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I have been asked to prove that the number of elements in a finite sigma algebra over a set $X$ is $2^n$ for some integer $n$. How do I go about this problem? I have no idea where to start. Thanks in advance for any ideas.

Do I need to prove that given a set $F$, $\sigma(F)$ is actually a power set of some set say $S$?

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+1, nice question! – Salech Alhasov Aug 12 '12 at 6:59

3 Answers 3

Let $\Sigma$ be the $\sigma$-algebra. Choose $x \in X$, and define $M_x = \cap_{M \in \Sigma, x \in M} M$. Clearly $M_x \neq \emptyset$, and $M_x \in \Sigma$.

Furthermore, the collection $F = \{M_x \} \subset \Sigma$ is a partition of $X$ (and finite, of course). To see this, suppose $M_x \cap M_y \neq \emptyset$. Then we must have $M_x = M_y$, or else either $M_x \setminus M_y $ or $M_y \setminus M_x $ would be strictly smaller sets contradicting the definition of either $M_x$ or $M_y$.

Furthermore, it is clear that if $M \in \Sigma$, then $M = \cup_{x \in M} M_x$, hence every element of $\Sigma$ is the (disjoint) union of members of $F$ (the empty set taken as the union of no members of $F$), hence $|\Sigma| = 2^{|F|}$.

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Note that a finite $\sigma$-algebra $\mathcal{A}$ has minimal nonempty elements. Show that every element of $\mathcal{A}$ is a union of these minimal elements.

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To clarify: If you manage to prove Arthur Fischer's claim. Then let $A_1$, ..., $A_n$ be these finitely many minimal elements. To each binary sequence $\sigma$ of length associated it with the set $\bigcup_{\sigma(n) = 1} A_n$. So there are $2^n$ elements in this sigma algebra. – William Aug 12 '12 at 5:34

Here, in page 4 you will find a proof and a efficiant algorithm finding $n$.

Good luck!

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The link is broken… – Dirk Jul 7 at 8:49
Thatt's still broken, but this contains a proof which yields the result in the finite case:… – BrianO Oct 12 at 23:20

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