Exhaustion by compact sets eventually include fixed compact set.

Question

Suppose that $\Omega \subseteq \mathbb R^n$ non-empty, open, and $n \ge 1$.

We have an increasing collection $(K_i)_{i \in \mathbb N}$ of compact subsets of $\Omega$ with union $\Omega$.

For $K \subset \Omega$ fixed and compact, is it necessarily the case that $K \subseteq K_i$ for $i$ sufficiently large?

I know that this is true when we have the constraint that $K_i \subseteq \mathrm{interior}(K_{i+1})$ for each $i \in \mathbb N$.

I am not convinced that this will be the case without this constraint, but I have had some trouble cooking up a counterexample.

I think one could arise if we took $K$ to have non-empty interior and $K_i$ to have empty interior for each $i$, by taking some pathological choice of $K_i$.

Answer 1

Take $\Omega=B((0,0),1)\subset \mathbb{R}^2$, $K_n$ the union of the closed annulus of inner radius $\dfrac{1}{n}$ and outer radius $1-\dfrac{1}{n}$ and $\{(0,0)\}$, which clearly are compact, increasing and whose union is $\Omega$.

If $K=\left\{\left(\dfrac{1}{n},0\right)\right\}_{n\geq 2}\cup \{(0,0)\},$ which is compact as it is a convergent sequence with its limit, then $K$ is never contained in any $K_i$.

Answer 2

Here's a counterexample: Let $\Omega = B_1(0) \subseteq \mathbb{R}^2$ be the open unit disk and let $K_i = \{r e^{2 \pi i \alpha} \mid r, \alpha \in [0, 1 - 1 / i]\}$ ($i \geq 1$). Let $K = \overline{B_{1 / 2}(0)}$ be the closed disk with radius $1 / 2$ centered at the origin. Clearly $\bigcup_{i \geq 1} K_i = \Omega$, seeing as $1 - 1 / i \overset{i \to \infty}{\longrightarrow} 1$, but $K \nsubseteq K_i$ for all $i$ (since there's "always a slice missing" from $K_i$).

In particular, you don't need to assume the $K_i$ to have empty interior to find a counterexample (Edit: As David Gao points out in the comments, it is in fact impossible for all the $K_i$ to have empty interior since the union of the interiors of the $K_i$ has to be dense in $\Omega$).