Does the set intersection operation has unit

Question

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?

Answer 1

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 ...).

Answer 2

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).

Answer 3

That works for all subsets of U.
That however, does not include the set PU nor any of its subsets which are inself yields another unuversal thing PPU.
The notion of a universal set thing is an armature concept.
As you saw above, any universal set allows for bigger and bigger universal sets. There is no set of all sets.
Assume U is the set of all sets. As {U} is a set, {U} in U, so U in {U} in U which contradicts the axiom of foundation.