Martin’s axiom is consistent
If is an uncountable strong limit cardinal such that for any , then it is consistent that and MA. This is shown by using finite support iterated forcing to construct a model of ZFC in which this is true. Historically, this proof was the motivation for developing iterated forcing.
The proof uses the convenient fact that holds as long as it holds for all partial orders smaller than . Given the conditions on , there are at most names for these partial orders. At each step in the forcing, we force with one of these names. The result is that the actual generic subset we add intersects every dense subset of every partial order.
will be constructed by induction with three conditions: for all , , and satisfies the ccc. Note that a partial ordering on a cardinal is a function from to , so there are at most of them. Since a canonical name for a partial ordering of a cardinal is just a function from to that cardinal, there are at most of them.
At each of the steps, we want to deal with one of these possible partial orderings, so we need to partition the steps in to steps for each of the cardinals less than . In addition, we need to include every name for any level. Therefore, we partion into for each cardinal , with each having cardinality and the added condition that implies . Then each name for a partial ordering of is assigned some index , and that partial order will be dealt with at stage .
Formally, given for , can be constructed and the names for partial orderings of each cardinal enumerated by the elements of . for some and , and so some canonical name for a partial order of has already been assigned to .
Since is a name, it is also a name, so can be defined as if satisfies the ccc and by the trivial partial order otherwise. Obviously this satisfies the ccc, and so does as well. Since is either trivial or a cardinal together with a canonical name, . Finally, .
Proof that holds for
Lemma: It suffices to show that holds for partial order with size
Suppose is a partial order with and let be dense subsets of . Define functions for with (obviously such elements exist since is dense). Let be a function such that whenever and are compatible. Then pick some element and let be the closure of under and with the same ordering as (restricted to ).
Since there are only functions being used, it must be that . If then and clearly , so each is dense in . In addition, is ccc: if is an antichain in and then are incompatible in . But if they were compatible in then would be an element of , so they must be incompatible in . Therefore is an antichain in , and therefore must have countable cardinality, since satisfies the ccc.
By assumption, there is a directed such that for each , and therefore holds in full. ∎
Now we must prove that, if is a generic subset of , some partial order with and are dense subsets of then there is some directed subset of intersecting each .
If then additional elements can be added greater than any other element of to make , and then since there is an order isomorphism into some partial order of , assume is a partial ordering of . Then let .
Take canonical names so that , and for each and:
For any there is a maximal antichain such that if then either or and another maximal antichain such that if then either or . These antichains determine the value of those two formulas.
Then, since and for , it must be that , so is regular. Then , so , and therefore the names and are also names.
Lemma: For any , is a generic subset of
First, it is directed, since if then there is some such that , and therefore and .
Also, it is generic. If is a dense subset of then is dense in , since if then there is some , but then is compatible with , so . Therefore there is some , and so . ∎
Since and are names, and , so
Then there must be some such that
Let be a maximal antichain of such that , and define as a name with for each and if where and . That is, when and otherwise. Then this is the name for a partial ordering of , and therefore there is some such that , and . Since , .
The relationship between Martin’s axiom and the continuum hypothesis tells us that . Since was less than in , and since adds at most elements, it must be that .
|Title||Martin’s axiom is consistent|
|Date of creation||2013-03-22 13:21:59|
|Last modified on||2013-03-22 13:21:59|
|Last modified by||mathcam (2727)|