# Understanding how to compute sigma algebra

I'm having trouble understanding the production rules for sigma algebra. I know there are the following requirements for the sigma algebra:

1. $$\emptyset \in \mathcal A$$
2. when $$A \in \mathcal A$$ the $$A^c \in \mathcal A$$
3. when $$A_{1},A_{2},A_{3},\dots \in \mathcal A$$ then $$\cup _{i=1}^{\infty }A_{i} \in \mathcal A$$

My problem is the 3rd rule, I assumed that this means that the union of all (inkl. generated) elements has to be a subset as well. But with the following example, this was not the case.

Given $$\Omega = \{1, 2, 3, 4\}$$ the minimal sigma algebra containing $$\{1\}, \{1,2\}$$ is $$\{\emptyset, \{1\}, \{1, 2\}, \{2, 3, 4\}, \{3, 4\}, \{1, 2, 3, 4\}, \{1, 3, 4\}, \{2\}\}$$

but why would $$\{1, 3, 4\}, \{2\}$$ be part of the sigma algebra when rule three just creates a union of all subsets?

You don't only use the rules on the sets you start with. Instead, we have to continue applying the rules to all of the new sets we find from previous applications. Note that $$\{2\}=\{1\}^c\cap\{1,2\}=(\{1\}\cup\{1,2\}^c)^c$$ and $$\{1,3,4\}=\{2\}^c.$$
Because $$A\cap B=(A^c\cup B^c)^c$$ $$A\setminus B=A\cap B^c$$ and hence including both unions and complements means that you also include intersections, differences and so on.
For example $$\{2\}=\{1,2\}\setminus\{1\}=\{1,2\}\cap\{1\}^c=(\{1,2\}^c\cup\{1\})^c$$
To convince yourself that the minimal sigma-algebra containing $$\{1\}$$ and $$\{1,2\}$$ is the one you wrote, simply try to remove one of the elements of the sigma-algebra and explain why it is not a sigma-algebra anymore.
Call $$\mathcal{A}$$ your sigma-algebra. In particular, $$\{1\}\in \mathcal{A}$$ and $$\{3,4\}\in \mathcal{A}$$ by rule 2, being it the complement of $$\{1,2\}\in \mathcal{A}$$, thus $$\{1\}\cup \{3,4\}=\{1,3,4\}\in \mathcal{A}$$ by rule 3. Now $$\{2\}\in \mathcal{A}$$ since $$\{1,3,4\}\in \mathcal{A}$$ and rule 2 again.