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I know that a sigma algebra generated by a subset a, $\sigma(a) = \{\emptyset,a,a^c,E\}$.

But what about $\sigma({a,b})$? Would it be $\{\emptyset,a,a^c, b, b^c,E\}$?

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A $\sigma$-algebra is, in particular, closed under finite intersections and unions. Your claimed example is not. –  Zhen Lin Dec 11 '11 at 1:45
    
You are missing, potentially, $a\cap b$, $a\triangle b$, $a-b$, $b-a$, $(a-b)\cup b^c$, and lots more... –  Arturo Magidin Dec 11 '11 at 1:54
    
thanks, so the only way to find \sigma({a,b}) is to check $a\cap b$...? –  Thomas Dec 11 '11 at 1:58
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$\sigma(\{a\})$ generally has 4 elements; $\sigma(\{a,b\})$ generally has 16 elements. –  GEdgar Dec 11 '11 at 2:05
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And more generally, if you start with $n$ sets, the $\sigma$-algebra generated by those $n$ sets will generally have $2^{2^n}$ elements, always has at most that many, and you can always find an example where it has exactly that many elements. –  Arturo Magidin Dec 11 '11 at 2:11

1 Answer 1

It's quite easy to describe the $\sigma$-algebra generated by a finite partition $\{A_1,\dots,A_N\}$ of $\Omega$: it consists of the sets $\bigcup_{j\in J}A_j$, where $j\subset [N]$.

In this particular case, consider $A_1=A\cap B$, $A_2=A^c\cap B$, $A_3:=A\cap B^c$, $A_4:=A^c\cap B^c$. These sets form a partition of $\Omega$.

Of course, we can describe in the same way the $\sigma$-algebra generated by a finite collection $S_1,\dots,S_n$.

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