A description/proof of that for topological space, compact subset of subspace is compact on base space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of subspace topology.
- The reader knows a definition of compact subset of topological space.
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.
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 subspace, \(T_1 \subseteq T\), any compact subset, \(S \subseteq T_1\), of \(T_1\) is compact on \(T\).
2: Proof
For any open cover, \(\{U'_\alpha\}\), of \(S\) on \(T\), \(\{U_\alpha = U'_\alpha \cap T_1\}\) is an open cover of \(S\) on \(T_1\). There is a finite subcover, \(\{U_j\}\), of \(\{U_\alpha\}\). The corresponding \(\{U'_j\}\) is a finite subcover of \(\{U'_\alpha\}\).
