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Prove that $X$ is Hausdorff if and only if the diagonal $\Delta$ is closed in $X\times X$.

This exercise appeared on a previous exam in my course, and also in Munkres. Here's my attempt:

First I want to show the implication starting with assuming that $X$ is Hausdorff. Consider $(X\times X)\backslash \Delta$, then for any tuple $(x_1, x_2) \in (X\times X)\backslash \Delta$ we know $x_1\neq x_2$. Then there are neighbourhoods $U,V$ of $x_1,x_2$ respectively such that $U\cap V = \emptyset$. The set $U\times V$ is a basis element of $X\times X$ so it's in the product topology. So any tuple $(x_1, x_2) \in (X\times X)\backslash \Delta$ lies in an open set of $X\times X$, thus $(X\times X) \backslash\Delta$ is open which makes $\Delta$ closed in $X\times X$.

Now I assume that $\Delta$ is closed in $X\times X$. Then $(X\times X)\backslash\Delta$ is open in $X\times X$. Unless I'm mistaken, every open set $A\times B$ in $X\times X$ will be a subset of $(X\times X)\backslash\Delta$. Then $A\cap B = \emptyset$ because otherwise we would have $(x_1,x_1)\in A\times B \subseteq (X\times X)\backslash\Delta$ for some $x_1\in X$, which is a contradiction. Thus $X$ is Hausdorff.

I'm simply wondering if my argument is correct.

Najib Idrissi
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Auclair
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  • What is the diagonal $\Delta$? I haven't seen this kind of terminology – Siddharth Bhat May 23 '16 at 11:39
  • Is $X\times X\backslash \Delta=(X\times X)\backslash \Delta$? – BigbearZzz May 23 '16 at 11:39
  • It looks fine...almost. I'd prefer to see for any two different $;x,y\in X;$ two disjoint open sets containing them. This is not very clear from what you did. – DonAntonio May 23 '16 at 11:39
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    @NajibIdrissi No duplicate: the asker here is asking about the correctness of his proof. – DonAntonio May 23 '16 at 11:40
  • The first proof is correct, the second one isn't. You want to show that $X$ is Hausdorff, so start by picking $x \neq y \in X$. Then continue from there. Read the linked duplicate to see a correct proof. – Najib Idrissi May 23 '16 at 11:40
  • @SiddharthBhat $\Delta:={(x,x)|x\in X }$ – BigbearZzz May 23 '16 at 11:40
  • @Joanpemo This question has been asked dozens of times. It's not necessary to add another one to the list. – Najib Idrissi May 23 '16 at 11:40
  • @SiddharthBhat Is a common term in set theory and in topology. It is the set $$\Delta:=\left{,(a,b)\in X\times X;/;;a=b,\right}$$ – DonAntonio May 23 '16 at 11:41
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    @NajibIdrissi Again: this particular question is not about the fact, but about how the asker did when trying to prove it. The question you linked ask about hte proof itself. Two different things. – DonAntonio May 23 '16 at 11:42
  • @Joanpemo And again, this particular question has been asked dozens of time, and adding a new one to the site is useless or even actively harmful. I answered in the comments about the correctness of the proof, and now we can mark this question as a duplicate. – Najib Idrissi May 23 '16 at 11:43
  • @Auclair, can you put the tag proof-verification in your question? Thank you. – Masacroso May 23 '16 at 11:45
  • @Masacroso Done, thanks for reminding me. – Auclair May 23 '16 at 11:46
  • @NajibIdrissi Can't you really understand that the asker is not asking how to prove the claim but in fact he's asking how is his proof ?? The question you linked asked about the claim: two different things! – DonAntonio May 23 '16 at 11:48
  • @NajibIdrissi I think I corrected the second part of the proof. Thanks for the hint. – Auclair May 23 '16 at 11:53
  • @Joanpemo I can understand that. Now calm yourself, please. There is little archival value in keeping this sort of proof-verification question, and it's better to direct future readers to the Q&A I linked. – Najib Idrissi May 23 '16 at 11:59

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