If $X$ is a compact and Hausdorff topological space,$(x_n)_{n}$ is a sequence in $X$, for any ultrafilter $\mathcal{F}$ on $\mathbb{N}$, I know the fact that $\lim_{\mathcal{F}}x_n$ exists and is unique.

I think that the limit may be different for two different ultrafilters on $\mathbb{N}$. Can anyone show me some examples? Thanks!


If $\mathcal{F}_m$ is the principal ultrafilter corresponding to $m\in\mathbb{N}$, then the limit of $(x_n)$ with respect to $\mathcal{F}_m$ is just $x_m$. So any sequence which is not constant gives an example.

A little less trivially, there are also easy examples with nontrivial ultrafilters. For instance, let $(y_n)$ and $(z_n)$ be two sequences in $X$ which converge to distinct points $y$ and $z$, respectively. Let $x_{2n}=y_n$ and $x_{2n+1}=z_n$. Then for any nonprincipal ultrafilter $\mathcal{F}$ which contains the even numbers, the limit of $(x_n)$ with respect to $\mathcal{F}$ will be $y$, while for any nonprincipal ultrafilter $\mathcal{F}$ which contains the odd nubmers, the limit of $(x_n)$ with respct to $\mathcal{F}$ will be $z$.

More generally, if $(x_n)$ is any sequence, then every accumulation point is the limit of $x_n$ with respect to some nonprincipal ultrafilter. So, if $(x_n)$ has more than one accumulation point, this gives nonprincipal ultrafilters for which it has different limits.

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    $\begingroup$ One should perhaps point out that $\beta\Bbb N$ is a compact Hausdorff space that has a sequence that has $2^{2^{\aleph_0}}$ accumulation points. And of course $\Bbb Q\cap[0,1]$ is a countable sequence in $[0,1]$ with $2^{\aleph_0}$ accumulation points. $\endgroup$ – Asaf Karagila Oct 6 '18 at 7:28
  • $\begingroup$ If the sequence $ (x_n)$ has only one accumulation point,for different free ultrafilter on $\mathbb{N}$, the limit has only one value.Is that true?@Eric Wofsey $\endgroup$ – mathrookie Oct 6 '18 at 8:34
  • $\begingroup$ @Asaf Karagila,if $(x_n)=\Bbb Q\cap[0,1]$,then there exist $2^{\aleph_0}$ free filters on $\mathbb{N}$ such that each limit is different.Is that correct? $\endgroup$ – mathrookie Oct 6 '18 at 8:46
  • $\begingroup$ @mathrookie: Yes. For each $r\in[0,1]$ choose a sequence of rationals which converges to it (e.g. cut the decimal expansion of $r$). Now enumerate the rationals as $q_n$, and take an ultrafilter which contains the indices of the sequence converging to $r$. $\endgroup$ – Asaf Karagila Oct 6 '18 at 9:25
  • $\begingroup$ @mathrookie: Yes, if $(x_n)$ has only one accumulation point, then that accumulation point must actually be its limit as a sequence (assuming your space is compact). So, it is also the limit with respect to every nonprincipal ultrafilter. $\endgroup$ – Eric Wofsey Oct 6 '18 at 14:43

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