Intersection of two sets that contain other sets as elements How would the intersection of $A=\{a, b, e, \{a, b, c, d\}, \{d, e\}\}$ and $B=\{a, b, c, f, \{a, d\}, \{d, e\}\}$ be defined? I've searched quite a few books but no luck so far.

Just to elaborate on my comment and put it into an answer:
The definition of intersection is always the same.  The intersection of two sets $X$ and $Y$ is the set of all things that are in both $X$ and $Y$.
In other words
$z \in X \cap Y \iff z \in X \wedge z \in Y$.  It may be that $z$ is a set itself, but from $X$'s perspective it is just an element or not.
So in this case we can just look at each element in $A$ and ask if it is in $B$ and if it is, it will be in $A \cap B$.


*

*$a \in A$ and $a \in B$ so $a \in A \cap B$

*$b \in A$ and $b \in B$ so $b \in A \cap B$

*$e \in A$ but $b \notin B$ so $b \notin A \cap B$

*$\{a,b,c,d\} \in A$ but $\{a,b,c,d\} \notin B$ so $\{a,b,c,d\} \notin A \cap B$

*$\{d,e\} \in A$ and $\{d,e\} \in B$ so $\{d,e\} \in A \cap B$


At this point we are done.  We have checked which elements of $A$ are in $B$, it doesn't matter that $B$ has extra elements; for something to be in $A \cap B$ it must be in both.
Thus, we see that 
$A \cap B = \{a,b, \{d,e\}\}$
