A description/proof of that topological sum of paracompact topological spaces is paracompact
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological sum.
- The reader knows a definition of paracompact topological space.
- The reader admits the proposition that the topological sum of any possibly uncountable number of Hausdorff topological spaces is Hausdorff.
Target Context
- The reader will have a description and a proof of the proposition that the topological sum of any possibly uncountable number of paracompact topological spaces is paracompact.
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 paracompact topological spaces, \(\{T_\alpha\vert \alpha \in A\}\) where \(A\) is a possibly uncountable indices set, the topological sum, \(T := \coprod_\alpha T_\alpha\), is paracompact.
2: Proof
\(T\) is a Hausdorff topological space as the topological sum of Hausdorff topological spaces, by the proposition that the topological sum of any possibly uncountable number of Hausdorff topological spaces is Hausdorff.
For any open cover, \(S := \{U_\beta \subseteq T \vert \beta \in B\}\) where \(B\) is any possibly uncountable indices set, of \(T\), \(S_\alpha := \{U_\beta \cap T_\alpha \vert \beta \in B\}\) is an open cover of \(T_\alpha\). As \(T_\alpha\) is paracompact, there is a locally finite refinement, \({S_\alpha}' := \{{U_{\alpha-\gamma}}' \subseteq U_\beta \cap T_\alpha \vert \gamma \in C\}\) where \(C\) is a possibly uncountable indices set. \(\cup_\alpha {S_\alpha}'\) is a locally finite refinement of \(S\), because \({U_{\alpha-\gamma}}'\) is open on \(T\), \(\cup_\alpha {S_\alpha}'\) covers \(T\), \({U_{\alpha-\gamma}}' \subseteq U_\beta\), and around any point, \(p \in T\), \(p \in T_\alpha\), and there is a neighborhood, \(N_p\), contained in \(T_\alpha\) that intersects only finite elements of \({S_\alpha}'\), because \(T_\alpha\) is open and disjoint from any \(T_\beta\) such that \(\beta \neq \alpha\).
3: Note
When \(T\) is any Hausdorff topological space that is the disjoint union of any possibly uncountable number of open paracompact subspaces, \(T\) is paracompact, because \(T\) is the topological sum of the subspaces.
The subspaces have to be open and disjoint in order to apply this proposition, because otherwise, \(T\) would not be any topological sum.
