Closure of c$_0$ in Uniform Topology

Question

The Uniform Topology is generated by the metric $d$ on the set $X=l^\infty$

$d(x,y)= \sup \{ |x_n -y_n|: n\in \mathbb{N}\} $

We now prove $\bar{ c_0} \subset c_0 $, let $x \in \bar{c_0}$ we have to say that $x\in c_0$

Since $x\in \bar{c_0}$ there exist a sequence $x_n$ in $ c_0$ such that $x_n \rightarrow x$

Let $\epsilon >0$, we should find a stage $k\in \mathbb{N}$ such that $\forall i>k, |x(i)|<\epsilon$

$x_n \rightarrow x \implies x_n(i)\rightarrow x(i)$,

I dont know how to go further. Can someone help with these?

Answer 1

It is more easy to prove that $l^\infty\setminus c_0$ is open. Let $(x_n)\in l^\infty\setminus c_0$ be any sequence. Then $x$ does not converges to $0$, that is there exists $\varepsilon>0$ such a set $A=\{n\in\Bbb N: |x_n|\ge\varepsilon\}$ is infinite. Let $y\in l^\infty$ be any sequence such that $d(x,y)<\varepsilon/2$ and $n\in A$ be any index. Then

$$|y_n|\ge |x_n|-|x_n-y_n|\ge\varepsilon -\varepsilon/2\ge \varepsilon/2,$$

so $y$ does not converges to $0$, that is $y\in l^\infty\setminus c_0$.