For the first talk, see here. The second talk took place on September 21.
We want to prove of Theorem 1, that if
, then II has a winning coding strategy in
.
The argument makes essential use of the following:
Coding Lemma. Let
be a poset such that for all
,
Suppose that
. Then there is a map
such that
Proof. Note that is infinite. We may then identify it with some infinite cardinal
. It suffices to show that for any partial ordering
on
as in the hypothesis, there is a map
such that for any
, there is a
with
such that
.
Well-order in type
, and call
this ordering. We define
by transfinite recursion through
. Given
, let
be the set of its
-predecessors,
.
Our inductive assumption is that for any pair , we have chosen some
with
, and defined
. Let us denote by
the domain of the partial function we have defined so far. Note that
. Since
has size
, it must meet
. Take
to be least in this intersection, and set
, thus completing the stage
of this recursion.
At the end, the resulting map can be extended to a map with domain
in an arbitrary way, and this function clearly is as required.
Back to the proof of . Fix a perfect information winning strategy
for II in
, and a set
cofinal in
of least possible size. Pick a
such that for all
we have
.
Given , let
. Now we consider two cases, depending on whether for some
we have
or not.
Suppose first that for all
. Then the Coding Lemma applies with
in the role of
, and
as chosen. Let
be as in the lemma.
We define as follows:
- Given
, let
be such that
, and set
.
- Given
with
, let
be such that
, and set
.
Clearly, is winning: In any run of the game with II following
, player II’s moves cover their responses following
, and we are done since
is winning.
The second case, when there is some with
, will be dealt with in the next talk.
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