The finitely supported sequences form an incomplete normed space with different standard completions

If $x\in c_0$, then $x^{(N)}\in c_{00}$ and $\mathrm{\parallel }x-x^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{n>N}\mathrm{\mid }{x}_n\mathrm{\mid }\to 0$. So $c_{00}$ is dense in $c_0$ for the supremum norm.

For $1\le p<\mathrm{\infty }$ and $x\in {\mathrm{\ell }}^p$, the truncations satisfy $\mathrm{\parallel }x-x^{(N)}{\mathrm{\parallel }}_p^p={\sum }_{n>N}\mathrm{\mid }{x}_n{\mathrm{\mid }}^p\to 0$, so $c_{00}$ is dense in ${\mathrm{\ell }}^p$ for the ${\mathrm{\ell }}^p$ norm.

The sequence $u=(1,1\mathrm{/}2,1\mathrm{/}3,\dots )$ lies in $c_0$ but not in $c_{00}$, while its truncations $u^{(N)}\in c_{00}$ converge to $u$ in the supremum norm by step 1.1. Hence $c_{00}$ is not complete for that norm. Since $c_0$ is Banach by [L1] and the standard inclusion is a linear isometry with dense image by step 1.1, [L2] says that this inclusion exhibits $c_0$ as a supremum-norm completion of $c_{00}$.

Because ${\mathrm{\ell }}^p$ is Banach by [L3] and the standard inclusion of $c_{00}$ is a linear isometry with dense image by step 1.2, [L2] says that this inclusion exhibits ${\mathrm{\ell }}^p$ as an ${\mathrm{\ell }}^p$-completion of $c_{00}$.