Question on Sigma Algebras and Non-Finite Sets

[BrainHurts]

Homework Statement

Find a set X such that $\mathcal{A}_1 \text{ and } \mathcal{A}_2$ are $\sigma$-algebras where both $\mathcal{A}_1 \text{ and } \mathcal{A}_2$ consists of subsets of X. We want to show that there exists such a collection such that $\mathcal{A}_1 \cup \mathcal{A}_2$ is not a $\sigma$ - algebra

The Attempt at a Solution

So here's what I'm thinking. I feel like for sure we need to fail the condition of Countable additivity.

I'm using a simple example like $X = \{1,2,3\}$ and I chose something $\mathcal{A}_1 = \left\{\emptyset,\{1,2,3\}, \{1\}, \{2,3\} \right\}$

and $\mathcal{A}_2 = \left\{\emptyset,\{1,2,3\}, \{2\}, \{1,3\} \right\}$

I have shown that both $\mathcal{A}_1$ and $\mathcal{A}_2$ are $\sigma$ algebras.

Am I on the right track here? Should I think of non-finite sets?

 

[Dick]

Homework Statement

Find a set X such that $\mathcal{A}_1 \text{ and } \mathcal{A}_2$ are $\sigma$-algebras where both $\mathcal{A}_1 \text{ and } \mathcal{A}_2$ consists of subsets of X. We want to show that there exists such a collection such that $\mathcal{A}_1 \cup \mathcal{A}_2$ is not a $\sigma$ - algebra

The Attempt at a Solution

So here's what I'm thinking. I feel like for sure we need to fail the condition of Countable additivity.

I'm using a simple example like $X = \{1,2,3\}$ and I chose something $\mathcal{A}_1 = \left\{\emptyset,\{1,2,3\}, \{1\}, \{2,3\} \right\}$

and $\mathcal{A}_2 = \left\{\emptyset,\{1,2,3\}, \{2\}, \{1,3\} \right\}$

I have shown that both $\mathcal{A}_1$ and $\mathcal{A}_2$ are $\sigma$ algebras.

Am I on the right track here? Should I think of non-finite sets?

No, no need for infinite sets. Can you show the union of those two is not a sigma algebra?

 

[BrainHurts]

OK I'm going to do all the four steps

$\mathcal{A}_1 \cup \mathcal{A}_2 = \{ \emptyset, \{ 1,2,3 \}, \{ 1 \}, \{ 2,3 \}, \{2 \} , \{ 1,3\} \}$

1) it is clear that $\emptyset, \{1,2,3\}$are in $\mathcal{A}_1 \cup \mathcal{A}_2$

2) so if $A \in \mathcal{A}_1 \cup \mathcal{A}_2$, then $A^c \in \mathcal{A}_1 \cup \mathcal{A}_2$.

I think this is satisfied, e.g. if $A = \{ 1 \}$ , then $A^c = \{ 2,3 \}$ and both are in $\mathcal{A}_1 \cup \mathcal{A}_2$

3) if $B_1, ... B_n \in \mathcal{A}_1 \cup \mathcal{A}_2$ then both

$\bigcup_{i=1}^n A_i$ and $\bigcap_{i=1}^n A_i$ are both in $\mathcal{A}_1 \cup \mathcal{A}_2$

I think this is it! I just came up with it now,

so if I take $A_1 = \{ 2,3 \}$ and $A_2 \{1,3\}$ then the intersection is $\{ 3 \}$ and that's not in $\mathcal{A}_1 \cup \mathcal{A}_2$. Is this right? so it fails the condition that $\bigcup B_i$ is not in $\mathcal{A}_1 \cup \mathcal{A}_2$

 

[BrainHurts]

Sorry I meant $\bigcap B_i$ is not in $\mathcal{A}_1 \cup \mathcal{A}_2$

 

[Dick]

Sorry I meant $\bigcap B_i$ is not in $\mathcal{A}_1 \cup \mathcal{A}_2$

Sure. You can get {3} by intersections or unions and complements of sets in $\mathcal{A}_1 \cup \mathcal{A}_2$ but it's not in $\mathcal{A}_1 \cup \mathcal{A}_2$.