I've been solving some problems from my Functional Analysis course and there's a part of this problem I'm not sure how to do. The problem goes like this:
Given $\Sigma$ a collection of subsets of a non-empty set $X$ that verifies:
- $X\in\Sigma$
- $A,B\in\Sigma \implies B\setminus A\in\Sigma$
- $A,B\in\Sigma, A\cap B=\emptyset \implies A\cup B\in\Sigma$
- $(A_n)_{n\in\mathbb{N}}\subseteq\Sigma \text{ decreasing sequence} \implies \bigcap_{n=1}^\infty A_n\in\Sigma$
Prove $\Sigma$ is a $\sigma$-algebra.
What I did is just prove the three different points of the $\sigma$-algebra definition:
First I prove $\emptyset\in\Sigma$. This is easy since (1) says that $X\in\Sigma$ and using (2), $X\setminus X=\emptyset\in\Sigma$.
Now to prove $A\in\Sigma$ implies $A^c\in\Sigma$, I use again (1) and (2), $X\setminus A = A^c\in\Sigma$.
These two properties were easy to prove, the problem is the last one:
- Now I need to prove that $\Sigma$ is closed under countable union, to be said, $(A_n)_{n\in\mathbb{N}}\subseteq \Sigma \implies\bigcup A_n\in\Sigma$. It's clear that I'll need to use hypothesis (3) and (4). From (4) I can conclude that $\Sigma$ is closed under countable unions of increasing sequences of sets, because $$\left[\bigcap_{n\in\mathbb{N}} A_n \right]^c = \bigcup_{n\in\mathbb{N}} A_n^c$$ and $(A_n^c)$ is an increasing sequence (because $(A_n)$ was a decreasing one). But as I said this is verified for increasing countable sequences, not in general. I guess now I must use hypothesis (3) to extend it to all countable sequences, but I don't see how.
How can I finish the prove of the third property? Is the work I did till that point correct? Any help or hint will be appreciated, thanks in advance.