A description/proof of that on \(T_1\) topological space, point is \(\omega\)-accumulation point of subset iff it is accumulation point of subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of \(T_1\) topological space.
- The reader knows a definition of \(\omega\)-accumulation point of subset.
- The reader knows a definition of accumulation point of subset.
Target Context
- The reader will have a description and a proof of the proposition that on any \(T_1\) topological space, any point is an \(\omega\)-accumulation point of any subset if and only if the point is an accumulation point 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 \(T_1\) topological space, \(T\), and any subset, \(S \subseteq T\), any point, \(p \in T\), is an \(\omega\)-accumulation point of \(S\) if and only if \(p\) is an accumulation point of \(S\).
2: Proof
Let us suppose that \(p\) is an accumulation point of \(S\). For any neighborhood, \(U_{p}\), of \(p\), there is a point, \(p_{1} \in U_{p} \cap S\) such that \(p_{1} \neq p\). There is a neighborhood, \(U_{p, 1}\), of \(p\) such that \(p_{1} \notin U_{p, 1}\), because \(T\) is a \(T_1\) space. There is a point, \(p_{2} \in U_{p} \cap U_{p, 1} \cap S\) such that \(p_{2} \neq p\). \(p_{1} \neq p_{2}\), because \(p_{1} \notin U_{p, 1}\). There is a neighborhood, \(U_{p, 2}\), of \(p\) such that \(p_{2} \notin U_{p, 2}\). There is a point, \(p_{3} \in U_{p} \cap U_{p, 1} \cap U_{p, 2} \cap S\) such that \(p_{3} \neq p\). \(p_{1}, p_{2} \neq p_{3}\), because \(p_{1}, p_{2} \notin U_{p, 1} \cap U_{p, 2}\), and so on. After all, there is a point, \(p_{i} \in U_{p} \cap S\), that is distinct from \(p, p_{1}, p_{2}, ..., p_{i - 1}\), for any natural number \(i\). So, \(U_{p} \cap S\) has at least countably infinite number of points excluding \(p\).
Let us suppose that \(p\) is an \(\omega\)-accumulation point of \(S\). Obviously, \(p\) is an accumulation point of \(S\).
