Property of a locally compact Hausdorff topological space Let $Y$ be a locally compact and Hausdorff topological space;  let $A$ be a compact subspace of $Y$, and $B$ an open subset of $Y$ conteining $A$.
How can I find an open set $U$ (open in $Y$) and a compact space $K$ such that $$A\subset U\subset K\subset B\subset Y?$$
 A: The key point here is that in a locally compact Hausdorff space, every neighbourhood of a point $x$ contains a compact neighbourhood of $x$. Since if $N$ is a neighbourhood of $x$, then the interior $\overset{\Large\circ}{N}$ is also a neighbourhood of $x$, we may assume that our neighbourhood is open.
So let $X$ be a locally compact Hausdorff space, $x\in X$, and $U$ an open neighbourhood of $x$. Let $K$ be a compact neighbourhood of $x$, as guaranteed by local compactness. Since $U$ is open, $C := K \setminus U$ is a compact set, and since $x\in U$, we have $x \notin C$. Since $X$ is Hausdorff, for every $y \in C$ there are disjoint open neighbourhoods $V_y$ of $x$ and $W_y$ of $y$. Since $C$ is compact, and $$C \subset \bigcup_{y\in C} W_y,$$
there are finitely many points $y_1,\dotsc, y_n \in C$ with
$$C \subset \bigcup_{k = 1}^n W_{y_k}.$$
Then
$$V := \overset{\Large\circ}{K} \cap U \cap \bigcap_{k = 1}^n V_{y_k}$$
is an open neighbourhood of $x$. We have $V \subset K$, so $\overline{V}$ is compact (and contained in $K$). By construction,
$$\overline{V} \cap \bigcup_{k = 1}^n W_{y_k} = \varnothing,$$
so $\overline{V}\cap C = \varnothing$, whence $\overline{V}\subset K \setminus C = K \cap U \subset U$.

Then cover $A$ by finitely many open sets whose closure is compact and contained in $B$.
