Locally compact and regular topogical space needs to be Hausdorff?

Question

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

Answer 1

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.