Generalizations of Martin’s Axiom and the well-met condition

Recall that Martin’s Axiom asserts that for every partial order P satisfying c.c.c., and for any family D of <20 many dense subsets of P, there exists a directed subset G of P such that GD for all DD.

Solovay and Tennenbaum proved that Martin’s Axiom is consistent with 20>1.

In [Sh:80], Shelah proved that the following weak generalization of Martin’s Axiom is consistent with 21>2 and CH. If P is a partial order satisfying:

  1. |P|<21;
  2. P is σ-closed;
  3. P is 2-stationary-c.c.: if {pαα<2} is an injective enumeration of conditions, then there exists a regressive map f:ω2ω2 and a club Dω2, so that {pααDEω1ω2,f(α)=i} is directed for all i<ω2;
  4. well-met condition: for every p,qP, if {rPpr & qr} is nonempty (i.e., p and q are compatible), then the set has a least element,

then for any family D of <21 many dense subsets of P, there exists a directed subset G of P such that GD for all DD.

Let us look at requirements (1)–(4).

Requirement (1) is a rather natural restriction (though, this restriction may sometime be waived via tail-catching, or by reflection arguments in the presence of large cardinals).

Requirement (2) has to do with the fact that any forcing axiom for meeting more than 1 many dense sets which will be compatible with CH, will have to be applicable only to forcing notions that do not add reals.

Requirement (3): Recall that Martin’s Axiom implies that the product of any two c.c.c. posets is again c.c.c., whereas there are ZFC examples of two 2-c.c. posets whose product is not anymore 2-c.c. (see here). So, plain 2-c.c. must be replaced by a stronger variation.

Today, I learned from Ashutosh Kumar that [Sh:1036] offers a nice and relatively simple justification of Requirement (4). Luckily, the example builds upon a theorem of Shelah which I already presented in a previous blog post (see also Section 2 of this survey paper, for motivation):

Theorem (Shelah). If CδδEω1ω2 is a sequence such that Cδ is a club in δ of order-type ω1 for each δ, then there exists a sequence of local colorings fδ:Cδ2δEω1ω2, such that for every global coloring f:ω2ω2, there exists δEω1ω2 for which {αCδf(α)fδ(α)} is not bounded in δ.

The very same proof shows that “not bounded” may be strengthened to “is stationary”. Thus, fix a sequence fδδEω1ω2 witnessing the latter. For simplicity, we may assume that min(Cδ)=1 for all δEω1ω2. Define a notion of forcing P, as follows. pP iff it is a partial function from Eω1ω2 to Pω1(ω2), such that:

  1. dom(p) is a countable subset of Eω1ω2;
  2. p(δ) is a closed bounded subset of Cδ;
  3. if δ1<δ2 are in dom(p) and αp(δ1)p(δ2), then fδ1(α)=fδ2(α);
  4. if δ1δ2 are in dom(p), then β<max(p(δ1)) for all βCδ1Cδ2.

A condition q extends p, iff

  1. dom(q)dom(p);
  2. q(δ) is an end-extension of p(δ) for all δdom(p).

Let D:={Dα,δα<δ,δEω1ω2}, where Dα,δ:={pPδdom(p) & max(p(δ))>α}.

Then |D|=2 and if G happens to be a directed subset of P such that GD for all DD, then for all δEω1ω2, cδ:={p(δ)pG} is a club subset of Cδ, and f:={fδcδδEω1ω2} is a function! In particular, {αCδf(α)fδ(α) or αdom(f)} is nonstationary for all δEω1ω2.

