1. ## sigma-algebra

Hi!

Trying to get a grasp on $\displaystyle \sigma-algebra$.
Letīs say I have a the set $\displaystyle \Omega = \left\{1,2,3}\right\}$.

Now, if I want to find the smallest $\displaystyle \sigma-algebra \, F$ that includes both 1 and 2. Would this be:

$\displaystyle F = \left\{\emptyset, \Omega, 1, 2, 3, \left\{1,2\right\}\right\}$ ?

The empty set is there, and also its complement $\displaystyle \Omega$. Both 1 and 2 are there and their complement 3. Also the union of 1 and 2 are is there.

Thanks!

2. Thats not quite right.

First, here is a result for $\displaystyle \sigma$ algebras that is quite useful to know:

If $\displaystyle X$ is a set and $\displaystyle \Sigma \subset \mathcal{P}\{X\}$ is a $\displaystyle \sigma$-algebra on $\displaystyle X$ then
$\displaystyle \Sigma$ is closed under countable intersections.

Proof:

Let $\displaystyle \bigcap_{n=1}^\infty A_n$ for $\displaystyle A_i \in \Sigma.$ Then note that for each $\displaystyle A_i \in \Sigma$ we have that $\displaystyle \bar{A}_i \in \Sigma$ where $\displaystyle \bar{A}_i$ is the complement of $\displaystyle A_i.$ It follows that $\displaystyle \bigcap_{n=1}^\infty A_n = X \setminus{ \bigcup_{n=1}^\infty \bar{A}_n } \in \Sigma. _\Box$

But other than that I can point out that $\displaystyle 1,2,3$ are not themselves sets and can not possibly be in the $\displaystyle \sigma$-algebra. However, I think you meant $\displaystyle \{1\},\{2\},\{3\}$, which is correct.

You are however, missing some sets in this. Consider this corollary to the previous result: if $\displaystyle A,B\in \Sigma$ then $\displaystyle A \setminus B \in \Sigma.$ This means that for example $\displaystyle \{2,3\} = \Omega \setminus\{1\}\in \Sigma$ along with a few other sets.

3. I think I understood your claim and proof. You say that if we have a sequence of sets in a $\displaystyle \sigma$-algebra $\displaystyle F$ their intersection must also lie in $\displaystyle F$. Because all the complements of the sets must lie in $\displaystyle F$, we get that the union of all complements must lie in $\displaystyle F$, and because of this the complement of the union of complements must lie in $\displaystyle F$. And this is equal to the intersection of our original sets.

I donīt think I understand your corollary though.

I updated:
$\displaystyle F= \left\{\emptyset,\Omega, \left\{1\right\}, \left\{2\right\}, \left\{3\right\}, \left\{1,2\right\}, \left\{2,3\right\}, \left\{1,3\right\} \right\}$

4. Originally Posted by ecnanif
I donīt think I understand your corollary though.

I updated:
$\displaystyle F= \left\{\emptyset,\Omega, \left\{1\right\}, \left\{2\right\}, \left\{3\right\}, \left\{1,2\right\}, \left\{2,3\right\}, \left\{1,3\right\} \right\}$
The corollary stating: If $\displaystyle A,B \in \Sigma, then A \setminus B \in \Sigma$ means
That since $\displaystyle \Omega, \{1\} \in \Sigma$, by the corollary this implies $\displaystyle \Omega \setminus \{1\} = \{2,3\} \in \Sigma$.
He was just showing you how you could get your missing sets in your $\displaystyle \sigma$-algebra. Your updated set is now correct.

5. Ah okay =)
Thank you!