2
$\begingroup$

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?

  • 0
    I think [this](http://math.stackexchange.com/q/136204/8348) is the question the OP is talking about.2012-12-15

2 Answers 2

3

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 $U \subseteq \bigcup_k A_k$.
0

(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 $\bigcup X_k= U$).