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To each pair $(S,\mathcal{X})$ where $S=(s_i)_{i\in\mathbb{N}}$ is a decreasing sequence of positive real numbers and $\mathcal{X}\subseteq\mathbb{R}$, we can associate the alternation game $A_S(\mathcal{X})$ as follows:

  • Players $1$ and $2$ jointly build an increasing sequence of natural numbers $$a_0<b_0<a_1<b_1<...,$$ subject to the rule that $$\sum_{i<a_0}s_i>\sum_{a_0\le i<b_0}s_i> ... >\sum_{b_k\le i< a_{k+1}}s_i>\sum_{a_{k+1}\le i<b_{k+1}}s_i>...$$

  • Player $1$ wins iff the alternating series $$\left(\sum_{i<a_0}s_i\right)-\left(\sum_{a_0\le i<b_0}s_i\right)+ ... +\left(\sum_{b_k\le i< a_{k+1}}s_i\right)-\left(\sum_{a_{k+1}\le i<b_{k+1}}s_i\right)+...$$ converges to an element of $\mathcal{X}$.

An old question of mine asked about the behavior of these games when we specifically take $S$ to be the harmonic sequence $s_i={1\over i}$, but turned out to be difficult to attack. I'd like to get a better understanding of why this sort of question might be hard, and this seems like a good starting point:

Is it consistent with $\mathsf{ZFC}$ that some $A_S(\mathcal{X})$ is undetermined?

A positive answer to this question would help explain potential wildness, but would also seem to require us to be able to "code into" the alternation games (a la the Banach game), and at present I don't see how to do that.

Noah Schweber
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1 Answers1

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If we take $s_i = 2^{-i}$, we should even get that $\mathrm{ZFC}$ proves the existence of some $\mathcal{X}$ with undetermined $A_{(2^{-i})_{i \in \mathbb{N}}}(\mathcal{X})$. The key parts are that different plays yield different reals (as nothing in the tail can overcome a difference in a prefix), and that each strategy is compatible with continuumsly many plays (as a move here doesn't constrain the future at all).

We should thus be able to do the standard trick of well-ordering both players strategies with the least possible order type, and then building $\mathcal{X}$ in stages to foil any particular winning strategy.

Arno
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