Does the set intersection operation has unit I'm trying to get better understanding of binary operations, and I came across this problem: namely on one online discussions I saw that set intersection as binary operation doesn't have a unit, however I think it has:
The set intersection operation for any set universe $U$ is defined on the $\mathcal P(U) \times \mathcal P(U)$ set. From the definition of unit for binary operation, wouldn't it be true that the $U$ set is the unit of this operation. Any set in $PU$ intersected with $U$ will give back the set, and it works in both directions.
Am I doing something wrong, or the discussion on that place was wrong?
 A: Your universe $U$ is not a set. You cannot intersect it with anything, and so your proof does not work (for this reason, it is also incorrect to write $\mathcal P(U)$, as the operator $\mathcal P$ - returning the set of all subsets of a given set - can only be applied to, well, sets themselves.
Hence, what was claimed in the other discussion is true: there exist no set $A$ such that for all sets $B$, $A \cap B = B$.
Edit: It is possible to consider some sort of set algebra, in which such things do make sense ($A \cap U = A$, and so on). This is however a choice that has to be made consciously; the typical ZF(C) axioms do not define those operations, and the "collection of all sets, $U$", is explicitly not a set on which you can perform those operations (power set, cardinality, intersection ...).
A: For a given set $U$, the intersection indeed has $U$ as the identity element, among the subsets of $U$.
On the other hand, seeing the intersection as an operation on all sets, its identity element would be the biggest set (the union of all sets), and such a thing doesn't exist (at least it's not a set).
