# Is the Wallman compactification of a locally compact $T_1$ space locally compact?

Let $$X$$ be a $$T_1$$ topological space which is locally compact (in the sense that every point has a local base of compact neighbourhoods). Let $$W(X)$$ be the Wallman compactification of $$X$$. Is $$W(X)$$ locally compact?

As a step towards this, is the remainder $$X^*= W(X)\setminus X$$ locally compact?

The remainder $$X^*$$ is certainly closed in $$W(X)$$, and hence compact. To see this, suppose that $$X$$ is non-compact and that $$K$$ is a compact neighbourhood of a point $$x\in X$$, with interior $$U$$. Let $$C$$ be the complement of $$U$$ in $$X$$. Then $$C$$ belongs to every free closed ultrafilter $$\cal F$$ on $$X$$, for otherwise, by the disjointness of ultrafilters, there is closed set $$D$$ belonging to $$\cal F$$ disjoint from $$C$$. Then $$D\subseteq U\subseteq K$$, and hence $$D$$ is compact, and thus $$\cal F$$ is a fixed ultrafilter. Hence the closure of $$C$$ in $$W(X)$$ is $$C\cup X^*$$. It follows that $$U$$ is open in $$W(X)$$, and thus that $$X^*$$ is closed.

• I see that the closure of $C$ in $W(X)$ covers $X^*$ but how does this imply the closure of $C$ in $W(X)$ is disjoint from $U$? BTW I like this Q but I cannot answer it. – DanielWainfleet Jun 17 '20 at 3:58
• My understanding is that to get a base for the closed sets in $W(X)$ you take the closed sets $C$ in $X$ and let $C^+$ be the set of ultraclosed filters (a better term than the one I used in the question) of which $C$ is a member. So for the closed set $C$ in the question, $C^+= C\cup X^*$. – user558840 Jun 17 '20 at 9:12