A description/proof of that disjoint union of closed sets is closed in disjoint union topology
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of disjoint union topology.
- The reader knows a definition of closed set.
- The reader admits the proposition that the disjoint union of any complements is the disjoint union of the whole sets minus the disjoint union of the subsets.
Target Context
- The reader will have a description and a proof of the proposition that the disjoint union of any closed sets (at most 1 from each constituent topological space) is closed in the disjoint union topology.
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 disjoint union topological space, \(T := \coprod_\alpha T_\alpha\), and any closed sets, \(\{C_\beta \subseteq T_\beta\}\) where \(\{\beta\} \subseteq \{\alpha\}\), \(\coprod_\beta C_\beta\) is closed on \(T\).
2: Proof
\(\coprod_\beta C_\beta = \coprod_\beta T_\beta \setminus U_\beta\) where \(U_\beta \subseteq T_\beta\) is open on \(T_\beta\). \(\coprod_\beta (T_\beta \setminus U_\beta) = \coprod_\beta T_\beta \setminus \coprod_\beta U_\beta\), by the proposition that the disjoint union of any complements is the disjoint union of the whole sets minus the disjoint union of the subsets. \(\coprod_\beta C_\beta = \coprod_\alpha T_\alpha \setminus \coprod_{\gamma \in \{\alpha\} \setminus \{\beta\}} T_\gamma \setminus \coprod_\beta U_\beta = \coprod_\alpha T_\alpha \setminus (\coprod_{\gamma \in \{\alpha\} \setminus \{\beta\}} T_\gamma \cup \coprod_\beta U_\beta)\). \(\coprod_{\gamma \in \{\alpha\} \setminus \{\beta\}} T_\gamma \cup \coprod_\beta U_\beta\) is open.
