A description/proof of that quotient space of compact topological space is compact
Topics
About: topological space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that any quotient space of any compact topological space is compact.
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 compact topological space, \(T_1\), any quotient space, \(T_2\), such that \(f: T_1 \rightarrow T_2\) where \(f\) is a quotient map, is compact.
2: Proof
Let \(\{U_\alpha\}\) be any open cover of \(T_2\) such that \(\cup_\alpha U_\alpha = T_2\). Each \(f^{-1} (U_\alpha)\) is open on \(T_1\) by the definition of quotient topology. \(\{f^{-1} (U_\alpha)\}\) covers \(T_1\), because \(f^{-1} (T_2) = T_1 = f^{-1} (\cup_\alpha U_\alpha) = \cup_\alpha f^{-1} (U_\alpha)\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets. So, \(\{f^{-1} (U_\alpha)\}\) is an open cover of \(T_1\). As \(T_1\) is compact, there is a finite subcover, \(\{f^{-1} (U_i)\}\). Is \(\{U_i\}\) a cover of \(T_2\)? Let us suppose that there was a \(p \in T_2\) such that \(p \notin U_i\) for each \(i\). Then, \(f^{-1} (p) \cap f^{-1} (U_i) = \emptyset\), because if \(p' \in f^{-1} (p) \cap f^{-1} (U_i)\), \(f (p') = p\) and \(f (p') \in U_i\), which would mean \(p \in U_i\), a contradiction. But as \(f\) is surjective, there would be \(p' \in f^{-1} (p)\) and \(p' \notin \cup_i f^{-1} (U_i)\), so, \(\{f^{-1} (U_i)\}\) would not be any open cover of \(T_1\), a contradiction. So, \(\{U_i\}\) is a finite subcover of \(\{U_\alpha\}\). As there is a finite subcover for any open cover, \(T_2\) is compact.
