# Showing a family is actually an algebra

• Jan 12th 2012, 09:39 AM
matt.qmar
Showing a family is actually an algebra
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 $X$ and a collection of subsets $U$ of $X$, and given that
(i) $X \in U$ and $\emptyset \in U$
(ii) $E \in U$ implies that the complement $X\setminus E \in U$
( $U$ is closed under complements)
(iii) $U$ is closed under finite (or, case 2: countable) intersections, (ie, $E_1, E_2, ..., E_N \in U \Rightarrow \bigcap_{n=1}^{N}E_n \in U}$)

I am trying to show that $U$ 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 $U$ is closed under finite unions to show that it is an algebra.

To do this, take $E_1, E_2, ..., E_N \in U$. Then since $U$ is closed under complements and also under finite intersections, $\bigcap_{n=1}^{N}X\setminus E_n$ $\in U$.
Then by Demorgans Law, $\bigcap_{n=1}^{N}X\setminus E_n = X\setminus {\bigcup_{n=1}^{N}X\setminus E_n$. Since the left-hand side is in $U$, the right-hand side will be as well. Also, Since, we are closed under complements, $X\setminus {\bigcup_{n=1}^{N}X\setminus E_n \in U \Rightarrow {\bigcup_{n=1}^{N}X\setminus E_n \in U$.

Here is where I am stuck. Can I show that ${\bigcup_{n=1}^{N}X\setminus E_n \in U \Rightarrow {\bigcup_{n=1}^{N}E_n \in U$? 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!
• Jan 12th 2012, 10:03 AM
Plato
Re: Showing a family is actually an algebra
You are using the wrong LaTeX tags.
[TEX]X\setminus E[/TEX] gives $X\setminus E$