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.
Best Answer
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 $l^{\infty}$.