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I know that in general, $L^{\infty}$($E$) is not separable, like for example, if $E$ = [$a$,$b$]. But wouldn't $L^{\infty}$($E$) be separable if $E$ = $\mathbb{Q}$, i.e. the set of rational numbers? Wait a second, I actually think maybe it could be separable if $E$ = $\mathbb{R}$, the set of real numbers, since the rationals are a dense subset of $\mathbb{R}$. Correct me if I'm wrong. How can I show that $L^{\infty}$($E$) is not separable if $E$ contains a nondegenerate interval? Wouldn't that just follow from the fact that $L^{\infty}$[$a$,$b$] is not separable?

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    What measure are you putting on $\mathbb{Q}$? With the induced measure from $\mathbb{R}$, the rationals have zero measure, so I don't understand what $L^{\infty}(\mathbb{Q})$ could sensibly mean. With the counting measure, this is the same as $L^{\infty}(\mathbb{N})$, and it is a straightforward exercise to determine whether this is separable. The answer to your last question is "yes," but can you see why?2011-04-18
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    @Qiaochu: Lebesgue measure2011-04-18
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    Well, for Lebesgue measure, $L^{\infty}(\mathbb{Q}) = 0$ is very separable :)2011-04-18

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$L^\infty$ is effectively never separable. Indeed, for any measure space $(X,\mu)$, $L^\infty(X,\mu)$ is either finite dimensional or non-separable, according to whether or not $X$ can be written as a finite union of atoms. I recommend proving this as an exercise.