Prove that for every $A \in P (U)$ there is a unique $B \in P(U)$ such that for every $C \in P (U)$, $C \setminus A = C \cap B$.

Question

I know that there exists such a set $(U\setminus A)$ for which $C \setminus A = C \cap B$. However I have trouble proving that it is unique.

What I am trying to do is prove that $\forall D\in P(U)(C\setminus A = C\cap D \Rightarrow D=U\setminus A)$. I first assume $C\setminus A = C\cap D$ and try to prove $D=U\setminus A$, but this leads to nowhere. Any suggestions on how to approach this problem ?

Possible duplicate of Uniqueness proof for $\forall A\in\mathcal{P}(U)\exists!B\in\mathcal{P}(U)(C\setminus A=C\cap B)$ — Heptapod, Aug 25 '17 at 09:29

score 2 · Accepted Answer · answered Mar 11 '15 at 14:32

Let $B,D$ be such that for every $C\in P(U)$ we have $C\setminus A=C \cap D$ and $C\setminus A=C \cap B$. Then we have $C\cap D=C\setminus A = C\cap B$ for every $C\in P(U)$. In particular for $C=D$ we get $D\cap D = D = D\cap B$ and thus $D\subset B$, now taking $C=B$ will imply $B\subset D$.

Prove that for every $A \in P (U)$ there is a unique $B \in P(U)$ such that for every $C \in P (U)$, $C \setminus A = C \cap B$.

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