Convergence of subnets Suppose that exists a subnet of a net $( x_{\mu})$ in $X$ that converge to $x \in X$, then $x$ is a cluster point of $( x_{\mu})$.
My attempts were: suppose $x$ is not a cluster point of $( x_{\mu})$, then exists $U$ open set, $x \in U$, and $\mu_{0}$ such that $\forall \mu \geq \mu_{0}$ implies $x_{\mu} \in X-U$, then I would like show that all subnets cannot converge for $x$, but I can not do this.
Thanks for help.
 A: There are three different definitions of subnet in fairly common use; their definitions are given in this question and answer. I suspect that you’re using one of the definitions given in the question; the hint below is written for the second of those but is easily adapted if you’re using one of the other definitions.
HINT: Let $\nu=\langle D,\le\rangle$ be a directed set, and let $\langle x_\mu:\mu\in D\rangle$ be a net in $X$ that has a subnet converging to $x$, but that $x$ is not a cluster point of $\nu$. As you say, the assumption that $x$ is not a cluster point of $\nu$ means that there are an open nbhd $U$ of $x$ and a $\mu_0\in D$ such that $x_\mu\in X\setminus U$ for all $\mu\in D$ such that $\mu_0\le\mu$.
However, we also know that $\nu$ has a subnet converging to $x$. This means that there are a directed set $\langle E,\preceq\rangle$ and a function $\varphi:E\to D$ such that


*

*$\varphi(e_0)\le\varphi(e_1)$ whenever $e_0,e_1\in E$ and $e_0\preceq e_1$,   

*$\varphi[E]$ is cofinal in $D$, meaning that for each $\mu\in D$ there is an $e\in E$ such that $\mu\le\varphi(e)$, and  

*the subnet $\nu\circ\varphi=\langle x_{\varphi(e)}:e\in E\rangle$ of $\nu$ converges to $x$ in $X$.


That last clause implies that there is an $e_0\in E$ such that $x_{\varphi(e)}\in U$ whenever $e_0\le e$; why?
Now get a contradiction by using $\mu_0$ and $e_0$ to show that there is some $\mu\in D$ such that $x_\mu\in X\setminus U$ and $x_\mu\in U$.
