MHB Is $\mathcal{A}$ a $\sigma$-algebra over $M$?

[mathmari]

Hey!

We have the set $M=\{1,2,3,4\}$. I want to check if the following are $\sigma$-algebras over $M$.

1. $\mathcal{A}=\{\emptyset, \{1,2\}, \{3,4\}\}$
2. $\mathcal{A}=\{\emptyset, M, \{1\}, \{2,3,4\}, \{1,2\}, \{3,4\}\}$
3. $\mathcal{A}=\{\emptyset, M, \{3\}, \{1,2,4\}\}$
4. $\mathcal{A}=\{\emptyset, M, \{1\}, \{2\}, \{1,2\}, \{3,4\}, \{2,3,4\}, \{1,2,4\}, \{1,3,4\}\}$

$\mathcal{A}$ is a $\sigma$-algebra over $M$ if the following hold:

• $M$ is in $\mathcal{A}$.
• If $A$ is in $\mathcal{A}$, then so is the complement of $A$.
• If $A_n$ is a sequence of elements of $\mathcal{A}$, then the union of the $A_n$s is in $\mathcal{A}$.

I have done the following:

1. $\mathcal{A}=\{\emptyset, \{1,2\}, \{3,4\}\}$
   
   • $M \notin \mathcal{A}$.
   
   Therefore, $\mathcal{A}$ is not a $\sigma$-algebra over $M$.
2. $\mathcal{A}=\{\emptyset, M, \{1\}, \{2,3,4\}, \{1,2\}, \{3,4\}\}$
   
   • $M \in \mathcal{A}$.
   • $\emptyset \in \mathcal{A} \rightarrow \emptyset^c=M\in \mathcal{A}$
     $M \in \mathcal{A} \rightarrow M^c=\emptyset\in \mathcal{A}$
     $\{1\}\in \mathcal{A} \rightarrow \{1\}^c=\{2,3,4\}\in \mathcal{A}$
     $\{2,3,4\} \in \mathcal{A} \rightarrow \{2,3,4\}^c=\{1\}\in \mathcal{A}$
     $\{1,2\} \in \mathcal{A} \rightarrow \{1,2\}^c=\{3,4\}\in \mathcal{A}$
     $\{3,4\} \in \mathcal{A} \rightarrow \{3,4\}^c=\{1,2\}\in \mathcal{A}$
     
   • Do we have to check if the union of all the elements of $\mathcal{A}$ is again in $\mathcal{A}$ ?
3. $\mathcal{A}=\{\emptyset, M, \{3\}, \{1,2,4\}\}$
   
   • $M\in \mathcal{A}$.
   • $\emptyset \in \mathcal{A} \rightarrow \emptyset^c=M\in \mathcal{A}$
     $M \in \mathcal{A} \rightarrow M^c=\emptyset\in \mathcal{A}$
     $\{3\} \in \mathcal{A} \rightarrow \{3\}^c=\{1,2,4\}\in \mathcal{A}$
     $\{1,2,4\} \in \mathcal{A} \rightarrow \{1,2,4\}^c=\{3\}\in \mathcal{A}$
     
   • Do we have to check if the union of all the elements of $\mathcal{A}$ is again in $\mathcal{A}$ ?
4. $\mathcal{A}=\{\emptyset, M, \{1\}, \{2\}, \{1,2\}, \{3,4\}, \{2,3,4\}, \{1,2,4\}, \{1,3,4\}\}$
   
   • $M \in \mathcal{A}$.
   • $\emptyset \in \mathcal{A} \rightarrow \emptyset^c=M\in \mathcal{A}$
     $M \in \mathcal{A} \rightarrow M^c=\emptyset\in \mathcal{A}$
     $\{1\}\in \mathcal{A} \rightarrow \{1\}^c=\{2,3,4\}\in \mathcal{A}$
     $\{2\} \in \mathcal{A} \rightarrow \{2\}^c=\{1,3,4\}\in \mathcal{A}$
     $\{1,2\} \in \mathcal{A} \rightarrow \{1,2\}^c=\{3,4\}\in \mathcal{A}$
     $\{3,4\}\in \mathcal{A} \rightarrow \{3,4\}^c=\{1,2\}\in \mathcal{A}$
     $\{2,3,4\} \in \mathcal{A} \rightarrow \{2,3,4\}^c=\{1\}\in \mathcal{A}$
     $\{1,2,4\} \in \mathcal{A} \rightarrow \{1,2,4\}^c=\{3\}\notin \mathcal{A}$
   
   So, $\mathcal{A}$ is not a $\sigma$-algebra.

Is what I have done correct? Could you give me a hint how we could check the last property? (Wondering)

 

[I like Serena]

[*] Do we have to check if the union of all the elements of $\mathcal{A}$ is again in $\mathcal{A}$ ?

Is what I have done correct? Could you give me a hint how we could check the last property? (Wondering)

Hey mathmari! (Smile)

Yep. It's all correct.

To check the last property formally we need to check every sequence of elements.
With 6 elements that is $2^6=64$ possible sequences. (Sweating)

However, it suffices if we check every combination of 2 elements.
We can break up the union of a sequence of n elements into a series of unions of 2 elements after all.
With 6 elements that is 6 x 5 = 30 possible sequences.
Furthermore, combinations that contain $\varnothing$ or $M$ trivially satisfy the requirement.
So that leaves us with 4 x 3 = 12 combinations to verify.

 

[mathmari]

However, it suffices if we check every combination of 2 elements.
We can break up the union of a sequence of n elements into a series of unions of 2 elements after all.

Why does this suffice to check every combination of 2 elements? (Wondering)

With 6 elements that is 6 x 5 = 30 possible sequences.

