Past exam of the options II course of the University of Oxford / 2013 / E3 / b

Let $a^{(m)}=(1,1/2,\dots ,1/m,0,0,\dots )$. Each has finite support, and in the supremum norm metric it converges to $a=(1,1/2,1/3,\dots )$, because $d(a^{(m)},a)=1/(m+1)$.

The limit is bounded but does not have finite support. Thus the finite-support subspace is not closed in $X$, and by part (a)(iii) it is not complete. Equivalently, the displayed sequence is Cauchy in that subspace but has no limit there.

Show that the subspace $C$ of convergent real sequences is closed in $X$. Suppose $a^{(m)}\in C$ converges uniformly in the coordinate index to $a\in X$. Given $ϵ>0$, choose $m$ so ${\sup }_j|a_j-a_j^{(m)}|<ϵ/3$. Since the fixed sequence $a^{(m)}$ converges, it is Cauchy, so for sufficiently large $j,k$ its coordinate difference is below $ϵ/3$. ThenThus $a$ is a Cauchy sequence of real numbers and converges by completeness of $\mathbb{R}$. Hence $a\in C$, so $C$ is closed. Part (a)(iii) proves the convergent-sequence subspace is complete.