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If $\omega$ is a free ultrafilter on $\mathbb{N}$,$(x_n)$ is a sequence of complex numbers,what is the precise definition of "$lim_{\omega}(x_n)$ does not converge to $x$"?

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    $\begingroup$ My guess (not confident enough to make it an answer) would be: $x_* \omega = \{ S \mid x^{-1}(S) \in \omega \}$ is an ultrafilter on $\mathbb{C}$, and we're requiring that ultrafilter not to have limit $x$. So, that would mean that the neighborhood filter $\mathcal{N}_x \not\subseteq x_* \omega$, i.e. there is some $\epsilon > 0$ such that $\{ n \in \mathbb{N} \mid |x_n - x| < \epsilon \} \notin \omega$. $\endgroup$ – Daniel Schepler Nov 16 '18 at 22:26
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    $\begingroup$ Since $\omega$ is often synonymous to $\Bbb N$, I would to nominate this to the prize of "worst notation ever". :-) $\endgroup$ – Asaf Karagila Nov 16 '18 at 23:16
  • $\begingroup$ Yes, write $\mathcal{F}$ for the filter, and use $\lim_{\mathcal{F}} x_n$ instead, as is more usual. $\endgroup$ – Henno Brandsma Nov 17 '18 at 5:45
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First know what $x =\lim_\omega(x_n)$ means, then look at the negation:

$$x = \lim_\omega(x_n) \text{ iff } \forall O \text{ open with } x \in O: \{n: x_n \in O\} \in \omega$$

So the negation of that is that there exists some open neighbourhood $O_x$ of $x$ such that $\{n: x_n \in O_x\} \notin \omega$, but as $\omega$ is an ultrafilter (so has the property $A \notin \omega \leftrightarrow \omega\setminus A \in \omega$ for all subset $A$ of $\omega$) this is equivalent to the fact that $\{n: x_n \in X\setminus O_x\} \in \omega$.

So e.g. if $X$ were compact and we'd assume no $x$ is a limit, we'd have an open cover of $X$ of such $O_x$, hence a finite cover of these sets: $O_{x_1},\ldots,O_{x_N}$, and then we'd have contradiction as

$$\cap_{i=1}^N \{n: x_n \in X\setminus O_{x_i}\} \in \omega$$ while the left hand side is empty, as the $O_{x_i}$ form a cover.

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