879: For Topological Space, Union of Closures of Subsets Is Contained in Closure of Union of Subsets

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description/proof of that for topological space, union of closures of subsets is contained in closure of union 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 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.

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Target Context

• The reader will have a description and a proof of 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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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:
$T$: $\in \{\text{ the topological spaces }\}$
$\{S_{\beta }\subseteq T|\beta \in B\}$: $B\in \{\text{ the possibly uncountable index sets }\}$
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Statements:
${\cup }_{\beta \in B}\overline{S_{\beta }}\subseteq \overline{{\cup }_{\beta \in B}{S}_{\beta }}$, where the overlines denote the closures.
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2: Natural Language Description

For any topological space, $T$, and any set of any subsets, $\{S_{\beta }\subseteq T|\beta \in B\}$, where $B$ is any possibly uncountable index set, ${\cup }_{\beta \in B}\overline{S_{\beta }}\subseteq \overline{{\cup }_{\beta \in B}{S}_{\beta }}$, where the overlines denote the closures.

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3: Proof

Whole Strategy: Step 1: for each $p\in {\cup }_{\beta \in B}\overline{S_{\beta }}$, take a $\beta \in B$ such that $p\in \overline{S_{\beta }}$, and list the 2 cases, Case 1: $p\in S_{\beta }$; Case 2: $p\notin S_{\beta }$ and $p$ is an accumulation point of $S_{\beta }$; Step 2: for Case 1, see that $p\in \overline{{\cup }_{\beta \in B}{S}_{\beta }}$; Step 3: for Case 2, see that $p\in \overline{{\cup }_{\beta \in B}{S}_{\beta }}$; Step 4: conclude the proposition.

Step 1:

For each $p\in {\cup }_{\beta \in B}\overline{S_{\beta }}$, $p\in \overline{S_{\beta }}$ for a $\beta \in B$.

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

Step 2:

When $p\in S_{\beta }$, $p\in {\cup }_{\beta \in B}{S}_{\beta }$, so, $p\in \overline{{\cup }_{\beta \in B}{S}_{\beta }}$.

Step 3:

When $p\notin S_{\beta }$ and $p$ is an accumulation point of $S_{\beta }$, for each neighborhood of $p$, $N_p\subseteq T$, $N_p\cap S_{\beta }\ne \mathrm{\varnothing }$, so, $N_p\cap ({\cup }_{\beta \in B}{S}_{\beta })\ne \mathrm{\varnothing }$, so, $p$ is an accumulation point of ${\cup }_{\beta \in B}{S}_{\beta }$. So, $p\in \overline{{\cup }_{\beta \in B}{S}_{\beta }}$.

Step 4:

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

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References

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