For every $A\in \mathcal {P}(U)$ there is a unique $B\in \mathcal{P}(U)$ such that for every $C\in \mathcal{P}(U)$, $C\cap A=C-B$

Question

Pls help me out with the proof:

For every $A\in \mathcal {P}(U)$ there is a unique $B\in \mathcal {P}(U)$ such that for every $C\in \mathcal {P}(U), C \cap A=C-B$. For the existence part, we have to figure out some $B$ which works for all $C$. What can we assign for $B$?

Here $\mathcal{P}(U)$ denotes the power set of $U$

Is $P(U)$ the power set of $U$, where $U$ is the universal set??? — Anurag A, May 11 '15 at 17:17

score 0 · Accepted Answer · answered May 11 '15 at 17:40

Since $B$ can only depend on $U$ and $A$, there is a natural choice to guess: $B = U \setminus A$. Indeed, this works. To see why it is unique, show that no matter what choice of $B$ we take, we must have that $A \cap B = \varnothing$ and $A \cup B = U$.

For every $A\in \mathcal {P}(U)$ there is a unique $B\in \mathcal{P}(U)$ such that for every $C\in \mathcal{P}(U)$, $C\cap A=C-B$

Here $\mathcal{P}(U)$ denotes the power set of $U$

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