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.