How can I solve this set equation $A=(X\cap A) \cup B$ Let A and B be subsets of E, and $\mathcal P(E)$ the powerset of E. We define the application $f:\mathcal P(E)\to \mathcal P(E) $  by the formula $ f(X)= (X \cap A) \cup B$
Now given the previous situation I am asked to solve in $\mathcal P(E)$ the equation in $X$ : $f(X)=A$ 
I have developed the equation $A=(X\cap A) \cup B$ , and arrived to $$A=(X\cup B)\cap (A \cup B)$$ I will have to end up with an epression for X, and this is where I am stuck. 
From the first equation I can both deduce that $(X \cap A) \subseteq A$ and $  B \subseteq A$ . And from the one I arrived to I deduce that $ A \subseteq (X \cup B)$ and $A \subseteq (A \cup B)$ and this last one is $\to B\subseteq A$ as seen before. So mainly I arrived to three expressions which are $$(X \cap A) \subseteq A; and; A \subseteq (X \cup B) ; and ;B \subseteq A$$
And I don-t know how to continue from there.... if you could help or orientate me a little bit if i was following the wrong path...
 A: I claim the following:
If $B\subseteq A$, then the solutions are precisely the subsets of $E$ which contain $A\setminus B$.
Otherwise (i.e., if $B\nsubseteq A$) there are no solutions.
A: First of all.  If $B \not \subset A$ there is no solution as $B \subset (A \cap X) \cup B = A$.
So $B = A \setminus B^c$ and $(A\cap X)\cup (A\setminus B^c)= A \cap (X \cup B)$.
If $x \in A\setminus B$ then $x \in A= A\cap (X \cup B)$ so $x \in X \cup B$ but $x \not \in B$ so $x \in X$.  So $A\setminus B \subset X$.
Let $X = A\setminus B$ then $(A \cap A\setminus B)\cup B = (A\setminus B) \cup (A \setminus B^c) = A$.  So $A\setminus B$ is the minimal solution.
Let $Y$ be any set, let $X = (A\setminus B) \cup Y$.  Then $X\cap A = (A\setminus B)\cup (A\cap Y)$ and $(X\cap A)\cup B = (A\setminus B)\cup (A\cap Y) \cup B = A \cup (A\cap Y) = A$.
So $(A\setminus B) \cup Y$ is also a solution. So any $X \supset A \setminus B$  will be a solution.
So for any $X$ , $(A\cap X) \cup B$ if and only if $X  \supset (A\setminus B)$ and $B \subset A$.
