1. ## Sigma-algebras

I am considering the set $\displaystyle \left\{ 1,2,3,4,5,6 \right\}$ and two systems (sets) on $\displaystyle X$ given by:

$\displaystyle \mathcal{G}_{1} =\left\{ \left\{ 1,2,3 \right\}, \left\{2,3,4,5 \right\}, \left\{ 1,6 \right\} \right\}$

$\displaystyle \mathcal{G}_{2} =\left\{ \left\{ 1,4,5 \right\}, \left\{2,3,6 \right\}, \left\{ 4,5,6 \right\} \right\}$

I have to show that $\displaystyle \sigma(\mathcal{G}_{1})=\sigma(\mathcal{G}_{2})$.

I would first like to show that $\displaystyle \mathcal{G}_{1} \subset \sigma(\mathcal{G}_{2})$.

Suggestions to proceed are appreciated.

2. ## Re: Sigma-algebras

Originally Posted by detalosi
I am considering the set $\displaystyle \left\{ 1,2,3,4,5,6 \right\}$ and two systems (sets) on $\displaystyle X$ given by:

$\displaystyle \mathcal{G}_{1} =\left\{ \left\{ 1,2,3 \right\}, \left\{2,3,4,5 \right\}, \left\{ 1,6 \right\} \right\}$

$\displaystyle \mathcal{G}_{2} =\left\{ \left\{ 1,4,5 \right\}, \left\{2,3,6 \right\}, \left\{ 4,5,6 \right\} \right\}$

I have to show that $\displaystyle \sigma(\mathcal{G}_{1})=\sigma(\mathcal{G}_{2})$.

I would first like to show that $\displaystyle \mathcal{G}_{1} \subset \sigma(\mathcal{G}_{2})$.

Suggestions to proceed are appreciated.
How to start? Let $A = \{1,4,5\}$ and $B = \{4,5,6\}$. Then how about $\{1,6\} = A\cup B - A\cap B$ with the usual meaning of $-$ for sets? So $\{1,6\} \in \sigma (\mathcal{G}_{2})$

3. ## Re: Sigma-algebras

I am asked to show the above without writing the sigma-algebras.

How do you see that {1,6} is in $\displaystyle \sigma(\mathcal{G}_{2})$?

4. ## Re: Sigma-algebras

Originally Posted by detalosi
I am asked to show the above without writing the sigma-algebras.

How do you see that {1,6} is in $\displaystyle \sigma(\mathcal{G}_{2})$?
Because A and B are in $\mathcal{G}_{2}$ and I showed you how to express {1,6} in terms of unions, intersections etc of them.

5. ## Re: Sigma-algebras

Yes, I see.

Is this correct:

By definition $\displaystyle A\in \sigma(\mathcal{G}_{2})$ since $\displaystyle A\in \mathcal{G}_{2}$ and $\displaystyle B\in \sigma(\mathcal{G}_{2})$ since $\displaystyle B\in \mathcal{G}_{2}$.
Therefore $\displaystyle (A \cup B - A \cap B) \in \sigma(\mathcal{G}_{2})$ due to closure of $\displaystyle \sigma$-algebras under countable unions, countable intersections and complements.

How do I conclude from here that $\displaystyle \sigma(\mathcal{G}_{1}) \subset \sigma(\mathcal{G}_{2})$ ?

6. ## Re: Sigma-algebras

Originally Posted by detalosi
Yes, I see.

Is this correct:

By definition $\displaystyle A\in \sigma(\mathcal{G}_{2})$ since $\displaystyle A\in \mathcal{G}_{2}$ and $\displaystyle B\in \sigma(\mathcal{G}_{2})$ since $\displaystyle B\in \mathcal{G}_{2}$.
Therefore $\displaystyle (A \cup B - A \cap B) \in \sigma(\mathcal{G}_{2})$ due to closure of $\displaystyle \sigma$-algebras under countable unions, countable intersections and complements.

How do I conclude from here that $\displaystyle \sigma(\mathcal{G}_{1}) \subset \sigma(\mathcal{G}_{2})$ ?
Walagaster showed you that one set in $\mathcal{G}_1$ is in $\sigma(\mathcal{G}_2)$. Once you show that every set of $\mathcal{G}_1$ is in the latter sigma-algebra, then you know that $\sigma(\mathcal{G}_1)$ is the smallest sigma-algebra containing $\mathcal{G}_1$. If $\mathcal{G}_1 \subset \sigma(\mathcal{G}_2)$ (as you show element-by-element), then $\sigma(\mathcal{G}_2)$ by definition is a superset (not necessarily proper) of $\sigma(\mathcal{G}_1)$. Use your definitions.