smallest sigma algebra possible roll of a die n times [closed]

Consider the sample space $\Omega = \left\{1,2,\cdots,6\right\}^n $ and $n$ associated with $n$ rolls of a die. Any outcome $\omega$ of the sample space $\Omega$ can be represented as $(ω_1, \dots ,ω_n)$. Consider the following three subsets of sample space $A = \left\{\omega \in \Omega : \omega_1 \leq 2\right\}$, $B = \left\{\omega \in \Omega : 3 \leq \omega_1 \leq 4 \right\}$, $C = \left\{\omega \in \Omega : \omega_1 \geq 5 \right\}$.

What is the smallest sigma-algebra $\sigma(\left\{A,B,C\right\})$ containing three subsets $A$, $B$ and $C$?

In this question, I have tried to do the smallest sigma-algebra all are disjoint? Here As all the following sets $A$, $B$, and $C$ are disjoint sets, and if we are going to construct $\sigma$-algebra for each set. Then the smallest sigma-algebra is like $\left\{\phi, A \cap B \cap C , \Omega\right\}$ Is it the answer? Actually, I am stuck somewhere.

We say that the smallest $\sigma$-algebra containing $A,B,C$ is the $\sigma$-algebra generated by $A,B$, and $C$; we “generate” it by starting with the sets $A,B,C$ and finding all sets that we can arrive at by taking complements, countable unions, and countable intersections. In this case, our work is fairly straightforward. Indeed, the sets $A,B,C$ are disjoint, and one can also observe that $A \cup B \cup C = \Omega$. We then have $A^\mathsf{C} = B \cup C$, $B^\mathsf{C} = A \cup C$, and $C^\mathsf{C} = A \cup B$, all of which we already know are in the $\sigma$-algebra by taking unions. We have $A \cup A^\mathsf{C} = \Omega$, and $\Omega^\mathsf{C} = \varnothing$. (The sets $\varnothing$ and its complement are necessarily in every $\sigma$-algebra.)

What we end up with is: $$\sigma(\{A,B,C\}) = \{\varnothing,\Omega,A,B,C,A \cup B,A\cup C, B \cup C\}$$ We have shown that all of these sets must be in the $\sigma$-algebra, and you can check that they are closed under the complement, union, and intersection operations, so this is the smallest $\sigma$-algebra as needed.

This is effectively an equivalent problem to finding the smallest $\sigma$-algebra on $\{1,2,3\}$ that contains the sets $\{1\}, \{2\}$, and $\{3\}$. Since $A,B,C$ are a partition of the set $\Omega$, there are no proper subsets of $A,B$, or $C$ that we need to worry about when we generate the $\sigma$-algebra.