If Λ is a open subset of a metric space and $K⊆Λ_ε:=\{x:d(x,Λ^c)>ε\}$ is compact, is there a compact $L⊆Λ_ε$ s.t. $B_δ(x)⊆L$ for all $x∈K$?

Question

Let

• $(M,d)$ be a locally compact metric space
• $\overline B_\delta(x)$ denote the closed $\delta$-ball of $x$ in $M$
• $\Lambda\subseteq M$ be open

Now, let $\varepsilon>\delta>0$, $$\Lambda_\varepsilon:=\left\{x\in M:d(x,\Lambda^c)>\varepsilon\right\}$$ and $K\subseteq\Lambda_\varepsilon$ be compact.

Are we able to find a compact $\tilde K\subseteq\Lambda_\varepsilon$ with $\overline B_\delta(x)\subseteq\tilde K$ for all $x\in K$?

Answer 1

See the answer of Cover a compact set with closed balls.

The essence of your question is the following:

If $V$ is open in $M$ and $K \subset V$ compact, are we able to find $\tilde{K} \subset V$ such that $\overline{B}_\delta(x) \subset \tilde{K}$ for all $x \in K$? Clearly, $\delta$ has to be sufficiently small.

Define the distance of two subsets $A,B \subset M$ by $d(A,B) = inf \lbrace d(a,b) \mid a \in A, b \in B \rbrace$. If $A$ is compact, $B$ is closed and $A \cap B = \emptyset$, then it is easy to see that $d(A,B) > 0$. Choose an open neighborhood $U$ of $K$ such that $\overline{U}$ is compact and contained in $V$. Then $\delta = d(K,M-U) > 0$ and $B = \lbrace x \in M \mid d(x,K) \le \delta \rbrace \subset \overline{U}$. Clearly $B$ contains all $\overline{B}_\delta(x)$, $x \in K$.