A description/proof of that for sequence on topological space, around point, there is open set that contains only finite points of sequence if no subsequence converges to point
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of sequence on topological space.
- The reader knows a definition of convergence of net with directed index set.
- The reader admits the proposition that any net is frequently in each neighborhood of any point on the topological space, if and only if the net has a subnet that converges to the point.
Target Context
- The reader will have a description and a proof of the proposition that for any sequence on any topological space, around any point, there is an open set that contains only some finite points of the sequence if no subsequence converges to the point.
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 sequence, \(s: N \rightarrow T\) where \(N\) is the natural numbers set, around any point, \(p \in T\), there is an open set, \(U_p, p \in U_p\), that contains only some finite points of \(s\), if \(s\) has no subsequence that converges to \(p\).
2: Proof
Suppose that \(s\) has no subsequence that converges to \(p\). Any sequence is a net and any subsequence is a subnet. By the proposition that any net is frequently in each neighborhood of any point on the topological space, if and only if the net has a subnet that converges to the point, there is a neighborhood of \(p\), \(N_p\), in which \(s\) is not frequently, which means that there is an \(n \in N\) such that there is no \(n' \in N, n \leq n'\) such that \(s (n') \in N_p\). So, \(N_p\) contains at most \(n\) points of \(s\). There is an open set, \(U_p, p \in U_p, U_p \subseteq N_p\), and \(U_p\) contains no more points of \(s\).
