Cartesian product with all elements I have two sets A and B with
$A = \{1,2,3\} \\
B = \{ A, B, C, D, E \}$
Now I need to get something similar to the Cartesian product. If my understanding is correct, the Cartesian product would return several sets with only 2 elements in each of them.
I need the combinations of all elements:
$1A\\
...\\
1ABCDE\\
..\\
123A\\
...\\
123ABCDE$
The order doesn't matter; each new set should contain each element only once. So no $11A$.
Is there a suitable operation?
 A: If I understand you correctly, you want the set of subsets of elements from both $A$ and $B$ that contain at least one element of $A$ and one element of $B$. If that is what you want, then you want
$$[\mathcal P(A\cup B)\setminus\mathcal P(A)]\setminus\mathcal P(B)$$
or
$$\mathcal P(A\cup B)\cap\mathcal P(A)^c\cap\mathcal P(B)^c$$
where $\mathcal P(X)$ is the power set of $X$, i.e. the set of all subsets of $X$, "$\setminus$" is the setminus operator, i.e. $X\setminus Y=X\cap Y^c$ is the set of all elements in $X$ that are not in $Y$, and "$^c$" is the complement operator.
I have never seen separate notation for what you want, but combining the powerset and setminus operators is pretty easy. My first expression would look a little simpler if you remove the brackets but I wanted to make my order of operations explicit. I think the order is more clear in the second expression, though the complement operator may be more iffy in the general theory of sets in ZF theory. (It is perfectly fine in NBG, with which I am more familiar.)
