((gik) =g[ik, (i + 1)k)), then
n i (n(gik) = g[ik, (i + 1)k)).
Then there is a countable set of functions
i 2ik 2k such that g 2,
if i ((gik) = g[ik, (i + 1)k)), then
i ((gik) = g[ik, (i + 1)k)).
Proof. This is a standard argument which we leave to the reader.
Lemma 3.3 LetP be a partial order of size , and assume is aPname for a function
i 2
ik 2k. Then there is a set of size of functionsi 2
ik 2k such that g 2, if i ((gik) = g[ik, (i + 1)k)),then
i ((gik) = g[ik, (i + 1)k)).
Proof. This is wellknown and trivial.
Using the first two of these three lemmata we see that if we iterate -2k-linked forcing over a model V containing a family F 2 such that
() for all countable sets of functions
i 2ik 2k there is g F such
that for all , i ((gik) = g[ik, (i + 1)k)),
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then F still satisfies () in the final extension. We also have
Lemma 3.4 If F satisfies (), then econst2 (k) |F|.
Proof. Simply note F is a witness for econst2 (k). For given a predictor : 2
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Proof. Let P, Q; < be a finite support iteration of ccc forcing such
that each factor Q is forced to be a -(2k 1)-linked forcing notion of sizeless than k for some k 2. Also guarantee we take care of all such forcingnotions by a bookkeeping argument. Then m(-(2k 1)-linked) k isstraightforward. In view of Corollary 3.6 it suffices to prove econst2 (k) kfor all k. So fix k. Note that by stage k of the iteration we have adjoined afamily F of size k satisfying () above with countable replaced by less thank (for example, let F be the collection of Cohen reals added at limit stagesof countable cofinality below k). Show by induction on the remainder ofthe iteration that F continues to satisfy this version of (). The limit stepis taken care of by Lemma 3.2 because no new reals appear at limit steps of
uncountable cofinality. For the successor step, in caseQ is -2
-linked forsome k, use Lemma 3.1, and in case it is not -2k-linked (and thus ofsize less than k), use Lemma 3.3. By Lemma 3.4, e
const2 (k) k follows.
By somewhat changing the above proof, we can dualize Kamos CON(vconst2 >cof(N)) (and thus answer a question of his, see [Ka2]), and reprove his resultas well.
Theorem 3.8 (a) econst2 < add(N) is consistent; in fact, given < = cof(N) is consistent; in fact, given regularuncountable and = > , there is a partial orderP forcingvconst2 =c = and cof(N) = .
Proof. (a) Let P, Q; < be a finite support iteration of ccc forcingsuch that
for even , Q is amoeba forcing,
for odd , Q is a subforcing of some Pk of size less than .
Guarantee that we go through all such subforcings by a bookkeeping argu-ment. Then econst2 is straightforward, as is add(N) = c = . Now notethat amoeba forcing is --linked (like random forcing). Therefore we canapply Lemmata 3.1, 3.2, and 3.3 for all k simultaneously, and see that thereis a family F of size which satisfies the appropriate modified version of ()(such a family is adjoined after the first stages of the iteration).
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(b) First add many Cohen reals. Then make a stage finite support
iteration of amoeba forcing. Again, cof(N) = is clear. vconst2 = c = follows from Lemmata 3.1 and 3.2 using standard arguments.
One can even strengthen Theorem 3.7 in the following way. Say a partialorder P satisfies property Kk if for all uncountable X P there is Y X uncountable such that any k many elements from Y have a commonextension. Property Kk is a weaker relative of -k-linkedness. Let m(Kk)denote the least cardinal such that MA fails for property Kk partial orders.
Lemma 3.9 Assume CH. Pk does not have property K2k . In fact no prop-ertyK2k partial order adds a predictor which k-constantly predicts all ground
model reals.
Proof. List all predictors as {; < 1}. Choose reals f 2 such that does not k-constantly predict f for . Let X = {f; < 1}.
Let P have property K2k . Also let be a Pname for a predictor. Assumethere are conditions p P such that p k-constantly predicts ffrom m onwards. Without loss m = m for all , and any 2
k many phave a common extension. Let T 2 m(-(2k 1)-linked) is
consistent. This, however, is easy, for the forcing Pk is Suslin ccc [BJ] while
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it is wellknown that iterating Suslin ccc forcing keeps numbers like m(-
(2k 1)-linked) small (it even keeps the splitting number s small).The results in this section are related to work of Blass [Bl2, Section 10].
We briefly sketch the connection. Fix k 2. Momentarily call a function : m(-k-linked).
We close this section with a few questions. We have no dual result forTheorem 2.2 so far.
Question 3.11 Is econst2 > sup{econst2 (k); k < } consistent?
Question 3.12 Can econst2 have countable cofinality?
By Theorem 3.7, one of these two questions must have a positive answer.In fact, in view of the proof of Theorem 3.8, econst2 must be
either max{k; k } (in case the set has a max),
or sup{k; k } or its successor (in case the set has no max)
in the model of Theorem 3.7.
References
[BJ] T. Bartoszynski and H. Judah, Set theory. On the structure of thereal line, A K Peters, Wellesley, Massachusetts, 1995.
[Bl1] A. Blass, Cardinal characteristics and the product of countably manyinfinite cyclic groups, J. Algebra 169 (1994), 512-540.
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[Bl2] A. Blass, Combinatorial cardinal characteristics of the continuum, in:
Handbook of Set Theory (A. Kanamori et al., eds.), to appear.
[Br] J. Brendle, Evasion and prediction III Constant prediction anddominating reals, to appear.
[GSh] M. Goldstern and S. Shelah, Many simple cardinal invariants, Arch.Math. Logic 32 (1993), 203-221.
[Kd1] M. Kada, A note on various classes of evasion numbers, unpublishedmanuscript.
[Kd2] M. Kada, Strategic Sacks property and covering numbers for predic-tion, to appear.
[Ka1] S. Kamo, Cardinal invariants associated with predictors, in: LogicColloquium 98 (S. Buss et al., eds.), Lecture Notes in Logic 13 (2000),280-295.
[Ka2] S. Kamo, Cardinal invariants associated with predictors II, J. Math.Soc. Japan 53 (2001), 35-57.
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