Show that $L^{\infty}$ space does not have a countable dense set. 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.
Let $f_s(x) = \chi_{[0,s]}(x)$. If $0 \le s &lt; t \le 1$, what is $\|f_s - f_t\|_\infty$?
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||&lt;\epsilon $ for any $\epsilon $>0(take $\epsilon=\dfrac{1}{2}$)

This is not possible as $D$ is countable