Claim 1. Dα,δ is dense for every δEω1ω2 and α<δ.
Proof. Given qPDα,δ, we distinguish two cases.
If δdom(q), then let α:=min(Cδ(max(q(δ){α})+1)), and set p:=(q(dom(p){δ})){(δ,q(δ){α})}.If δdom(q), let t:={CβCγβγ in dom(q){δ}}. As t and dom(q) are countable, it is easy to construct an injective sequence αββdom(q){δ}βdom(q){δ}Cβ such that αβ>sup(q(β)(tβ)) for all βdom(q), and αδ>sup(α(tδ)).
Finally, let pP be such that dom(p)=dom(q){δ} and, p(β)={q(β){αβ},βdom(q){αδ},otherwise.◻

Claim 2. P is σ-closed.
Proof. Given an increasing sequence pnn<ω in P, define pP such that dom(p):=n<ωdom(pn) and p(δ) is the closure of {pn(δ)n<ω,δdom(pn)} for all δdom(p). To see that pP, we only need to verify that if δ1<δ2 are in dom(p) and αp(δ1)p(δ2), then fδ1(α)=fδ2(α). Fix a large enough N<ω such that δ1 and δ2 are in dom(pN). Put β:=sup(Cδ1Cδ2), then β<δ1<δ2, and by clause (4) above, β<max(pN(δi)) for i<2. By αCδ1Cδ2, we have αβ<max(pN(δi)) for i<2. As pn(δi) end-extends  pN(δi) whenever Nn<ω and i<2, we infer that αpN(δ1)pN(δ2), and so by Clause (3), fδ1(α)=fδ2(α). ◻

Notation. For every qP, denote gq:={fδq(δ)δdom(q)}.

Claim 3. CH entails that P has size 2 and satisfies the 2-stationary-c.c.
Proof. It is a basic counting argument that shows that CH entails a bijection ψ:ω2P. Next, given a sequence of conditions pααEω1ω2, we first extend each pα to a condition qα satisfying αdom(qα), and max(qα(δ))>α for all δdom(qα) above α.  As gqα is a countable partial function from ω2 to 2, we infer from CH and the Engelking-Karlowicz theorem the exisetnce of a sequence φi:ω22i<ω1 such that for each αEω1ω2, there is iα<ω1 with gqαφiα. Let D be the set of all δ<ω2 such that for all α<δ, {pPdom(p)α}ψ[δ]. Then D is a club, and for all δDEω1ω2, {pPdom(p)δ}ψ[δ]. Define a function h:Eω1ω2ω2 by stipulating h(δ):=sup({dom(qα)αEω1δ}), and let E:={δDh[δ]δ}. Finally, define a regressive function f:EEω1ω2ω1×ω2 by letting f(δ):=(iδ,ψ1(qδδ)).

To see that {pααEEω1ω2,f(α)=i} is directed for all iω1×ω2, fix arbitrary δ1<δ2 in EEω1ω2 with f(δ1)=f(δ2).  As qδ1δ1=qδ2δ2 and δ1δ2E, we have qδ1(dom(qδ1)dom(qδ2))=qδ2(dom(qδ1)dom(qδ2)).

Set p:=qδ1qδ2. Then p(α) is a closed bounded subset of Cα for all αdom(p). Also, gp is a function, since gpφiδ1=φiδ2. So, p satisfies Clauses (1)–(3) above. Now, as in the end of the proof of Claim 1, it is easy to find pP with dom(p)=dom(p) such that p(α) is an end-extension of p(α) by a singleton for all αdom(p). Clearly, p is stronger than qδ1 and qδ2, let alone pδ1 and pδ2. ◻

Note that any condition pP admits c many pairwise incompatible extensions, hence, CH is also necessary for P to satisfy the 2-c.c.

Claim 4. P does not satisfy the well-met condition.
Proof. By iterating the pressing down lemma, we can find a stationary subset S of Eω1ω2 along with ordinals α0<α1<α2 such that for all δS, the first three elements of Cδ are {α0,α1,α2}. Pick δ0<δ1 from S. Let pi:={(δi,{αi})} for i<2. Then p0 and p1 are compatible elements of P, but due to Clause (4) above, p0p1 is not in P, and so the set {rPp0r & p1r} does not admit a least element. ◻

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