Open Projection and Closed Graph Implies Quotient is Hausdorff

Theorem
Let $\RR \subseteq X \times X$ be an equivalence relation on a topological space $\struct {X, \tau}$.

Let $X / \RR$ be the quotient space.

Let $p$ denote the quotient mapping.

Let:
 * $\RR$ be closed in $X \times X$
 * $p$ be an open mapping.

Then $X / \RR$ is Hausdorff.

Proof
Let $\eqclass x \RR, \eqclass y \RR \in X / \RR$ such that $\eqclass x \RR \ne \eqclass y \RR$.

Then $\tuple {x, y} \notin \RR$.

Since $\RR$ is closed, $\paren {X \times X} \setminus \RR$ is open.

Let:
 * $\BB = \set { U \times V : U, V \in \tau}$

By Natural Basis of Product Topology, $\BB$ is a basis for the product topology on $X \times X$.

Then choose $U, V \in \tau$ such that $\tuple {x, y} \in U \times V \subseteq \paren {X \times X} \setminus \RR$.

Thus $x \in U$, so:
 * $\eqclass x \RR = \map p x \in p \sqbrk U$

Similarly, $y \in V$, so:
 * $\eqclass y \RR = \map p y \in p \sqbrk V$

Since $p$ is open, $p \sqbrk U$ and $p \sqbrk V$ are open.

that $p \sqbrk U \cap p \sqbrk V \ne \O$.

Then there exist $z \in U$ and $w \in V$ such that $\eqclass z \RR = \eqclass w \RR$.

Then $z \RR w$, and so $\tuple {z, w} \in \RR$.

However, $\tuple {z, w} \in U \times V$, contradicting $U \times V \subseteq \paren {X \times X} \setminus \RR$.

Thus:
 * $p \sqbrk U \cap p \sqbrk V = \O$

Hence the result.

Also see

 * Hausdorff Space iff Diagonal Set on Product is Closed