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 $x\otin K_{n}$ . 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 $x\otin\\partial K_{n}$ and we have a contradiction. $K_{n}$ is therefore closed.
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$K_{n}\\subset\\operatorname{int}K_{n+1}$
Suppose $x\\in K_{n}$ and $x\otin\\operatorname{int}K_{n+1}$ . 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 $x\\in\\overline{B_{\\frac{1}{n+1}}(y)}\\subset B_{\\frac{1}{n}}(y)\\implies x\otin K%_{n}$ and we have a contradiction.
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$\\bigcup_{n=1}^{\\infty}K_{n}=A$
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 $x\otin B_{\\frac{1}{n}}(y)$ 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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$K_{n}\\subset\\operatorname{int}K_{n+1}$ because $K_{k,n}\\subset\\operatorname{int}K_{k,n+1}$ and $\\operatorname{int}(A\\cup B)\\subset\\operatorname{int}A\\cup\\operatorname{int}B$
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$\\bigcup_{n=1}^{\\infty}K_{n}=A$
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.

单词	ProofOfExhaustionByCompactSetsFormathbbRn
释义	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 $x\otin K_{n}$ . 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 $x\otin\\partial K_{n}$ and we have a contradiction. $K_{n}$ is therefore closed. • $K_{n}\\subset\\operatorname{int}K_{n+1}$ Suppose $x\\in K_{n}$ and $x\otin\\operatorname{int}K_{n+1}$ . 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 $x\\in\\overline{B_{\\frac{1}{n+1}}(y)}\\subset B_{\\frac{1}{n}}(y)\\implies x\otin K%_{n}$ 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 $x\otin B_{\\frac{1}{n}}(y)$ 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.
随便看	BorelCantelliLemma Borel-Cantelli lemma BorelFunctionalCalculus Borel functional calculus BorelGroupoid Borel groupoid BorelGspace Borel G-space BorelIsomorphism Borel isomorphism BorelMeasure Borel measure BorelMorphism Borel morphism Borelsigmaalgebra BorelSpace Borel space BorelSubalgebra Borel subalgebra BorelSubgroup Borel subgroup Borel σ-algebra BornologicalSpace bornological space BorsukUlamTheorem

proof of exhaustion by compact sets for ℝn

proof of exhaustion by compact sets for $\\mathbb{R}^{n}$