saharon shelah and juris steprans- maximal chains in ^omega-omega and ultrapowers of the integers

8/3/2019 Saharon Shelah and Juris Steprans- Maximal Chains in ^omega-omega and Ultrapowers of the Integers

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MAXIMAL CHAINS IN AND ULTRAPOWERS OFTHE INTEGERS

SAHARON SHELAH AND JURIS STEPRANS

Abstract. Various questions posed by P. Nyikos concerning ul-trafilters on and chains in the partial order (,


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2 SAHARON SHELAH AND JURIS STEPRANS

If U is an ultrafilter then the integers modulo U will refer to theultrapower of the integers with respect to U and will be denoted by/U. If U is an ultrafilter then C will be said to be unboundedmodulo U if, letting [f]U represent the equivalence class of f in/U, the set {[f]U : f C} is unbounded in the linear order /U.The least ordinal which can be embedded cofinally in a linear orderingL is denoted by cof(L) cof( /U) will be an important invariant of

U in the following discussion.For reference, here are Nyikos axioms (throughout C refers to a

maximal chain of nondescending functions in and U refers to anultrafilter)

Axiom 1 (U)(C)(C is unbounded modulo U)

Axiom 2 (C)(U)(C is unbounded modulo U) Axiom 3 (U)(C)(C is unbounded modulo U) Axiom 4 (U)(C)(C is unbounded modulo U) Axiom 5 (C)(U)(C is unbounded modulo U) Axiom 5.5 (U)(cof( /U) = b) Axiom 6 (C)(U)(C is unbounded modulo U) Axiom 6.5 (C)(U)(cof(C) = cof( /U))

Various implications and non-implications between these axioms areestablished in [5]. As well, it is observed that Axiom 2 is equivalent tothe equality b = d.

2. Non-monotone functions

The definition of chains as


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MAXIMAL CHAINS IN AND ULTRAPOWERS OF THE INTEGERS 3

Fortunately, many of these axioms turn out to be equivalent andothers are simply false. The following simple observation of Nyikos can

be used to see this.

Lemma 2.1. There is a mapping : such that

(f) is strictly increasing for every f f (f) for every f if f g and f = g then (f) m such that

Mi=0 g(i) >

mi=0(f(i) g(i)). Hence

(f)(j) < (g)(j) for all j M.Finally observe that if f(n) < g(n) and f is nondecreasing then

f(i) < g(n) for each i n. Hence (f)(n) f(n)(n + 1) + n h(n))}

and observe that V is open. From the Open Colouring Axiom it followsthat there are only two possibilities.

The first is that there is a partition = nXn such that [X]2V = for each n . In this case there must be some n such thatXn is cofinal in . Choose such that X = X and let

f(n) = min{h(n) : }. Now, if > and n X then thereis some such that n X and hence n dom(f). Moreover, if and n X then g(n) h(n) and so g(n) f(n). So f is thedesired function.

The second possibility is that there is X []1 such that [X]2 V.Since 2 it is possible to choose some such that X . It isthen possible to choose M , g : M and Y [X]1 such that

if Y then X \ M X if Y and n X \ M then g(n) g(n) h(n) g M = g

Then if and {, } [Y]2 and n X X it must be thateither n M or n < M. In the first case it follows that n X andso g(n) g(n) h(n). In the second case it may be concluded thatg(n) = g(n) = g(n) h(n). It follows that {, } / V which is acontradiction.

Theorem 2.3. The conjunction of Axiom 2 and the Open ColouringAxiom implies Axiom 5(, ).

Proof: To begin, recall that it was shown in [5] that Axiom 2implies that b = d. Hence it is possible to choose a (, )-chain

{d : d} which is also a dominating family in . Also, if C is any(, )-chain then C is of the form {g : d}. Define E(, ) ={n : g(n) d(n)}.

