TITLE: $H^{\infty}$ is not separable
QUESTION [5 upvotes]: Consider the Hardy space, $(H^{\infty},\|•\|{\tiny{\infty}})$ where, $$  \|f\|{\tiny{\infty}} = \sup\{f(z) : z\in\mathbb{D} \}<\infty,$$ 
  for a $f\in H^{\infty}.$ Prove that $H^{\infty}$ is not separable.

My attempt:
I tried to find, $\phi^{i}$, $i\in I,$ functions, holomorphic, bounded over the unit disk, such that, $$B = \{\phi^{i} : i\in I\}$$ has uncountably many members and for every $i,j\in I$ with $i \neq j$, there exists a M>0 : $\|\phi^{i} - \phi^{j}\|{\tiny{\infty}}>M$.
But I cant find a spesific example that works. I tried for example $\phi^{t}(z) = e^{it}f(z)$ where $f\in H^{\infty}$. Its not easy, but there must be an elementary way to solve it, without using heavy theorems ( like interpolation theorems) or algebraic methods.

REPLY [5 votes]: We have a nice family of inner functions:
For $\zeta \in \partial \mathbb{D}$, let
$$f_{\zeta}(z) = \exp \biggl( \frac{z+\zeta}{z - \zeta}\biggr).$$
For $\zeta_1 \neq \zeta_2$, consider the radial limits of $f_{\zeta_k}$ at $\zeta_1$ to see $\lVert f_{\zeta_1} - f_{\zeta_2}\rVert_{\infty} \geqslant 1$.