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Suppose $\mathcal{F}$ is any free ultrafilter on $\beta\mathbb{N}\setminus\mathbb{N}$,$(x_n)$ is a sequence of complex numbers.

My question is:What is the deference between the limit along $\mathcal{F}$ ,$lim_{\mathcal{F}}x_n$ and $lim_{n\to \infty}x_n$?If $lim_{n\to\infty}x_n\not \to 0$,can we conclude that $lim_{\mathcal{F}}x_n\not\to 0$?

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    $\begingroup$ The sequence $(0,1,0,1,0,1,0,1,0,1,\dots)$ doesn't converge. It converges with respect to every ultrafilter $\mathcal F$ on $\mathbb N$. You should figure out what it converges to, in terms of $\mathcal F$. And then you should probably try to prove that every bounded sequence of complex numbers converges with respect to every ultrafilter on $\mathbb N$. $\endgroup$ – Andreas Blass Dec 7 '18 at 2:23
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No, I'll use Andreas' elementary example from the comments: the sequence $x_n = 0$ for $n$ even, $x_n= 1$ for $n$ odd, does not converge in $\mathbb{R}$ in the usual topology.

However, if $\mathcal{F}$ is a free ultrafilter on $\mathbb{N}$, then either $E= \{2n: n \in \mathbb{N}\}$ or $O = \{2n+1: n \in \mathbb{N}\}$ is in $\mathcal{F}$ (as $O \cup E = \mathbb{N}$ and we have an ultrafilter). If the former, then for any neighbourhood $U$ of $0$ we have that $\{n : x_n \in U\} \supseteq E$ so $\{n : x_n \in U\} \in \mathcal{F}$, and so $\lim_\mathcal{F} x_n = 0$ and if the latter we similarly see that $\lim_\mathcal{F} x_n = 1$. In any case, the non-convergent sequence $(x_n)_n$ does converge along any free ultrafilter. So the proposed implication does not hold.

In fact one can show that any bounded sequence in the reals or complex numbers has a limit along any ultrafilter whether it converges (usually) or not.

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