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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?

Eric Wofsey
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Jonathan Schnell
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  • While it is true that the cardinality is smaller, that's not what your statement means (which is even clearer if you look at wikipedia's statement). –  Jul 14 '21 at 15:29
  • The Wikipedia article says that the co-countable $\sigma$-algebra is distinct from the power set, not that it has a smaller cardinality. (I don't know about the cardinality, I'm just pointing out that you seem to have misunderstood the statement.) – saulspatz Jul 14 '21 at 15:30

1 Answers1

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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 ^_^

Chris Grossack
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  • Thank you for the answer. I was hoping that there could be a simpler proof, showing the fact that every uncountable set $X$ has covering $X = A \cup A^c$ where $A,A^c$ uncountable. – Jonathan Schnell Jul 15 '21 at 12:25
  • @JonathanSchnell -- unfortunately, no such simpler proof exists. As I mentioned in my last comment, without some kind of argument relying on the axiom of choice, the theorem fails. Indeed amorphous sets are uncountable sets which cannot even be written as a disjoint union of two infinite sets! It's consistent with $\mathsf{ZF}$ that these sets exist (and they make your desired result false), so you're going to need to use choice in some fundamental way. – Chris Grossack Jul 15 '21 at 20:18
  • Oh, okey many thanks ;-) – Jonathan Schnell Jul 16 '21 at 09:15