**Proof.**
Let $F$ be a closed subset of $X$, if $y\in F$ then $\{ y\} \subset F$, so $\overline{\{ y\} } \subset \overline{F} = F$ as $F$ is closed. Thus for all specialization $x$ of $y$, we have $x\in F$.

Let $x, y\in X$ such that $x\in \overline{\{ y\} }$ and let $T$ be a subset of $X$. Saying that $T$ is stable under specialization means that $y\in T$ implies $x\in T$ and reciprocally saying that $T$ is stable under generalization means that $x\in T$ implies $y\in T$. Therefore (3) is proven using contraposition.

The second property follows from (1) and (3) by considering the complement.
$\square$

## Comments (0)