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If $G\subseteq2^{\mathbb{N}}$ is comeager then exist is a partition $\mathbb{N}=A_0\cup A_1$, $A_0\cap A_1=\emptyset$, and sets $B_i \subseteq A_i$ for $i \in \{0,1\}$, such that for $A \subseteq \mathbb{N}$, if either $A\cap A_0=B_0$ or $A\cap A_1=B_1$, then $A \in G$.

I am very interested is able to resolve this I do not want to solve it.

How obtaining a partition of $\mathbb {N}$. A suggestion, thanks

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Suppose $F = \bigcup_k W_k$ is an increasing union of closed nowhere dense sets in $2^{\mathbb{N}}$ and $F$ is closed under the bit-flip operation $x \mapsto \overline{x}$.

Claim: There exist $x_F \in 2^{\mathbb{N}}$ and an increasing sequence $\langle n_k : k \in \mathbb{N} \rangle$ such that for every $x \in F$, for all but finitely many $k$, $x \upharpoonright [n_k, n_{k+1}) \neq x_F \upharpoonright [n_k, n_{k+1})$.

Now let $A_0 = \bigcup_k [n_{2k}, n_{2k+1})$, $B_0 = A_0 \cap x_F$. Define $B_1$ analogously. It is clear that for every $A \in F$, $A \cap A_i \neq B_i$.

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  • $\begingroup$ Interesting... (however, I guess it doesn't fully answer the question as $F$ may not be closed under bit-flip) $\endgroup$ Commented Mar 2, 2015 at 18:25
  • $\begingroup$ We can just close the original nowhere dense sets under bit flips - This still gives us closed nowhere dense sets. $\endgroup$
    – Ashutosh
    Commented Mar 2, 2015 at 18:36

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