An inner-product space need not be complete

The pairing is an inner product on $c_{00}$: linearity in the first argument and conjugate symmetry are finite-sum algebra, and $⟨x,x⟩={\sum }_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2=0$ forces every coordinate $x_k$ to vanish; the induced length is the ${\mathrm{\ell }}^2$ norm of the finitely supported sequence.

Let $u^{(N)}$ be the sequence with $u_k^{(N)}=1\mathrm{/}(k+1)$ for $k<N$ and $u_k^{(N)}=0$ for $k\ge N$; each $u^{(N)}$ lies in $c_{00}$, and for $M>N$ one has $\mathrm{\parallel }{u}^{(M)}-u^{(N)}{\mathrm{\parallel }}^2={\sum }_{N\le k<M}1\mathrm{/}(k+1{)}^2={\sum }_{N<m\le M}1\mathrm{/}{m}^2$, a difference of partial sums of the convergent $p$-series, which tends to $0$ as $N,M\to \mathrm{\infty }$ by [A1]; hence $(u^{(N)})$ is Cauchy in the ${\mathrm{\ell }}^2$ norm.

Suppose $v\in c_{00}$ were a limit of $(u^{(N)})$ in the induced norm; then for each fixed $k$, Cauchy–Schwarz applied to $v-u^{(N)}$ and the $k$-th coordinate vector gives $\mathrm{\mid }{v}_k-u_k^{(N)}\mathrm{\mid }\le \mathrm{\parallel }v-u^{(N)}\mathrm{\parallel }$, so $v_k={\lim }_N{u}_k^{(N)}=1\mathrm{/}(k+1)$ for every $k$, and $v$ has infinitely many nonzero coordinates, contrary to finite support.

Hence the Cauchy sequence $(u^{(N)})$ in the inner-product space $c_{00}$ has no limit there, so $c_{00}$ is not complete for its induced norm, and the statement that every inner-product space is complete is false.