188: Closure of Subset Is Union of Subset and Accumulation Points Set of Subset

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A description/proof of that closure of subset is union of subset and accumulation points set of subset

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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 knows a definition of closure of subset.
• The reader admits a local characterization of closure: for any topological space and any subset, any point on the topological space is on the closure of the subset if and only if its every neighborhood contains a point on the subset.

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Target Context

• The reader will have a description and a proof of 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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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 subset, $S\subseteq T$, the closure of $S$, $\bar{S}$, is the union of $S$ and the accumulation points set of $S$, $ac(S)$, which is $\bar{S}=S\cup ac(S)$.

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2: Proof

For any point, $p\in \bar{S}$, if $p\in S$, $p\in S\cup ac(S)$, otherwise, by a local characterization of closure: for any topological space and any subset, any point on the topological space is on the closure of the subset if and only if its every neighborhood contains a point on the subset, every neighborhood of $p$ contains a point of $S$, but as $p$ is not any point on $S$, every neighborhood of $p$ contains a point of $S$ excepting $p$, so, $p$ is an accumulation point of $S$, so, $p\in S\cup ac(S)$.

For any point, $p\in S\cup ac(S)$, if $p\in S$, $p\in \bar{S}$, otherwise, $p$ is an accumulation point of $S$, so, every neighborhood of $p$ contains a point of $S$ by the definition of accumulation point, so, by a local characterization of closure: for any topological space and any subset, any point on the topological space is on the closure of the subset if and only if its every neighborhood contains a point on the subset, $p\in \bar{S}$.

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References

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