1. The problem statement, all variables and given/known data Let U be any set. (a) Prove that for every A ∈ P (U) there is a unique B ∈ P (U) such that for every C ∈ P (U), C \ A = C ∩ B. Remark: P(U) = the Power set of U, i.e. A ∈ P (U) then A⊆U 2. Relevant equations 3. The attempt at a solution The question's form is as follows: ∀A(A∈P(U)⇒∃!B(Q(B))) Where Q(B) = B∈P(U)∧∀C(C∈P(U)⇒C\A=C∩B) I first let A be an arbitrary element, then A∈P(U)⇒∃!B(Q(B)) Suppose A⊆U, then I want to prove ∃!B(Q(B)) Existence: I want to prove ∃B(Q(B)) Uniqueness: I want to prove that, ∀r∀s(Q(r)∧(Q(s)⇒r=s) I start with existence, To find a B such that Q(B) is true, I consider all subsets of U and their relation to A in order to find a suitable B using the hint C\A=C∩B Let u be an arbitrary subset of U Since both A and the subsets of U are both subsets of U, then there are two possibilities: either A∩u=∅∨A∩u≠∅ In the first case u\A is just u, therefore B can be any set such that u⊆B since u∩B=u =u\A In the second case u\A = some subset of u, say u’ since u∩A≠∅. We want to find the choices for B such that u∩B = u’. Therefore B can be any set that does not contain u∩A and contains u’; For example: U\A, or U\(u∩A) Even considering these choices for B, I cannot narrow it down to a choice where B satisfies Q(B) since the possibilities for B in both cases are case-specific and do not apply to both cases. This is where I’m stuck, any hints or even the solution would be much appreciated. I feel like I’m overlooking something simple and that I might be over-complicating things.