Determine the two $\sigma$-algebras of subsets of $X$ generated by 
Let $A$ be a fixed subset of a set $X$. Determine the two $\sigma$-algebras of subsets of $X$ generated by
  
  
*
  
*$\{A\}$
  
*$\{B: A\subseteq B \subseteq X\}$
  

I managed to find the first one, being:


*

*$\{\emptyset, ~X,~ A, ~X\setminus A\}$


Is this correct?
I am, however, not sure how to determine the $\sigma$-algebra generated by (2).
 A: For starters, you know the $\sigma$-algebra in (2) will contain $\emptyset$, $X$, and all sets $B$ such that $A \subseteq B \subseteq X$. (Note: there's not one fixed $B$. The set in (2) consists of all sets $B$ containing $A$.) Let's call those sets $B$ with $A\subseteq B \subseteq X$ sets of type 1.
Since $\sigma$-algebras are closed under taking complements, you also need the complement of each set of type 1. If $A \subseteq B \subseteq X$, what can you say about $C:=X\setminus B$? That is, how would you related it to $A$ or $X\setminus A$? Do you get all sets of that form? Call these sets of type 2.
Next you need to consider countable unions and intersections. If you take a countable union of sets of type 1, you get another set of type 1, so that's already in the $\sigma$-algebra. Same thing if you take a countable intersection them. (You should verify these claims!) You should check that the same is true for taking countable unions/intersections of sets of type 2.
You could also take a countable union that consists of sets of type 1 and type 2. What would such a set look like? I claim these are already in the $\sigma$-algebra. Do you see what type they are? You could also take a countable intersection consisting of sets of type 1 and type 2. Again, I claim these are already in the $\sigma$-algebra.
Thus $\{\emptyset,X\}\cup \{\text{sets of types 1 and 2}\}$ is a $\sigma$-algebra, and it's the smallest one containing the generating set.
