---
tag: X117TDBR51
kind: definition
name: Convergence along an ultrafilter
msc: [54A20, 40J05]
defines: QG9QRZAFTG
uses: [J64Y3JC0RE, F6QDHKE6J3, 0CQHFPGJ45, 0ZCY0Z80B6, 0TVJZYB6SH, 0S1FJ8VEX9, DT5PH9QVQ7, SYF2ND6J97]
sources:
  - url: https://en.wikipedia.org/wiki/Ultrafilter_on_a_set
---

## Definition

Let $S$ be a set and let $\mathcal{U}$ be an [[ultrafilter|J64Y3JC0RE]] on $S$.
Let $X$ be a [[topological space|0CQHFPGJ45]], let $f \colon S \to X$ be a map, and let $x \in X$.
The map $f$ *converges to $x$ along $\mathcal{U}$* if
$$\set{s \in S : f(s) \in V} \in \mathcal{U}$$
for every [[neighbourhood|0ZCY0Z80B6]] $V$ of $x$.
When this limit is unique it is written $\lim_{\mathcal{U}} f$ or $\mathcal{U}\text{-}\lim_{s} f(s)$.

## Remarks

This is the [[filter limit|F6QDHKE6J3]] 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|0TVJZYB6SH]], then every $f \colon S \to X$ converges along $\mathcal{U}$ to some point; this is closely related to [[compactness via ultrafilters|DT5PH9QVQ7]].
If $X$ is [[Hausdorff|0S1FJ8VEX9]], 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|SYF2ND6J97]]: it gives $(1, 0, 1, 0, \dots)$ the value $0$ or $1$, while every Banach limit gives $1/2$.