Let $l^\infty$ be the set of bounded sequences. Let $f : \mathbb{N} \to \mathbb{R}_+$ be a function such that $\lim_{n \to +\infty} f(n) = 0$.
I want to show the set $A = \{x \in l^\infty : |x_n| \leq f(n) $ $\forall n \in \mathbb{N} \}$ is sequentially compact.
For an arbitrary sequence $(x^j \in A)_{j \in \mathbb{N}}$, we know that $|x_n^j| \leq f(n)$ for all $j \in \mathbb{N}$.
I am guessing a convergent subsequence might converge to the sequence of all zeros, but I am not able to construct such a convergent subsequence.