$l^\infty(\mathbb{N}; \mathbb{R})$ is the set of all bounded sequences in $\mathbb{R}$. Let $d(a, b) = \sup_n |a_n - b_n|$ for $a, b \in l^\infty$. Then, $(l^\infty(\mathbb{N} ; \mathbb{R}), d)$ is a metric space.
I want to show that the closed ball $B_r(0) = \{x \in l^\infty(\mathbb{N}, \mathbb{R}) : d(0, x) \leq r \}$ is bounded, closed but not compact. I already showed it is bounded and closed.
Since this is a metric space, the notion of compactness and sequential compactness coincide. Therefore, I need to find a sequence $(x_n \in B_r(0))_{n \in \mathbb{N}}$ which does not have a convergent subsequence.
I am considering a sequence $x_n = \left (r \cos(\frac{2 \pi}{n} k) \right )_{k \in \mathbb{N}}$ so that:
$x_1 = r, r, r, ...$
$x_2 = -r, r, -r, -r ...$
and so on ...
Perhaps if I can show that for $m \neq n$, we have $\sup_k |x_m - x_n| = 2 r$ ?
So I tried writing $\sup_k |x_m - x_n| = r \sup_k \big | \cos(\frac{2 \pi k}{m}) - \cos(\frac{2 \pi k}{n}) \big |$ ?
I cannot proceed. I am not sure if this is a right sequence to consider.
A much easier example to work with is $$ x_{n,k}= \begin{cases} 1 & n=k \\0 & else \end{cases} $$
Then, for $n\neq m$, $\sup_{k} |x_{n,k}-x_{m,k}|=1$ (simply by considering $k=n$ and $k=m$ respectively), so the sequence cannot have any Cauchy sub-sequences. Also, clearly, each sequence is an element of the closed unit ball.