Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 2 / b / i / Solution

The source prints $m\ne 0$ in the zero case. Since the coordinates are indexed by positive integers, that overlaps its $m=n$ case and does not define a sequence. We use the intended coordinate vectors, with $e_m^{(n)}=1$ for $m=n$ and $e_m^{(n)}=0$ for $m\ne n$.

Every $e^{(n)}$ belongs to $X$, because its entries are bounded by one and its coordinate sequence eventually vanishes. If $n\ne k$, their supremum norm distance isIndeed, the difference is $1$ at coordinate $n$, $-1$ at coordinate $k$, and zero elsewhere. No subsequence is a Cauchy sequence, since arbitrarily late distinct terms remain distance one apart. Every convergent sequence is a Cauchy sequence, so no convergent subsequence exists. Therefore $\boxed{X\text{ is not sequentially compact}}$.