A description/proof of that closure of subset is union of subset and accumulation points set of 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.
- The reader admits a local characterization of closure: for any topological space and any subset, any point on the topological space is on the closure of the subset if and only if its every neighborhood contains a point on the subset.
Target Context
- The reader will have a description and a proof of the proposition that the closure of any subset is the union of the subset and the accumulation points set of the 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, \(T\), and any subset, \(S \subseteq T\), the closure of \(S\), \(\overline{S}\), is the union of \(S\) and the accumulation points set of \(S\), \(ac (S)\), which is \(\overline{S} = S \cup ac (S)\).
2: Proof
For any point, \(p \in \overline{S}\), if \(p \in S\), \(p \in S \cup ac (S)\), otherwise, by a local characterization of closure: for any topological space and any subset, any point on the topological space is on the closure of the subset if and only if its every neighborhood contains a point on the subset, every neighborhood of \(p\) contains a point of \(S\), but as \(p\) is not any point on \(S\), every neighborhood of \(p\) contains a point of \(S\) excepting \(p\), so, \(p\) is an accumulation point of \(S\), so, \(p \in S \cup ac (S)\).
For any point, \(p \in S \cup ac (S)\), if \(p \in S\), \(p \in \overline{S}\), otherwise, \(p\) is an accumulation point of \(S\), so, every neighborhood of \(p\) contains a point of \(S\) by the definition of accumulation point, so, by a local characterization of closure: for any topological space and any subset, any point on the topological space is on the closure of the subset if and only if its every neighborhood contains a point on the subset, \(p \in \overline{S}\).
