# sigma-algebra

• Jun 28th 2010, 10:37 AM
ecnanif
sigma-algebra
Hi!

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

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

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

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

Thanks!
• Jun 28th 2010, 11:02 AM
gmatt
Thats not quite right.

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

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

Proof:

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

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

You are however, missing some sets in this. Consider this corollary to the previous result: if $A,B\in \Sigma$ then $A \setminus B \in \Sigma.$ This means that for example $\{2,3\} = \Omega \setminus\{1\}\in \Sigma$ along with a few other sets.
• Jun 28th 2010, 11:24 AM
ecnanif
I think I understood your claim and proof. You say that if we have a sequence of sets in a $\sigma$-algebra $F$ their intersection must also lie in $F$. Because all the complements of the sets must lie in $F$, we get that the union of all complements must lie in $F$, and because of this the complement of the union of complements must lie in $F$. And this is equal to the intersection of our original sets.

I donīt think I understand your corollary though.

I updated:
$F= \left\{\emptyset,\Omega, \left\{1\right\}, \left\{2\right\}, \left\{3\right\}, \left\{1,2\right\}, \left\{2,3\right\}, \left\{1,3\right\} \right\}$
• Jun 28th 2010, 12:50 PM
MattMan
Quote:

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

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

The corollary stating: If $A,B \in \Sigma, then A \setminus B \in \Sigma$ means
That since $\Omega, \{1\} \in \Sigma$, by the corollary this implies $\Omega \setminus \{1\} = \{2,3\} \in \Sigma$.
He was just showing you how you could get your missing sets in your $\sigma$-algebra. Your updated set is now correct.
• Jun 28th 2010, 11:19 PM
ecnanif
Ah okay =)
Thank you!