What is the largest possible compact set that can fit inside an open set?

Question

If $U$ is open and $C \subset U$ is compact, there is a compact set $D \subset U$ such that $C \subset \text{interior}(D)$. Now, by inductive reasoning, this means that there is actually a chain (maybe infinite?) of compact sets $C, D, E, F, .....$ such that each one is a subset of the next bigger compact set. Then, what is the biggest possible compact subset? For example, suppose $C=\{1,\frac{1}{2}, \frac {1}{3},....,0 \}$ is the harmonic sequence and $U=(-1,2)$. Then how do we construct $D$? I suspect that the choice of $D, E, ...$ etc. is not unique. In fact we could keep adding elements from the interval $(1,2)$ to $C$ thus forming $D$. Is it right? Thanks.

Answer 1

If $U$ is open and $C \subset U$ is compact, there is a compact set $D \subset U$ such that $C \subset \operatorname{int}(D)$.

This is not always true, you need extra assumptions on the space $X$ containing $U$ and $C$. In fact, it is true if and only iof $X$ is locally compact which means that for each $x \in X$ and each open neighborhood $U$ of $x$ there exists a compact $K \subset X$ such that $x \in \operatorname{int}(K)$ and $K \subset U$. Note that if $X$ is Hausdorff, then $X$ is locally compact iff each point has a compact neighborhood.

1. Your condition implies that $X$ is locally compact (take $C= \{x\}$).

2. If $X$ is locally compact, choose compact $K_x \subset U$ for each each $x \in C$ such that $x \in \operatorname{int}(K_x)$. Finitely many $U_i = \operatorname{int}(K_{x_i})$ cober $C$. Then $D = \bigcup K_{x_i}$ is compact, $C \subset \bigcup U_i \subset \operatorname{int}(D)$ and $D \subset U$.

Considering locally compact $X$, we can state:

An open $U \subset X$ contains a largest compact subset if and only if $U$ is itself compact. (For example $U =[0,1] \subset X = [0,1] \cup [2,3])$.

If $U$ is compact, we may take $D = U$. Then $C \subset \operatorname{int}(D) = U$.

If there is a largest compact $D^* \subset U$ such that $C \subset \operatorname{int}(D^*)$, then $D^* = U$. To see this, let $x \in U$. Take a compact $D \subset U$ such that $C \subset \operatorname{int}(D)$. Then also $D' = D \cup \{x\}$ is a compact subset of $U$ with $C \subset \operatorname{int}(D')$. But we must have $D' \subset D^*$ because $D^*$ is maximal. Thus $D^*$ contains all $x \in U$.