A description/proof of that open set minus closed set is open
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 open set minus any closed set is open.
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 open set, \(U \subseteq T\), and any closed set, \(C \subseteq T\), the open set minus the closed set, \(U \setminus C\), is open on \(T\).
2: Proof
For any point, \(p \in U \setminus C\), \(p \in U\) and \(p \notin C\), which means that \(p \in T \setminus C\) while \(T \setminus C\) is open. There are open sets, \(p \in U_1 \subseteq U\) and \(p \in U_2 \subseteq T \setminus C\). \(p \in U_1 \cap U_2\) while \(U_1 \cap U_2\) is open and \(U_1 \cap U_2 \subseteq U\) and \(U_1 \cap U_2 \subseteq T \setminus C\), which means that \(U_1 \cap U_2 \cap C = \emptyset\), so, \(U_1 \cap U_2 \subseteq U \setminus C\). As at any point, \(p \in U \setminus C\), there is an open set, \(U_1 \cap U_2\), such that \(p \in U_1 \cap U_2 \subseteq U \setminus C\), by the local criterion for openness, \(U \setminus C\) is open.
