accumulation points and convergent subnets

Proposition.

Let $X$ be a topological space and ${\mathrm{(}{x}_{\alpha }\mathrm{)}}_{\alpha \mathrm{\in }A}$ a net in $X$. A point $x\mathrm{\in }X$is an accumulation point of $\mathrm{(}{x}_{\alpha }\mathrm{)}$ if and only if some subnet of $\mathrm{(}{x}_{\alpha }\mathrm{)}$ 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 $\beta \ge {\beta }_1$, as well as ${\beta }_2\in B$ such that ${\alpha }_{\beta }\ge \alpha$ for $\beta \ge {\beta }_2$. Finally, because $B$ is directed, there exists $\beta \in B$ such that $\beta \ge {\beta }_1$ and $\beta \ge {\beta }_2$; we then have ${\alpha }_{\beta }\ge \alpha$ 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 $\alpha \ge \gamma$ 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 $({\gamma }^{\prime },U^{\prime })\ge (\gamma ,U)$, then ${\alpha }_{({\gamma }^{\prime },U)}\ge {\gamma }^{\prime }\ge \gamma$ and $x_{{\alpha }_{({\gamma }^{\prime },U^{\prime })}}\in U^{\prime }\subseteq U$.∎