# $X$ is Hausdorff iff the diagonal is a closed subset of $X^2$ [duplicate]

This question already has an answer here:

Let $(X,\tau)$ be a topological space. Consider $X^2$ with the product topology. Show that $X$ is Hausdorff iff the diagonal $D = \{(x,y) \in X^2 \mid x=y\}$ is a closed subset of $X^2$.

## marked as duplicate by Najib Idrissi, J.-E. Pin, John B, Dan Rust, Harish Chandra RajpootMar 2 '16 at 12:09

• This has been asked a billion times. – Najib Idrissi Mar 2 '16 at 11:04
• Can you please answer just one more time .I think its bit different from that one if not so can you share the answer – user319276 Mar 2 '16 at 11:25
• Click the link. Read the (multiple) answers. – Dan Rust Mar 2 '16 at 11:27
• I Could not find the link – user319276 Mar 2 '16 at 11:31
• It is the first comment to this question by Najib Idrissi. – Dan Rust Mar 2 '16 at 11:34

Hint: Convince yourself that for $x \neq y$, finding distinct open neighborhoods $U_x$ and $U_y$ is the same as finding an open neighborhood of $(x,y) \in X \times X$, which does net meet the diagonal.

• The downvoter might explain him/herself? – MooS Mar 2 '16 at 11:23
• Sure: At almost 12k reputation, one would think you would recognize such obvious duplicates and not answer them. Especially if it's to give a hint answer like that. – Najib Idrissi Mar 2 '16 at 12:32