8-10

Prove thatL^\infty(E) is not separable for any E where \vert E\vert>0.
Definition
A metric space is separable if it has countable dense subset.
Proof
Definef_{E_j}(x)=\chi_{E_j}(x)for E_j\subset E such that j\in\mathbb{R}. We can do this, because Eis uncountable since it has positive measure, and so clearly it has uncountably many subsets. Choose any two sets E_1, E_2, we have\Vert f_{E_1}-f_{E_2}\Vert_\infty=1Let \{g_j\} be countable dense subset of L^\infty(\mathbb{R}), this implies that for anyf, and any \epsilon, there exist a g_j such that\Vert g_j-f\Vert_\infty<\epsilonLet consider balls of radius \dfrac{1}{3} for each f_{E_j}, then we have g_j for each f_{E_j}. Yet be noticed that g_i\neq g_j since each ball has diameter at most 2/3, and so it can only contain at most one of the elements of \{g_j\}. This leads to the contradiction that \{g_j\} is countable set since \{f_{E_j}\} is uncountable. Hence L^\infty(E) is not separable for any E where \vert E\vert>0.