Does there exist an infinite $\sigma$-algebra that contains no nonempty member which has no nonempty proper measurable subset? We know that an infinite $\sigma$-algebra contains an infinite sequence of disjoint sets.
Does there exist an infinite $\sigma$-algebra that contains no nonempty member which has no nonempty proper measurable subset?
 A: Put the topology of pointwise convergence of nets on the set of all functions $f:\mathbb R\to\{0,1\}$. In other words, the product topology on the product of continuum-many copies of $\{0,1\}$, where the latter has the discrete topology. Consider the $\sigma$-algebra of all Borel sets in this space.  An open set is a set of all functions whose values at finitely many points are specified.  An intersection of countably many open sets is a set of all functions whose values at countably many points are specified.
In other words, an intersection of countably many open sets is a set of the form
$\{f : \forall x\in A\  f(x)=g(x)\}$ for some fixed function $g$ and some countable set $A\subseteq\mathbb R$.  This will always have a proper subset that is a non-empty Borel set: just add some more members to $A$, but not uncountably many.
Does every Borel set have a subset that is the intersection of countably many Borel sets?  On that point I'm rusty; maybe this won't work if that's not true.
If one can show that a set containing only one function $f:\mathbb R\to \{0,1\}$ is never a Borel set in this space, I think that's enough.
