Box topology

**aharonidan** · March 28th 2011, 11:37 PM

Hello math experts,
Let $X=R^{N}$ be a topological space with the standard box topology.
Show that the collection of sequences that converge to 0 is an open-closed set in $X$ .

thanks

**Drexel28** · March 29th 2011, 12:25 AM

Originally Posted by aharonidan

Hello math experts,
Let $X=R^n$ be a topological space with the standard box topology.
Show that the collection of sequences that converge to 0 is an open-closed set in $X$ .

thanks

I'm confused. Presumably you mean that $X=\mathbb{R}^\mathbb{N}$ . I'll help you with one half. Suppose that $(a_n)\notin Z$ where $Z$ is the set of all null sequences. Then, there exists some $\varepsilon>0$ such that for every $N\in\mathbb{N}$ one has that there is some $n\geqslant N$ for which $|a_n|>\varepsilon$ . So, let $\displaystyle O=\prod_{n\in\mathbb{N}}A_n$ where $A_n=\mathbb{R}-(-\varepsilon,\varepsilon)$ if $|a_n|>\varepsilon$ and $\mathbb{R}$ otherwise. This is clearly open in the box topology since it's the product of open sets, but clearly $\displaystyle O\cap Z=\varnothing$ . Thus $X-Z$ is open etc. etc.

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