295: For Topological Space, Compact Subset of Subspace Is Compact on Base Space

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A description/proof of that for topological space, compact subset of subspace is compact 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 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 compact subset of any subspace is compact 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, $T$, and any subspace, $T_1\subseteq T$, any compact subset, $S\subseteq T_1$, of $T_1$ is compact on $T$.

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2: Proof

For any open cover, $\{U_{\alpha }^{\prime }\}$, of $S$ on $T$, $\{U_{\alpha }=U_{\alpha }^{\prime }\cap T_1\}$ is an open cover of $S$ on $T_1$. There is a finite subcover, $\{U_j\}$, of $\{U_{\alpha }\}$. The corresponding $\{U_j^{\prime }\}$ is a finite subcover of $\{U_{\alpha }^{\prime }\}$.

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References

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