Find the least sigma-field $\mathcal{F}$ such that $A \in \mathcal{F}$ and $B \in \mathcal{F}$ Let $A,B \subset X$ and let $A \not= \emptyset$ and $B \not= \emptyset$ such that $A \cap B = \emptyset$. Find the least sigma-field $\mathcal{F}$ such that $A \in \mathcal{F}$ and $B \in \mathcal{F}$.What will change when $A \cap B \not= \emptyset$ ?
My attempt:
I think that $\mathcal{F} = \left\{\emptyset,X,A,B,X \setminus A, X \setminus B , A \cup B, X \setminus (A \cup B) \right\} $
What will change when $A \cap B \not= \emptyset$ ?
I think that 
$\mathcal{F} = \left\{\emptyset,X,A,B,X \setminus A, X \setminus B , A \cup B, X \setminus (A \cup B), A \cap B, X \setminus (A \cap B) \right\} $
Am I right? I will grateful for check. 
 A: When faced with something like this, I sometimes resort to this process to ensure that I have accounted for all the possibilities:
Draw a Venn diagram, and label the individual areas 1, 2, 3 and 4. Then list the possibilities.

So I use 234 as an abbreviation for $A\cup B$, etc.
Now the 16 sets you need are all possible combinations: $1, 2, 3, 4, 12, 13, \dots 234, 1234$ together with the empty set.
Listing the sets you require in a table, you get:
$$\begin{eqnarray}
~ & ~ & \emptyset & ~ & X & ~ & 1234 \\
23 & ~ & A & ~ & X\setminus A & ~ & 14 \\
34 & ~ & B & ~ & X\setminus B & ~ & 12 \\
3 & ~ & A\cap B & ~ & X\setminus (A\cap B) & ~ & 124 \\
234 & ~ & A\cup B & ~ & X\setminus (A\cup B) & ~ & 1 \\
2 & ~ & A \setminus B & ~ & X\setminus (A \setminus B) & ~ & 134 \\
4 & ~ & B \setminus A & ~ & X\setminus (B \setminus A) & ~ & 123 \\
24 & ~ & A \Delta B & ~ & X\setminus (A \Delta B) & ~ & 13 \\
\end{eqnarray}
$$
Where $\Delta$ is the symmetric difference $(A\setminus B) \cup (B\setminus A)$.
You should be able to work out which ones you have missed from here.
