384: Closed Set on Closed Topological Subspace Is Closed on Base Space

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A description/proof of that closed set on closed topological subspace is closed on base space

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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 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 closed set on any closed topological subspace is closed on the base space.

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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, $T_1$, and any closed topological subspace, $T_2\subseteq T_1$, of $T_1$, any closed set, $C\subseteq T_2$, on $T_2$, is closed on $T_1$.

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2: Proof

As $C$ is closed on $T_2$, $T_2\setminus C$ is open on $T_2$. For any point, $p\in T_2\setminus C$, there is an open neighborhood, $U_p\subseteq T_2\setminus C$, of $p$ on $T_2$. By the definition of subspace topology, $U_p=U_p^{\prime }\cap T_2$ where $U_p^{\prime }\subseteq T_1$ is open on $T_1$. But $U_p^{\prime }=U_p^{\prime }\cap (T_2\cup (T_1\setminus T_2))=(U_p^{\prime }\cap T_2)\cup (U_p^{\prime }\cap (T_1\setminus T_2))\subseteq (T_2\setminus C)\cup (T_1\setminus T_2)$, because $U_p^{\prime }\cap T_2\subseteq T_2\setminus C$ and $(U_p^{\prime }\cap (T_1\setminus T_2))\subseteq T_1\setminus T_2$. As $(T_2\setminus C)\cup (T_1\setminus T_2)=T_1\setminus C$, $U_p^{\prime }\subseteq T_1\setminus C$, which means that around any point on $T_2\setminus C$, there is an open neighborhood on $T_1$ that is contained in $T_1\setminus C$. As $T_2$ is closed on $T_1$, $T_1\setminus T_2$ is open on $T_1$. For any point, $p\in T_1\setminus T_2$, there is an open neighborhood, $U_p\subseteq T_1\setminus T_2\subseteq T_1\setminus C$, of $p$ on $T_1$, which means that around any point on $T_1\setminus T_2$, there is an open neighborhood on $T_1$ that is contained in $T_1\setminus C$. As $T_1\setminus C=(T_2\setminus C)\cup (T_1\setminus T_2)$, around any point on $T_1\setminus C$, there is an open neighborhood on $T_1$ that is contained in $T_1\setminus C$, which means that $T_1\setminus C$ is open on $T_1$, so, $C$ is closed on $T_1$.

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References

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