# Find smallest $\sigma$-algebra with intersections

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?

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 $$\{4\}=\{3,4\}\setminus (\{2,3\}$$.
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.