I have a question regarding the countable/co-countable $\sigma$-algebra. I will write first its definition:
Example. Let $X$ be an uncountable infinite set. Then
$$ \mathcal{A} = \{A \subseteq X\mid A\text{ is at most countable or } A^c\text{ is at most countable}\} $$
is a $\sigma$-algebra, which is strictly smaller than $\mathcal{P}(X) = 2^X$. (stated in Wikipedia)
Now to prove that $\mathcal{A}$ is a $\sigma$-algebra is quite easy, but how can we prove that $\mathcal{A} \subsetneq \mathcal{P}(X)$?
With $X=\mathbb{R}$ we can write $B=(-\infty,0]$ with $B^c=(0,+\infty)$, both are uncountable, with
$$ B\in \mathcal{P}(X) \quad \text{ and }\quad B\notin \mathcal{A} $$
So $\mathcal{A} \subsetneq\mathcal{P}(X)$.
But with a general $X$ how can we do it?
Best Answer
Welcome to MSE!
You have the right idea -- we want to find some subset $B$ of $X$ so that $B$ and $B^c$ are both uncountable. Intuitively it's clear that such a decomposition should exist, but it's hard to see how to get your hands on one without something like the order structure of $\mathbb{R}$.
One way to do this is with a bit of cardinal arithmetic. It turns out that for any infinite cardinal, $\lambda + \lambda = \lambda$. So using this fact, we can find bijections
$$X \cong \lambda \cong \lambda + \lambda$$
Since $\lambda + \lambda$ is (by definition) the size of the disjoint union of two sets of size $\lambda$, this gives us the desired decomposition (do you see how?).
As an aside, you're going to have to do something tricky like this, as without AC there are uncountable sets whose countable/cocountable algebra is the whole powerset algebra. See here.
I hope this helps ^_^