Continuity of a function on $\ell^\infty$ I want to prove the set $c_{0}$ is closed in $\ell^{\infty}$. But I saw what could be done with the following function: $f: \ell^\infty \to \mathbb{R} $ defined by $x \mapsto \limsup_n | x_n | $. But I can't prove that $f$ is continuous. Can someone help me? Thanks.
 A: If $(x_n)_{n\geqslant 1}\in\ell^\infty$ and $(y_n)_{n\geqslant 1}\in\ell^\infty$, we have
$$
\limsup_{n}\left\lvert x_n\right\rvert\leqslant \limsup_{n}\left(\left\lvert x_n-y_n\right\rvert+\left\lvert y_n\right\rvert\right)
\leqslant  \limsup_{n} \left\lvert x_n-y_n\right\rvert+\limsup_{n} \left\lvert y_n\right\rvert
$$
and similarly
$$
\limsup_{n}\left\lvert y_n\right\rvert\leqslant\limsup_{n} \left\lvert x_n-y_n\right\rvert+\limsup_{n} \left\lvert x_n\right\rvert
$$
hence
$$
\left\lvert \limsup_{n}\left\lvert x_n\right\rvert-\limsup_{n}\left\lvert y_n\right\rvert \right\rvert\leqslant  \limsup_{n} \left\lvert x_n-y_n\right\rvert$$
and I let you conclude.
A: A direct approach: Let $x=(x_n)_n\in l^{\infty}\setminus c_0.$ Then there exists $r>0$ such that the set $S=\{n:|x_n|>r\}$ is infinite. The open ball $B(x,r/2)$ in $l^{\infty}$ is disjoint from $c_0$. Because if $(y_n)_n\in B(x,r/2)$ then the set  $T= \{n: |y_n|>r/2\}$ is infinite (because $T\supset S$), so the sequence $(y_n)_n $ cannot converge to $0.$
