# Nested sequence of compact subsets covering an open set in $\mathbb{R}^n$

Let $$A$$ be an open set in $$\mathbb{R}^n$$. I would like to prove the following result:

There exists a sequence of compact sets $$\{D_i\}$$ with the following properties:

(a) Each $$D_i$$ is a subset of $$A$$.

(b) $$D_i \subset Int(D_{i+1}) \ \forall i$$, where $$Int()$$ denotes interior.

(c) $$\bigcup_{i=1}^\infty D_i = A$$.

(This result is used without proof in Lemma 16.2 of "Analysis on Manifolds" by James Munkres.)

Constructing a sequence with the first 2 properties is easy. We start with an arbitrary compact set $$D_1 \subset A$$ with non-empty interior (e.g. a closed ball centered at some point of A). We then use the following fact: every compact set contained inside an open set $$A$$ can be enclosed in the interior of another compact set contained inside $$A$$. This allows us to construct a sequence $$\{D_i\}$$ satisfying (a) and (b). However I don't know how to ensure that their union is $$A$$.

• We can write $A$ as a countable union of open balls. Can you see what to do from there? – Prototank Oct 25 '18 at 14:39

Take$$D_n=\left\{a\in A\,\middle|\,\|a\|\leqslant n\wedge d(a,\partial A)\geqslant\frac1n\right\}.$$