357: For Subset of Topological Space, Closure of Subset Minus Subset Has Empty Interior

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A description/proof of that for subset of topological space, closure of subset minus subset has empty interior

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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
• 3: Note

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Starting Context

• The reader knows a definition of closure of subset.
• The reader knows a definition of interior of 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 its subset, the closure of the subset minus the subset has the empty interior.

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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 interior of $\bar{S}\setminus S$ is empty, which is $int(\bar{S}\setminus S)=\mathrm{\varnothing }$.

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2: Proof

For any point, $p\in \bar{S}\setminus S$, $p$ is an accumulation point of $S$, and so, any neighborhood of $p$, $N_p\subseteq T$, intersects $S$, which means that $N_p$ is not contained in $\bar{S}\setminus S$. So, there is no open set contained in $\bar{S}\setminus S$.

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3: Note

When $S$ is open on $T$, $\bar{S}\setminus S$ is closed on $T$, by the proposition that any closed set minus any open set is closed, so, $\overline{\stackrel{―}{S}\setminus S}=\bar{S}\setminus S$, so, $int(\overline{\stackrel{―}{S}\setminus S})=\mathrm{\varnothing }$, which means that $\bar{S}\setminus S$ is nowhere dense.

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References

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