$X$ - regular, $A \subset X$ - closed $\Rightarrow \ \ X/A$ - quotient space is Hausdorff

Question

Could you tell me how to prove that if $X$ is a regular space , $ \ A \subset X$ is closed, then $ X/A$ (quotient space) is Hausdorff?

I know what the topology of a quotient space looks like. If $q: X \rightarrow X/A$, then $U \subset X/A$ is open $\iff \ \ q^{-1}(U)$ is open. But I don't know how to use it here.

Could you help me?

Thank you.

Answer 1

You basically just do it.

First, note that the quotient mapping works as follows: $$ q(x) = \begin{cases} x, &\text{if }x \notin A \\ A, &\text{if }x \in A. \end{cases}$$

So let's pick two distinct points $x,y \in X / A$. What follows is a basic outline, with the details left to be filled in.

• If neither point is $A$, then $x,y \in X \setminus A$ (which is open in $X$), and we can use the Hausdorffness of $X$ (and a parenthetical observation above) to help separate them in $X / A$.
• If one of them, say $y$, is $A$, then we know that $x \in X \setminus A$, and we can instead turn to the regularity of $X$ to help separate these points in $X / A$.