Exercise 20, § 9

One says that a separated space X is minimal if its topology $\mathfrak{T}$ is such that every topology on X, separated and less fine than $\mathfrak{T}$, is necessarily identical to $\mathfrak{T}$; then $\mathfrak{T}$ is necessarily semi-regular (I, p. 103, exer. 20). Prove that in order that a separated space X be minimal, it is necessary and sufficient that every filter base on X formed of open sets, and having a single adherent point, be convergent; X is then absolutely closed (exer. 19). (If $\mathfrak{B}$ is a filter base on X formed of open sets and having no adherent point, consider, for a point $a \in X$, the filter base $\mathfrak{B}'$ formed by the unions $V \cup A$, where A ranges over $\mathfrak{B}$ and V the set of open neighbourhoods of $a$; deduce a contradiction.)