215: Subset on Topological Subspace Is Closed iff There Is Closed Set on Base Space Whose Intersection with Subspace Is Subset

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A description/proof of that subset on topological subspace is closed iff there is closed set on base space whose intersection with subspace is 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 subspace topology.
• The reader knows a definition of closed set.

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Target Context

• The reader will have a description and a proof of the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is 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 its subspace, $T_1\subseteq T$, any subset, $S\subseteq T_1$, is closed on $T_1$ if and only if there is a closed set, $C\subseteq T$, such that $S=C\cap T_1$.

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2: Proof

Suppose that there is a $C$. $T_1\setminus S=T_1\setminus (C\cap T_1)=(T\setminus C)\cap T_1$, because for any point, $p\in T_1\setminus (C\cap T_1)$, $p\notin C\cap T_1$, $p\notin C$, $p\in T\setminus C$, so, $p\in (T\setminus C)\cap T_1$; for any point, $p\in (T\setminus C)\cap T_1$, $p\in T\setminus C$, $p\notin C$, $p\notin C\cap T_1$, $p\in T_1\setminus (C\cap T_1)$. So, $T_1\setminus S$ is open on $T_1$, and $S$ is closed on $T_1$.

Suppose that $S$ is closed on $T_1$. $T_1\setminus S$ is open on $T_1$, so, $T_1\setminus S=U\cap T_1$ where $U\subseteq T$ is open on $T$. $S=(T\setminus U)\cap T_1$, because for any point, $p\in S$, $p\in T\setminus U$, because otherwise, $p\in U$ then $p\in T_1\setminus S$, a contradiction, $p\in (T\setminus U)\cap T_1$; for any $p\in (T\setminus U)\cap T_1$, $p\in S$, because otherwise, $p\notin S$, $p\in T_1\setminus S$, $p\in U\cap T_1$, $p\in U$, a contradiction. $C$ can be take to be $T\setminus U$.

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References

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