-1
$\begingroup$

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 $X^2$.

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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.

$\endgroup$
  • 2
    $\begingroup$ The downvoter might explain him/herself? $\endgroup$ – MooS Mar 2 '16 at 11:23
  • $\begingroup$ 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. $\endgroup$ – Najib Idrissi Mar 2 '16 at 12:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.