# Prove $\Delta_X$ is closed iff X is Hausdorff [duplicate]

I'm attempting a topology proof and think my proof is correct but I'm not 100% sure. The problem is as follows: Define $\Delta_X=\{(x,x)\in X\}$. Prove that $X$ is Hausdorff if and only if $\Delta_X$ is a closed set. The reverse proof is fine but for my proof of the forward one I attemopted as follows.

Assume $\Delta_X$ is closed, then $D=(\Delta_X)^C$ is open. Let $p_1,p_2\in D$ then $p_1 \neq p_2$. Take a union of open sets $U=\cup_{\alpha \in A}U_{\alpha}$ such that $p_1,p_2 \in U$. If $p_i \in U_{\alpha_i}$ with $U_{\alpha_1} \cap U_{\alpha_2} = \varnothing$ then we're done. If not, then $p_1,p_2$ are contained in the same open set. Write this set as a union of open sets such that $p_1$ and $p_2$ are not contained in the same open set. Therefore, $X$ is Hausdorff.

I'm not sure if this is correct but if it isn't any comments on where it is wrong and hints at improving it would be greatly appreciated!

## marked as duplicate by Mario Carneiro, Joey Zou, R_D, Cameron Williams, user91500Oct 5 '16 at 6:20

You're supposed to prove that given $p_1,p_2\in \color{red}{X}$ with $p_1 \neq p_2$, there are disjoint open sets containing $p_1$ and $p_2$, respectively. If $p_1$ and $p_2$ are distinct points of $X$, then $(p_1,p_2)\notin \Delta_X$. So there is a basic open set $U \times V$ of $X \times X$ containing $(p_1,p_2)$ which is disjoint from $\Delta_X$. Now show that $U$ is disjoint from $V$.
• I think I see now. So to prove $U$ and $V$ are disjoint, you could assume they're not, therefore they have point in common which is a contradiction since $U \times V$ is disjoint from $\Delta_X$ ? – Crunch Oct 4 '16 at 21:55