317: Intersection or Finite Union of Closed Sets Is Closed

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A description/proof of that intersection or finite union of closed sets is closed

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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 closed set.
• The reader admits the proposition that for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets.
• The reader admits the proposition that for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets.

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Target Context

• The reader will have a description and a proof of the proposition that the intersection of any possibly uncountable number of closed sets or the union of any finite number of closed sets is closed.

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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, the intersection of any possibly uncountable number of closed sets, ${\cap }_{\alpha }{C}_{\alpha }$ where $C_{\alpha }\subseteq T$ closed where $\alpha$ is any possibly uncountable indices set, or the union of any finite number of closed sets, ${\cup }_i{C}_i$ where $C_i\subseteq T$ closed where $i$ is any finite indices set, is closed.

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2: Proof

The complement of the intersection, $T\setminus {\cap }_{\alpha }{C}_{\alpha }$, equals $T\setminus {\cap }_{\alpha }(T\setminus U_{\alpha })$ where $U_{\alpha }:=T\setminus C_{\alpha }$ open. But ${\cap }_{\alpha }(T\setminus U_{\alpha })=T\setminus {\cup }_{\alpha }{U}_{\alpha }$ by the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets. As ${\cup }_{\alpha }{U}_{\alpha }$ is open, $T\setminus {\cup }_{\alpha }{U}_{\alpha }$ is closed, so, $T\setminus {\cap }_{\alpha }{C}_{\alpha }$ is open.

The complement of the union, $T\setminus {\cup }_i{C}_i$, equals $T\setminus {\cup }_i(T\setminus U_i)$ where $U_i:=T\setminus C_i$ open. But ${\cup }_i(T\setminus U_i)=T\setminus {\cap }_i{U}_i$ by the proposition that for any set, the union of the complements of any subsets is the complement of the intersection of the subsets. As ${\cap }_i{U}_i$ is open, $T\setminus {\cap }_i{U}_i$ is closed, so, $T\setminus {\cup }_i{C}_i$ is open.

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References

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