A description/proof of that for normal topological space, collapsed topological space by closed subset is normal
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of normal topological space.
- The reader knows a definition of closed set.
- The reader knows a definition of collapsed topological space by subset.
- The reader admits the proposition that the preimage of any closed set under any continuous map is a closed set.
Target Context
- The reader will have a description and a proof of the proposition that for any normal topological space, the collapsed topological space by any closed subset is normal.
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 normal topological space, \(T\), and any closed subset, \(S \subseteq T\), the collapsed topological space, \(T/S\), is normal.
2: Proof
Let \(f\) be the quotient map, \(f: T \rightarrow T/S\), and \(C_1, C_2 \in T/S\) be any closed subsets on \(T/S\) such that \(C_1 \cap C_2 = \emptyset\). \(f^{-1} (C_i)\) is closed on \(T\) as \(f\) is continuous, by the proposition that the preimage of any closed set under any continuous map is a closed set. \(f^{-1} (C_1) \cap f^{-1} (C_2) = \emptyset\). As \(T\) is normal, there are some open sets, \(U_1, U_2 \subseteq T\), around \(f^{-1} (C_1), f^{-1} (C_2)\) such that \(U_1 \cap U_2 = \emptyset\).
If \(S \notin C_1, C_2\), \(f^{-1} (C_i) \subseteq T \setminus S\), so, \(U'_i := U_i \cap (T \setminus S)\), open on \(T\), still contains \(f^{-1} (C_i)\), while \(U'_1 \cap U'_2 = \emptyset\). \(f (U'_i)\) is open on \(T/S\), because \(f^{-1} (f (U'_i)) = U'_i\) is open on \(T\). \(f (U'_i)\) contains \(C_i\). \(f (U'_1) \cap f (U'_2) = U'_1 \cap U'_2 = \emptyset\).
If \( S \notin C_1\) and \(S \in C_2\), \(f^{-1} (C_1) \subseteq T \setminus S\) and \(S \subseteq f^{-1} (C_2)\), and \(U_1 \cap S = \emptyset\). \(f (U_i)\) is open on \(T/S\), because \(f^{-1} (f (U_i)) = U_i\) is open on \(T\). \(f (U_i)\) contains \(C_i\). \(f (U_1) \cap f (U_2) = \emptyset\).
If \(S \in C_1\) and \(S \notin C_2\), the situation is symmetric with the above case.
