Why is the spectrum of $L^\infty[0,1]$ inseparable?

Question

In the Gelfand correspondance any measurable subset of $[0,1]$ corresponds to a clopen subset of $\text{Spec}(L^\infty[0,1])$, so clearly the space contains a bunch of clopens. I'm not sure if this is relevant and remain stuck on the problem. The text I read said that it was essentially due to the Borel measure on $[0,1]$ being atomless, so this might be a more general result.

Answer 1

Let $X=\operatorname{Spec}(L^\infty[0,1])$. A point of $X$ corresponds to an ultrafilter on the Boolean algebra of measurable subsets of $[0,1]$ modulo null sets. In particular, any such ultrafilter must contain sets of arbitrarily small positive measure (this follows from atomlessness: we can keep splitting elements of the ultrafilter into two subsets of half the measure, and one of the halves must be in the ultrafilter).

Now suppose $\{x_n:n\in\mathbb{N}\}$ is a countable subset of $X$. For each $n$, we can pick a set $A_n$ in the ultrafilter corresponding to $x_n$ such that $\mu(A_n)<\epsilon_n$, for any $\epsilon_n>0$ that we choose. In particular, we can choose a sequence $(\epsilon_n)$ such that $\sum_n\epsilon_n<1$. Now consider the measurable set $A=[0,1]\setminus\bigcup_n A_n$. By our choice of $(\epsilon_n)$, $A$ has positive measure. However, $A$ is not in the ultrafilters corresponding to any of our $x_n$. Thus the set of points of $X$ whose corresponding ultrafilters contain $A$ is a nonempty open subset of $X$ which does not contain any $x_n$. That is, $\{x_n\}$ is not dense in $X$.