355: Complement of Open Dense Subset Is Nowhere Dense

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A description/proof of that complement of open dense subset is nowhere dense

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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 nowhere dense subset.
• The reader knows a definition of dense subset.
• The reader admits the proposition that for any subset of any topological space, the interior of the complement of the subset is the complement of the closure 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 its any open dense subset, the complement of the subset is nowhere dense.

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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 its any open dense subset, $S\subseteq T$, which means that $\bar{S}=T$, the complement of $S$, $T\setminus S$, is nowhere dense, which means that $int(\overline{T\setminus S})=\mathrm{\varnothing }$.

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2: Proof

$int(T\setminus S)=T\setminus \bar{S}$, by the proposition that for any subset of any topological space, the interior of the complement of the subset is the complement of the closure of the subset. As $\bar{S}=T$, $int(T\setminus S)=\mathrm{\varnothing }$. As $S$ is open, $T\setminus S$ is closed, and $\overline{T\setminus S}=T\setminus S$, so, $int(\overline{T\setminus S})=int(T\setminus S)=\mathrm{\varnothing }$.

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References

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