# Finding the smallest $\sigma$-algebra generated by $\mathcal{U}=\{\{1,3,5\},\{1,5,7,9\}\}$

Let $$\Omega=\{1,3,5,7,9\}$$ and $$\mathcal{U}=\{\{1,3,5\},\{1,5,7,9\}\}$$. I am trying to find the smallest $$\sigma$$-algebra, $$\sigma_\mathcal{U}$$, generated by $$\mathcal{U}$$.

By considering the partition $$\{3\}, \{1,5\}, \{7,9\}$$, I found that $$\sigma_\mathcal{U}=\{\emptyset,\{1,3,5,7,9\},\{3\},\{1,5\},\{7,9\},\{1,3,5\},\{1,5,7,9\},\{3,7,9\}\}.$$ I know that the $$\sigma_\mathcal{U}$$ generated by $$\mathcal{U}$$ is the power set of my partition and should therefore contain eight elements.

Is my logic correct? I'm somewhat of a novice at measure theory.

Your logic is indeed correct. A $$\sigma$$-algebra containing $$\mathcal U$$ would have to contain $$\mathcal V = \{\{3\}, \{1,5\}, \{7,9\}\}$$ and hence the power set $$\mathcal P(\mathcal V)$$.
As the smallest $$\sigma$$-algebra containing $$\mathcal U$$ is the intersect of all $$\sigma$$-algebras containing $$\mathcal U$$, it is equal to $$\sigma_{\mathcal U}$$.