Next, let {M : d} be a sequence of elementary submodels of(H(c+), ) such that

|M | < d for each d {g : d} M and {d : d} M for each d M


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M M for each d

and let () = M d. Define F to be the filter generated by

{E((), ( + 1)) : d and is odd }

and observe that if F is a proper filter then C will be cofinal in modulo U for any ultrafilter extending F.

Hence it suffices to show that F is proper. To this end let F be thefilter generated by

{E((), ( + 1)) : and is odd }

and prove by induction that each F is proper. Moreover, it will beshown by induction that F I(C) = . If = 0, is odd or is alimit then there is nothing to do so suppose that = + 1, where

is odd, and that F is a proper filter such that F I(C) = .Notice that F M because is odd. Therefore it suffices to show

that for each B I(C)+ there is some d such that E((), )

B I(C)+ the reason being that the elementarity of M+1 willguarantee that E((), ( + 1)) B I(C)+ for each B I(C)+.Elementarity also assures that it may as well be assumed that B M.But if there is some B I(C)+ such that E((), ) B I(C) foreach c then it is possible to find h such that

dom(h) = B E((), ) for each d h(n) g(n) for every n B E((), ) and for each d

g dom(h)

h for each cIt follows that {(h, g BE((), )) : d} satisfies the hypothesisof Lemma 2.2. Since the Open Colouring Axiom is being assumed,there is a function f such that g f B E((), ) for each d. It follows that for each d there are only finitely many n Bsuch that g(n) > max{d(n), f(n)} contradicting that B / I(C).

Notice that it is shown in [8] that the Proper Forcing Axiom impliesthe hypothesis of Theorem 2.3. Moreover it is a Corollary that Ax-iom 5(, ) does not imply Axiom 1 because it is easy to check that

Martins Axiom and hence the Proper Forcing Axiom impliesthat Axiom 1 fails. In particular, it is possible to inductively define a(


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Axiom 5(


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Proof: In [5] it is shown that Axiom 2 is equivalent to the equalityb = d. Since b is regular it follows that d is regular. Moreover, ifC

is an unbounded (,N)-chain then cof(C) = d. Since C consists ofnondecreasing functions it is clear that {c A : c C} is unboundedfor each infinite set A. Hence, by Lemma 2.3, it follows that there isan ultrafilter U such that C is unbounded modulo U.

3. Oracle Chain Conditions and Locally Cohen PartialOrders

It will be shown that there is a model of set theory where Axiom 6.5

fails. This answers the first two questions in Problem 5 of [5]. C.Laflamme has remarked that in some models of NCF (see [2] for anoverview of this area) Axiom 6.5 fails as well because it is possible toprovide a classification of chains in these models. The restriction tochains does not play an important role in this theorem and, in fact,the theorem is slightly stronger than required at least formally because of this.

Theorem 3.1. There is a model where cof( /U) = for everyultrafilter U but every unbounded subset of has an unbounded subsetof size 1.

Proof: The plan of the proof is to start with a model V in which1 and 2(1) in other words, the trapping of subsets of2 occursat ordinals of cofinality 1 in 2 both hold. In this model a finitesupport iteration {(P,Q) : 2} will be constructed along with asequence of oracles [6] {M : 2} more precisely, M is a P-name for an oracle. The oracles will be chosen so that if{g : 1} isa P-name, guessed by the 2(1) sequence, for an unbounded subsetof then M is chosen so that ifQ is any partial order satisfying theM-chain condition then forcing with Q does not destroy the unbound-edness of {g : 1}. Provided that P2/P satisfies the M-chain

condition, it will follow that every unbounded subset of has cofi-nality 1 because every unbounded subset is reflected at some initialstage by the 2(1) sequence. The rest of the result will follow onceit is shown how to construct Q satisfying the M-chain condition andadding an upper bound to any given sequence from some ultrapowerof the integers.

The construction of {(P,Q) : 2} and {M : 2} is, ofcourse, done by induction. If is a limit then P is simply the direct


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limit of{P : }. The construction ofM and Q does not dependon whether or not is a limit.

