Show that a $\sigma$-field is either finite or uncountable. After looking around a bit I see that this is a standard claim, but I'm wondering if there's any way to argue what I had thought of during a quiz.
I thought to assume that $\mathscr{F}$ was in fact countable, so that there is some bijection $f:\mathscr{F} \rightarrow \mathbb{N}$. If this is the case we can associate every element of the $\sigma$-field to some integer; i.e index them as a sequence $(F_n)_{n \geq 1}$ which contains every element of the $\sigma$-field. But then because $\sigma$-fields are closed under countable unions, we have that $\cup_{n=1}^\infty F_n$ is an element of $\mathscr{F}$ that wasn't in the sequence. Contradiction?
Is there anything that can be done here to salvage this type of argument as it's all I could think of when I encountered this. Also I'd really appreciate if someone could tell me where this idea takes a wrong turn. Thanks in advance for the help!
 A: There's no easy way to make an argument like the one you've described work.  Assuming you have a countably infinite $\sigma$-field and enumerating it at $(F_n)_{n\geq 1}$ is a good start.  The problem is that it could happen that $\bigcup_{n=1}^{\infty}F_n = F_k$ for some $k$.  In fact, if your $\sigma$-field $\mathcal{F}$ is on a set $X$ then you will have $X \in \mathcal{F}$, and hence there will be some $k$ such that $F_k = X = \bigcup_{n=1}^{\infty}F_n$.
You could try to fix this problem by taking a union (or intersection) of some subcollection of $\mathcal{F}$, rather than taking the union of all the sets.  You again run into the problem of trying to figure out how to choose the subfamily so that the union (or intersection) you take can't be on the list already.  In the abstract it's hard to think of how you'd pick such a clever subcollection, but I think you could view the standard proof (eg, If $S$ is an infinite $\sigma$ algebra on $X$ then $S$ is not countable) as showing that such a choice is possible.
