Locally compact preregular spaces are completely regular

Question

First some definitions:

A space is called locally compact if each point has at least one compact nbhd.

Two points $a$ and $b$ are called topologically distinguishable if there is some open set containing $a$ and not $b$, or containing $b$ and not $a$. For example, in a $T_0$ space any two points are topologically distinguishable.

A space is called preregular (or $R_1$) if any two topologically distinguishable points are separated by disjoint nbhds. It's a generalization of both Hausdorff spaces and regular spaces.

It is a well-known result that every locally compact Hausdorff space is completely regular. Now wikipedia as well this answer from Tyrone mention that the same holds if Hausdorff is replaced by regular, or even more generally by preregular.

Theorem: Every locally compact preregular space is completely regular.

Can anyone provide a proof?

Answer 1

Consider the equivalence relation on your space, say we call it $X$, according to which $x R y$ iff $x$ and $y$ are indistinguishable, and let $Y=X/R$.

Observe that every open subset of $X$ is saturated under $R$, almost by definition! Therefore the quotient map $q:X\to Y$ is an open map.

It then follows that $Y$ is locally compact. Moreover the preregular hypothesis now easily implies that $Y$ is Hausdorff.

If $C$ is a closed subset of $X$, and $x\in X\setminus C$, then $q(C)$ is closed and $q(x)\notin q(C)$.

The classical result applied to $Y$ now provides a function $f:Y\to[0,1]$ such that $f(q(x))=1$, and $f=0$ on $q(C)$.

The composition $f\circ q$ now satisfies the desired properties.