302: For Topological Space, Subspace Subset That Is Compact on Base Space Is Compact on Subspace

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A description/proof of that for topological space, subspace subset that is compact on base space is compact on subspace

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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 compact subset of topological space.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, any subspace subset that is compact on the base space is compact on the subspace.

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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 subspace, $T_1\subseteq T$, any subset, $S\subseteq T_1$, that is compact on $T$ is compact on $T_1$.

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2: Proof

For any open cover, $\{U_{\alpha }\}$, of $S$ on $T_1$, $U_{\alpha }=U_{\alpha }^{\prime }\cap T_1$ where $U_{\alpha }^{\prime }$ is open on $T$. $\{U_{\alpha }^{\prime }\}$ is an open cover of $S$ on $T$. There is a finite subcover, $\{U_j^{\prime }\}$, of $\{U_{\alpha }^{\prime }\}$. The corresponding $\{U_j\}$ is a finite subcover of $\{U_{\alpha }\}$, because as $S\subseteq {\cup }_j{U}_j^{\prime }$, $S\subseteq ({\cup }_j{U}_j^{\prime })\cap T_1={\cup }_j(U_j^{\prime }\cap T_1)={\cup }_j{U}_j$.

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References

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