A description/proof of that for regular topological space, collapsed topological space by closed subset is Hausdorff
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of regular topological space.
- The reader knows a definition of closed set.
- The reader knows a definition of collapsed topological space by subset.
- The reader knows a definition of Hausdorff topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any regular topological space, the collapsed topological space by any closed subset is Hausdorff.
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 regular topological space, \(T\), and any closed subset, \(S \subseteq T\), the collapsed topological space, \(T/S\), is Hausdorff.
2: Proof
Let \(f\) be the quotient map, \(f: T \rightarrow T/S\), and \(p_1, p_2 \in T/S\) be any points on \(T/S\) such that \(p_1 \neq p_2\).
If \(p_1, p_2 \in T \setminus S\), as \(T\) is regular and so Hausdorff, there are some open sets, \(U_1, U_2 \subseteq T\), around \(p_1, p_2\) such that \(U_1 \cap U_2 = \emptyset\). \(U'_i := U_i \cap (T \setminus S)\) is open on \(T\) and contains \(p_i\), and \(U'_1 \cap U'_2 = \emptyset\). \(f (U'_i) = U'_i\) is open on \(T/S\), because \(f^{-1} (f (U'_i)) = U'_i\) is open on \(T\). \(f (U'_i)\) contains \(p_i\). \(f (U'_1) \cap f (U'_2) = U'_1 \cap U'_2 = \emptyset\).
If \(p_1 \in T \setminus S\) and \(p_2 = S\), as \(T\) is regular, there are some open sets, \(U_1, U_2 \subseteq T\), around \(p_1, S\) such that \(U_1 \cap U_2 = \emptyset\). \(f (U_i)\) is on \(T/S\), because \(f^{-1} (f (U_i)) = U_i\) is open on \(T\). \(f (U_i)\) contains \(p_i\). \(f (U_1) \cap f (U_2) = \emptyset\).
If \(p_1 = S\) and \(p_2 \in T \setminus S\), the situation is symmetric with the above case.
