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Let $X$ be a topological space. Let $\bar{X}$ denote a compactification of $X$ (i.e., $\bar{X}$ is a compact Hausdorff space such that $X$ is an open dense subspace of $\bar{X}$). Notice that in particular this implies that $X$ is Hausdorff. Define the diagonal $\triangle_X$ of $X$ by $$\triangle_X=\{(x,x) \in X \times X \mid x \in X\}.$$

It is a well known fact that if $X$ is a topological space, then $X$ is Hausdorff if and only if its diagonal is closed. However, I am interested in the following:

Question: Is $\triangle_X$ closed in $\bar{X} \times \bar{X}?$

Any help would be greatly appreciated.

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First, minor issues: your context makes it more appropriate to say a compactification, instead of the compactification. Furthermore, it is not usual to require that $X$ is open in $\overline{X}$, but that will not make a difference.

That said, the response is that $\Delta_X$ is never closed in $\overline{X} \times \overline{X}$ (as long as $\overline{X}$ is not $X$, of course).

One way to see this is simply by handling the dense hypothesis: pick $p$ a point of $\overline{X}$ which is not in $X$. Given a neighbourhood of the form $U_1 \times U_2$ of $(p,p)$, since $X$ is dense in $\overline{X}$ it follows that there exists a $q$ in $X$ such that $q \in U_1 \cap U_2$, and thus $(q,q) \in U_1 \times U_2$. Thus $(p,p)$ is a limit point of $\Delta_X$ which is not in $\Delta_X$.

Another way is the following: suppose that $\Delta_X$ is closed in $\overline{X} \times \overline{X}$. Then it is compact, since the later is compact. Since the mapping $\pi_1: \overline{X} \times \overline{X} \to \overline{X}$ is continuous, $\pi_1(\Delta_X)=X$ is compact, a contradiction.

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  • $\begingroup$ Exceptional answer. Thank you. $\endgroup$ – Pawel Aug 4 '18 at 16:54
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This is not true. Consider $X$ to be the open interval $(0,1)$, and $\overline X=[0,1]$. Then $\Delta_X=\{(t,t):t\in(0,1)\}$ is not a closed subset of $\overline X\times \overline X=[0,1]\times[0,1]$, as it has a limit point $(0,0)\notin\Delta_X.$

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  • $\begingroup$ Simple and elegant counterexample. +1 $\endgroup$ – Pawel Aug 4 '18 at 16:54
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Certainly not, unless $X$ is compact. Because $\Delta$ is homeomorphic to $X$, and $\Delta$ closed in $\overline X\times\overline X$ implies $\Delta$ compact.

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