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Basically i tried to prove that $A\cup B= \Omega \Rightarrow B= A^{C}$. so then the $\sigma(C)= \{\emptyset, \Omega, A, A^{C}\}$. is right?.

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  • $\begingroup$ $B=A^c$ just if they are disjoint (and its immediate if they really are) $\endgroup$ – Robson Sep 13 '18 at 16:04
  • $\begingroup$ $A\cup B=\Omega$ does not imply $B=A^C$, so you will not be able to prove this. $\endgroup$ – Mike Earnest Sep 13 '18 at 16:09
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how about $\{A,B,A^c,B^c,A^c\cup B^c, A\cap B, \Omega,\emptyset\}$

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The $\sigma$-álgebra generated by a set $C$ is by definition the least $\sigma$-álgebra that contains $C$.

Observe that the following set is a $\sigma$-álgebra and is in fact the least one: $$\mathcal{A}=\{\emptyset,\Omega,A,B,A^c,B^c,A\cap B,A\cap B^c,B\cap A^cA^c\cap B^c\}$$ Note that here it isn't of our preocupation if some of the elements are write twice or more.

You can see that $\mathcal{A}$ satisfy all properties for being a $\sigma$-álgebra.

And it is the least one because if $\mathcal{A'}$ is another $\sigma$-álgebra such that $A,B\in \mathcal{A'}$, then by the properties of a $\sigma$-álgebra we have that all elements of $\mathcal{A}$ are elements of $\mathcal{A'}$ too, i.e., $\mathcal{A}\subset\mathcal{A'}$.

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