For the second talk (and a link to the first one), see here. The third talk took place on September 28.

In the second case, we fix an with . We can clearly assume that is infinite, and it easily follows that . This is because any can be coded by the pair , and there are only many possible values for the second coordinate.

In particular, is infinite, and we can fix a partition of into countably many pieces, each of size . Recall that we are assuming that and have fixed a set cofinal in of smallest possible size. We have also fixed a perfect information winning strategy for II, and an with for all .

For each , fix a surjection .

We define as follows:

- Given , let
,

and

such that ,

and set .

- Suppose now that , that , and that there is an such that , , and . Let
be such that ,

and

and set .

- Define in other cases.

A straightforward induction shows that is winning. The point is that in a run of the game where player II follows :

- Player II’s moves code the part that lies outside of of player II’s moves in a run of the game following where I plays sets covering the sets in the original run. For this, note that at any inning there is a unique index such that player II’s move covers , is disjoint from , and meets in a set that is neither empty nor all of , and this codes the inning of the game, and the piece of player II’s move in codes the history of the run played so far.
- is eventually covered completely, so in particular the parts inside of player II’s responses in the run are covered as well.

This completes the proof of Theorem 1.

By way of illustration, consider the case where is the ideal of finite sets of some set . Then whether II has a winning coding strategy turns into the question of when it is that . This certainly holds if or if . However, it fails if .

This example illustrates how player II really obtains an additional advantage when playing in rather than just in . To see that this is the case, consider the same as above with . This is an instance of the *countable-finite game*. We claim that II has a winning coding strategy in this case. To see this, consider a partition of into countably many sets with . For each , pick a winning coding strategy for the countable-finite game on , and define a strategy in so that for each it simulates a run of the game with II following , as follows: In inning , II plays on for ; player I’s moves in the “-th board” are the intersection with of I’s moves in , and I’s first move occurred at inning . (II can keep track of in several ways, for example, noticing that, following the proof of Theorem 1 produces coding strategies that never play the empty set.)

Note that this strategy is not winning in , the difference being that there is no guarantee that (for any ) the first moves of I in the -st board are going to be covered by II’s responses. On the other hand, the strategy is winning in , since, no matter how late one starts to play on the -th board, player I’s first move covers I’s prior moves there (and so, II having a winning coding strategy for the game that starts with this move, will also cover those prior moves).

The first place where this argument cannot be continues is when . However, suffices to see that player I has a winning strategy in in this case, and so we can continue. This illustrates the corollary stated in the first talk, that suffices to guarantee that II always has a winning coding strategy in .

The natural question is therefore how much one can weaken the assumption, and trying to address it leads to Theorem 2, which will be the subject of the next (and last) talk.

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