The sequence spaces c_0 and ell-infinity

Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with absolute value in the real case and modulus $\mathrm{\mid }a+ib\mathrm{\mid }=\sqrt{a^2+b^2}$ in the complex case. A scalar sequence here is a function $x:\mathbb{N}\to \mathbb{K}$, including index zero. Let

${\mathrm{\ell }}^{\mathrm{\infty }}=\{x=(x_n{)}_{n\in \mathbb{N}}:\text{there is }C\ge 0\text{ with }\mathrm{\mid }{x}_n\mathrm{\mid }\le C\text{ for all }n\},c_0=\{x\in {\mathrm{\ell }}^{\mathrm{\infty }}:x_n\to 0\},$

both equipped with $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{n\in \mathbb{N}}\mathrm{\mid }{x}_n\mathrm{\mid }$. Thus $c_0$ is a specified linear subspace of the bounded-sequence space ${\mathrm{\ell }}^{\mathrm{\infty }}$.

For each bounded sequence, the nonempty real set $\{\mathrm{\mid }{x}_n\mathrm{\mid }:n\in \mathbb{N}\}$ has a finite supremum by The Cauchy-sequence reals have the least-upper-bound property. This proves that the displayed norm exists, rather than presuming existence in the definition of boundedness. Its uniqueness requires no choice of supremum witnesses.

Here $x_n\to 0$ means that for every real $\epsilon >0$ there is $N\in \mathbb{N}$ such that $\mathrm{\mid }{x}_n\mathrm{\mid }<\epsilon$ for all $n\ge N$. Addition and scalar multiplication are coordinatewise. The scalar triangle inequality makes bounded sequences and null sequences linear spaces and gives the triangle inequality for the displayed supremum norm. Absolute homogeneity follows coordinatewise, and a zero supremum forces every coordinate to vanish.