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.