289: Closed Set Minus Open Set Is Closed

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A description/proof of that closed set minus open set is closed

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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 closed set.
• The reader admits the local criterion for openness.

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Target Context

• The reader will have a description and a proof of the proposition that any closed set minus any open set is closed.

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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, $T$, any closed set, $C\subseteq T$, and any open set, $U\subseteq T$, the closed set minus the open set, $C\setminus U$, is closed on $T$.

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2: Proof

For any point, $p\in T\setminus (C\setminus U)$, $p\notin C\setminus U$. 1): $p\notin C$ or 2): $p\in C\cap U$.

For 1), $p\in T\setminus C$, open. There is an open neighborhood, $U_p\subseteq T\setminus C$, of $p$. $U_p\subseteq T\setminus (C\setminus U)$? For any point, $p^{\prime }\in U_p$, $p^{\prime }\notin C$, so, $p^{\prime }\notin C\setminus U$. So, yes, $U_p\subseteq T\setminus (C\setminus U)$.

For 2), there is an open neighborhood, $U_p\subseteq U$, of $p$. $U_p\subseteq T\setminus (C\setminus U)$? For any point, $p^{\prime }\in U_p$, $p^{\prime }\notin C\setminus U$. So, yes, $U_p\subseteq T\setminus (C\setminus U)$.

By the local criterion for openness, $T\setminus (C\setminus U)$ is open, so, $C\setminus U$ is closed.

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References

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