You are here
Homeproof of exhaustion by compact sets for $\mathbb{R}^n$
Primary tabs
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 $0<r<\frac{1}{n}$ such that $x\in B_{r}(y)$. But $B_{r}(y)\cap K_{n}=\{\}$ because $B_{{\frac{1}{n}}}(y)\cap K_{n}=\{\}$ and $0<r<\frac{1}{n}\implies B_{r}(y)\subset B_{{\frac{1}{n}}}(y)$. 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 $\frac{1}{n}<r$ 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.
Mathematics Subject Classification
5300 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections