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.