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.
Could someone please help me with this?
Best Answer
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.