880: For Locally Finite Set of Subsets of Topological Space, Closure of Union of Subsets Is Union of Closures of Subsets

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description/proof of that for locally finite set of subsets of topological space, closure of union of subsets is union of closures of subsets

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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: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of locally finite set of subsets of topological space.
• The reader knows a definition of closure of subset of topological space.
• 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.
• The reader admits the proposition that for any topological space and any set of any subsets, the union of the closures of the subsets is contained in the closure of the union of the subsets.

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Target Context

• The reader will have a description and a proof of the proposition that for any locally finite set of subsets of any topological space, the closure of the union of the subsets is the union of the closures of the subsets.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$
$\{S_{\beta }\subseteq T|\beta \in B\}$: $\in \{\text{ the locally finite sets of subsets of }T\}$, where $B\in \{\text{ the possibly uncountable index sets }\}$
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Statements:
$\overline{{\cup }_{\beta \in B}{S}_{\beta }}={\cup }_{\beta \in B}\overline{S_{\beta }}$
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2: Natural Language Description

For any topological space, $T$, and any locally finite set of subsets, $\{S_{\beta }\subseteq T|\beta \in B\}$, where $B$ is any possibly uncountable indices set, $\overline{{\cup }_{\beta \in B}{S}_{\beta }}={\cup }_{\beta \in B}\overline{S_{\beta }}$.

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3: Proof

Whole Strategy: Step 1: see that ${\cup }_{\beta \in B}\overline{S_{\beta }}\subseteq \overline{{\cup }_{\beta \in B}{S}_{\beta }}$; Step 2: see that $\overline{{\cup }_{\beta \in B}{S}_{\beta }}\subseteq {\cup }_{\beta \in B}\overline{S_{\beta }}$.

Step 1:

${\cup }_{\beta \in B}\overline{S_{\beta }}\subseteq \overline{{\cup }_{\beta \in B}{S}_{\beta }}$, by the proposition that for any topological space and any set of any subsets, the union of the closures of the subsets is contained in the closure of the union of the subsets.

Step 2:

Let us see that $\overline{{\cup }_{\beta \in B}{S}_{\beta }}\subseteq {\cup }_{\beta \in B}\overline{S_{\beta }}$.

Let us define $S:={\cup }_{\beta \in B}{S}_{\beta }$.

For each $p\in \bar{S}$, $p\in S$ or ($p\notin S$ and $p$ is an accumulation point of $S$), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

If $p\in S$, $p\in S_{\beta }\subseteq \overline{S_{\beta }}$ for an $\beta \in B$, so, $p\in {\cup }_{\beta \in B}\overline{S_{\beta }}$.

If $p\notin S$ and $p$ is an accumulation point of $S$, for any neighborhood of $p$, $N_p\subseteq T$, $N_p\cap S\ne \mathrm{\varnothing }$. In fact, for a fixed $\beta \in B$ (which may depend on $p$), for any neighborhood of $p$, $N_p\subseteq T$, $N_p\cap S_{\beta }\ne \mathrm{\varnothing }$, because if there was a neighborhood of $p$, $N_{p,\beta }$, such that $N_{p,\beta }\cap S_{\beta }=\mathrm{\varnothing }$ for each $\beta \in B$, while there was a $N_p^{\prime }$ that intersected only finite number of $S_{\beta }$ s where $\beta \in B$, denoted by $\{S_j|j\in J\}$ where $J\subseteq B$ was a nonempty finite index set (if it was empty, $N_p^{\prime }\cap S=\mathrm{\varnothing }$, a contradiction against $p$'s being an accumulation point of $S$), $N_p^{\prime }\cap {\cap }_{j\in J}{N}_{p,j}$ would be a neighborhood of $p$ and would not intersect any $S_{\beta }$ such that $\beta \in B$, then it would not intersect $S$, a contradiction against $p$'s being an accumulation point of $S$.

So, $p$ is an accumulation point of $S_{\beta }$ for a $\beta \in B$, so, $p\in \overline{S_{\beta }}$. So, $p\in {\cup }_{\beta \in B}\overline{S_{\beta }}$.

So, $\bar{S}=\overline{{\cup }_{\beta \in B}{S}_{\beta }}\subseteq {\cup }_{\beta \in B}\overline{S_{\beta }}$.

So, $\overline{{\cup }_{\beta \in B}{S}_{\beta }}={\cup }_{\beta \in B}\overline{S_{\beta }}$.

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References

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