226: For Set Plus Set as an Element, Open Sets That Are Subsets of Set and Subsets Whose Complements Are Finite Is Topology

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A description/proof of that for set plus set as an element, open sets that are subsets of set and subsets whose complements are finite 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, for the set plus the set as an element, the set of subsets which are subsets of the set and the subsets whose complements are finite 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, $S$, for the set, $S^{\prime }=S\cup \{S\}$, the set of subsets, $O$, such that $S_{\alpha }\in O$ if and only if $S_{\alpha }\subseteq S$ or $S^{\prime }\setminus S_{\alpha }$ is finite constitutes a topology.

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2: Proof

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

Let us denote $U_{\alpha }$ for any $U_{\alpha }\subseteq S$ and $V_{\beta }$ for any $V_{\beta }\subseteq S^{\prime }$ such that $S^{\prime }\setminus V_{\beta }$ is finite.

Is $({\cup }_{\alpha }{U}_{\alpha })\cup ({\cup }_{\beta }{V}_{\beta })$ open for any $\{\alpha \}$ and any non-empty $\{\beta \}$? $S^{\prime }\setminus (({\cup }_{\alpha }{U}_{\alpha })\cup ({\cup }_{\beta }{V}_{\beta }))\subseteq S^{\prime }\setminus {\cup }_{\beta }{V}_{\beta }={\cap }_{\beta }{S}^{\prime }\setminus V_{\beta }$, 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, and it is finite.

Is $({\cup }_{\alpha }{U}_{\alpha })\cup ({\cup }_{\beta }{V}_{\beta })$ open for any $\{\alpha \}$ and the empty $\{\beta \}$? ${\cup }_{\alpha }{U}_{\alpha }$ is a subset of $S$.

Is $({\cap }_i{U}_i)\cap ({\cap }_j{V}_j)$ open for any non-empty finite $\{i\}$ and any finite $\{j\}$? It is a subset of $S$.

Is $({\cap }_i{U}_i)\cap ({\cap }_j{V}_j)$ open for the empty $\{i\}$ and any finite $\{j\}$? $S^{\prime }\setminus {\cap }_j{V}_j={\cup }_j{S}^{\prime }\setminus V_j$, 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. It is finite.

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References

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