The problem is: Let $∼$ be an equivalence relation on a regular space $X$. If the natural projection $X \to X \, /∼$ is closed, show that $G(∼)$ is closed in $X \times X$. $G(∼) = \{(x, x' ) \, |\, x, x' \in X \, \text{and} \,x ∼ x' \, \}$.
I know that the product of regular spaces is regular, to see that $G(∼)$ is closed I am trying to see that its complement is open, I take an open in the complement of $G$ but I do not come to anything. Also try to project $(x, x') \in G(∼)$ into each set $X \, /∼$.
Can I do this?
Some clue
Hint: If $x\not\sim y$, then $x\not\in \pi[\{y\}]$, where $\pi:X\to X/\sim$ is the canonical projection. Since regular spaces are Hausdorff, singletons are closed, and hence $\pi[\{y\}]$ is closed. Thus, $x$ and $\pi[\{y\}]$ can be separated by open sets. Apply the same principle to $y$ and $\pi[\{x\}]$.