Question asked on the structure of open sets in $\mathbb{R}^n$

Question

There was a question asked: An open subset $U\subseteq R^n$ is the countable union of increasing compact sets. There Davide gave an answer. Can anyone tell me how the equality holds, and the motivation behind this construction?

Answer 1

Let's first fix a $k$ and consider Davide's set (which I will slightly rewrite) $$X_k := U \cap A_k \cap B_k$$ where $$\begin{gather}A_k := \{x : \lVert x\rVert\leq k\} \\ B_k := \{x : d(x,U^c)\geq k^{-1}\} = \{ x : B ( x; k^{-1} ) \subseteq U\}.\end{gather}$$ One thing to note is that as $B_k \subseteq U$, we actually have that $$X_k = A_k \cap B_k.$$ It is quite easy to show that $A_k$ and $B_k$ are individually closed, and that $A_k$ is bounded, so $X_k$ must be closed and bounded, i.e., compact.

To see that $B_k$ is closed, note that if $x \notin B_k$, then we may take $y \in B ( x ; k^{-1} ) \cap U^c$, and let $\delta = \frac{k - d(x,y)}{2}$. Given any $z \in B ( x; \delta )$, we have that $$d ( z , y ) \leq d ( z,x) + d (x,y) < \delta + d(x,y) = \frac{k+d(x,y)}{2} < k$$ and so $B ( z;k^{-1}) \not\subseteq U$. Therefore $B ( x;\delta ) \subseteq B_k^c$.

It is also easy to show that $X_k \subseteq X_{k+1}$, and so we have an increasing sequence of compact subsets of $U$. It is also clear that $\bigcup_k X_k \subseteq U$, so we need only show the reverse inclusion.

If $x \in U$, then since $U$ is open there is an $k_0 \in \mathbb{N}$ such that $B ( x ; k_0^{-1} ) \subseteq U$. Also, there is a $k_1 \in \mathbb{N}$ such that $\| x \| \leq k_1$. Letting $k = \max \{ k_0 , k_1 \}$ it follows that $x \in A_k \cap B_k = X_k$.

The basic idea of the construction is, I think, as follows:

• The sets $B_k$ consist of those points of $U$ which are "far away" from the boundary of $U$. As $U$ is open, every point in $U$ is some positive distance from the boundary of $U$, and so there must be a $k$ such that $d ( x , U^c ) \geq k^{-1}$. We actually have that $U = \bigcup_k B_k$. However, if $U$ is itself an unbounded set, it could be that certain of the $B_k$ are unbounded, and so it does not suffice to only consider these sets.
• The sets $A_k$ are there to ensure that the given set is bounded. As $\mathbb{R}^n = \bigcup_k A_k$, we also have that $U \subseteq \bigcup_k A_k$.

Answer 2

(answer for the "motivation" part of the question)

I stumbled on two applications of that property although I'm not well-versed enough to develop so I'll just cite "Topological spaces, distributions and kernels", François Treves: (Lemma 10.1 p.87)

1. the property/lemma is used to proove that $C^k(U)$ with some topology defined p.86 is metrizable.
2. in Example II: Spaces of test functions p.131-133, it is used to proove that $C^k(U)$ (and other spaces) are inductive limits of the $C^k(X_k)$ ($X_k$ such that $\bigcup X_k= U$).

Edits: now that I see this question several years later, I would say that the motivation is tautologically to "approximate" the open subset $U$ with countably many compacts. As a consequence,

3. in integration theory this can be used to show density of $\mathcal{C}(\mathbb{R}^n)$ in $L^1(\mathbb{R}^n)$ with the L1 norm. I don't know the exact reason why but there is a notion of inner and outer regularity of the Lebesgue measure, and one of the two involves compact subsets. One of the possible proofs uses that characteristic functions of compact subsets can be approximated by a sequence of continuous functions and then one uses these characteristic functions to approximate more general functions, ultimately approximating any $f\in L^1$.
4. and for topological manifolds (ultimatedly this will also be related to integration, more precisely to existence of partitions of unity), there is a notion of paracompactness which is surprisingly related to the notion of $\sigma$-compacts (it has never clear to me whether this is equivalent to countable at infinity but...), cf. Topology and Geometry, Glen E. Bredon, Thm 12.11 p.38 or Topologie Générale, N. Bourbaki, Théorème 5 "p.82 on the pdf file" or p. I.70.