A description/proof of that for regular topological space, neighborhood of point contains closed neighborhood
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of regular topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any regular topological space, any neighborhood of any point contains a closed neighborhood.
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, \(T\), any point, \(p \in T\), and any neighborhood of \(p\), \(p \in N_p \subseteq T\), \(N_p\) contains a closed neighborhood.
2: Proof
There is an open set, \(U_p \subseteq N_p, p \in U_p\). \(T \setminus U_p\) is closed and does not contain \(p\). So, by the definition of regular topological space, there are some disjoint neighborhoods, \(N_1, N_2\), such that \(p \in N_1 \subseteq T\) and \(T \setminus U_p \subseteq N_2 \subseteq T\), \(N_1 \cap N_2 = \emptyset\) where we take \(N_1\) and \(N_2\) as open neighborhoods, which is always possible (see the Note of the definition article). In fact, \(\overline{N_1} \cap N_2 = \emptyset\) where \(\overline{N_1}\) is the closure of \(N_1\), because \(\overline{N_1} = N_1 \cup ac (N_1)\) where \(ac (N_1)\) is the set of all the accumulation points of \(N_1\), but any accumulation point of \(N_1\) does not belong to \(N_2\), because for any point, \(p' \in N_2\), as \(N_2\) is open, there is an open set, \(U_{p'}\), \(p' \in U_{p'} \subseteq N_2\), which does not contain any point of \(N_1\), so, \(p'\) is not any accumulation point of \(N_1\). So, \(\overline{N_1} \subseteq T \setminus N_2 \subseteq T \setminus (T \setminus U_p) = U_p\), because \(T \setminus U_p \subseteq N_2\). So, \(\overline{N_1}\) is a closed neighborhood contained in \(U_p\), which is contained in \(N_p\).
