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For a cardinal $\kappa$ I'll use the phrase "$(\kappa,\kappa)$-game" to mean "two-player, perfect-information, deterministic game on $\kappa$ of length $\kappa$."

Say that a cardinal $\kappa$ is LOI (= limited open indeterminacies) iff for every tree $T\subseteq\kappa^{<\kappa}$ the corresponding "$\kappa$-open" game $G_T$ has the property that there is a set $\Sigma_T$ of strategies for (at least) one of players $1$ and $2$ such that

  • every strategy for the other player is beaten by some element of $\Sigma_T$, and

  • $\vert \Sigma_T\vert<2^\kappa$.

(To be clear, the game $G_T$ is the $(\kappa,\kappa)$-game in which on turn $\alpha<\kappa$ player $parity(\alpha)$ plays an ordinal $\beta_\alpha$, and player $1$ wins iff the sequence $(\beta_\alpha)_{\alpha<\kappa}$ eventually leaves $T$.)

LOI is a "near-determinacy" principle for (a not-too-small class of) games of uncountable length which is not trivially trivialized by the failure of determinacy for games of length $\omega_1$. For example, if $\kappa$ is inaccessible then recasting an undetermined game of length $\lambda<\kappa$ as a game of length $\kappa$ by just adding "dummy moves" does result in an undetermined game on $\kappa$ but does not result in a failure of LOI at $\kappa$.

I'm curious whether, in fact, this does trivialize:

Question: is the existence of an uncountable LOI cardinal consistent with $\mathsf{ZF}$?

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  • $\begingroup$ In case it's useful, I have a tentative argument that the clopen version of LOI is equivalent to LOI itself, but I'm still checking the details. $\endgroup$ Commented Sep 5, 2022 at 0:27
  • $\begingroup$ Can we also say that $\kappa$ is LOL if the requirement for it to be LOI is ridiculous? $\endgroup$
    – Asaf Karagila
    Commented Sep 5, 2022 at 9:13
  • $\begingroup$ All $κ$ with $κ^{<κ}<2^κ$ satisfy this property since if player 2 (i.e. closed) does not have a winning strategy, then each strategy is defeated by some play, and there are $≤κ^{<κ}$ plays winning for player 1. $\endgroup$ Commented Sep 7, 2022 at 21:49
  • $\begingroup$ @DmytroTaranovsky Oh, yes, that's a good point. $\endgroup$ Commented Sep 7, 2022 at 22:08

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