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Consider X a topological space, that is locally compact and regular. Is X necessarily a Hausdorff space? Thank you in advance.

Guilherme N.
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  • No, see for instance an indiscrete topological space with at least two points. However, a regular space is Hausdorff if and only if it's T0, regardless of local compactness. – Sassatelli Giulio May 30 '23 at 15:58

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Any $T_0$ regular space is Hausdorff. Given points $x, y\in X$ such that $x\ne y$, either $x$ is not in the closure of $\{y\}$, or vice versa. In either case, due to regularity you can separate the points by disjoint open neighborhoods.

If you drop the $T_0$ assumption - it's possible to create artificial counter-examples. Start from any locally compact regular Hausdorff space $X$, take some point $x_0$ in $X$, and double it by replacing it by two distinct points. Now define the topology with the same open sets as before - either containing both points or containing neither point. This topology doesn't differentiate between the two points in any way, and otherwise has the same properties as before.

Chad K
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