Proposition 14

Let $X$ be a regular space, $R$ a closed equivalence relation on $X$. Then the graph $C$ of $R$ in $X \times X$ is closed.

Let $(a, b)$ be a point of $X \times X$ in the closure of $C$, and let $V$ (resp. $W$) be a closed neighbourhood of $a$ (resp. a neighbourhood of $b$) in $X$; then there is a point $(x, y) \in C \cap (V \times W)$. Since $x \in V$, $y$ belongs to the saturation $S$ of $V$ with respect to $R$; hence each neighbourhood $W$ of $b$ meets $S$. By hypothesis $S$ is closed, and therefore $b \in S$. Now let $B$ be the saturation of $\{b\}$ with respect to $R$, then each closed neighbourhood $V$ of $a$ meets $B$; since by hypothesis $B$ is closed and $X$ is regular, it follows that $a \in B$ and therefore that $(a, b) \in C$. This completes the proof.