a) Give an example of a bounded closed subset of $ A = \{(x_n) \in \ell^1: \sum_{n\geq1} x_n = 1\}$ which is not compact. The metric we consider on A is induced by the normal norm on $\ell^1$.
b) Show that the set $B = \{(x_n)\in \ell^1: \sum_{n\geq1} n|x_n| = 1\}$ is compact in $\ell^1$.
My try: a)The sequences in $\hat{A} = \{e_i = (0,\ldots 0 , 1 ,0 \ldots, ),\; \forall i \in \mathbb{N}\}$ has no converging subsequences. Is it a closed subset? It seems like it contains all its boundary points (A is not vector space). b) If we take a sequence $(x^k) \in B$ we also have $(x^k) \in \ell^\infty$. But since all elements in $\ell^\infty$ has subsequences that converges, this is true for $(x^k)$?
Edited: Davids method seems to work, both Im trying to get it to work with the following hint.
Hint: You can use without proof the diagonalization process to conclude that every bounded sequence $(x_n) \in\ell^\infty$ has a subsequence $x^{n_k}$ that converges in each component. Moreover, sequences in $\ell^1$ are obviously bounded in their $\ell^1$ norm.
So what I need to show is that the convergence of a sequence does not end up outside B?