L^p space related question for p= infinity

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Homework Statement


We are asked to exhibit a measurable set E such that L^p(E) is separable for p= infinity. Also, we have to show that L^infinity(E) is not separable if E contains a nondegenerate interval.

Homework Equations


A normed linear space X is separable provided there is a countable subset that is dense in X.
In my real analysis class, we showed that L^infinity[a,b] is not separable but then my professor said that L^infinity of a countable set will be separable.

The Attempt at a Solution


Going by what my professor said, I'm thinking that L^infinity(Q) would work here, but I'm having a little trouble starting out on how to show that this is separable. Any ideas? I would really appreciate some help here. For the second part of the problem, I think it just follows from the fact that L^infinity[a,b] is not separable, but I'm struggling to figure out why. Well, wouldn't [a,b] itself consist of disjoint sets of nondegenerate intervals? Then again, maybe this question is related to something different...
 
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smanalysis said:
Going by what my professor said, I'm thinking that L^infinity(Q) would work here, but I'm having a little trouble starting out on how to show that this is separable. Any ideas?

OK, you COULD proceed this way, but you're making it hard on yourself. What about L^\infty(\{0\}), so simply take a singleton. Isn't that a lot easier??

I would really appreciate some help here. For the second part of the problem, I think it just follows from the fact that L^infinity[a,b] is not separable, but I'm struggling to figure out why.

Let's fix some notations, let E be our set that contains the nondegenerate interval [a,b]. You could begin by showing that there exists an embedding L^\infty([a,b])\rightarrow L^\infty(E).

Well, wouldn't [a,b] itself consist of disjoint sets of nondegenerate intervals?

Yes, of course, but I don't see how this would help you...
 
I would listen to micromass' advice. Especially since I don't think L^infinity(Q) is separable. Why don't you try to show that it isn't? It's a pretty simple Cantor diagonal type argument.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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