Sigma algebra question Let $Σ_1$ and $Σ_2$ be two sigma-algebras on the same set $X$, such that their union $Σ_1 ∪Σ_2$ is also a sigma-algebra. Prove that for $A,B ⊆ X$ such that $A ∈ Σ_1 \setminus Σ_2$ and $B ∈ Σ_2 \setminus Σ_1$ it always holds that $A∩B \neq\emptyset$.

I cant see how it isn't equal to the empty set. $A$ is in sigma $1$ but not in $2$. And $B$ is in $2$ but not in $1$ so they must be disjoint so their intersection must have nothing in common so it must be the empty set.
EDIT: $A,B \in \Sigma_1 \cup \Sigma_2=\Sigma$ so $(X \setminus A),(X \setminus B) \in \Sigma$ so $(X \setminus A) \cap (X \setminus B)=\emptyset \in \Sigma$ then $(X \setminus A)^c \cap (X \setminus B)^c=A \cap B=\emptyset ^c=X \in \Sigma$. Is this correct?
 A: Suppose $A,B ⊆ X$ such that $A ∈ Σ_1 \setminus Σ_2$ and 
$B ∈ Σ_2 \setminus Σ_1$. 
Since $Σ_1 ∪Σ_2$ is a $\sigma$-algebra and $A,B\in Σ_1 ∪Σ_2$, we have $A \cup B\in Σ_1 ∪Σ_2$. So $A \cup B\in Σ_1 $ or $A \cup B\in Σ_2$. 
Suppose  $A \cup B\in Σ_1 $. Let us prove $A\cap B \notin Σ_1$.  In fact, if  $A\cap B \in Σ_1$, we would have 
$$B= ((A \cup B)\setminus A)\cup (A\cap B) \in  Σ_1 $$ 
Contradiction, as $B ∈ Σ_2 \setminus Σ_1$. So $A\cap B \notin Σ_1$ and in particular $A\cap B \neq \emptyset$. 
Now, Suppose  $A \cup B\in Σ_2 $. Let us prove $A\cap B \notin Σ_2$.  In fact, if  $A\cap B \in Σ_2$, we would have 
$$A= ((A \cup B)\setminus B)\cup (A\cap B) \in  Σ_2 $$ 
Contradiction, as $A ∈ Σ_1 \setminus Σ_2$. So $A\cap B \notin Σ_2$ and in particular $A\cap B \neq \emptyset$. 
So, in both cases we have $A\cap B \neq \emptyset$.
Important Remark: In fact, the statement "for $A,B ⊆ X$ such that $A ∈ Σ_1 \setminus Σ_2$ and $B ∈ Σ_2 \setminus Σ_1$ it always holds that $A∩B \neq\emptyset$" is vacuously true, because of the following result: 
Let $\Sigma_1$ and $\Sigma_2$ be two $\sigma$-algebras 
on the same set $X$, such that their union 
$\Sigma_1 \cup\Sigma_2$ is also a $\sigma$-algebra. Then either $\Sigma_1 \subseteq \Sigma_2$ or $\Sigma_2 \subseteq \Sigma_1$. In other other, there is no  $A,B \subseteq X$ such that $A \in \Sigma_1 \setminus \Sigma_2$ and 
$B \in \Sigma_2 \setminus \Sigma_1$. 
Proof: Suppose there is $A,B \subseteq X$ such that $A \in \Sigma_1 \setminus \Sigma_2$ and $B \in \Sigma_2 \setminus \Sigma_1$. 
Since $Σ_1 ∪Σ_2$ is a $\sigma$-algebra and $A,B\in Σ_1 ∪Σ_2$, we have $A \setminus B\in Σ_1 ∪Σ_2$. So $A \setminus B\in Σ_1 $ or $A \setminus B\in Σ_2$ 
If  $A \setminus B\in Σ_1$ then $A\cap B= A\setminus (A\setminus B) \in \Sigma_1$. 
If  $A \setminus B\in Σ_2$ then, since $A= (A \setminus B)\cup (A\cap B)$ and $A \notin \Sigma_2$, we must have that $A\cap B \notin \Sigma_2$. Since $A\cap B\in Σ_1 ∪Σ_2$, we have that $A\cap B \in \Sigma_1$. 
So we have proved that $A\cap B \in \Sigma_1$.
Now, from $B \setminus A\in Σ_1 ∪Σ_2$, in a similar way we can prove that $A\cap B \in \Sigma_2$.  So we have $A\cap B \in \Sigma_1 \cap \Sigma_2$
Finally, note that $A \cup B\in Σ_1 ∪ Σ_2$. 
If $A \cup B\in Σ_1$ then 
$B= ((A \cup B)\setminus A)\cup (A\cap B) \in  Σ_1 $.
Contradiction, as $B ∈ Σ_2 \setminus Σ_1$.
If $A \cup B\in Σ_2$ then 
$A= ((A \cup B)\setminus B)\cup (A\cap B) \in  Σ_2 $. 
Contradiction, as $A ∈ Σ_1 \setminus Σ_2$.
So there is no  $A,B \subseteq X$ such that $A \in \Sigma_1 \setminus \Sigma_2$ and $B \in \Sigma_2 \setminus \Sigma_1$. So either $\Sigma_1 \subseteq \Sigma_2$ or $\Sigma_2 \subseteq \Sigma_1$.
A: The key issue is that the sigma algebras are required to have a union which is also a sigma algebra.   What is required for that to happen?
A: I don't see how you prove $(X\setminus A)\cap (X\setminus B)=\emptyset$.
Clearly $A\cup B\in\Sigma_1\cup\Sigma_2$. W.l.o.g. $A\cup B\in \Sigma_1$. Suppose $A\cap B=\emptyset$, then $(A\cup B)\setminus A = B\in\Sigma_1$.
A: Some people are addressing the question directly; here I will address instead what appears to be a misunderstanding involved.
Suppose $X= \{a,b,c,d\},\quad$ $\Sigma_1 = \Big\{ \{a,b\},\  \{c,d\} \Big\},\quad$ and $\Sigma_1 = \Big\{ \{a,c\},\  \{b,d\} \Big\}$.
Then we have


*

*$\{a,b\}\in\Sigma_1\setminus\Sigma_2$, and

*$\{a,c\}\in\Sigma_2 \setminus\Sigma_1$, and

*the intersection of those two sets is $\{a\}$, which is not empty.


(In this case the union is not a sigma-algebra, but I think this example may nonetheless show what is wrong in the logic that led to the mistaken conclusion.)
