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I was able to show that when $p ≥ 1$, the $L^p$ space on the interval $[0,1]$ has a countable dense set.

However, when $p$ is infinite, how to prove that $L^p$ space on the interval $[0,1]$ does not have a countable dense set? I can't find some way to approach.

Watson
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Keith
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    Try to find an uncountable subset such that the minimal distance between two elements is non-zero. – Ben Grossmann Apr 08 '15 at 15:46
  • For example: ${1,0}^{\Bbb N}$ is such a subset of $\ell^\infty$. – Ben Grossmann Apr 08 '15 at 15:47
  • A roundabout, but interesting, way would be the following. Suppose $L^\infty$ is separable. Then the closed unit ball in $L^1$ is weak-* compact. Now present a counterexample to this (a sequence in the closed unit ball of $L^1$ with no weakly convergent subsequence). The classic example of such a sequence is $n \chi_{[0,1/n]}$. – Ian Apr 08 '15 at 15:54

2 Answers2

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Let $f_s(x) = \chi_{[0,s]}(x)$. If $0 \le s < t \le 1$, what is $\|f_s - f_t\|_\infty$?

Umberto P.
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Consider all those elements $e_i$ whose terms are either $0$ or $1$ .They all belong to $L^\infty$ and they are uncountable having cardinality $c$

$||e_i-e_j||=1$

Now if we a countable dense set $D$ say then we should have for each $e_i$ an element $d_i$ such that $||e_i-d_i||<\epsilon $ for any $\epsilon $>0(take $\epsilon=\dfrac{1}{2}$)

This is not possible as $D$ is countable

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    You have presented an answer for $\ell^\infty$, that is, $L^\infty(\mathbb{N},c)$, where $c$ is the counting measure. The OP is asking about the case $L^\infty([0,1],m)$ where $m$ is the Lebesgue measure. Of course using something like binary expansion you can translate between the two, but still, this is not an answer to the question as it was asked. – Ian Apr 08 '15 at 15:55