Let $\mathcal{M}$ be an infinite $\sigma$-algebra.
$\mathbf{a.} \ \mathcal{M}$ contains an infinite sequence of disjoint sets
I am aware of the solution discussed here. In fact when my professor first gave me this problem, that is very similar to my approach.
My question is about a possible alternate solution that I think is a little nicer. This is by no means a formal argument but an outline of what I could make into a formal argument.
Since $\mathcal{M}$ is a $\sigma$-algebra, it is closed under countable unions and complements (and intersections). So it's easy to show that it is also closed under countable symmetric differences. We can then put a ring structure on $\mathcal{M}$ with addition as symmetric difference and multiplication as intersection.
We claim that $\mathcal{M}$ is also a boolean ring. The result then follows from the fact that an infinite boolean ring contains infininitely many zero divisors.
The main idea here being that if two sets are disjoint, they're zero-divisors in a ring-theoretic sense. Is this a valid approach to the problem?