For the third talk (and a link to the second one), see here. The fourth talk took place on October 12.
We want to show the following version of Theorem 2:
Theorem. Suppose is a singular strong limit cardinal of uncountable cofinality. Then the following are equivalent:
- For each ideal on , player II has a winning coding strategy in .
Since has uncountable cofinality, option 2 above is equivalent to saying that the instance of corresponding to holds.
Before we begin the proof, we need to single out some elementary consequences in cardinal arithmetic of the assumptions on . First of all, since is singular strong limit, then for any cardinal , we have that
We are ready to address the Theorem.
Proof. We use Theorem 1. If option 1. fails, then there is an ideal on with .
Note that , and . Moreover, if , then since, otherwise,
So and then, by Hausdorff, in fact , and option 2. fails.
Suppose option 2. fails and let , so and . We use to build an ideal on with .
For this, we use that there is a large almost disjoint family of functions from into . Specifically:
Lemma. If is singular strong limit, there is a family with and such that for all distinct , we have that .
In my notes, I have a proof of a general version of this result, due to Galvin and Hajnal, see Lemma 12 here; essentially, we list all functions , and then replace them with (appropriate codes for) the branches they determine through the tree . Distinct branches eventually diverge, and this translates into the corresponding functions being almost disjoint.
Pick a family as in the lemma, and let be a subfamily of size . Let . We proceed to show that and use to define an ideal on as required.
First, obviously . Since and , it follows that , or else , since is strong limit.
Clearly, is an ideal. We claim that . First, each singleton with is in , so . Define by . Since the functions in are almost disjoint, it follows that is 1-1. Let be the image of . By construction, is cofinal in . But then
where the first inequality follows from noticing that any has size at most . It follows that , as claimed.
Finally, we argue that , which completes the proof. For this, consider a cofinal , and a map such that for all , we have .
Since is cofinal in , it follows that is cofinal in . But this gives the result, because
and we are done.