Let $l^{\infty}$ be the vector space of all bounded sequences $x=(x_n)$ of real numbers with the norm $||x||=\sup_{n\in\mathbb{N}}|x_n|$ and $l^{\infty}$ is complete.
I am trying to show that the set $S=\{(x_n)\in l^{\infty}:|x_n|\leq2^{-n}\forall n\in\mathbb{N}\}$ is a compact subset of $l^{\infty}$.
Here is how I approach the problem (where I used the sequential compactness definition):
We need to show that every sequence of $S$ has a subsequence which converges to an element of $S$.
Let $(y_n)$ be an arbitrary sequence in $S$. Since $l^{\infty}$ is complete, then every Cauchy sequence in $S$ has a limit in $S$.
Let $(y_k)$ be a subsequence of $(y_n)$. Then $|y_k|\leq2^{-k}$ by definition of $S$. Then as $k\to\infty$, $2^{-k}\to0$, then $y_k\to0$ and $0\in S$ since $|0|\leq2^{-n}$. So the subsequence $(y_k)$ converges to an element in $S$. Hence $S$ is a compact subset of $l^{\infty}$.
Some of my doubts on my proof:
I did not really use the fact that $l^{\infty}$, does this information really needed?
I am not very sure how to show there exists a convergent subsequence, the way I did it is just assume an arbitrary subsequence and show that it converges. Honestly I doubt whether my approach is correct?
Is there anything that I need to fix in my proof? Is there any mistake, if so how can I write a better proof? Could anyone please lend some help.
Thanks!
You are working with sequence of the a sequences set, so that your $(y_n)$ is actually $\forall n, y_n = (y_{n,k})_k \in S$.
In this case, the information is written as : $(y_n) \subset S$ then, $\forall n,k \geq 1, |y_{n,k}| \leq 2^{-k}$.
A good method to extract subsequences is to use Cantor diagonal method.
This link should help you :)
Compact subsets in $l_\infty$ (converse of my last question)