# limit of a sequence along ultrafilter

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.
• 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. – Asaf Karagila Oct 6 '18 at 7:28
• 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 – mathrookie Oct 6 '18 at 8:34
• @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? – mathrookie Oct 6 '18 at 8:46
• @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 $r$. – Asaf Karagila Oct 6 '18 at 9:25
• @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. – Eric Wofsey Oct 6 '18 at 14:43