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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?

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Will you please link to the answer you are asking about? (Note that you have enough reputation to post comments on all posts, and in some cases it may suffice to comment to ask directly for clarification of answers.) – Jonas Meyer Dec 15 '12 at 5:44
@JonasAn open subset U subset of R^n is the countable union of increasing compact sers – Ester Dec 15 '12 at 6:02
I think this is the question the OP is talking about. – arjafi Dec 15 '12 at 7:18
@Arthur: Thank you. – Jonas Meyer Dec 15 '12 at 7:29
up vote 2 down vote accepted

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 $U \subseteq \bigcup_k A_k$.
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(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 $\bigcup X_k= U$).
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