A description/proof of that closed set on closed topological subspace is closed on base space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of subspace topology.
- The reader knows a definition of closed set.
Target Context
- The reader will have a description and a proof of the proposition that any closed set on any closed topological subspace is closed 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, \(T_1\), and any closed topological subspace, \(T_2 \subseteq T_1\), of \(T_1\), any closed set, \(C \subseteq T_2\), on \(T_2\), is closed on \(T_1\).
2: Proof
As \(C\) is closed on \(T_2\), \(T_2 \setminus C\) is open on \(T_2\). For any point, \(p \in T_2 \setminus C\), there is an open neighborhood, \(U_p \subseteq T_2 \setminus C\), of \(p\) on \(T_2\). By the definition of subspace topology, \(U_p = U_p' \cap T_2\) where \(U_p' \subseteq T_1\) is open on \(T_1\). But \(U_p' = U_p' \cap (T_2 \cup (T_1 \setminus T_2)) = (U_p' \cap T_2) \cup (U_p' \cap (T_1 \setminus T_2)) \subseteq (T_2 \setminus C) \cup (T_1 \setminus T_2)\), because \(U_p' \cap T_2 \subseteq T_2 \setminus C\) and \((U_p' \cap (T_1 \setminus T_2)) \subseteq T_1 \setminus T_2\). As \((T_2 \setminus C) \cup (T_1 \setminus T_2) = T_1 \setminus C\), \(U_p' \subseteq T_1 \setminus C\), which means that around any point on \(T_2 \setminus C\), there is an open neighborhood on \(T_1\) that is contained in \(T_1 \setminus C\). As \(T_2\) is closed on \(T_1\), \(T_1 \setminus T_2\) is open on \(T_1\). For any point, \(p \in T_1 \setminus T_2\), there is an open neighborhood, \(U_p \subseteq T_1 \setminus T_2 \subseteq T_1 \setminus C\), of \(p\) on \(T_1\), which means that around any point on \(T_1 \setminus T_2\), there is an open neighborhood on \(T_1\) that is contained in \(T_1 \setminus C\). As \(T_1 \setminus C = (T_2 \setminus C) \cup (T_1 \setminus T_2)\), around any point on \(T_1 \setminus C\), there is an open neighborhood on \(T_1\) that is contained in \(T_1 \setminus C\), which means that \(T_1 \setminus C\) is open on \(T_1\), so, \(C\) is closed on \(T_1\).