Given P, use the results of pages 124 to 127 of [6] to find a P-namefor a single oracle N such that ifQ satifies the N-chain condition thenit satisfies the M chain condition for each . Let C be the setguessed by the 2(1) sequence at . If C is not a P-name for anunbounded subset of then let M = N. Otherwise, use Lemma2.1 on page 122 of [6] to find an oracle M such that ifQ satisfies theM-chain condition then the subset C remains unbounded afterforcing with Q. The use of Lemma 2.1 requires checking that ifC is an unbounded chain then adding a Cohen real will not destroy itsunboundedness. This is a result of the folklore which can be found in[7]. Then use the results of pages 124 to 127 of [6] to find a single oracleM such that any Q which satisfies the M-chain condition will alsosatisfy the M-chain condition and the N-chain condition.

Suppose that the 2(1) sequence has also trapped a filter U which is an ultrafilter in the intermediate generic extension by P and an increasing sequence {f : 1} in the reduced power of theintegers modulo U. (So it is being assumed that, by some coding, the2(1) sequence traps triples of sets the first component of thetriple at is a candidate for C in the construction ofM while thesecond and third components are candidates for the ultrafilter U andthe sequence {f : 1}.) The only thing left to do is to construct

Q satisfying the M-chain condition and adding an upper bound for{f : 1} in the reduced power modulo U.

Let M = {M : 1}. The partial order Q is constructed

by induction on 1 in VP it will be similar to the forcing which

adds a dominating real but with extra side conditions. In particular,a sequence of partial functions {S : 1}

is constructed byinduction on 1 and Q

is defined to be the set of all pairs (F, ) such

that F : k is a finite partial function and []


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which belongs to M remains predense in Q+1 . This, of course, will

ensure that Q = Q1 satisfies the M-chain condition.

Suppose that {S : } have been constructed. Let A be the set ofall of the dense open subsets ofQ which belong to M this includes

all those dense open subsets ofQ which belong to M some .Choose h to be some function which dominates all members ofM; in other words, if g M then g h. Let {(Ai, (Fi, i)) :i } enumerate A Q . Now choose, by induction on , integers{Ki : i } such that Ki < Ki+1 and K0 = 0. Given Ki, defineFij Fj for each j i such that if dom(Fj) Ki then dom(F

ij ) = Ki,

(Fj, j) (Fij , j) and Fij (k) h(k) if k dom(F

ij \ Fj). Now choose

(Fij , ij) Aj such that (F

ij ,

ij) (F

ij , ). Let Ki+1 be such that

dom(Fij ) Ki+1 for each j i.Let Xm = i[K2i+m, K2i+m+1) for m 2 and note that there exists

m 2 such that Xm U . Let S = h Xm ; the reason being thatU is an ultrafilter in VP and X0 VP . To see that this definitionof S ensures that every dense open subset of Q

which belongs to

M remains predense in Q+1 let (F, ) Q

+1 and let D M be

dense open in Q . It follows that (F, \ {}) Q . To simplify

notation assume that m = 1. Choose j such that dom(F) K2j andsuch that (D, (F, \ {}) = (Ak, (Fk, k)) for some k 2j. Since

dom(F2jk ) K2j it follows that F2jk (n) h(n) if n dom(F

2jk \ Fk).

Moreover [K2j , K2j+1) domS = . Hence (F2jk , 2jk ) is compatiblewith (F, ).

The methods of the previous theorem can also be used to show thatit is consistent that Axiom 6 holds but Axiom 5.5 fails. In establishingthis it will be helpful to introduce the following definition.

Definition 3.1. A partial order (P, ) will called locally Cohen if foreveryX [P]0 there is Y [P]0 such thatX Y andY is completelyembedded in P in other words, if A Y is a maximal antichain inthe partial order (Y, Y Y) then it is also maximal in (P, ).

