Set of all sequences converging to $0$ forms a Banach Space.

Question

Let $c_0 =\{(x_n)\in \mathbb{R}:x_{n}\to 0\},$ then show that $c_0\subset l^{\infty}$ and that $c_0$ forms a Banach Space.

Since convergent sequences are bounded clearly $c_0\subset l^{\infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{\infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{\infty}\to 0$ as $n\to \infty.$ Then we want to show that $u_{k}\in c_{0}.$ Clearly for a given $\epsilon>0$ there exists $N$ such that for all $n\geq N$ we have that $$||u_{k}^{n}-u_{k}||<\epsilon$$ for all $k\in \mathbb{N}.$ Therefore for each $k\in \mathbb{N}$ we have
$$|u_{k}^{n}-u_{k}|\leq ||u_{k}^{n}-u_{k}||<\epsilon.$$ If we send $k\to \infty$ we get that $$|u_k|\leq \epsilon$$ and therefore $\lim_{k\to \infty}u_k=0.$ This show that $c_0$ is closed and hence complete.

Is this argument correct?

Answer 1

Your argument is indeed correct, but it is simpler to prove that ${c_0}^\complement$ is open in $\ell^\infty$. This can be done by noting that if $(u_n)_{n\in\mathbb N}\in{c_0}^\complement$, then there are two possibilites:

1. $(u_n)_{n\in\mathbb N}$ converges to $l\neq0$. Then the open ball centered at $(u_n)_{n\in\mathbb N}$ with radius $\lvert l\rvert$ is contained in ${c_0}^\complement$.
2. $(u_n)_{n\in\mathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $\varepsilon>0$ such that$$(\forall N\in\mathbb{N})(\exists m,n\in\mathbb{N}):m,n\geqslant N\wedge\lvert u_m-u_n\rvert\geqslant\varepsilon.$$Then the open ball centered at $(u_n)_{n\in\mathbb N}$ with radius $\frac\varepsilon3$ is contained in ${c_0}^\complement$.