388: For Hausdorff Topological Space and 2 Disjoint Compact Subsets, There Are Disjoint Open Subsets Each of Which Contains Compact Subset

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A description/proof of that for Hausdorff topological space and 2 disjoint compact subsets, there are disjoint open subsets each of which contains compact subset

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of Hausdorff topological space.
• The reader knows a definition of compact subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any Hausdorff topological space and its any 2 disjoint compact subsets, there are some disjoint open subsets each of which contains one of the compact subsets.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any Hausdorff topological space, $T$, and any disjoint compact subsets, $S_1,S_2\subseteq T$, such that $S_1\cap S_2=\mathrm{\varnothing }$, there are some open subsets, $U_1,U_2\subseteq T$, such that $S_i\subseteq U_i$ and $U_1\cap U_2=\mathrm{\varnothing }$.

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2: Proof

For any point, $p_{2,\alpha }\in S_2$, for each point, $p_{1,\beta }\in S_1$, there are some disjoint open neighborhoods, $U_{p_{1,\beta }}\cap U_{p_{2,\alpha ,\beta }}=\mathrm{\varnothing }$. $\{U_{p_{1,\beta }}\}$ is an open cover of $S_1$ and has a finite subcover, $\{U_{p_{1,i}}\}$. There is the corresponding $\{U_{p_{2,\alpha ,i}}\}$. Let us define $U_{p_{1,\alpha }}:={\cup }_i{U}_{p_{1,i}}$ and $U_{p_{2,\alpha }}:=\cap U_{p_{2,\alpha ,i}}$, where $U_{p_{1,\alpha }}$ is indexed with $\alpha$, because it depends on $p_{2,\alpha }$. $U_{p_{i,\alpha }}$ is open and $U_{p_{2,\alpha }}$ is nonempty (because $p_{2,\alpha }$ is contained) open. $S_1\subseteq U_{p_{1,\alpha }}$ and $U_{p_{1,\alpha }}\cap U_{p_{2,\alpha }}=\mathrm{\varnothing }$, because for any $p\in U_{p_{2,\alpha }}$, $p\in U_{p_{2,\alpha ,i}}$ for each $i$ and $p\notin U_{p_{1,i}}$ for each $i$.

$\{U_{p_{2,\alpha }}\}$ where $\alpha$ is moved around is an open cover of $S_2$ and has a finite subcover, $\{U_{p_{2,i}}\}$, and there is the corresponding, $U_{p_{1,i}}$. Let us define $U_1:={\cap }_i{U}_{p_{1,i}}$ and $U_2:={\cup }_i{U}_{p_{2,i}}$, both open. $S_1\subseteq U_1$ and $S_2\subseteq U_2$. $U_1\cap U_2=\mathrm{\varnothing }$, because for any $p\in U_1$, $p\in U_{p_{1,i}}$ for each $i$ and $p\notin U_{p_{2,i}}$ for each $i$.

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References

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