My question is regarding Exercise 1-22, which states:
If U is open and C $\subset$ U is compact, show that there is a compact set D such that C $\subset$ interior D and D $\subset$ U.
I could prove this result assuming U is a subset of Euclidean space. Is my assumption correct? Is this true for general spaces?
It holds for every locally compact $T_2$ topological space $(X,\mathcal{T})$ that
If $S \subseteq T \subseteq X$ with $S$ compact and $T$ open, there exists a pre-compact open $V$ with $S \subseteq V \subseteq V^- \subseteq T$.
This is a immediate consequence that $(X,\mathcal{T})$ has a compact local basis. Indeed, for each $x \in S$, there exists a compact neighbourhood $U(x) \ni x$ with $U(x) \subseteq T$. Then $\left(U(x)^\circ\right)_{x \in S}$ forms an open cover of $S$ for which there is a finite sub-cover $(U_j^\circ)_{j=0}^{m-1}$. Define $V = \bigcup_{j=0}^{m-1} U_j^\circ$. Then $S\subseteq V$, and $V^- \subseteq \bigcup_{j=0}^{m-1} U_j$ is a subset of a compact set, and hence compact. Therefore $V$ is pre-compact.