354: Complement of Empty-Interior Especially Nowhere Dense Subset Is Dense

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A description/proof of that complement of empty-interior especially nowhere dense subset is 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 closure of the complement of the subset is the complement of the interior 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 empty-interior especially nowhere dense subset, the complement of the subset is 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 empty-interior especially nowhere dense subset, $S\subseteq T$, which means that $intS=\mathrm{\varnothing }$ especially $int\bar{S}=\mathrm{\varnothing }$, the complement of $S$, $T\setminus S$, is dense, which means that $\overline{T\setminus S}=T$.

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2: Proof

$\overline{T\setminus S}=T\setminus intS$, by the proposition that for any subset of any topological space, the closure of the complement of the subset is the complement of the interior of the subset. As $intS=\mathrm{\varnothing }$, $\overline{T\setminus S}=T$. When $int\bar{S}=\mathrm{\varnothing }$, $intS=\mathrm{\varnothing }$, because as $\bar{S}$ does not contain any open subset of $T$, the possibly smaller $S$ does not contain any open subset of $T$.

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References

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