Equivalence relations on a regular space

Question

I read the next proposition in Bourbaki. I didnt understand why each closed neighborhood $V$ of $a$ meets $B$ and why $a\in B$.

Answer 1

Let $\varphi:X\to X/R$ be the canonical map.

• $V$ meets $B=\varphi^{-1}(\{\varphi(b)\})$ because $b\in S=\varphi^{-1}(\varphi(V))$.
• $a\in B$ because: 1) $X$ being regular, the closed neighborhoods $V$ of $a$ form a neighborhood basis for $a$, and each of them meets $B$, hence $a\in\overline B$; 2) $B$ is closed, as the saturation of a closed subset in $X$ (the subset $\{b\}$: for Bourbaki, Hausdorff is mandatory in the definition of the notion of regular space) by a closed relation.