# Prove $(l^{\infty},d_{\infty})$ is complete (metric space)

I need to prove that the space of the limited sequences $$l^{\infty}$$, with the metric $$d_{\infty}\left((x_{n}),(y_{n})\right)=\sup\{|x_{n}-y_{n}| : n\in\mathbb{N}\}$$ is complete.

I saw this question in other posts, but always talking about measure theory, and I'm working only with metric spaces.

Basically, I need to show that every Cauchy sequence of elements in $$l^{\infty}$$ is convergent. Let $$(x_{n})_{n\in\mathbb{N}}=\left((x_{1,n})_{n\in\mathbb{N}},(x_{2,n})_{n\in\mathbb{N}},\dots,(x_{i,n})_{n\in\mathbb{N}},\dots\right)$$ a Cauchy sequence in $$l^{\infty}$$. So, for all $$i\in\mathbb{N},$$ $$(x_{i,n})_{n\in\mathbb{N}}$$ is a Cauchy Sequence. I only know that $$(x_{i,n})_{n\in\mathbb{N}}$$ is limited. How can I prove that $$((x_{i,n})_{n\in\mathbb{N}}$$ is convergent? Proving that, I can finish the question.

Cauchy property of the sequence says given $$\epsilon >0$$ there exists $$m$$ such that $$\sup_i|x_{i,n} -x_{i,k}| <\epsilon$$ for all $$n,k \geq m$$. You have already observed that $$\lim_{n\to \infty} x_{i,n}$$ exists for all $$i$$. Call this limit $$x^{(i)}$$. We have $$|x_{i,n} -x_{i,k}| <\epsilon$$ whenever $$n,k \geq m$$. Let $$k \to \infty$$ in this to get $$|x_{i,n} -x^{(i)}| \leq\epsilon$$ whenever $$n \geq m$$. Take sup over $$i$$ to get $$\sup_i|x_{i,n} -x^{(i)}| \leq\epsilon$$ whenever $$n \geq m$$. This implies $$(x^{(i)}) \in l^{\infty}$$ and $$x_n$$ converges to $$(x^{(i)})$$ in the norm of $$l^{\infty}$$.
• Kavi, why $(x_{i,n})$ converges for all $i$? O only know that this is a limited Cauchy sequence. Does it implies that the sequence is convergent? – Mateus Rocha Oct 25 '18 at 19:12
• Oh, so $l^{\infty}$ is the space of limited real sequences? I thought that was the space of any limited sequences. – Mateus Rocha Oct 25 '18 at 23:27
• @MateusRocha In standard notation $l^{\infty}$ is the space of all bounded real sequences or the space of all bounded complex sequences. Both $\mathbb R$ and $\mathbb C$ are complete. BTW, you should say 'bounded' instead of 'limited'. – Kavi Rama Murthy Oct 25 '18 at 23:33