196: Open Set Intersects Subset if It Intersects Closure of Subset

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A description/proof of that open set intersects subset if it intersects closure 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.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, any open set intersects any subset if it intersects the closure 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, $S\subseteq T$, any open set, $U$, intersects $S$ if $U$ intersects the closure of $S$, $\bar{S}$.

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2: Proof

Suppose that $U\cap \bar{S}\ne \mathrm{\varnothing }$ and $U\cap S=\mathrm{\varnothing }$. $S\subseteq (T\setminus U)$ and $\bar{S}\cap (T\setminus U)\subset \bar{S}$. So, $S\subseteq \bar{S}\cap (T\setminus U)\subset \bar{S}$, a contradiction, because $\bar{S}\cap (T\setminus U)$ would be a closed set that contains $S$, which (the closed set) would be smaller than the closure of $S$. So, $U\cap S\ne \mathrm{\varnothing }$ if $U\cap \bar{S}\ne \mathrm{\varnothing }$.

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References

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