Generated Sigma-Algebra Preserves Finiteness

Theorem
Let $X$ be a set, and let $A_1, \ldots, A_n \subseteq X$ be subsets of $X$.

Then $\sigma \left({\left\{{A_1, \ldots, A_n}\right\}}\right)$ is a finite set, where $\sigma$ denotes generated $\sigma$-algebra.

Proof
Proceed by induction on $n$, that is, on the number of generators.

Basis for the Induction
The case $n = 0$ is verified as follows:


 * $\sigma \left({\varnothing}\right) = \left\{{\varnothing, X}\right\}$

from Sigma-Algebra Generated by Empty Set.

Thus $\sigma \left({\varnothing}\right)$ is finite.

This is the basis for the induction.

Induction Hypothesis
Fix $n \in \N$.

Assume that $\sigma \left({\left\{{A_1, \ldots, A_n}\right\}}\right)$ is finite.

This is the induction hypothesis.

Induction Step
This is the induction step:

For brevity, write $\Sigma_n$ for $\sigma \left({A_1, \ldots, A_n}\right)$.

Observe that:


 * $\left\{{A_1, \ldots, A_{n+1}}\right\} \subseteq \Sigma_n \cup \left\{{A_{n+1}}\right\} \subseteq \Sigma_{n+1}$

Hence Condition on Equality of Generated Sigma-Algebras applies to yield:


 * $\Sigma_{n+1} = \sigma \left({\Sigma_n \cup \left\{{A_{n+1}}\right\}}\right)$

From Sigma-Algebra Extended by Single Set, it follows that:


 * $\sigma \left({\Sigma_n \cup \left\{{A_{n+1}}\right\}}\right) = \left\{{\left({E_1 \cap A_{n+1}}\right) \cup \left({E_2 \cap A_{n+1}^c}\right): E_1, E_2 \in \Sigma_n}\right\}$

Thus there is a surjection $f: \Sigma_n \times \Sigma_n \to \Sigma_{n+1}$, defined by:


 * $f \left({E_1, E_2}\right) := \left({E_1 \cap A_{n+1}}\right) \cup \left({E_2 \cap A_{n+1}^c}\right)$

By Cardinality of Surjection, it now follows that:


 * $\left\vert{\Sigma_{n+1}}\right\vert \le \left\vert{\Sigma_n \times \Sigma_n}\right\vert$

and the latter equals $\left\vert{\Sigma_n}\right\vert^2$ by Cardinality of Cartesian Product.

By the induction hypothesis, this is necessarily finite.

Therefore, $\Sigma_{n+1}$ is also finite.

The result follows by the Principle of Mathematical Induction.