The real square-summable sequence space is complete

For each fixed $n$, $\mathrm{\mid }{x}_n^{(k)}-x_n^{(j)}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(k)}-x^{(j)}{\mathrm{\parallel }}_2$, so $(x_n^{(k)}{)}_k$ is Cauchy in $\mathbb{R}$. By [F2] let $x_n={\lim }_k{x}_n^{(k)}$; each coordinate limit is unique, so these values define a sequence $x$. To see $(x^{(k)})$ is bounded, choose $K$ so that $\mathrm{\parallel }{x}^{(k)}-x^{(j)}{\mathrm{\parallel }}_2<1$ for all $k,j\ge K$; then $\mathrm{\parallel }{x}^{(k)}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{x}^{(K)}{\mathrm{\parallel }}_2+1$ for $k\ge K$, and the finitely many earlier norms have a finite maximum. Thus $\mathrm{\parallel }{x}^{(k)}{\mathrm{\parallel }}_2\le M$ for some finite $M$. For any finite $F\subset \mathbb{N}$, ${\sum }_{n\in F}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2={\lim }_k{\sum }_{n\in F}\mathrm{\mid }{x}_n^{(k)}{\mathrm{\mid }}^2\le M^2$. Taking the supremum over finite $F$ gives ${\sum }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\le M^2$, hence $x\in {\mathrm{\ell }}^2$.

Let $\epsilon >0$. Choose $K$ such that $\mathrm{\parallel }{x}^{(k)}-x^{(j)}{\mathrm{\parallel }}_2<\epsilon \mathrm{/}2$ for all $k,j\ge K$. Fix $k\ge K$. For every finite $F\subset \mathbb{N}$, coordinate convergence and continuity of finite sums give ${\sum }_{n\in F}\mathrm{\mid }{x}_n^{(k)}-x_n{\mathrm{\mid }}^2={\lim }_{j\to \mathrm{\infty }}{\sum }_{n\in F}\mathrm{\mid }{x}_n^{(k)}-x_n^{(j)}{\mathrm{\mid }}^2\le {\epsilon }^2\mathrm{/}4$. Taking the supremum over finite $F$ yields $\mathrm{\parallel }{x}^{(k)}-x{\mathrm{\parallel }}_2\le \epsilon \mathrm{/}2<\epsilon$. Thus $x^{(k)}\to x$ in ${\mathrm{\ell }}^2$. Every Cauchy sequence converges, proving completeness and, by [F3], the Hilbert-space claim.