# An algebra on a finite set is a $\sigma$-algebra on the finite set.

I am learning probability theory, and it is my desire to prove the following statement: an algebra on a finite set is a $$\sigma$$-algebra on the finite set.

Let $$\Omega$$ be finite and $$\mathcal{A}$$ be an algebra on $$\Omega$$. By definition, we have the following facts: $$\mathcal{A} \subseteq 2^{\Omega}$$ (power set); $$\Omega \in \mathcal{A}$$; $$\mathcal{A}$$ is $$\setminus$$-closed; $$\mathcal{A}$$ is $$\cup$$-closed.

To prove that $$\mathcal{A}$$ is a $$\sigma$$-algebra on $$\Omega$$, we only need to prove $$\mathcal{A}$$ is closed under countable unions. This requires me to go from finite unions to infinite unions.

Let $$S$$ be a sequence on $$\mathbb{N}$$ such that for any $$i \in \mathbb{N}$$, $$S_{i} \in \mathcal{A}$$. Then it is my desire to prove that $$\bigcup_{i \in \mathbb{N}} S_{i} \in \mathcal{A}$$. How to prove it using the given facts?

• What is wrong with saying $\mathcal A$ is finite (it is a subset of $2^{\Omega}$) so any countable union of elements of $\mathcal A$ is a finite union of elements of $\mathcal A$ when you remove duplicates. Commented Aug 19, 2023 at 15:47
• Well, if $\Omega$ is finite, can you actually have infinitely many distinct sets in your countable collection?
– Mark
Commented Aug 19, 2023 at 15:47
• @Henry Ha, seems to be the case :) Commented Aug 19, 2023 at 15:49

If you have finitely many sets to work with, any uncountable union must be actually a finite union: Since the Number of elements in the truncated union $$\bigcup_{i=0}^n A_i$$ is non decreasing in function of $$n$$, there must be finitely many indicies $$n_1,…n_k$$ where the number of elements strictly increased. Otherwise the countable union would be an infinite set. Now we see, that $$\bigcup_{i=0}^\infty A_i=A_{n_1} \cup …\cup A_{n_k}$$, since these sets are the only sets contributing to new elements in the unions. Therefore we represented the infinite union as a finite one.