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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

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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.

  1. 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.

  2. 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 $\subseteq$.

  3. 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.

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  • $\begingroup$ well the elements of P(A-B) will be the subsets made of elements strictly belonging to A and are disjoint from B? $\endgroup$ – Sev Dec 15 '14 at 7:01
  • $\begingroup$ Yes, that is correct. What does it tell you about $A$? $\endgroup$ – Asaf Karagila Dec 15 '14 at 7:07
  • $\begingroup$ that A is a subset of A-B so all members of A belong to A-B and consequently none in B? $\endgroup$ – Sev Dec 15 '14 at 7:20
  • $\begingroup$ Yeah, that's the main point here. $\endgroup$ – Asaf Karagila Dec 15 '14 at 7:21
  • $\begingroup$ so are the hints you gave me from the definition of a power set? I am confused in how to formalize the proof $\endgroup$ – Sev Dec 15 '14 at 7:24
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We will prove the contrapositive:

If $A \cap B \neq \emptyset$, then $\mathcal P(A - B) \neq \mathcal P(A)$.

Suppose that there is some element $x \in A \cap B$ so that $x \in A$ and $x \in B$. Then $x \notin A - B$ (otherwise, if $x \in A - B$, then $x \notin B$, contradicting the fact that $x \in B$). This implies that $\{x\} \subseteq A$ and $\{x\} \not\subseteq A - B$ so that $\{x\} \in \mathcal P(A)$ while $\{x\} \notin \mathcal P(A - B)$. So $\mathcal P(A - B) \neq \mathcal P(A)$, as desired. $~~\blacksquare$

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Think only about the set $A$ itself:

$A \in \mathcal P(A) $ is a subset of $A\setminus B \subseteq A$ iff $A\setminus B = A$ iff $A \cap B=\emptyset$

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