316: Closure of Union of Finite Subsets Is Union of Closures of Subsets

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A description/proof of that closure of union of finite 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: Description
• 2: Proof

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Starting Context

• 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 the closure of the union of any finite number of 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: Description

For any topological space, $T$, and any finite number of subsets, $S_i\subseteq T$, $\overline{{\cup }_i{S}_i}={\cup }_i\overline{S_i}$ where the over lines denote the closures.

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2: Proof

$S:={\cup }_i{S}_i$.

While $\bar{S}$ is the smallest closed set that contains $S$ by the definition of closure of set, ${\cup }_i\overline{S_i}$ is a closed set that contains $S$. So, $\bar{S}\subseteq {\cup }_i\overline{S_i}$.

For any $p\in {\cup }_i\overline{S_i}$, $p\in \overline{S_i}$ for an $i$. $p\in S_i$, or $p\notin S_i$ and $p$ is an accumulation point of $S_i$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset. If $p\in S_i$, $p\in S$, so, $p\in \bar{S}$. If $p\notin S_i$ and $p$ is an accumulation point of $S_i$, for any neighborhood, $N_p\subseteq T$, of $p$, $N_p\cap S_i\ne \mathrm{\varnothing }$. If $p\in S$, $p\in \bar{S}$, otherwise, $p\notin S$ and $N_p\cap S\ne \mathrm{\varnothing }$, so, $p$ is an accumulation point of $S$, so, $p\in \bar{S}$. So, ${\cup }_i\overline{S_i}\subseteq \bar{S}$.

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References

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