Closed sets with filters

Question

I am studying topological vector spaces, and, being at the begin of the textbook (Sevres) I frequently encounter rephrasing and generalisations of known results about metric spaces in terms of filters. In particular, it is known that in a metric space

A is a closed set in a metric space M IFF any sequence contained in A, which converges to a point x, converges in A (i.e. x belongs to A)

I was trying to prove something as:

"A is a closed set in a TVS E IFF any filter on A converging to a point x, converges in A (i.e. x belongs to A)"

Left to right implication sounds ok, but I can't prove the opposite, without adding the hypothesis of Hausdorff space on E (indeed satisfied by metrizable spaces)

Is this condition necessary? Is there any more general criterion?

Answer 1

This holds true in any topological space:

$A$ is closed in $X$ iff for every filterbase $\mathcal{F}$ on $A$ (i.e. consisting of subsets of $A$) that converges to $x$ in $X$ we have that $x \in A$.

Proof sketch: suppose $A$ is closed and we have such a filterbase $\mathcal{F}$ on $A$, then for every neighbourhood $O$ of $x$ we have that $O$ contains some $F \in \mathcal{F}$ and thus $O \cap A \neq \emptyset$; this implies $x \in \overline{A}=A$. For the reverse, suppose $A$ satisfies the filterbase property and let $x \in \overline{A}$, then $\{O \cap A: O \text{ (open) neighbourhood of } x\}$ is a filterbase on $A$ that converges to $x$, so by the property $x \in A$ and so $A$ is closed.

No metric or Hausdorffness is needed, which is why filters are popular.