A description/proof of that closure of union of finite subsets is union of closures of subsets
Topics
About: topological space
The table of contents of this article
Starting Context
- 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 the closure of the union of any finite number of subsets is the union of the closures of the subsets.
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 finite number of subsets, \(S_i \subseteq T\), \(\overline{\cup_i S_i} = \cup_i \overline{S_i}\) where the over lines denote the closures.
2: Proof
\(S := \cup_i S_i\).
While \(\overline{S}\) is the smallest closed set that contains \(S\) by the definition of closure of set, \(\cup_i \overline{S_i}\) is a closed set that contains \(S\). So, \(\overline{S} \subseteq \cup_i \overline{S_i}\).
For any \(p \in \cup_i \overline{S_i}\), \(p \in \overline{S_i}\) for an \(i\). \(p \in S_i\), or \(p \notin S_i\) and \(p\) is an accumulation point of \(S_i\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset. If \(p \in S_i\), \(p \in S\), so, \(p \in \overline{S}\). If \(p \notin S_i\) and \(p\) is an accumulation point of \(S_i\), for any neighborhood, \(N_p \subseteq T\), of \(p\), \(N_p \cap S_i \neq \emptyset\). If \(p \in S\), \(p \in \overline{S}\), otherwise, \(p \notin S\) and \(N_p \cap S \neq \emptyset\), so, \(p\) is an accumulation point of \(S\), so, \(p \in \overline{S}\). So, \(\cup_i \overline{S_i} \subseteq \overline{S}\).
