Assume that $\mathcal P(A-B)= \mathcal P(A)$. Prove that $A\cap B = \varnothing$.
What I did:
I tried proving this directly and I got stuck.
Let $X$ represent a nonempty set, and let $X\in\mathcal P(A-B)$.
By definition of power set and set difference:
$X\subseteq A$ and $X\nsubseteq B$.
By definition of a subset and subset negation:
Let $y$ be an arbitrary element of $X$ such that:
$(\forall y)(y \in X \rightarrow y \in A)\land (\exists y)(y \in X \land y \nsubseteq B)$
This is where I get stuck, how do I confirm I have no common elements with the fact that there is at least one element they don't share?
I also tried proving the contrapositive by assuming that the intersection is not disjoint, where I let $X$ be a subset in the intersection then $X$ must belong to both $A$ and $B$ and that's where I got stuck.
I apologize for bad formatting first time poster
Best Answer
HINT: $A\in\mathcal P(A)$, so $A\in\mathcal P(A-B)$. What can you say about all the elements in $\mathcal P(A-B)$ and $B$?
Now that we covered the approach to the correct answer, let me give you a short critique about your efforts.
If $X\in\mathcal P(A-B)$ it is not just that $X\nsubseteq B$, it is that $X\cap B=\varnothing$, this is a much stronger fact, and we do use it here, as seen above.
In the explicit writing that $X\subseteq A$ and $X\nsubseteq B$, you say that there is some $y\in X$ such that $y\nsubseteq B$. That might be a typo, and you meant $y\notin B$, but nonetheless you should be wary about mixing $\in$ and $\subseteq$.
As before, you write in the last attempt that you try to take $X$ in the intersection of $A$ and $B$, then it is a subset of both, this is also not true. It is an element of both. And a subset of the intersection is a subset of both $\mathcal P(A)$ and $\mathcal P(B)$. But that's mainly what you can say about that.