Does Regular imply $T_{3.5}$?

Question

Prove or disprove: If a topological space is Hausdorff and locally compact then it is $T_{3.5}.$

I managed to prove that $T$ is regular. But how will I prove that it is $T_{3.5}$?

Answer 1

Let $C=\overline C\subset T$ and $p\in T\setminus C.$

Since T is locally compact there exists open $U\subset T\setminus C$ with $p\in U,$ such that $\overline U$ is compact. Since $T$ is regular there exists open $V$ with $p\in V\subset \overline V\subset U.$

The subspace $\overline U$ is a compact Hausdorff space so it is a normal space. Now $\{x\}$ and $\overline U \setminus V$ are disjoint closed subsets of $\overline U.$ By Urysohnn's Lemma there exists a continuous $f:\overline V\to [0,1]$ with $f(p)=1 $ and $f(\overline U\setminus V)\subset \{0\}.$

Extend $f$ to the domain $T$ by letting $f(T\setminus \overline U)\subset \{0\}.$ To finish, verify that $f:T\to [0,1$ is continuous.

Answer 2

A Hausdorff locally compact space $X$ has a one-point (Aleksandroff) Hausdorff compactification $\alpha X$ that it is a subspace of.

$\alpha X$ is $T_4$ (being compact and Hausdorff) and so $T_{3\frac12}$ (by Urysohn's lemma, essentially). Subspaces of $T_{3\frac12}$ spaces are still $T_{3\frac12}$ hence so is $X$.