open equivalence relation and closed graph of it

Question

I need to prove that if $\sim$ is an open equivalence relation on a topological space S and $R = \{(x,y)\in S\times S : x\sim y\}$ is a close subset of $S\times S$ then $\Delta = \{(x,x)\in S\times S\}$ is a close subset of $S\times S$. I tried to apply ideas from the theory and exercises with similar requests, like that the quotient map is an open map but failed to solve that.

Answer 1

The result is false.

Let $S=\{0,1,2\}$ with the topology $\tau=\big\{\varnothing,\{0,1\},\{2\},S\big\}$, and let $\sim$ be the equivalence relation on $S$ whose equivalence classes are $\{0,1\}$ and $\{2\}$ clearly

$$R=(\{0,1\}\times\{0,1\})\cup\{\langle 2,2\rangle\}\;.$$

$R$ is a closed subset of $S\times S$, and the quotient map $q:S\to S/\!\!\sim$ is open, but $S$ is not Hausdorff, so $\Delta$ is not closed in $S\times S$. (Specifically, neither $\langle 0,1\rangle$ nor $\langle 1,0\rangle$ has an open nbhd in $S\times S$ that is disjoint from $\Delta$.)

Added: Apparently you wanted to assume the result that a space is Hausdorff if and only if its diagonal is closed and prove the following result:

Suppose $\sim$ is an open equivalence relation on a topological space $S$. Then the quotient space $S/\!\!\sim$ is Hausdorff if and only if the graph $R$ of $\sim$ is closed in $S\times S$.

Let $q:S\to S/\!\!\sim$ be the quotient map. Suppose that $q(x)$ and $q(y)$ are distinct points of $S/\!\!\sim$. Then $x\not\sim y$, so $\langle x,y\rangle\in(S\times S)\setminus R$. $R$ is closed in $S\times S$, so there are open $U,V\subseteq S$ such that $\langle x,y\rangle\in U\times V\subseteq(S\times S)\setminus R$. Clearly $U$ and $V$ are disjoint open nbhds of $x$ and $y$, respectively, in $S$. The map $q$ is open, so $q[U]$ and $q[V]$ are disjoint open nbhds of $q(x)$ and $q(y)$, respectively, in $S/\!\!\sim$. Thus, $S/\!\!\sim$ is Hausdorff if $R$ is closed.

For the other direction assume that $S/\!\!\sim$ is Hausdorff; you need to show that $R$ is closed. Let $\Delta=\{\langle q(x),q(x)\rangle:x\in S\}$, the diagonal in $(S/\!\!\sim)\times(S/\!\!\sim)$. By hypothesis this is closed, and $q$ is continuous, so $q^{-1}[\Delta]$ is closed in $S\times S$. But $q^{-1}[\Delta]$ is ... ?