Why isn't $B = \{x \in l_{\infty} : \| x \|_{\infty} \leq 1 \}$ a compact metric space?

Question

I have thought about an example of Cauchy sequence for which every subsequence has no limit in $B$, but I wasn't able to find one.

Answer 1

The space $B$ is not a compact metric space because if $e_n$ is the element of $\ell^\infty$ which consists only of zeros, except for the $n$^th term, which is a $1$, then the sequence $(e_n)_{n\in\mathbb N}$ has no convergent subsequence (the distance between any two distinct $e_n$'s is $1$).

However, it is complete, so you will not find in it a Cauchy sequence that doesn't converge.

Answer 2

Let $u_n:= (0,0,,,,0,1,0,0,...)$ where $1$ is on the $n$-th place. Then $u_n \in B$ for all $n$.

Suppose that $B$ is compact, then $(u_n)$ contains a convergent subsequence $(u_{n_k})$.

Hence $(u_{n_k})$ is Cauchy. But for $k \ne j$ we have $n_k \ne n_j$, thus

$$||u_{n_k}-u_{n_j}||_{\infty}=1,$$

a contradiction.