How do we get that number of possible sequences? I got stuck right now. (Wondering)

 

[I like Serena]

Why does this suffice to check every combination of 2 elements?

Let's take a look at the sequence $(A_1, A_2, A_3, ..., A_n)$.
Then we need that $A_1\cup A_2\cup A_3\cup ...\cup A_n \in \mathcal{A}$ yes?

Or written otherwise that $((((A_1\cup A_2)\cup A_3)\cup ...)\cup A_n) \in \mathcal{A}$.
But we also need that $A_1\cup A_2\in\mathcal A$, since that's also a possible sequence.
Every parenthesized expression of effectively 2 elements must be in $\mathcal A$.

In other words, if we pick each combination of 2 elements and verify that it's in $\mathcal A$, it follows that each sequence that is built from more than 2 elements is also in $\mathcal A$.

How do we get that number of possible sequences? I got stuck right now.

In how many ways can we pick 2 elements from $\mathcal A$ if it has 6 elements? (Wondering)

 

[mathmari]

Let's take a look at the sequence $(A_1, A_2, A_3, ..., A_n)$.
Then we need that $A_1\cup A_2\cup A_3\cup ...\cup A_n \in \mathcal{A}$ yes?

Or written otherwise that $((((A_1\cup A_2)\cup A_3)\cup ...)\cup A_n) \in \mathcal{A}$.
But we also need that $A_1\cup A_2\in\mathcal A$, since that's also a possible sequence.
Every parenthesized expression of effectively 2 elements must be in $\mathcal A$.

In other words, if we pick each combination of 2 elements and verify that it's in $\mathcal A$, it follows that each sequence that is built from more than 2 elements is also in $\mathcal A$.

I got it! (Nerd)

Furthermore, combinations that contain $\varnothing$ or $M$ trivially satisfy the requirement.

We use here the fact that $\emptyset \cup S=S$ and $M\cup S=M$, for any set $S$, right? (Wondering)
So, we have the following:

2. $\mathcal{A}=\{\emptyset, M, \{1\}, \{2,3,4\}, \{1,2\}, \{3,4\}\}$

• $M \in \mathcal{A}$.
• $\emptyset \in \mathcal{A} \rightarrow \emptyset^c=M\in \mathcal{A}$
  $M \in \mathcal{A} \rightarrow M^c=\emptyset\in \mathcal{A}$
  $\{1\}\in \mathcal{A} \rightarrow \{1\}^c=\{2,3,4\}\in \mathcal{A}$
  $\{2,3,4\} \in \mathcal{A} \rightarrow \{2,3,4\}^c=\{1\}\in \mathcal{A}$
  $\{1,2\} \in \mathcal{A} \rightarrow \{1,2\}^c=\{3,4\}\in \mathcal{A}$
  $\{3,4\} \in \mathcal{A} \rightarrow \{3,4\}^c=\{1,2\}\in \mathcal{A}$

$\{1\}\cup \{2,3,4\}=M\in \mathcal{A}$
$\{1\}\cup \{1,2\}=\{1,2\}\in \mathcal{A}$
$\{1\}\cup \{3,4\}=\{1,3,4\}\notin \mathcal{A}$

Therefore, $\mathcal{A}$ is not a $\sigma$-algebra.

3. $\mathcal{A}=\{\emptyset, M, \{3\}, \{1,2,4\}\}$

• $M\in \mathcal{A}$.
• $\emptyset \in \mathcal{A} \rightarrow \emptyset^c=M\in \mathcal{A}$
  $M \in \mathcal{A} \rightarrow M^c=\emptyset\in \mathcal{A}$
  $\{3\} \in \mathcal{A} \rightarrow \{3\}^c=\{1,2,4\}\in \mathcal{A}$
  $\{1,2,4\} \in \mathcal{A} \rightarrow \{1,2,4\}^c=\{3\}\in \mathcal{A}$
• Do we have to check if the union of all the elements of $\mathcal{A}$ is again in $\mathcal{A}$ ?

$\{3\}\cup \{1,2,4\}=M\in \mathcal{A}$

So, $\mathcal{A}$ is a $\sigma$-algebra. Is everything correct? (Wondering)

In how many ways can we pick 2 elements from $\mathcal A$ if it has 6 elements? (Wondering)

Aren't there $\binom{6}{2}=\frac{6!}{2!\cdot 4!}=\frac{4!\cdot 5\cdot 6}{2\cdot 4!}=5\cdot 3=15$ ways? (Wondering)

 

[I like Serena]

We use here the fact that $\emptyset \cup S=S$ and $M\cup S=M$, for any set $S$, right?

Yep.
And actually we also have $S\cup S=S$ for any set $S$. (Nerd)

Is everything correct?

Yep. (Nod)

Aren't there $\binom{6}{2}=\frac{6!}{2!\cdot 4!}=\frac{4!\cdot 5\cdot 6}{2\cdot 4!}=5\cdot 3=15$ ways?

Indeed.
As I wrote it, I took 6 choices for the 1st element and 5 choices for the 2nd element, giving us 6 x 5 = 30 ways.
Using $S\cup T=T\cup S$, we can reduce that by a factor 2 giving the 15 ways that you found. (Mmm)

 

[mathmari]

Yep.
And actually we also have $S\cup S=S$ for any set $S$. (Nerd)
Yep. (Nod)
Indeed.
As I wrote it, I took 6 choices for the 1st element and 5 choices for the 2nd element, giving us 6 x 5 = 30 ways.
Using $S\cup T=T\cup S$, we can reduce that by a factor 2 giving the 15 ways that you found. (Mmm)

I understand! Thank you so much! (Mmm)