Is it true that if a given equivalence relation is closed, then each equivalence class is closed?

Question

Let $X$ be a topological space and $R$ a closed equivalence relation on $X$. Then is it true that each equivalence class (under the relation $R$) is closed subset in $X$. Here by a closed equivalence relation $R$, we mean that $R$ is closed subset of $X \times X$ under the product topology.

Answer 1

Suppose $[x]$ is an equivalence class for some $x\in X$. To show [x] is closed, we need show $X\backslash [x]$ is open. It is enough to show that for arbitrary $y\in X\backslash [x]$ there is a neighborhood $V\ni y$ with $V\cap [x]=\emptyset$. To this end, suppose by contradiction that there is $y\notin [x]$ such that every open $V\ni y$ contains an element $x_V\in[x]$.

Since $R$ is closed, $X\times X\backslash R$ is open, and contains $(x,y)$. Therefore by definition of product topology there is an open "rectangle" $U \times V\subseteq X\times X\backslash R$, with $(x,y)\in U\times V$. Then if $x_V$ is as described above, we have $x\in U$, $x_V\in V$, and so $(x,x_V)\in U\times V\subseteq X\times X\backslash R$, contradicting the fact that $x_V\in [x]$.