315: For Locally Finite Open Cover of Topological Space, Closure of Union of Open Sets Is Union of Closures of Open Sets

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A description/proof of that for locally finite open cover of topological space, closure of union of open sets is union of closures of open sets

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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 locally finite cover of topological space.
• The reader knows a definition of closure of subset.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any locally finite open cover of any topological space, the closure of the union of any possibly uncountable open sets in the cover is the union of the closures of the open sets.

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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, $T$, any locally finite open cover, $\{U_{\alpha }\subseteq T|\alpha \in A\}$ where $A$ is a possibly uncountable indexes set, and any possibly uncountable open sets, $\{U_{\beta }|\beta \in B\}$ where $B\subseteq A$, $\overline{{\cup }_{\beta \in B}{U}_{\beta }}={\cup }_{\beta \in B}\overline{U_{\beta }}$.

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2: Proof

$U:={\cup }_{\beta \in B}{U}_{\beta }$.

For any $p\in \bar{U}$, $p\in U$, or $p\notin U$ and $p$ is an accumulation point of $U$ by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset. If $p\in U$, $p\in U_{\beta }$ for a $\beta \in B$, so, $p\in {\cup }_{\beta \in B}\overline{U_{\beta }}$. If $p\notin U$ and $p$ is an accumulation point of $U$, for any neighborhood, $N_p\subseteq T$, of $p$, $N_p\cap U\ne \mathrm{\varnothing }$. In fact, $N_p\cap U_{\beta }\ne \mathrm{\varnothing }$ for a fixed $\beta \in B$, because if there was a neighborhood, $N_{p-\beta }$, of $p$ such that $N_{p-\beta }\cap U_{\beta }=\mathrm{\varnothing }$ for each $\beta \in B$, while there is a neighborhood of $p$, $N_p$, that intersects only finite number of $U_{\alpha }$s where $\alpha \in A$, denoted by $\{U_{\gamma }|\gamma \in C\}$ where $C$ is a finite indexes set, $C^{\prime }:=C\cap B$. $C^{\prime }\ne \mathrm{\varnothing }$ because if $C^{\prime }=\mathrm{\varnothing }$, $N_p\cap U_{\beta }=\mathrm{\varnothing }$ for every $\beta \in B$, then, $N_p\cap U=\mathrm{\varnothing }$, a contradiction against $p$'s being an accumulation point of $U$; $N_p\cap {\cap }_{{\gamma }^{\prime }\in C^{\prime }}{N}_{p-{\gamma }^{\prime }}$ would be a neighborhood of $p$ and would not intersect any $U_{\beta }$ such that $\beta \in B$, then it would not intersect $U$, a contradiction against $p$'s being an accumulation point of $U$. So, $p$ is an accumulation point of $U_{\beta }$ for a $\beta \in B$, so, $p\in \overline{U_{\beta }}$. So, $p\in {\cup }_{\beta \in B}\overline{U_{\beta }}$.

For any $p\in {\cup }_{\beta \in B}\overline{U_{\beta }}$, $p\in \overline{U_{\beta }}$ for a $\beta \in B$. $p\in U_{\beta }$, or $p\notin U_{\beta }$ and $p$ is an accumulation point of $U_{\beta }$. If $p\in U_{\beta }$, $p\in U$, so, $p\in \bar{U}$. If $p\notin U_{\beta }$ and $p$ is an accumulation point of $U_{\beta }$, for any neighborhood, $N_p$, of $p$, $N_p\cap U_{\beta }\ne \mathrm{\varnothing }$, so, $N_p\cap U\ne \mathrm{\varnothing }$, so, $p$ is an accumulation point of $U$. So, $p\in \bar{U}$.

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References

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