54A20-AccumulationPointsAndConvergentSubnets.tex

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\pmtitle{accumulation points and convergent subnets}
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\begin{document}
\begin{prop*}
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 $x$.
\end{prop*}
\begin{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 $x_{\alpha_{(\gamma^\prime,U^\prime)}}\in U^\prime\subseteq U$.
\end{proof}
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