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Let $A$ be set of finite number of subsets of set $\Omega$. How many members are there in the sigma algebra generated by $A$ ?

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I have just fixed Tex, putting '$A$' instead of $A$ (as these on are mathematical symbols, not exactly works). –  Davide Giraudo Sep 23 '12 at 14:07
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It depends. For example, let $A_1$ and $A_2$ be sets, neither of which is the empty set or $\Omega$. Suppose that $A_1$ is a proper subset of $A_2$. Then the $\sigma$-algebra generated by $A_1$ and $A_2$ has $8$ sets. By appropriate choices, we can achieve $4$ and even $2$. There are also choices of $A_1$ and $A_2$ that give a $\sigma$-algebra of size $16$, which is the maximum possible if $n=2$.

In general, let $A_1, A_2,\dots,A_n$ be our sets. Consider the collection $\mathbb{K}$ of all non-empty sets of the shape $$A_1^{e_1}\cap A_2^{e_2}\cap \cdots\cap A_n^{e_n},\tag{$1$}$$ where for each $i$, $e_i$ is either the empty word or the word consisting of the complement symbol. So we are saying "yes" or "no" to the $A_i$ in all possible ways.

If $\mathbb{K}$ has $k$ elements, then the $\sigma$-algebra generated by the $A_i$ has $2^k$ elements. For the elements of the $\sigma$-algebra are the union of any collection of elements of $\mathbb{K}$. In a sense, the elements of $\mathbb{K}$ are the atoms of the $\sigma$-algebra.

For a $\sigma$-algebra of maximum cardinality, assume that the $2^{n}$ expressions in $(1)$ all give non-empty sets. Then our $\sigma$-algebra has $2^{(2^n)}$ elements.

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