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I was asked to find the smallest $\sigma$-algebra ($\sigma(Z)$) generated by the following set $Z$={{1,2},{2,3},{3,4}} where the sample space is $\Omega$={1,2,3,4}. The thing is that the middle element of $Z$ is causing me a lot of trouble. The smallest $\sigma(Z)$ I could find was the following:

$\sigma(Z)$={$\emptyset$,$\Omega$,{1},{2},{3},{4},{1,2},{2,3},{3,4},{1,3},{1,4},{2,4},{1,2,3},{2,3,4},{1,3,4},{1,2,4}}

I know that the $\sigma$-algebra has to be closed under complements and unions, and that is how I got to that set. Is this right? Is there a faster way to solve this kind of problems?

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2 Answers 2

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The moment you get $\{1\},\{2\},\{3\},\{4\}$ in the sigma algebra you can conclude that all subsets of $\{1,2,3,4\}$ are also in it. (Because any subset of $\{1,2,3,4\}$ is a finite union of singletons). In this case $\{1\}=\{1,2,\}\setminus \{2,3\}$, $\{2\}=\{1,2\}\cap \{2,3\}$, $\{3\}=\{2,3,\}\setminus (\{1,2\}$ and $\{4\}=\{3,4\}\setminus (\{2,3\}$.

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There is a faster way : Since $\{1\},\{2\},\{3\},\{4\}\in \sigma (Z)$, then $\sigma (Z)$ is the power set of $\{1,2,3,4\},$ and what you found is correct.

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