A description/proof of that set of neighborhood bases at all points determines topology
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of neighborhood basis at point.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, any set of neighborhood bases at all points determines the topology.
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, \(T\), any set of neighborhood bases at all points, \(\{B_p\}\) where \(p \in T\), determines the topology, which means that there cannot be any different topology for the points set of \(T\), for which \(\{B_p\}\) is a set of neighborhood bases at all points.
2: Proof
Let \(T_1\) denote the topology of \(T\) and suppose that there is another topology, \(T_2\), for the points set of \(T\), for which \(\{B_p\}\) is a set of neighborhood bases at all points. Let us think of any open neighborhood, \(U_p\), of any point \(p \in T\), for \(T_1\). For any \(p' \in U_p\), \(U_p\) is a neighborhood of \(p'\) for \(T_1\), so, there is a neighborhood, \(U_{p'}\), for \(T_1\) from \(B_{p'}\) such that \(U_{p'} \subseteq U_p\), by the definition of neighborhood basis. But \(U_{p'}\) is also a neighborhood for \(T_2\) by the supposition. So, with respect to \(T_2\), as there is a neighborhood at any point of \(U_p\) contained in \(U_p\), \(U_p\) is open, which means that any open set for \(T_1\) is open also for \(T_2\). By symmetry, any open set for \(T_2\) is open also for \(T_1\). So, \(T_1 = T_2\).
