1

Let $S$ be any set. By $\mathcal{l}^{\infty}(S)$ we denote the space of all bounded functions $S\longrightarrow \mathbb{C}$ endowed with the $\sup$-norm. It can be shown that each normed space $X$ can be isometrically embedded into $\mathcal{l}^{\infty}(S)$ for some $S$. Really, if $X$ is some normed space, consider the set $S$ of all bounded linear functionals $\{T: X\longrightarrow \mathbb{C}: \|T\|=1\}$ whose norm is $1$. Now we build an embedding $F$ as follows: $F(x): T\mapsto Tx\in \mathbb{C}$.
It is easy to see that $F$ is linear, that $T(x)$ is a bounded function $S\longrightarrow\mathbb{C}$, and that $F$ respects the norm (i.e. $\|x\|=\sup\{|Tx|: T\in X^{*}\text{ and }\|T\|=1\}$).

Well, how to prove that each separable space can be isometrically embedded into $\mathcal{l}^\infty =\mathcal{l}^\infty(\mathbb{N})$? We need something the same, really?

Michael Hardy
  • 1
  • I changed \mathrm{sup} to \sup. That is standard. When one writes a\sup_b c, it appears as $\displaystyle a\sup_b c$ in a "displayed" context and $a\sup_b c$ in an "inline" context. By contrast a\mathrm{sup}c appears as $\displaystyle a\mathrm{sup}c$, without proper spacing before and after $\sup$, and a\mathrm{sup}_bc appears as $\displaystyle a\mathrm{sup}_bc$, with the subscript not directly below $\sup$ even in a "displayed" setting and also without proper spacing. – Michael Hardy Nov 09 '13 at 21:07
  • Have you considered trying to use the separability of the embedded space in some way? E.g., using, in some way, the existence of a countable basis of the embedded space? – Stromael Nov 09 '13 at 21:08

0 Answers0