270: Topological Space Is Compact Iff for Every Collection of Closed Subsets for Which Intersection of Any Finite Members Is Not Empty, Intersection of Collection Is Not Empty

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A description/proof of that topological space is compact iff for every collection of closed subsets for which intersection of any finite members is not empty, intersection of collection is not empty

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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 compact topological space.
• The reader knows a definition of closed set.

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Target Context

• The reader will have a description and a proof of the proposition that any topological space is compact if and only if for its every collection of closed subsets for which the intersection of any finite members is not empty, the intersection of the collection is not empty.

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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 topological space, $T$, $T$ is compact if and only if for its every collection, $S:=\{C_{\alpha }\}$, of closed subsets for which the intersection of any finite members is not empty, the intersection, $\cap C_{\alpha }$, of the collection is not empty.

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2: Proof

Let us suppose that $\cap C_{\alpha }\ne \mathrm{\varnothing }$. For any open cover, $\{U_{\alpha }\}$, let us think of $S:=\{T\setminus U_{\alpha }\}$. If there was no finite subset, $\{T\setminus U_i\}$, such that ${\cap }_i(T\setminus U_i)=\mathrm{\varnothing }$, ${\cap }_{\alpha }(T\setminus U_{\alpha })\ne \mathrm{\varnothing }$, by the supposition. then, there would be a point, $p\in {\cap }_{\alpha }(T\setminus U_{\alpha })$, so, $p\in T\setminus U_{\alpha }$ for every $\alpha$, $p\notin U_{\alpha }$ for every $\alpha$, so, $\{U_{\alpha }\}$ would not be an open cover, a contradiction, so, there is a finite subset, $\{T\setminus U_i\}$, such that ${\cap }_i(T\setminus U_i)=\mathrm{\varnothing }$. Then, for any point, $p\in T$, if $p\in T\setminus U_i$ for an $i$, $p\notin T\setminus U_j$ for a $j\ne i$, then, $p\in U_j$; if $p\notin T\setminus U_i$ for an $i$, $p\in U_i$; so, anyway, $\{U_i\}$ covers $T$. So, there is a finite subcover for any open cover.

Let us suppose that $T$ is compact. Let us suppose that $\cap C_{\alpha }=\mathrm{\varnothing }$. $\{T\setminus C_{\alpha }\}$ would cover $T$, because for any point, $p\in T$, $p\notin C_{\alpha }$ for an $\alpha$, so, $p\in T\setminus C_{\alpha }$. There would be a finite subcover, $\{T\setminus C_i\}$, but for any point, $p\in T$, $p\in T\setminus C_i$ for an $i$, $p\notin C_i$, so, $p\notin {\cap }_i{C}_i$, so, no point would belong to ${\cap }_i{C}_i$, so, ${\cap }_i{C}_i=\mathrm{\varnothing }$, a contradiction.

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References

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