Set theory seminar – Marion Scheepers: Coding strategies (III)

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 X\in J with {}|J(X)|<|J|. We can clearly assume that S is infinite, and it easily follows that {}|{\mathcal P}(X)|=|J|. This is because any Y\in J can be coded by the pair (Y\cap X,X\cup(Y\setminus X)), and there are only {}|J(X)| many possible values for the second coordinate.

In particular, X is infinite, and we can fix a partition X=\bigsqcup_n X _n of X into countably many pieces, each of size {}|X|. Recall that we are assuming that \text{cf}(\left< J\right>,{\subset})\le|J| and have fixed a set H cofinal in \left< J\right> of smallest possible size. We have also fixed a perfect information winning strategy \Psi for II, and an f:\left< J\right>\to H with A\subseteq f(A) for all A.

For each n, fix a surjection f:{\mathcal P}(X_n)\setminus\{\emptyset,X_n\}\to{}^{<\omega}H.

We define F:J\times\left< J\right>\to J as follows:

  1. Given O\in \left< J\right>, let

    A=\Psi(\left<f(O)\right>)\setminus X,


    B\in{\mathcal P}(X_0)\setminus\{\emptyset, X_0\} such that f_0(B)=\left< f(O)\right>,

    and set F(\emptyset,O)=A\cup B.

  2. Suppose now that (T,O)\in J\times\left< J\right>, that T\ne\emptyset , and that there is an n such that \bigcup_{j<n} X_j\subseteq T, T\cap X_n\ne\emptyset,X_n, and T\cap\bigcup_{k>n}X_k=\emptyset. Let

    B\in{\mathcal P}(X_{n+1})\setminus\{\emptyset,X_{n+1}\} be such that f_{n+1}(B)=f_n(T\cap X_n){}^\frown\left< f(O)\right>,


    A=\Psi(f_{n+1}(B))\setminus X

    and set F(T,O)=A\cup\bigcup_{j\le n}X_j\cup B.

  3. Define F(T,O)=\emptyset in other cases.

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

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

This completes the proof of Theorem 1. \Box

By way of illustration, consider the case where J is the ideal of finite sets of some set S. Then whether II has a winning coding strategy turns into the question of when it is that \text{cf}({\mathcal P}_{\aleph_1}(S))\le|S|. This certainly holds if {}|S|={\mathfrak c} or if {}|S|<\aleph_\omega. However, it fails if {}|S|=\aleph_\omega.

This example illustrates how player II really obtains an additional advantage when playing in WMG(J) rather than just in RG(J). To see that this is the case, consider the same J as above with {}|S|=\aleph_\omega. 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 S into countably many sets S_n with {}|S_n|=\aleph_n. For each n, pick a winning coding strategy \sigma_n for the countable-finite game on S_n, and define a strategy in WMG(J) so that for each n it simulates a run of the game WMG(J\cap{\mathcal P}(S_n) with II following \sigma_n, as follows: In inning n, II plays on S_i for i\le n; player I’s moves in the “i-th board” are the intersection with S_i of I’s moves in WMG(J), and I’s first move occurred at inning i. (II can keep track of n 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 RG(J), the difference being that there is no guarantee that (for any i) the first i moves of I in the (i+1)-st board are going to be covered by II’s responses. On the other hand, the strategy is winning in WMG(J), since, no matter how late one starts to play on the i-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 |S|=\aleph_{\omega_1}. However, {\sf GCH} suffices to see that player I has a winning strategy in RG(J) in this case, and so we can continue. This illustrates the corollary stated in the first talk, that {\sf GCH} suffices to guarantee that II always has a winning coding strategy in WMG(J).

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


4 Responses to Set theory seminar – Marion Scheepers: Coding strategies (III)

  1. […] the third talk (and a link to the second one), see here. The fourth talk took place on October […]

  2. […] second case, when there is some with , will be dealt with in the next talk. 43.614000 […]

  3. […] the third talk (and a link to the second one), see here. The fourth talk took place on October […]

  4. […] second case, when there is some with , will be dealt with in the next talk. 43.614000 -116.202000 Like this:LikeBe the first to like […]

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