If $A,B,C$ form a partition of $\Omega$ then describe the smallest $\sigma$-algebra containing the sets $A,B$ and $C$. Question:

If $A,B,C$ form a partition of $\Omega$ then describe the smallest $\sigma$-algebra containing the sets $A,B$ and $C$.

This seems like a straightforward question but it is giving me a really hard time.
At first I thought the set is $\mathcal{F}=\{ \emptyset , \Omega \}$. However, this obviously doesn't include $A,B$ and $C$. So I thought intuitively that the set should just be $\mathcal{F} = \{ \emptyset , \Omega, A,B,C\}$. Now this makes sense as it satisfies the definition of the $\sigma$-algebra but I don't have a solid idea on how to prove that this is indeed the smallest possible set. I feel as if I'm missing some kind of fine detail that'll just make everything click but at the moment this all is just really confusing.
Any help would be greatly appreciated, thank you. 
 A: $\mathcal{F} = \{ \emptyset , \Omega, A,B,C\}$ is a nice try. But unfortunately, it is still not a $\sigma$-algebra, because the complement of $A$, $A^c=\Omega-A= B\cup C$ (here, we used the assumption that $A, B, C$ form a partition of $\Omega$) is not in $\mathcal{F}$. Similarly, $B^c= A\cup C$, and $C^c=A\cup B$ are not in $\mathcal{F}$. 
So we have to enlarge $\mathcal{F}$ further by including them. 
Therefore, we try $\mathcal{F}=\{ \emptyset , \Omega, A,B,C, B\cup C,A\cup C, A\cup B\}$.
This is indeed a $\sigma$-algebra, which I invite you to prove it rigorously by showing that it satisfies the three properties of the $\sigma$-algebra: 
(i) $\Omega\in \mathcal{F}$ (trivial!)
(ii) If $X\in\mathcal{F}$, then the complement $X^c=\Omega-X$ is still in $\mathcal{F}$ (you just have to check finitely number of times, since $\mathcal{F}$ is finite)
(iii) If $X_1,..., X_k\in\mathcal{F}$, then $X_1\cup \cdots X_k\in\mathcal{F}$
(again you just have to check finitely number of times, since $\mathcal{F}$ is finite)
