A description/proof of that 2 continuous maps into Hausdorff topological space that disagree at point disagree on neighborhood of point
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of continuous map.
- The reader knows a definition of neighborhood of point.
Target Context
- The reader will have a description and a proof of the proposition that any 2 continuous maps from any topological space into any Hausdorff topological space that (the maps) disagree at any point disagree on a neighborhood of the point.
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, \(T_1\), any Hausdorff topological space, \(T_2\), and any continuous maps, \(f_1, f_2: T_1 \to T_2\), if \(f_1 (p) \neq f_2 (p)\) at any point, \(p \in T_1\), there is a neighborhood, \(N_p\), of \(p\) such that \(f_1 (p') \neq f_2 (p')\) for each point, \(p' \in N_p\).
2: Proof
Let us suppose that \(f_1 (p) \neq f_2 (p)\). There are some disjoint open neighborhoods, \(U_{f_1 (p)}, U_{f_2 (p)} \in T_2\), of \(f_1 (p), f_2 (p)\), respectively, such that \(U_{f_1 (p)} \cap U_{f_2 (p)} = \emptyset\), because \(T_2\) is Hausdorff. As \(f_i\) is continuous, there is a neighborhood, \(N_{p, i}\), of \(p\) such that \(f_i (N_{p, i}) \subseteq U_{f_i (p)}\). \(N_p := N_{p, 1} \cap N_{p, 2}\) is a neighborhood, and \(f_i (N_p) \subseteq U_{f_i (p)}\). \(f_1 (N_p) \cap f_2 (N_p) = \emptyset\), which means that \(f_1 (p') \neq f_2 (p')\) for each \(p' \in N_p\).
