Convergence of filters

Remarks

For a map $f \colon S \to X$ from a set $S$ and a filter $\mathcal{G}$ on $S$, the image filter is $f(\mathcal{G}) = \set{B \subseteq X : f^{-1}(B) \in \mathcal{G}}$. One says that $f(s) \to x$ along $\mathcal{G}$ if $f(\mathcal{G}) \to x$.

Example: let $(x_n)$ be a sequence in $X$, and let $\mathcal{F}$ be the family of sets $B \subseteq X$ that contain $x_n$ for all large $n$. Then $\mathcal{F}$ is a filter, and $\mathcal{F} \to x$ if and only if the sequence converges to $x$, since both mean that each neighbourhood of $x$ contains $x_n$ for all large $n$.

Example: the neighbourhood filter $\mathcal{N}(x)$ converges to $x$, and so does the principal filter $\set{B : x \in B}$.

Non-example: in $\R$, let $\mathcal{F}$ be the filter of sets that contain an interval $[n, \infty)$. For each $x \in \R$, the neighbourhood $(x - 1, x + 1)$ contains no such interval, so $\mathcal{F}$ does not converge to $x$.

A space is Hausdorff if and only if every filter on it has at most one limit.