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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?

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2 Answers 2

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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.

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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 $\mathcal{F}$.

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