A description/proof of that intersection of closure of subset and open subset is contained in closure of intersection of subset and open subset
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 closure of subset.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the intersection of the closure of any subset and any open subset is contained in the closure of the intersection of the subset and the open subset.
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 subset, \(S \subseteq T\), and any open subset, \(U \subseteq T\), \(\overline{S} \cap U \subseteq \overline{S \cap U}\) where the over lines denote the closures.
2: Proof
For any point, \(p \in \overline{S} \cap U\), \(p \in \overline{S}\) and \(p \in U\). If \(p \in S\), \(p \in \overline{S \cap U}\). If \(p \notin S\), \(p\) is an accumulation point of \(S\), so, for any open neighborhood, \(U_p \subseteq T\), of \(p\), \(U_p \cap S \neq \emptyset\). \(U_p \cap (S \cap U) \neq \emptyset\)? As \(p \in U\), \(U_p \cap U \neq \emptyset\), open, and is an open neighborhood of \(p\), so, \(U_p \cap (S \cap U) = (U_p \cap U) \cap S \neq \emptyset\). So, \(p\) is an accumulation point of \(S \cap U\), so, \(p \in \overline{S \cap U}\).
