225: Open Sets Whose Complements Are Finite and Empty Set Is Topology

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A description/proof of that open sets whose complements are finite and empty set is topology

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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 topological space.
• The reader admits 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.
• The reader admits the proposition 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 for any set, the set of subsets which are subsets whose complements are finite and the empty set constitutes a topology.

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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 set, $S$, the set of subsets, $O$, such that $S_{\alpha }\in O$ if and only if $S\setminus S_{\alpha }$ is finite or $S_{\alpha }=\mathrm{\varnothing }$ constitutes a topology.

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2: Proof

$S\in O$ as $S\setminus S=\mathrm{\varnothing }$ is finite. $\mathrm{\varnothing }\in O$.

Is ${\cup }_{\alpha }{S}_{\alpha }$ open for any $\{\alpha \}$? $S\setminus {\cup }_{\alpha }{S}_{\alpha }={\cap }_{\alpha }S\setminus S_{\alpha }$, 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 $S\setminus S_{\alpha }$ is finite, ${\cap }_{\alpha }S\setminus S_{\alpha }$ is finite.

Is ${\cap }_i{S}_i$ open for any finite $\{i\}$? $S\setminus {\cap }_i{S}_i={\cup }_{i}S\setminus S_i$, by the proposition for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets. As $S\setminus S_i$ is finite and $\{i\}$ is finite, ${\cup }_{i}S\setminus S_i$ is finite.

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References

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