= Solution
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 $\epsilon>0$, choose $m$ so $\sup_j|a_j-a_j^{(m)}|<\epsilon/3$. Since the fixed <sequence> $a^{(m)}$ converges, it is Cauchy, so for sufficiently large $j,k$ its coordinate difference is below $\epsilon/3$. Then
$$
|a_j-a_k|\leq|a_j-a_j^{(m)}|+|a_j^{(m)}-a_k^{(m)}|+|a_k^{(m)}-a_k|<\epsilon.
$$
Thus $a$ 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 \b[the convergent-sequence subspace is complete].
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