The notion of locally Cohen partial orders has already been isolatedand investigated by W. Just in [4] who refers to locally Cohen partialorders as harmless. The motivation of Just was that any locally Cohenforcing satisfies the oracle chain condition for every oracle.

Let S() be the canonical partial order for adding a scale of length in with finite conditions. To be precise, a condition p belongs to


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S() if and only if p : p np is a function and p []


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in the model V[G, H] where G is generic over P. There is no problemin doing this because S() is locally Cohen and hence satisfies the M

chain condition for each indeed, S() satisfies any oracle chaincondition. It is therefore easy to use Claim 3.3 on page 127 of [6] toobtain M exactly as in the proof of Theorem 3.1.

Let G be P2 generic over V[H]. Exactly as in the proof of Theorem3.1, it can be shown that the cofinality of any ultrapower of the integersis 2 while b = 1 in V[G, H]. On the other hand, the fact that{c : 2} is unbounded follows from genericity and the fact thatS( + 1) = S() D({c : }) so c is not dominated by anyfunction from V[G, H] where G is the restriction of G to P.

Corollary 3.1. Axiom 6 does not imply Axiom 5.5.

To see that the model constructed in Theorem 3.2 is a model ofAxiom 6 but not of Axiom 5.5 observe first that Axiom 5.5 fails becauseb = 1 while the cofinality of any ultrapower of the integers is 2.On the other hand, there is (, )-chain of length 2 d. UsingLemma 2.1 it is possible to construct from this a (


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possible to use subsets of 2 to code such names for sets of reals. If Xis some name in a suitable partial order for a subset of [ ]0 then

c(X) will denote the subset of 2 which codes it while if X 2 thend(X) will denote the name it codes. The details of the coding will notbe important. Define a partial order on 2 by if and only ifc(D) = c(D ).

Now construct P as a finite support iteration of {P : 2} suchthat P+1 = P C D where C adds a Cohen real, A : 2 andD is some partial order which has yet to be defined. At the same time,construct a partial function : 2 2 so that if is in the domainof then 1 P d(D) is an ultrafilter and () is the minimumordinal such that () / {() : } and such that g() is a Pname.

Given P, define D by p D if and only if

p = (fp, p)

fp

p []


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It remains to be shown that P satisfies the countable chain conditionand that b = 1 after forcing with P. Both these facts will follow once

it has been shown that P is locally Cohen. To this end, it is worthobserving that P has a dense set of conditions which are somewhatdetermined a condition p will be said to be somewhat determined ifthe support of p is p [2]


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q P h0 = H0 and h1 = H1 q P ga() I0 = G for some G : I0 (if / W this can

be ignored) q P = q P A

1 {k} d(D)

That it is possible to arrange for the first two clauses follows from thefact that ga() is a P name and so any information about it can beobtained without changing h0 or h1. To satisfy the last clause, usethe fact that , which follows because is no longer maxinal in ( + 1)2.

Now define W = (W \ {}) {} and observe that W is a set ofmaximal elements in ( ). Define v = v W {(, k + 1mod 2)} and a = a W {(, ())} and observe that both a and vare still functions of the right type. Then use the induction hypothesison to find q q which is somewhat determined and such that thisis witnessed by n(q) and, such that for each W there is M() and G() : M() such that q P ga() M() = G() and,

moreover, hp,0 (i) = v() provided that i n(q) \ M(). Without loss

of generality, n(q) I0 and n(q) I1.Then let p = (q, (h0, h

1, ) where h

0 : n(q

) 2 is the extensionof H0 to n(q

) such that h0(i) = v() if i n(q) \ I0 and h1 is the

extension of H1 such that

h1(i) = max({f(i) : and }{G(())(i) : and })

for i n(q) \ I1. Notice that maximum is taken over actual integersrather than names for integers. The definition also respects the re-quirements of extension in the partial order P. Defining M() = I0and G() = G satisfies the extra induction hypothesis.

