Completeness of countable square-summable sequences

Let $(x^{(k)}{)}_{k\ge 1}$ be Cauchy in ${\mathrm{\ell }}^2({\mathbb{N}}_0,\mathbb{C})$. For each fixed $n$, $\mathrm{\mid }{x}_n^{(k)}-x_n^{(\mathrm{\ell })}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(k)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_2$, so [F2] gives a unique coordinate limit $x_n$. The sequence is norm-bounded: choose $k_0$ so that $\mathrm{\parallel }{x}^{(k)}-x^{(k_0)}{\mathrm{\parallel }}_2<1$ for $k\ge k_0$, bound those terms by $1+\mathrm{\parallel }{x}^{(k_0)}{\mathrm{\parallel }}_2$, and enlarge the bound to include the finitely many earlier norms. Call the resulting bound $C$. For every finite $F\subseteq {\mathbb{N}}_0$, finite-sum continuity gives ${\sum }_{n\in F}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2={\lim }_k{\sum }_{n\in F}\mathrm{\mid }{x}_n^{(k)}{\mathrm{\mid }}^2\le C^2$. Taking the supremum over finite $F$ shows $x\in {\mathrm{\ell }}^2$.

Given $\delta >0$, choose $k_1$ so $\mathrm{\parallel }{x}^{(k)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_2<\delta \mathrm{/}2$ whenever $k,\mathrm{\ell }\ge k_1$. Fix $k\ge k_1$. For each finite $F$, coordinate convergence and the finite sum imply ${\sum }_{n\in F}\mathrm{\mid }{x}_n^{(k)}-x_n{\mathrm{\mid }}^2={\lim }_{\mathrm{\ell }\to \mathrm{\infty }}{\sum }_{n\in F}\mathrm{\mid }{x}_n^{(k)}-x_n^{(\mathrm{\ell })}{\mathrm{\mid }}^2\le (\delta \mathrm{/}2{)}^2$. Taking the supremum over finite $F$ gives $\mathrm{\parallel }{x}^{(k)}-x{\mathrm{\parallel }}_2\le \delta \mathrm{/}2<\delta$. Thus every Cauchy sequence converges in this norm; [F3] gives the Hilbert-space conclusion. The coordinate limits are uniquely determined, so no choice principle is used.