A description/proof of that compactness of topological subset as subset equals compactness as subspace
Topics
About: topological space
The table of contents of this article
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.
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.
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.
Main Body
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.
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 \(S\), \(U_\alpha = U'_\alpha \cap S\) where \(U'_\alpha\) is an open set on \(T\). \(\{U'_\alpha\}\) is an open cover of \(S\) on \(T\), so, there is a finite subcover, \(\{U'_j\}\). Then, \(\{U'_j \cap S\} = \{U_j\}\) 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 \(S\), \(\{U'_\alpha\}\), on \(T\), \(\{U_\alpha = U'_\alpha \cap S\}\) is an open cover of \(S\) on \(S\), so, there is a finite subcover, \(U_j\). Then, the corresponding \(\{U'_j\}\) is a finite subcover of \(\{U'_\alpha\}\).
