# 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 $K_{n}=A\backslash C_{n}$.

• $K_{n}$ 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 $0 such that $x\in B_{r}(y)$. But $B_{r}(y)\cap K_{n}=\{\}$ because $B_{\frac{1}{n}}(y)\cap K_{n}=\{\}$ and $0. This implies that $x\notin\partial K_{n}$ and we have a contradiction. $K_{n}$ is therefore closed.

• $K_{n}\subset\operatorname{int}K_{n+1}$

Suppose $x\in K_{n}$ and $x\notin\operatorname{int}K_{n+1}$. 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 $x\in K_{n}\implies x\in A$ we must have $x\in\overline{B_{\frac{1}{n+1}}(y)}$. But $x\in\overline{B_{\frac{1}{n+1}}(y)}\subset B_{\frac{1}{n}}(y)\implies x\notin K% _{n}$ and we have a contradiction.

• $\bigcup_{n=1}^{\infty}K_{n}=A$

Suppose $x\in A$, since $A$ is open there must exist $r>0$ such that $B_{r}(x)\subset A$. Considering $n$ such that $\frac{1}{n} 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 $A_{k}$.

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

• $K_{n}\subset\operatorname{int}K_{n+1}$ because $K_{k,n}\subset\operatorname{int}K_{k,n+1}$ and $\operatorname{int}(A\cup B)\subset\operatorname{int}A\cup\operatorname{int}B$

• $\bigcup_{n=1}^{\infty}K_{n}=A$

First find $k$ such that $x\in A_{k}$. This will always be possible since all it requires is that $k>|x|$. Finally since $n>k\implies K_{k,n}\subset K_{n,n}$ by construction the argument for the bounded case is directly applicable.

Title proof of exhaustion by compact sets for $\mathbb{R}^{n}$ ProofOfExhaustionByCompactSetsFormathbbRn 2013-03-22 15:51:38 2013-03-22 15:51:38 cvalente (11260) cvalente (11260) 5 cvalente (11260) Proof msc 53-00