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To prove: A space $X$ is Hausdorff if and only if the set $\{(x,x)~|~x\in X\}$ is closed in $X\times X$.

If $X$ is Hausdorff then if a sequence $\{x_n\}_{n=1}^{\infty}$ of elements of $X$ converges in $X$, then the limit is unique. So a sequence in the given set, $\{(x_n,x_n)\}_{n=1}^{\infty}$ has to go to the point $(x,x)$ if $\lim_{n\rightarrow\infty}x_n=x$, and it will not converge if the sequence $\{x_n\}_{n=1}^{\infty}$ doesn't. Hence the given set is closed.

How can I show the converse.

QED
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1 Answers1

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Define:$$\Delta:=\{\langle x,x\rangle\mid x\in X\}$$

Let it be that $u,v\in X$ with $u\neq v$ or equivalently $\langle u,v\rangle\in\Delta^{\complement}$.

$\Delta^{\complement}$ is an open set in $X\times X$ so open sets $U,V$ must exists with $\langle u,v\rangle\in U\times V\subseteq\Delta^{\complement}$.

Now observe that $U\cap V=\varnothing$ while $u\in U$ and $v\in V$.

drhab
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  • Can any open set of $X\times X$ be represented as $U\times V$ for $U,V$ open in$X$. – QED Nov 09 '17 at 15:51
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    No. The collection ${U\times V\mid U,V\in\tau_X}\subseteq\tau_{X\times X}$ is not the topology (so $\neq\tau_{X\times X}$) but serves as a base of the that topology. – drhab Nov 09 '17 at 15:54