# Let $\Omega$ be a finite set. Let $\mathcal{F}\subset\mathcal{P}(\Omega)$ be an algebra. Show that $\mathcal{F}$ is a $\sigma$-algebra.

Let $$\Omega$$ be a finite set. Let $$\mathcal{F}\subset\mathcal{P}(\Omega)$$ be an algebra. Show that $$\mathcal{F}$$ is a $$\sigma$$-algebra.

MY ATTEMPT

Since $$\mathcal{F}$$ is an algebra, $$\Omega\in\mathcal{F}$$. Moreover, if $$A\in\mathcal{F}$$, then $$A^{c}\in\mathcal{F}$$. Finally, if $$A,B\in\mathcal{F}$$, then $$A\cup B\in\mathcal{F}$$.

Now we have to prove that the countable union of sets in $$\mathcal{F}$$ does belong to $$\mathcal{F}$$.

Here it is the sketch of my attempt to prove it: since there are finitely many subsets of $$\Omega$$, the countable union has to have finitely many different sets in its composition. Consequently, such union is a finite union of subsets of $$\Omega$$, which clearly belongs to $$\mathcal{F}$$ since it is an algebra.

However I am not sure if it is a good approach or how to formalize it.

Your sketch is pretty much a proof already. If you wanted to be more precise about it: suppose you have a countable family $$(A_i)_{i\in I}$$ of subsets of $$\Omega$$, then this is equivalently a function $$f:I\to2^\Omega$$. Since $$2^\Omega$$ is finite, so is the image $$fI$$, so write $$fI = \{B_1,\dots,B_n\}$$ and now $$\bigcup_{i\in I}A_i = B_1\cup\dots\cup B_n$$. As $$\mathcal F$$ is an algebra, it is closed under binary union, so by induction this union will be in $$\mathcal F$$ also.
Your explanation is good enough. Not sure how much formal you want the proof to be. If you want to make it very formal, it is possible. Let $$(A_n)_{n=1}^\infty$$ be a sequence of elements in $$\mathcal{F}$$. Define an equivalence relation on $$\mathbb{N}$$ like this: $$n\equiv m$$ iff $$A_n=A_m$$. The set of equivalence classes $$\mathbb{N}/\equiv$$ is finite, for example because $$\mathcal{F}$$ is finite and you can define a surjective function $$\rho: \mathcal{F}\to\mathbb{N}/\equiv$$ by $$\rho(A)=[m]$$ if $$A$$ belongs to the sequence $$(A_n)$$ and $$A=A_m$$, and $$\rho(A)=[1]$$ if $$A$$ does not belong to the sequence. This map is a well defined surjection, so $$\mathbb{N}/\equiv$$ is a finite set.
Ok, so there are finitely many equivalence classes $$S_1,...,S_k$$. Let $$b_i$$ be the smallest natural number which belongs to $$S_i$$. Then using a two sided inclusion we can easily show that $$\cup_{i=1}^\infty A_i=\cup_{i=1}^k A_{b_i}$$, so the union is actually a union of finitely many elements from $$\mathcal{F}$$.