A description/proof of that for topological space, subspace subset that is compact on base space is compact on subspace
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 subspace subset that is compact on the base space is compact on the 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 subspace, \(T_1 \subseteq T\), any subset, \(S \subseteq T_1\), that is compact on \(T\) is compact on \(T_1\).
2: Proof
For any open cover, \(\{U_\alpha\}\), of \(S\) on \(T_1\), \(U_\alpha = U'_\alpha \cap T_1\) where \(U'_\alpha\) is open on \(T\). \(\{U'_\alpha\}\) is an open cover of \(S\) on \(T\). There is a finite subcover, \(\{U'_j\}\), of \(\{U'_\alpha\}\). The corresponding \(\{U_j\}\) is a finite subcover of \(\{U_\alpha\}\), because as \(S \subseteq \cup_j U'_j\), \(S \subseteq (\cup_j U'_j) \cap T_1 = \cup_j (U'_j \cap T_1) = \cup_j U_j\).
