Power sets of sets with nonempty intersection (Set Theory) I started learning about Set Theory and I ran into the following unproven lemma:
There exist sets $A$ and $B$ with $A\cap B=\emptyset$ such that the power set of $A$ is equal to the power set of $A\setminus B$.
The statement is: P(A) = P(A/B) $\Leftrightarrow$ $A$ and $B$ are disjoint.
I tried every possible pair of sets I could come-up with, yet I cannot find a concrete example.
I would really appreciate it if someone could point me in the right direction.
Thanks
 A: Are you sure that this is supposed to be correct.
When $A$ and $B$ are sets with $A\cap B\neq\emptyset$, then there is $x\in A$ with $x\in B$.
Then $\{x\}\in\mathcal{P}(A)$. But $\{x\}\notin\mathcal{P}(A\setminus B)$. Since $\{x\}$ can not be a subset of $A\setminus B$.
Suppose otherwise, and $\{x\}\subseteq A\setminus B$. Then $x\in A\setminus B$, but $x\in A$ and $x\in B$. So $x\notin A\setminus B$. Which is a contradiction.
Edit: On your update.
When the statement is to show that $\mathcal{P}(A)=\mathcal{P}(A\setminus B)\Leftrightarrow~ A\cap B=\emptyset$, then we can prove this as follows:
$\Rightarrow$.
Let $\mathcal{P}(A)=\mathcal{P}(A\setminus B)$.
Suppose $A\cap B\neq\emptyset$. Then the proof given above can be applied here, to yield a contradiction.
You might want to work out the differences, or details yourself.
$\Leftarrow$:
Let $A\cap B=\emptyset$. Show $\mathcal{P}(A)=\mathcal{P}(A\setminus B)$.
This is trivial, as $A\cap B=\emptyset$ implies that $A\setminus B=A$.
Indeed:
$A\setminus B\subseteq A$ holds always.
For: $A\subseteq A\setminus B$ let $x\in A$. Since $A$ and $B$ are disjoint, we have that $x\notin B$. So $x\in A\setminus B$.
So we have shown that $A=A\setminus B$ (under the assumption that $A\cap B=\emptyset$.)
