290: Compactness of Topological Subset as Subset Equals Compactness as Subspace

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A description/proof of that compactness of topological subset as subset equals compactness as 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 topological space.
• The reader knows a definition of compact topological space.
• The reader knows a definition of compact topological subset.

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Target Context

• The reader will have a description and a proof of the proposition that the compactness of any topological subset as a subset equals the compactness as a 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 subset, $S\subseteq T$, $S$ is compact as a subset if and only $S$ is compact as a topological subspace.

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2: Proof

Let us suppose that $S$ is compact as a subset. Any open cover of $S$ on $T$ has a finite subcover. For any open cover of $S$, $\{U_{\alpha }\}$, on $S$, $U_{\alpha }=U_{\alpha }^{\prime }\cap S$ where $U_{\alpha }^{\prime }$ is an open set on $T$. $\{U_{\alpha }^{\prime }\}$ is an open cover of $S$ on $T$, so, there is a finite subcover, $\{U_j^{\prime }\}$. Then, $\{U_j^{\prime }\cap S\}=\{U_j\}$ is a finite subcover of $\{U_{\alpha }\}$.

Let us suppose that $S$ is compact as a subspace. Any open cover of $S$ on $S$ has a finite subcover. For any open cover of $S$, $\{U_{\alpha }^{\prime }\}$, on $T$, $\{U_{\alpha }=U_{\alpha }^{\prime }\cap S\}$ is an open cover of $S$ on $S$, so, there is a finite subcover, $U_j$. Then, the corresponding $\{U_j^{\prime }\}$ is a finite subcover of $\{U_{\alpha }^{\prime }\}$.

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References

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