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Let $\mathscr{U}$ be a free ultrafilter on the positive integers $\mathbf{N}$ and fix $U \in \mathscr{U}$ such that $U$ is not cofinite (thanks J.D.Hamkins for the correction.)

Consider the natural bijection between $(0,1]$ and the infinite $\{0,1\}$-sequences with infinitely many ones (written in base $2$).

Define also the sequence $x=(x_n)$ by $x_n=0$ if $n \in U$ and $x_n=1$ otherwise, so that $x$ is $\mathscr{U}$-convergent to $0$.

Hence, for each $\omega \in (0,1]$, we can consider the subsequence $x \upharpoonright \omega:=(x_{n_k})$, where the index $n_k$ is taken iff there is a $1$ in the corresponding representation of $\omega$.

Question. Is it possible that $$ S:=\left\{\omega \in (0,1]:\, (x\upharpoonright \omega)\, \text{ is }\mathscr{U}-\text{convergent to }0 \right\} $$ is not a meager set?

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    $\begingroup$ Could you describe the intended subsequence a bit more clearly? Should the subscript $n$ be $k$? Is the point that you are re-indexing the subsequence, but then still using the same ultrafilter? $\endgroup$
    – Joel David Hamkins
    Commented Apr 15, 2018 at 18:43
  • $\begingroup$ I try with an example: if $\omega=1/8$ then its representation in base $2$ with infinitely many ones is $0,00011111...$. Hence the subsequence defined by $x \upharpoonright 1/8$ is $(x_{n+3})$. Similarly, $x \upharpoonright 1/3$ is the subsequence $(x_{2n})$. (Yes, it is just a reindexing of the sequence, using the same ultrafilter at the end.) $\endgroup$
    – Paolo Leonetti
    Commented Apr 15, 2018 at 18:56
  • $\begingroup$ If $U=\mathbf{N}$, then all the subsequences are the constant $0$ sequence, in which case $S$ is everything, hence not meager. $\endgroup$
    – Joel David Hamkins
    Commented Apr 15, 2018 at 18:59
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    $\begingroup$ I'd expect that, now that you've excluded cofinite $U$, your set $S$ will never be meager; in fact it should never have the Baire property, and neither it nor its complement should have a non-meager Borel subset. Unfortunately, I won't have any time soon to think carefully about this, so I hope someone else will come along with an answer. (Also, it shouldn't matter that $U\in\mathcal U$, only that $U$ is neither finite nor cofinite.) $\endgroup$
    – Andreas Blass
    Commented Apr 16, 2018 at 1:17
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    $\begingroup$ Fix $A=\{a_1,a_2,\ldots\} \in \mathscr{U}$ and consider the subsequence $(x_{a_k})$. It would be sufficient to prove that $\{k: x_{a_k} =0\} \in \mathscr{U}$, that is, $\{k: a_k \in U\} \in \mathscr{U}$. By construction $A \cap U \in \mathscr{U}$. So, it would be enough that, if $S=\{s_1,s_2,\ldots\}$ and $Z=\{z_1,z_2,\ldots\}$ satisfy $s_n \le z_n$ for all $n$ and $Z \in \mathscr{U}$ then $S \in \mathscr{U}$. Does there exist an ultrafilter $\mathscr{U}$ with this property? $\endgroup$
    – Paolo Leonetti
    Commented Apr 17, 2018 at 16:18

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