Proposition 6

Let $\mathfrak{B}$ be a filter base on a subset A of a topological space X. Then every cluster point of $\mathfrak{B}$ in X belongs to $\overline{A}$; and conversely every point of $\overline{A}$ is a limit point of a filter on A.

The first assertion is trivial; on the other hand, if $x \in \overline{A}$, the trace on A of the neighbourhood filter of x in X is a filter on A which evidently converges to x.