Find $\sigma$-field $\sigma(\mathscr A)$ generated by $\mathscr A$ Let $\Omega =\{1,2,3,4\}$, and let $\mathscr A = \{\{1\},\{2\}\}$.
Find $\sigma$-field $\sigma(\mathscr A)$ generated by $\mathscr A$.
My answer is $\sigma(\mathscr A) = \mathscr A \cup \mathscr A^c = \{\{1\},\{2\}\} \cup \{\{3\},\{4\}\} = \{\{1\},\{2\},\{3\},\{4\}\}$. Because 1. $\Omega \subset \sigma(\mathscr A)$, 2. $A \in \sigma(\mathscr A)$ and $A^c \in \sigma(\mathscr A)$ 3. $\cup^\infty_{n=1}A_n \in \{\{1\},\{2\},\{3\},\{4\}\} \in \sigma(\mathscr A)$.
Can anyone help me to check if it is right? And will there be many other solution?
 A: By definition the $\sigma$-field/-algebra generated by a set is the minimal $\sigma$-algebra that contains the elements of the generating set. If you want to generate a $\sigma$-algebra from a subset, you need to consider all sets that can be generated by a countable sequence of intersections and complements (and hence unions). So if you write all ways of taking unions, intersections and complements of elements in your generating set, then you'll get your $\sigma$-algebra.
One observation to make this simpler for your specific example is that $3$ and $4$ are not in the generating set, which means these elements aren't necessarily "distinguishable". If this $\sigma$-algebra were describing events in an experiment, then maybe your collection instrument just outputs $3+$ rather than either $3$ or $4$. Therefore the minimal $\sigma$-algebra's elements will always include both $3$ and $4$ together or neither of them.
Using this you get:
$\{\emptyset,\{1\},\{2\},\{3,4\},\{1,3,4\},\{2,3,4\},\{1,2\},\Omega\}$
