A description/proof of that subset on topological subspace is closed iff there is closed set on base space whose intersection with subspace is subset
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 closed set.
Target Context
- The reader will have a description and a proof of the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
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 its subspace, \(T_1 \subseteq T\), any subset, \(S \subseteq T_1\), is closed on \(T_1\) if and only if there is a closed set, \(C \subseteq T\), such that \(S = C \cap T_1\).
2: Proof
Suppose that there is a \(C\). \(T_1 \setminus S = T_1 \setminus (C \cap T_1) = (T \setminus C) \cap T_1\), because for any point, \(p \in T_1 \setminus (C \cap T_1)\), \(p \notin C \cap T_1\), \(p \notin C\), \(p \in T \setminus C\), so, \(p \in (T \setminus C) \cap T_1\); for any point, \(p \in (T \setminus C) \cap T_1\), \(p \in T \setminus C\), \(p \notin C\), \(p \notin C \cap T_1\), \(p \in T_1 \setminus (C \cap T_1)\). So, \(T_1 \setminus S\) is open on \(T_1\), and \(S\) is closed on \(T_1\).
Suppose that \(S\) is closed on \(T_1\). \(T_1 \setminus S\) is open on \(T_1\), so, \(T_1 \setminus S = U \cap T_1\) where \(U \subseteq T\) is open on \(T\). \(S = (T \setminus U) \cap T_1\), because for any point, \(p \in S\), \(p \in T \setminus U\), because otherwise, \(p \in U\) then \(p \in T_1 \setminus S\), a contradiction, \(p \in (T \setminus U) \cap T_1\); for any \(p \in (T \setminus U) \cap T_1\), \(p \in S\), because otherwise, \(p \notin S\), \(p \in T_1 \setminus S\), \(p \in U \cap T_1\), \(p \in U\), a contradiction. \(C\) can be take to be \(T \setminus U\).
