Recall that Martin’s Axiom asserts that for every partial order satisfying c.c.c., and for any family of many dense subsets of , there exists a directed subset of such that for all .
Solovay and Tennenbaum proved that Martin’s Axiom is consistent with .
In [Sh:80], Shelah proved that the following weak generalization of Martin’s Axiom is consistent with and CH. If is a partial order satisfying:
- ;
- is -closed;
- is -stationary-c.c.: if is an injective enumeration of conditions, then there exists a regressive map and a club , so that is directed for all ;
- well-met condition: for every , if is nonempty (i.e., and are compatible), then the set has a least element,
then for any family of many dense subsets of , there exists a directed subset of such that for all .
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 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 -c.c. posets whose product is not anymore -c.c. (see here). So, plain -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 is a sequence such that is a club in of order-type for each , then there exists a sequence of local colorings , such that for every global coloring , there exists for which is not bounded in .
The very same proof shows that “not bounded” may be strengthened to “is stationary”. Thus, fix a sequence witnessing the latter. For simplicity, we may assume that for all . Define a notion of forcing , as follows. iff it is a partial function from to , such that:
- is a countable subset of ;
- is a closed bounded subset of ;
- if are in and , then ;
- if are in , then for all .
A condition extends , iff
- ;
- is an end-extension of for all .
Let , where
Then and if happens to be a directed subset of such that for all , then for all , is a club subset of , and is a function! In particular, is nonstationary for all .
Claim 1. is dense for every and .
Proof. Given , we distinguish two cases.
If , then let , and set If , let . As and are countable, it is easy to construct an injective sequence such that for all , and .
Finally, let be such that and,
Claim 2. is -closed.
Proof. Given an increasing sequence in , define such that and is the closure of for all . To see that , we only need to verify that if are in and , then . Fix a large enough such that and are in . Put , then , and by clause (4) above, for . By , we have for . As end-extends whenever and , we infer that , and so by Clause (3), .
Notation. For every , denote
Claim 3. CH entails that has size and satisfies the -stationary-c.c.
Proof. It is a basic counting argument that shows that CH entails a bijection . Next, given a sequence of conditions , we first extend each to a condition satisfying , and for all above . As is a countable partial function from to , we infer from CH and the Engelking-Karlowicz theorem the exisetnce of a sequence such that for each , there is with . Let be the set of all such that for all , . Then is a club, and for all , . Define a function by stipulating , and let . Finally, define a regressive function by letting .
To see that is directed for all , fix arbitrary in with . As and , we have
Set . Then is a closed bounded subset of for all . Also, is a function, since . So, satisfies Clauses (1)–(3) above. Now, as in the end of the proof of Claim 1, it is easy to find with such that is an end-extension of by a singleton for all . Clearly, is stronger than and , let alone and .
Note that any condition admits many pairwise incompatible extensions, hence, CH is also necessary for to satisfy the -c.c.
Claim 4. does not satisfy the well-met condition.
Proof. By iterating the pressing down lemma, we can find a stationary subset of along with ordinals such that for all , the first three elements of are . Pick from . Let for . Then and are compatible elements of , but due to Clause (4) above, is not in , and so the set does not admit a least element.