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We have a set A with finite elements, where A is a subset of Power set of Omega. prove that total number of elements in the sigma algebra generated by A is at most 2^2^|Omega|

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closed as off-topic by Care Bear, Tomás, Daniel Fischer, le gâteau au fromage, Hakim Jul 20 at 0:12

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What have you tried? –  Nate Eldredge Sep 21 '12 at 22:01

1 Answer 1


Suppose that $\Sigma$ is a $\sigma$-algebra generated by $A$, then $\Sigma \subseteq \wp(A)$; and since $A \subseteq \wp(\Omega)$, so consequently $|\Sigma|\le 2^{|A|}$ and $|A|\le 2^{|\Omega|}$. Hence $|\Sigma|\le 2^{2^{|\Omega|}}$


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