A description/proof of that topological space is normal iff for closed set and its containing open set there is closed-set-containing open set whose ~
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 closure of subset.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is normal if and only if for any closed set and its any containing open set, there is a closed-set-containing open set whose closure is contained in the former open set.
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
Any topological space, \(T\), is normal if and only if for any closed set, \(C \subseteq T\), and any open set, \(U_1 \subseteq T\), such that \(C \subseteq U_1\), there is an open set, \(U_2 \subseteq T\), such that \(C \subseteq U_2 \subseteq \overline{U_2} \subseteq U_1\) where \(\overline{U_2}\) is the closure of \(U_2\).
2: Proof
Suppose that \(T\) is normal. Suppose that there are C and \(U_1\). \(C_2 := T \setminus U_1\) is closed and C and \(C_2\) are disjoint. There are some open sets, \(U_2, U_3 \subseteq T\), that are disjoint such that \(C \subseteq U_2\) and \(C_2 \subseteq U_3\). \(C_4 := T \setminus U_3\) is closed and \(C_2 = T \setminus U_1 \subseteq U_3 = T \setminus C_4\), so, \(C_4 \subseteq U_1\). As \(U_2\) and \(U_3\) are disjoint, \(U_2 \subseteq T \setminus U_3 = C_4\). So, \(U_2 \subseteq C_4 \subseteq U_1\), but as \(\overline{U_2}\) is the intersection of all the closed sets that contain \(U_2\), \(U_2 \subseteq \overline{U_2} \subseteq C_4\). So, \(C \subseteq U_2 \subseteq \overline{U_2} \subseteq U_1\).
Suppose that for C and \(U_1\), there is \(U_2\) such that \(C \subseteq U_2 \subseteq \overline{U_2} \subseteq U_1\). For any disjoint closed sets, \(C_1, C_2 \subseteq T\), \(U_1 := T \setminus C_2\) is open and \(C_1 \subseteq U_1\). So, there is an open set, \(U_2 \subseteq T\), such that \(C_1 \subseteq U_2 \subseteq \overline{U_2} \subseteq U_1\). As \(\overline{U_2}\) is closed, \(U_3 := T \setminus \overline{U_2}\) is open and \(T \setminus U_3 = \overline{U_2} \subseteq U_1 = T \setminus C_2\), which implies \(C_2 \subseteq U_3\). As \(U_3 = T \setminus \overline{U_2} \subseteq T \setminus U_2\), \(U_2\) and \(U_3\) are disjoint.
