proof of exhaustion by compact sets for ${\mathbb{R}}^{n}$
First consider $A\subset {\mathbb{R}}^{n}$ to be a bounded open set and designate the open ball centered at $x$ with radius $r$ by ${B}_{r}(x)$
Construct ${C}_{n}={\bigcup}_{x\in \partial A}{B}_{\frac{1}{n}}(x)$, where $\partial A$ is the boundary of $A$ and define ${K}_{n}=A\backslash {C}_{n}$.

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${K}_{n}$ is compact.
It is bounded since ${K}_{n}\subset A$ and $A$ is by assumption^{} bounded. ${K}_{n}$ is also closed. To see this consider $x\in \partial {K}_{n}$ but $x\notin {K}_{n}$. Then there exists $y\in \partial A$ and $$ such that $x\in {B}_{r}(y)$. But ${B}_{r}(y)\cap {K}_{n}=\{\}$ because ${B}_{\frac{1}{n}}(y)\cap {K}_{n}=\{\}$ and $$. This implies that $x\notin \partial {K}_{n}$ and we have a contradiction^{}. ${K}_{n}$ is therefore closed.

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${K}_{n}\subset \mathrm{int}{K}_{n+1}$
Suppose $x\in {K}_{n}$ and $x\notin \mathrm{int}{K}_{n+1}$. This means that for all $y\in \partial A$, $x\in \overline{{B}_{\frac{1}{n+1}}(y)}\vee x\in {\mathbb{R}}^{n}\backslash A$. Since $x\in {K}_{n}\u27f9x\in A$ we must have $x\in \overline{{B}_{\frac{1}{n+1}}(y)}$. But $x\in \overline{{B}_{\frac{1}{n+1}}(y)}\subset {B}_{\frac{1}{n}}(y)\u27f9x\notin {K}_{n}$ and we have a contradiction.

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${\bigcup}_{n=1}^{\mathrm{\infty}}{K}_{n}=A$
Suppose $x\in A$, since $A$ is open there must exist $r>0$ such that ${B}_{r}(x)\subset A$. Considering $n$ such that $$ we have that $x\notin {B}_{\frac{1}{n}}(y)$ for all $y\in \partial A$ and thus $x\in {K}_{n}$.
Finally if $A$ is not bounded consider ${A}_{k}=A\cap {B}_{k}(0)$ and define ${K}_{n}={\bigcup}_{k=1}^{n}{K}_{k,n}$ where ${K}_{k,n}$ is the set resulting from the previous construction on the bounded set ${A}_{k}$.

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${K}_{n}$ will be compact because it is the finite union of compact sets.

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${K}_{n}\subset \mathrm{int}{K}_{n+1}$ because ${K}_{k,n}\subset \mathrm{int}{K}_{k,n+1}$ and $\mathrm{int}(A\cup B)\subset \mathrm{int}A\cup \mathrm{int}B$

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${\bigcup}_{n=1}^{\mathrm{\infty}}{K}_{n}=A$
First find $k$ such that $x\in {A}_{k}$. This will always be possible since all it requires is that $k>x$. Finally since $n>k\u27f9{K}_{k,n}\subset {K}_{n,n}$ by construction the argument for the bounded case is directly applicable.
Title  proof of exhaustion by compact sets for ${\mathbb{R}}^{n}$ 

Canonical name  ProofOfExhaustionByCompactSetsFormathbbRn 
Date of creation  20130322 15:51:38 
Last modified on  20130322 15:51:38 
Owner  cvalente (11260) 
Last modified by  cvalente (11260) 
Numerical id  5 
Author  cvalente (11260) 
Entry type  Proof 
Classification  msc 5300 