Finding the $\sigma$-algebra generated by $\mathcal{U}=\{\{1,2,3\},\{3,4,5\}\}$. In my study of $\sigma$-algebra, I am having difficulty understanding the concept of the smallest $\sigma$-algebra.
Here, the set $\mathcal{F}\in P(\Omega)$ (where $P(\Omega)$ denotes the power set of $\Omega$) is a $\sigma$-algebra if $\mathcal{F}$ has the following properties:
\begin{align}
(1):& \ \ \emptyset\in\mathcal{F} \\
(2):& \ \ A\in\mathcal{F}\implies A^c\in\mathcal{F} \\
(3):& \ \ A_1,A_2,...\in\mathcal{F}\implies \bigcup_{i=1}^\infty A_i\in\mathcal{F}
\end{align}
Now, given a family $\mathcal{U}$ of subsets of $\Omega$, there is a smallest $\sigma$-algebra $\sigma_{\mathcal{U}}$ containing $\mathcal{U}$. This $\sigma$-algebra is the intersection of all $\sigma$-algebra containing you.
To help improve my understanding, I considered the following example. Let $\Omega=\{1,2,3,4,5\}$ and let $\mathcal{U}=\{\{1,2,3\},\{3,4,5\}\}=\{A,B\}$. Then,
\begin{align}
\sigma_{\mathcal{U}}&=\{\underbrace{\emptyset}_{\text{by $(1)$}},\underbrace{\Omega,A,B,A^c,B^c}_{\text{by $(2)$}},\underbrace{A\cup B,A^c\cup B, A\cup B^c, A^c\cup B^c}_\text{by $(3)$},\underbrace{(A\cup B)^c,(A^c\cup B)^c,(A\cup B^c)^c,(A^c\cup B^c)^c}_\text{by $(2)$}\} \\ 
&=\{\emptyset,\{1,2,3,4,5\},\{1,2,3\},\{3,4,5\},\{4,5\},\{1,2\}\}
\end{align}
Is this correct?
 A: Assuming that $A$ equals $\{1,2,3\}$, and not $\{1,2\}$, the sigma-algebra generated by $\{A, B\}$ is the power set of the three sets $\{1,2\}$, $\{3\}$, $\{4,5\}$, hence it has eight elements. (You are missing the set $\{3\}$ and its complement $\{1,2,4,5\}$.)
This follows from the result: If the universe $\Omega$ is countable, then every sigma-algebra on $\Omega$ can be generated by a partition of $\Omega$.
So to determine $\sigma({\mathcal U})$ for your example, your first step is to find the partition that corresponds to $\mathcal U$: that is, for every $\omega\in\Omega$ find the smallest member of $\sigma({\mathcal U})$ that contains $\omega$. Check that there is no way to separate $1$ and $2$ using unions and/or intersections from $\mathcal U$, and similarly you cannot find a way to separate $4$ from $5$. On the other hand, the element $3$ is covered by the singleton set $\{3\} = \{1,2,3\}\cap\{3,4,5\}$. The three sets $\{1,2\}$, $\{3\}$, $\{4,5\}$ are disjoint and their union is $\Omega$, so that's your partition.
