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

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 $\boxed{X\text{ is not sequentially compact}}$.

Yes. Let $(x^{(k)}{)}_{k\ge 1}$ be any sequence in $A$, where $x^{(k)}=(x_m^{(k)}{)}_{m\ge 1}$. At coordinate $m$, the values lie in the compact interval $[-1/m,1/m]$. By part (a)(ii) we can successively extract a convergent subsequence for coordinate one, a further one for coordinate two, and so on. Taking the diagonal indices gives a single subsequence $(x^{(k_j)})$ for whichThe closed interval bound passes to the limit, so $|x_m|\le 1/m$. Hence $x_m\to 0$ and $x=(x_m)\in A$.

Coordinate convergence alone would not imply supremum norm convergence. Here the common tail bound supplies the missing estimate. Given $ϵ>0$, choose $M$ with $2/(M+1)<ϵ$. For $m>M$,For the finitely many $m\le M$, coordinate convergence makes all differences less than $ϵ$ once $j$ is large. Therefore $d(x^{(k_j)},x)<ϵ$, andThis is an instance of compactness from uniformly vanishing sequence tails. The common $1/m$ bound is exactly what the coordinate vectors in part (i) lack.