Convergence along an ultrafilter

Remarks

This is the filter limit of $f$ along the filter $\mathcal{U}$.

Example: for $a \in S$ and the principal ultrafilter $\mathcal{U}_a = \set{A \subseteq S : a \in A}$, every $f$ converges to $f(a)$ along $\mathcal{U}_a$. Indeed, for each neighbourhood $V$ of $f(a)$ the set $\set{s : f(s) \in V}$ contains $a$.

Non-example: let $\mathcal{U}$ be a non-principal ultrafilter on $\N$ and $f(n) = n$ in $\R$. For any $x \in \R$, the set $\set{n : \abs{n - x} < 1}$ is finite, and a non-principal ultrafilter contains no finite set. So $f$ converges to no point along $\mathcal{U}$.

If $X$ is compact, then every $f \colon S \to X$ converges along $\mathcal{U}$ to some point; this is closely related to compactness via ultrafilters. If $X$ is Hausdorff, the limit is unique. For a non-principal ultrafilter $\mathcal{U}$ on $\N$, the map $x \mapsto \lim_{\mathcal{U}} x_n$ on bounded real sequences is linear and multiplicative, and its value lies between $\liminf_n x_n$ and $\limsup_n x_n$. It is not a Banach limit: it gives $(1, 0, 1, 0, \dots)$ the value $0$ or $1$, while every Banach limit gives $1/2$.