accumulation points and convergent subnets
Proposition.
Let be a topological space and a net in . A point is an accumulation point of if and only if some subnet of converges to .
Proof.
Suppose first that is a subnet of converging to . Given an open subset of containing and , we may select such that for , as well as such that for . Finally, because is directed, there exists such that and ; we then have and , so that is frequently in , whence is an accumulation point of . Conversely, suppose that is an accumulation point of , let be the set of open neighborhoods of in , directed by reverse inclusion, and let , directed in the natural way. For each pair , select such that and ; is then a subnet of that converges to , for given and , if , then and . ∎
Title | accumulation points and convergent subnets |
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Canonical name | AccumulationPointsAndConvergentSubnets |
Date of creation | 2013-03-22 18:37:40 |
Last modified on | 2013-03-22 18:37:40 |
Owner | azdbacks4234 (14155) |
Last modified by | azdbacks4234 (14155) |
Numerical id | 9 |
Author | azdbacks4234 (14155) |
Entry type | Theorem |
Classification | msc 54A20 |
Related topic | Net |
Related topic | Neighborhood |
Related topic | DirectedSet |
Related topic | CompactnessAndConvergentSubnets |