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
Also, since the cofinality of is uncountable, we have Hausdoff’s result that if
, then
. I have addressed both these computations in my lecture notes for Topics in Set Theory, see here and here.
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.
Now define
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.