Hello!

I am struggling with just the very last stage of the solution and any suggestions would be much appriciated!! Thanks!

I am trying to show that given a set $\displaystyle X$ and a collection of subsets $\displaystyle U$ of $\displaystyle X$, and given that

(i) $\displaystyle X \in U$ and $\displaystyle \emptyset \in U$

(ii) $\displaystyle E \in U$ implies that the complement $\displaystyle X\setminus E \in U $

($\displaystyle U $ is closed under complements)

(iii) $\displaystyle U $ is closed under finite (or, case 2: countable) intersections, (ie, $\displaystyle E_1, E_2, ..., E_N \in U \Rightarrow \bigcap_{n=1}^{N}E_n \in U} $)

I am trying to show that $\displaystyle U $ is an algebra (or, case 2: a sigma algebra)

Handling case 1, closed under finite intersections, I think I have come up with a proof using some set theory but I was hoping someone might be able to verify.

So, firstly, I think all we need to show is that $\displaystyle U$ is closed under finite unions to show that it is an algebra.

To do this, take $\displaystyle E_1, E_2, ..., E_N \in U$. Then since $\displaystyle U $ is closed under complements and also under finite intersections, $\displaystyle \bigcap_{n=1}^{N}X\setminus E_n$ $\displaystyle \in U $.

Then by Demorgans Law, $\displaystyle \bigcap_{n=1}^{N}X\setminus E_n = X\setminus {\bigcup_{n=1}^{N}X\setminus E_n$. Since the left-hand side is in $\displaystyle U$, the right-hand side will be as well. Also, Since, we are closed under complements, $\displaystyle X\setminus {\bigcup_{n=1}^{N}X\setminus E_n \in U \Rightarrow {\bigcup_{n=1}^{N}X\setminus E_n \in U$.

Here is where I am stuck. Can I show that $\displaystyle {\bigcup_{n=1}^{N}X\setminus E_n \in U \Rightarrow {\bigcup_{n=1}^{N}E_n \in U $? Because then I would be done (we would have shown it is closed under finite unions). And I believe a similar argument would work for countable unions and sigma algebras?

Again, any help appriciated!! Thank you!