Q) Let $T$ be a unit circle in the complex plane. Show that $L^{\infty}(T)$ is not separable. I know this has been answered before but wanted to understand a proof.
Proof: $L^{\infty}(T)\cong L^{\infty}([0,2\pi])$. Will show that $L^{\infty}([0,2\pi])$ is not separable. Let
$$S = \{f\in L^{\infty}([0,2\pi]) | f = 1_{[0,r]},0<r<2\pi\}$$
1) $S$ is an uncountable subset of $L^{\infty}([0,2\pi])$ and if $f,g \in S$ with $f\neq g$, then $||f-g||_{\infty} = 1$. Suppose $L^{\infty}([0,2\pi])$ is separable. Thus there exists a countable dense set, say $M$ in $L^{\infty}([0,2\pi])$.
2) But $\cup_{m\in M} B_{0.4}(m)$ contains $L^{\infty}([0,2\pi])$ (Why?) where $B_{0.4}(m)$ is an open ball of radius $0.4$ around $m$.
3) Since each $B_{0.4}(m)$ contains at most one element of $S$ (Why?), therefore $\cup_{m\in M} B_{0.4}(m)$ contains at most countable elements of $S$. $S$ being uncountable, so $S$ is not a subset of $\cup_{m\in M} B_{0.4}(m)$, a contradiction. Therefore $L^{\infty}([0,2\pi])$ is not separable.
May I know why $2)$ and $3)$ are true? Thanks.
2) Since $M$ is dense in $L^{\infty}$, for $f\in L^{\infty}$, there is a $m\in M$ such that $\|m-f\|_{L^{\infty}}<0.4$, this also means that $f\in B_{0.4}(m)$.
3) If there were $f\ne g$, $f,g\in S$, then $1=\|f-g\|_{L^{\infty}}\leq\|f-m\|_{L^{\infty}}+\|m-g\|_{L^{\infty}}<1/4+1/4=1/2$, a contradiction.