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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.

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  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Jun 17 '20 at 3:58
  • $\begingroup$ 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, $C^+= C\cup X^*$. $\endgroup$ – user558840 Jun 17 '20 at 9:12

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