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 $l^{\infty}$.

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – Mateus Rocha Oct 25 '18 at 19:12
  • $\begingroup$ @MateusRocha The real line is complete. Any Cauchy sequence of real numbers converges. $\endgroup$ – Kavi Rama Murthy Oct 25 '18 at 23:09
  • $\begingroup$ Oh, so $l^{\infty}$ is the space of limited real sequences? I thought that was the space of any limited sequences. $\endgroup$ – Mateus Rocha Oct 25 '18 at 23:27
  • 1
    $\begingroup$ @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'. $\endgroup$ – Kavi Rama Murthy Oct 25 '18 at 23:33
  • $\begingroup$ Ok, thanks! My main language is not english. That is the reason for that mistakes. $\endgroup$ – Mateus Rocha Oct 25 '18 at 23:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.