Proposition 15

Let $X$ be a regular space, $F$ a closed subset of $X$, $R$ the equivalence relation on $X$ obtained by identifying all the points of $F$ [in other words, the equivalence relation whose equivalence classes are $F$ (if $F \neq \emptyset$*) and the sets* $\{x\}$ where $x \in CF$*]. Then the quotient space $X/R$ is Hausdorff.

Let $M$ and $N$ be two distinct equivalence classes in $X$. If each of them consists of a single point in the complement of $F$, then there exist two disjoint open neighbourhoods of $M$ and $N$ in the Hausdorff subspace $CF$; these are neighbourhoods of $M$ and $N$ in $X$ and are saturated with respect to $R$. If $M = F$ (so that $F \neq \emptyset$) and $N = \{ b \}$ where $b \notin F$, then since $X$ is regular there is an open neighbourhood of $b$ and an open neighbourhood of $F$ which do not intersect; these neighbourhoods are saturated with respect to $R$, and the proposition is proved.

Note that the quotient space $X/R$ is not necessarily regular (Chapter IX, § 4, Exercise 14).