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
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