A description/proof of that for disjoint subset and open set, closure of subset and open set are disjoint
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.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any disjoint subset and open set, the closure of the subset and the open set are disjoint.
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, \(T\), and any disjoint subset, \(S \subseteq T\), and open set, \(U \subseteq T\), \(S \cap U = \emptyset\), the closure of \(S\), \(\overline{S}\), and \(U\) are disjoint, which is \(\overline{S} \cap U = \emptyset\).
2: Proof
Suppose that \(\overline{S}\) and \(U\) were not disjoint. There would be a point, \(p \in \overline{S}, U\). By the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, \(p\) would be on \(S\) or would be an accumulation point of \(S\), but the former is impossible, because \(S\) and \(U\) are disjoint, so, \(p\) would be an accumulation point of \(S\). As \(p \in U\), there would be an open set around \(p\), \(U_p\), such that \(p \in U_p \subset U\), but \(U_p\) would not contain any point of \(S\), so, \(p\) would not be any accumulation point of \(S\), a contradiction.
