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Rudin asserts that $L^{p}(\mathbb{S}^{1})$ is separable for $1\le p<\infty$, but not for $p=\infty$. I am wondering why this is true.

For any $f\in L^{p}(\mathbb{S}^{1})$ we should be able to approximate it via the standard trigonometric series, hence lead to separability if we let the coefficents be rational. But I am wondering why for $L^{\infty}$ we cannot use the following system $f_{n}=\sum^{2^{k}}_{n=1}a_{n}\chi(E_{n})$ where $E_{n}$ is the characteristic function on intervals of length $[\frac{\pi}{2^{k}}]$ and $a_{n}\in \mathbb{Q}$. Then every function in $L^{\infty}$ should be able to be approximated by some $f_{n}$ with $n,k$ large enough. The whole set of $f_{n,k}$ would have same cardinality as the algebraic numbers, hence should be countable.

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    Right, that doesn't successfully approximate it in $L^\infty$ norm.2012-12-28

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In $L^\infty(S^1)$ you can found an uncountable $1-$separated set of functions, so $L^\infty$ isn't separable. An example $ \left\lbrace \chi_I\quad \big\vert \quad \forall I\subset S^1\right\rbrace $ indeed $ \forall I, I'\quad I\neq I'\quad \rightarrow \quad \Vert \chi_I - \chi_{I'}\Vert_{\infty}=1. $

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    I see. So simple functions would not work for $L^{\infty}$ norms. I will think about how to prove it.2012-12-28