Given a set (universal set) $Ω = \{1,2,3,4\}$, the smallest possible σ-algebra on $\Omega$ is this set $S = \{\{1,2,3,4\},\emptyset\}$, is it? wiki gives this definition of sigma-algebra

Let X be some set, and let ${\mathcal {P}}(X)$ represent its power set. Then a subset ${\displaystyle \Sigma \subseteq {\mathcal {P}}(X)}$ is called a σ-algebra if it satisfies the following three properties:
  
  
*
  
*X is in Σ, and X is considered to be the universal set in the following context.
  
*Σ is closed under complementation: If A is in Σ, then so is its complement, X \ A.
  
*Σ is closed under countable unions: If $A_1, A_2, A_3, ...$ are in Σ, then so is $A = A_1 ∪ A_2 ∪ A_3 ∪ …$ .
  
  
  since {X, ∅} satisfies condition (3), it follows that {X, ∅} is the smallest possible σ-algebra on X.

Given this set (universal set) $\Omega = \{1,2,3,4\}$, and then its power set 
\begin{equation*} 
{\mathcal {P}}(\Omega)=
\left\{\emptyset, \left\{1\right\}, \left\{2\right\}, \left\{3\right\}, \left\{4\right\}, \left\{1, 2\right\}, \left\{1, 3\right\}, \left\{1, 4\right\}, \left\{2, 3\right\}, \left\{2, 4\right\}, \left\{3, 4\right\}, \left\{1, 2, 3\right\}, \left\{1, 2, 4\right\}, \left\{1, 3, 4\right\}, \left\{2, 3, 4\right\}, \left\{1, 2, 3, 4\right\}\right\}
\end{equation*}
so, the smallest possible σ-algebra on $\Omega$ is this set $S = \{\{1,2,3,4\},\emptyset\}$, which has only 2 elements, is my understanding right?
 A: That's right. In general, given any set X, the smallest $\sigma$ -algebra on $Χ$ is $\{\emptyset, X\}$, having two elements, and the largest $\sigma$ -algebra on $Χ$ is $\mathcal{P}(X)$, where $\mathcal{P}(X)$ is the power set of $X$. 
When $X=\{1,2,3,4\}$, the smallest $\sigma$ -algebra on $Χ$ is $\{\emptyset, X\}$ , having exactly two elements, and the largest $\sigma$ -algebra on $Χ$ is $\mathcal{P}(X)$, having exactly the $|\mathcal{P}(X)|=2^4=16$ elements that you described above.
Going a little further, for any family $\mathcal{F}$ of subsets of $Χ$, an interesting example is to define $$\sigma(\mathcal{F}):=\bigcap\{\mathcal{B}\mid B \hspace{3mm}\sigma -algebra \text{ on } X \text{ and } \mathcal{B}\supseteq  \mathcal{F} \}$$ 
Then you can show that $\sigma(\mathcal{F})$ is the smallest $\sigma$ -algebra on $Χ$ that contains $\mathcal{F}$. Of course, when $\mathcal{F}=\{\emptyset\}$, then  $\sigma(\mathcal{F})=\{\emptyset, X\}$, as we descrived above.
A: Yes.
This corresponds to the case where the only two measurable sets are the empty set and $\Omega$ itself.
