# Convergence/Sequences/Box Topology

Let $\mathbb{R}^\omega$ be the countable product of $\mathbb{R}$. Make it a topological space using the box topology. Show that the sequence $\{(1/n, 1/n, ....)$ | $n \in \mathbb{Z}_+\}$ does not converge to $(0,0,...)$.

I know that a given sequence $\{x_n$ | $n \in \mathbb{Z}_+\}$ such that each $x_n$ is in $X$, $\{x_n\}$ converges to x if for any open set $U \subset X$ containing $x$ there is an $N \in \mathbb{Z}_+$ such that $x_n \in U$ whenever $n>N$. Is it because there can exist a $x_1 \in U$; however there is no $N \in \mathbb{Z}_+$ such that $1>N$?

Consider the following neighbourhood of $(0,0,\ldots)$ $$(-1,1)\times(-1/2,1/2)\times\cdots\times(-1/n,1/n)\times\cdots$$