Sigma Algebra

A collection of subsets of $S$ is called a Sigma Algebra denoted by $\mathcal{B}$, if it satisfies:

1. \(\emptyset \in \mathcal{B}\)
2. \(A \in \mathcal{B} \implies A^\complement \in \mathcal{B}\)
3. \(A_1,A_2,A_3,… \in \mathcal{B} \implies \bigcup_{i} A_{i}\)

Which can be interpreted as:

1. The empty set is an element of $\mathcal{B}$
2. If $A$ is in $\mathcal{B}$ then the complement of $A$ (i.e. $A^\complement$) is also in $\mathcal{B}$
3. For any set of $A$s in $\mathcal{B}$, the union of all those $A$s is also in $\mathcal{B}$

If $S$ is empty, then we don’t have much to speak of. If $S$ is not empty, there is always the trivial sigma algebra $\mathcal{B}=\{\emptyset,S\}$. If $S$ has some structure, then $\mathcal{B}=\mathcal{P}(S)$¹ is also a sigma algebra.

I wonder how many sigma algebras there are for a set of size $n$?

Example:

Let $\left\vert{S}\right\vert=3$ with $S=\{1,2,3\}$

$\mathcal{B}_1=\{\emptyset,S\}$
$\mathcal{B}_2=\{\emptyset,\{1\},\{2,3\},S\}$
$\mathcal{B}_3=\{\emptyset,\{2\},\{1,3\},S\}$
$\mathcal{B}_4=\{\emptyset,\{3\},\{1,2\},S\}$
$\mathcal{B}_5=\{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},S\}=\mathcal{P}(S)$

So when $n=3$, there are 5 sigma algebras total. I believe in general, the answer is related to Bell numbers:

$B_n=\sum_{k=0}^{k=n} {n\brace k}$, where ${n\brace k}$ are the Stirling numbers of the second kind.

To help substantiate the claim, we find the third bell number:

$$B_{3}=0+1+3+1=5$$

Of course this doesn’t constitute a proof. But it does satisfy my suspicions :)

¹ Read as the power set of $S$, which is the set of all subsets of $S$.