Trouble with a proof exercise in Set Theory regarding subsets and Power Sets Question as posed: Let U be any set. Prove that 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\setminus A=C \cap B $.
Proof (so far): Let U be an arbitrary set. Let A be an arbitrary set. Suppose $A\in\mathcal{P}(U)$. Let B=$U\setminus A$ Let C be an arbitrary set as well. Suppose $C\in\mathcal{P}(U)$. Let x be an arbitrary object. Suppose $x\in C\setminus A$. Then $x\in C$ and $x\notin A$. Note $x\in C$ so $x\in U$. Thus $x\in C$, $x\in U$, and $x\notin A$. Thus $x \in C \cap (U \setminus A)$ and so $x \in C \cap B$. Now suppose instead that $x \in C \cap B$ first (as opposed to $x\in C\setminus A$ first as we did before). Note that $C\cap B = C \cap (U\setminus A)=U \cap (C\setminus A) \subseteq (C\setminus A) $. Thus $x\in C\setminus A$. Thus B is a set such that for every $A\in\mathcal{P}(U)$ and $C\in\mathcal{P}(U)$ (as A and C were both arbitrary sets and x was an arbitrarily chosen object), that $C\setminus A = C \cap B$. Therefore clearly $U\setminus A$ is a set that exists and which fills the condition being looked for in our claim.
Now we will prove B is unique. Suppose for all sets $A\in\mathcal{P}(U)$, there is another arbitrary set D such that $D\in\mathcal{P}(U)$ such that for all sets C where $C\in\mathcal{P}(U)$, $C\setminus A = C \cap D$. But, $C\cap B = C\setminus A$. Thus $C\cap D = C\cap B$.......
The issue with proceeding: I have no idea on how to go from the fact that the two intersections are equal to proving that D=B. I am not looking for so much of a complete answer as just a bone that will get me from point Alice to point Bob. I have tried every which way that I can think of, but keep coming up short.
 A: Hint: If $B \ne D$, then they differ on some element of $U$. Use that element to construct a counter-example to the claim that $D$ has the same property as $B$.
A: This is not a direct answer to your question, but I'd like to demonstrate a 'calculational' approach to set theory questions, where we first expand the set theory definitions to go from the set theory level to the level of logic, then use the laws of logic simplify, and take it from there.
(Note that I am using notations I learned from Dijkstra et al.; see, e.g., EWD1300.)$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\P}[1]{\mathcal{P}(#1)}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
Written a bit more formally, you are asked to prove
$$
\tag{0}
\langle \forall A : A \in \P{U} : \langle \exists! B : B \in \P{U} : \langle \forall C : C \in \P{U} : C \setminus A \;=\; C \cap B \rangle \rangle \rangle
$$
Now, the simplest way to prove a statement of the form $\;\langle \exists! z :: R \rangle\;$, is to rewrite $\;R\;$ as something of the form $\;z = \ldots\;$.

Therefore, we rewrite the rightmost part inside the $\;\exists! B\;$ quantification to $\;B = \ldots\;$, under the assumption of the leftmost part $\;B \in \P{U}\;$ or equivalently $\;B \subseteq U\;$, as follows:
$$\calc
    \tag{1}
    \langle \forall C : C \in \P{U} : C \setminus A \;=\; C \cap B \rangle
\op=\hint{definition of $\;\P{\cdot}\;$; definition of equality of sets}
    \langle \forall C : C \subseteq U : \langle \forall x :: x \in C \setminus A \;\equiv\; x \in C \cap B \rangle \rangle
\op=\hint{merge quantifications; definitions of $\;\setminus, \cap\;$}
    \langle \forall x,C : C \subseteq U : x \in C \land x \not \in A \;\equiv\; x \in C \land x \in B \rangle
\op=\hint{logic: extract common conjunct out of $\;\equiv\;$}
    \langle \forall x, C : C \subseteq U \land x \in C : x \not \in A \;\equiv\; x \in B \rangle
\op=\hints{logic: $\;\langle \forall z : Q : R \rangle\;$ is equivalent to $\;\langle \exists z :: Q \rangle \then R\;$,}\hint{if $\;R\;$ does not contain $\;z\;$}
    \langle \forall x : \langle \exists C :: C \subseteq U \land x \in C \rangle : x \not \in A \;\equiv\; x \in B \rangle
\op=\hint{set theory: simplify}
    \langle \forall x : x \in U : x \not \in A \;\equiv\; x \in B \rangle
    \tag{*}
\op=\hints{logic: move common conjunct back into $\;\equiv\;$}\hints{-- since we want to bring $\;U\;$ and $\;A\;$ together,}\hint{and $\;U\;$ and $\;B\;$, to exploit our assumptions}
    \langle \forall x :: x \in U \land x \not \in A \;\equiv\; x \in U \land x \in B \rangle
\op=\hints{RHS: simplify using assumption $\;B \subseteq U\;$;}\hint{LHS: definition of $\;\setminus\;$ -- preparing for the next step}
    \langle \forall x :: x \in U \setminus A \;\equiv\; x \in B \rangle
\op=\hint{definition of equality of sets}
    U \setminus A \;=\; B
\endcalc$$
(This calculation may look daunting, but really up until $\ref{*}$ all we've done is expand definitions and take steps to simplify, based on the shape of the formula after each step.)
This proves that $\;U \setminus A\;$ is the unique $\;B\;$ that satisfies $\ref 1$, assuming $\;B \subseteq \P{U}\;$.
And since this $\;B\;$ also satisfies $\;B \in \P{U}\;$ (since $\;U \setminus A \subseteq U\;$), we've proven that
$$
\langle \exists! B : B \in \P{U} : \langle \forall C : C \in \P{U} : C \setminus A \;=\; C \cap B \rangle \rangle
$$
for all sets $\;U\;$ and $\;A\;$.

Note that we've actually proven a more general result than $\ref{0}$, since we did not need the assumption $\;A \subseteq \P{U}\;$.
Finally, for me the most important feature of this proof is that coming up with the set $\;U \setminus A\;$ did not require any real insight, it was not pulled like a rabbit out of a magician's hat.  Instead, it followed directly from the shape of the formulas, the desire to simplify, and the wish to apply the assumptions that were given.
