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.