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Here I described the Gale-Stewart game on Baire space $\mathbb{N}^{\mathbb {N}}$.

If the set A is determined (one of the players has a winning strategy), what can we say about the determinacy of its complement $\mathbb{N}^{\mathbb {N}}\setminus A$?

I would appreciate any help.

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  • $\begingroup$ Nothing in general: Let $A = \{0y: y \in \mathbb{N}^{\mathbb{N}}\} \cup \{nx: x \in \mathbb{N}^{\mathbb{N}} \setminus X, n \geq 1\}$ where $X \subseteq \mathbb{N}^{\mathbb{N}}$ is undetermined. Then $A$ is determined (player I has a winning strategy) but its complement isn't. $\endgroup$ – hot_queen Jun 19 '18 at 18:07
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    $\begingroup$ One thing you quickly understand in the context of descriptive set theory is that very little can be said in general from the determinacy of individual sets, while there is better chance of positive results when what is assumed is the determinacy of all sets of certain complexity. $\endgroup$ – Andrés E. Caicedo Jun 20 '18 at 20:33

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