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Let $\omega$ be a nonprincipal ultrafilter on $\mathbb N$.

A standard construction gives an $\omega$-limit, say $x_\omega$, for any bounded sequence $(x_n)$ of real numbers. Namely, there is unique real value $x_\omega$ such that
$$\{\,n\in\mathbb N\mid |x_\omega-x_n|<\varepsilon\,\}\in \omega$$ for any $\varepsilon>0$.

Clearly $x_\omega$ is a partial limit of $x_n$ [i.e., $x_\omega$ is a limit of a subsequence of $(x_n)$].

Question. Is it always possible to choose subsequence $(x_n)$, $n\in J$ converging to $x_\omega$ and such that $J\in\omega$?

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up vote 15 down vote accepted

No, only if $\omega$ is a p-point.

If $(A_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_k$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0.

To find a non-p-point, partition $\mathbb N$ into countably many infinite sets $A_k$, let $I$ be the ideal of sets meeting each $A_k$ only finitely often (except for finitely many $k$), and let $\omega$ be dual to any maximal ideal extending $I$. Then each $A_k$ is in $I$, hence null modulo $\omega$. The partition $(A_k)$ witnesses that $\omega$ is not a p-point.

Note that ZFC does not prove that there are p-points, but existence of p-points follows from CH (or weaker assumptions).

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