To see that P is locally Cohen let X [P]0. Let M be a countableelementary submodel of (H(3),P, {D : 2}, , X). It suffices toshow that P M is completely embedded in P. To see that this is so,

let A PM be a maximal antichain in PM and let p P; withoutloss of generality p can be assumed to be somewhat determined and,moreover, it may be assumed that this is witnessed by n(p). Let p be

defined so that dom(p) = dom(p)M and p() = (hp,0 , hp,1 ,

pM)for dom(p). Note that p M P. Hence there is q A andq P M such that q q and q p without loss of generality itmay be assumed that q is determined and this is witnessed by n(q). Itmust be shown that p and q are compatible.


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As in the proof that S() is locally Cohen, for dom(p) \ dom(q)extend hp,1 to h

1 by defining

h1 (m) = max({hq,1 (m) : and dom(q)} . . .

. . . {Gq()(m) : and dom(q) and m Mq() and A(m) = k(q , , )})

for m n(q) \ n(p), recalling that Gq(), Mq() and k(q , , ) arewitnesses to the fact that q is somewhat determined. This will certainlyassure that if and are in the domain of q and thenhq,1 (m) h

1 (m) h

q,1 (m) the fact that h

1(m) h

q,1 (m) follows

from the defintion of the third coordinates in p. Also, g()(m) h1 (m) if A(m) = k(p,,) will be true if m M() by construction.

Next, if dom(p) \ M, dom(p) M and then defineh0 (m) = 1 + k(p,,) mod 2. If this can be done then, ifm n(p),it is not necessary for hq,1 (m) to be greater than g()(m). If there isno dom(p) M such that , do not extend hp,0 at all. Noticethat in this last case it is still possible that there is some dom(q)such that . However, because it is only necessary for hq,1 (m) tobe greater than g()(m) in case q,, this will cause no problemsbecause q, M if M.

What must be checked, though, is that no conflict arises as a resultof this definition of h0 . After all, it is conceivable that and but k(q,, ) = k(q,, ). To see that this does not happen,

suppose that ,

, k(p,,) = k(p,,

) and {,

} M. Itfollows that if

= sup{ : and and dom()}

then M. Hence and so there is some such that , and dom() such that . Hence A1 {0} is measuredby the ultrafilter d(D). Since d(D) d(D) and d(D) d(D) itfollows that k(p,,) = k(p,,).

4. Open Questions

Table 1 of implications and non-implications summarizes the knownresults about the axioms discussed in this paper. The key to under-standing Table 1 is that

if there is a in the entry in the row headed by Axiom R andthe column headed by Axiom C then Axiom R implies AxiomC


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implies Axiom C is a trivial implication by trivial is meant some-thing which can be deduced by considering the quantifiers in the rele-

vant axioms. An N in that entry means that either the implication ornon-implication can be found in [5]. The enumeration of the followinglist corresponds to the numbered entries in Table 2. So, for example,the second entry of this list refers to Theorem 2.4 because this is thereason there is an in Table 1 in the row corresponding to Axiom 2and the column corresponding to Axiom 5(N).

(1) Theorem 2.1(2) Theorem 2.4(3) The fact that Axiom 4 does not imply Axiom 5(


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7. J. Steprans, Combinatorial consequences of adding Cohen reals, Submitted toProceedings of the 1991 Jerusalem Conference Set Theory of the Reals, 19.

8. S. Todorcevic, Partition problems in topology, vol. 84, American MathematicalSociety, Providence, 1989.

Institute of Mathematics, Hebrew University, Jerusalem, Givat Ram,

Israel and Department of Mathematics, Rutgers University, New Brunswick,

New Jersey

Department of Mathematics, York University, 4700 Keele Street,

North York, Ontario, Canada M3J 1P3

saharon shelah and juris steprans- maximal chains in ^omega-omega and ultrapowers of the integers

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