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In question 2 of this exam we see that topological space $X$ is Hausdorff if and only if the diagonal $\Delta=\{(x,x)\in X\times X\}$ is a closed. I came across this theorem in a similar assignment question and I am happy with the proof and validity of this theorem.

However, an equivalent definition of closed is that a subspace contains all its limit points. As far as I am concerned a sequence $x_i\in X$ converges exactly when $(x_i,x_i)\in X\times X$ converges and if $x_i\rightarrow x$ then surely $(x_i,x_i)\rightarrow (x,x)$ but this would imply that $\Delta$ is closed and hence every topological space would be Hausdorff. Clearly this is not the case.

My only thoughts are that perhaps some of the definitions I have used do not work for example a limit being well defined but so far my checks seem to fail to provide an adequate issue with my blatant contradiction.

What's gone wrong?

andrew
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    A sequence can have, a priori, multiple limits. Take a space $X$ with more than two points and with topology ${\emptyset,X}$, then every point is the limit of every sequence. – Alessandro Feb 04 '22 at 16:15
  • @Alessandro so uniqueness of a limit is the Hausdorff property I am using in this "proof"? As I understand it non-Hausdorff spaces can have unique limits this cannot be the issue I have here. – andrew Feb 04 '22 at 16:22
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    The other error in your argument (other than what @Alessandro already pointed out) is that you think of limits of sequences only. A set is closed iff is contains all limits of filters (or nets) in the set. A non-Hausdoff space can have unique limits of all sequences, but not of all filters. – Andreas Blass Feb 04 '22 at 16:37

2 Answers2

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As already said in the comments, if you're space is not Hausdorff, then sequences can have a priori multiple limit points. As also pointed out in the comments by @Andreas Blass, your space might not be Hausdorff, but such that every sequence has a unique limit (you can find an example here). However, the following hold:

A space is Hausdorff if and only if every net has a unique limit

You can find a proof here

Alessandro
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If in a non-Hausdorff space $X$ the sequence $(x_n)$ has limits $a$ and $b$ with $a \neq b$ (this can happen!) then $(x_n,x_n)$ is a sequence in $\Delta$ that converges to $(a,b)$ (and also to $(b,a)$), both of which are not in $\Delta$. So your argument fails.

There are also non-Hausdorff spaces where sequence limits are unique (US spaces) but then another flaw is shown: for closedness in general spaces it is not enough to consider closedness under sequential limits, but filters or nets must be used.

Henno Brandsma
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