# accumulation points and convergent subnets

###### Proposition.

Let $X$ be a topological space  and $(x_{\alpha})_{\alpha\in A}$ a net in $X$. A point $x\in X$ is an accumulation point  of $(x_{\alpha})$ if and only if some subnet of $(x_{\alpha})$ converges  to $x$.

###### Proof.

Suppose first that $(x_{\alpha_{\beta}})_{\beta\in B}$ is a subnet of $(x_{\alpha})$ converging to $x$. Given an open subset $U$ of $X$ containing $x$ and $\alpha\in A$, we may select $\beta_{1}\in B$ such that $x_{\alpha_{\beta}}\in U$ for $\beta\geq\beta_{1}$, as well as $\beta_{2}\in B$ such that $\alpha_{\beta}\geq\alpha$ for $\beta\geq\beta_{2}$. Finally, because $B$ is directed, there exists $\beta\in B$ such that $\beta\geq\beta_{1}$ and $\beta\geq\beta_{2}$; we then have $\alpha_{\beta}\geq\alpha$ and $x_{\alpha_{\beta}}\in U$, so that $(x_{\alpha})$ is frequently in $U$, whence $x$ is an accumulation point of $(x_{\alpha})$. Conversely, suppose that $x$ is an accumulation point of $(x_{\alpha})$, let $N$ be the set of open neighborhoods of $x$ in $X$, directed by reverse inclusion, and let $B=A\times N$, directed in the natural way. For each pair $(\gamma,U)\in B$, select $\alpha_{(\gamma,U)}\in B$ such that $\alpha\geq\gamma$ and $x_{\alpha_{(\gamma,U)}}\in U$; $(x_{\alpha_{(\gamma,U)}})_{(\gamma,U)\in B}$ is then a subnet of $(x_{\alpha})$ that converges to $x$, for given $U\in N$ and $\gamma\in A$, if $(\gamma^{\prime},U^{\prime})\geq(\gamma,U)$, then $\alpha_{(\gamma^{\prime},U)}\geq\gamma^{\prime}\geq\gamma$ and $x_{\alpha_{(\gamma^{\prime},U^{\prime})}}\in U^{\prime}\subseteq U$. ∎

Title accumulation points and convergent subnets AccumulationPointsAndConvergentSubnets 2013-03-22 18:37:40 2013-03-22 18:37:40 azdbacks4234 (14155) azdbacks4234 (14155) 9 azdbacks4234 (14155) Theorem msc 54A20 Net Neighborhood   DirectedSet CompactnessAndConvergentSubnets