A description/proof of that closed set minus open set is closed
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of closed set.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that any closed set minus any open set is closed.
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 closed set, \(C \subseteq T\), and any open set, \(U \subseteq T\), the closed set minus the open set, \(C \setminus U\), is closed on \(T\).
2: Proof
For any point, \(p \in T \setminus (C \setminus U)\), \(p \notin C \setminus U\). 1): \(p \notin C\) or 2): \(p \in C \cap U\).
For 1), \(p \in T \setminus C\), open. There is an open neighborhood, \(U_p \subseteq T \setminus C\), of \(p\). \(U_p \subseteq T \setminus (C \setminus U)\)? For any point, \(p' \in U_p\), \(p' \notin C\), so, \(p' \notin C \setminus U\). So, yes, \(U_p \subseteq T \setminus (C \setminus U)\).
For 2), there is an open neighborhood, \(U_p \subseteq U\), of \(p\). \(U_p \subseteq T \setminus (C \setminus U)\)? For any point, \(p' \in U_p\), \(p' \notin C \setminus U\). So, yes, \(U_p \subseteq T \setminus (C \setminus U)\).
By the local criterion for openness, \(T \setminus (C \setminus U)\) is open, so, \(C \setminus U\) is closed.
