Let $\ell^{\infty}$ be the space of bounded sequences of real numbers, endowed with the norm $\|\mathbf x\|_\infty=\sup_{n\in N}|x_n|$, where $\mathbf x=(x_n)_{n\in\Bbb N}$.
Prove that the closed unit ball of $ \ell^{\infty}, B'(\mathbf 0,1)={x \in \ell^{\infty} ; \|\mathbf x\|_{\infty} \le 1}$}, is not compact.
So basically I think I want to show that the implication that a closed and bounded subset of a metric space is compact is not necessarily true. In order to do this I think I will show it is not sequentially compact as the question tells me I am allowed to use the equivalence between compactness and sequential compactness in the setting of metric spaces. However, I am unsure if this approach is correct and do not know what sequence to use to show that the sequential compactness fails. Any help, hints or explanations are appreciated!
Best Answer
Hint: Show that there is a sequence of points $e_i$ on the unit sphere such that for $i\neq j$, $\|e_i-e_j\|_\infty=1$. Conclude that this sequence does not have a convergent subsequence, and therefore the unit ball is not compact.