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My first instinct with this proof is to assume the opposite of the hypothesis, as in a proof by contradiction.

My work is as follows:

Suppose $(A-B) \cap (B-A) \neq \emptyset$.

Consider an $x \in (A-B) \cap (B-A)$.

If $x \in (A-B) \land x \in (B-A)$

$(x \in A \land x \notin B) \land (x \in B \land x \notin A)$

$(x \in A \land x\notin A) \land (x \in B \land x \notin B)$

These are contradictions. Hence such an element $x$ does not exist.

$\implies (A-B) \cap (B-A) = \emptyset$.

My issue is that it is apparent that assuming that the set is not empty is apparently not good practice? Is there a better way to approach this proof?

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  • $\begingroup$ What a problem do it directly? $\endgroup$ – BBVM Jun 4 '16 at 2:27
  • $\begingroup$ I posted another approach to the problem and I personally don't think your solution is 'not a good practice'. $\endgroup$ – zxcvber Jun 4 '16 at 2:30
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    $\begingroup$ Assuming a set is non empty is only a "bad habit" if you expect people to accept it un questioning. A proof by contradiction is the exact opposite. You assume it ONLY to tear it down. Nothing wrong with that! What is wrong is assuming it is non _ empty, come to a conclusion, and never verify that it was nonempty. $\endgroup$ – fleablood Jun 4 '16 at 2:39
  • $\begingroup$ Well, I don't recommend this but $A \subset A-B$ and $B-A \subset A^$ so $A-B \cap B-A \subset A\capA^c=\emptyset $. But to answer your question there is nothing wrong with saying "let x \in A" if you are willing to accept that it might turn out that x doesn't exist. $\endgroup$ – fleablood Jun 4 '16 at 2:46
  • $\begingroup$ "Assuming the set is not empty is not good practice" depends on what you're doing with the set. Here you're using a proof by contradiction to show that the set is empty. In this case, absolutely nothing wrong with assuming the set is not empty. In fact, it's required to do what you need to do. If it makes you feel better you can start by pointing out that there are only two possibilities: the set is empty or it isn't. If it's empty, we're done. If it's not, then contradiction. This covers all possible cases. $\endgroup$ – user307169 Jun 4 '16 at 3:04
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The contradictory method works, although one of the other options available is to prove that

$$x \in (A-B) \implies x \notin (B-A).$$

This would implies that these two sets ($(A-B)$ and $(B-A)$) are disjoint and thus $$(A-B) \cap (B-A) = \emptyset.$$

If we see that $$x \in (A-B)$$ $$\implies x \in A \wedge x \notin B$$ $$\implies x \notin B$$ $$\implies x \notin B-A,$$

This claim concludes.

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Just another solution;

Since $A-B=A\cap B^{C}$ (here, the $^{C}$ is for complementary set),

$$(A-B)\cap(B-A)=(A\cap B^{C})\cap(B\cap A^{C})$$

Since the operation $\cap$ is commutative, the equation above can be written as

$$A\cap B^{C}\cap B \cap A^{C}$$

The result is of course, a null set.

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To do a proof by contradiction you assume to intersection isn't empty. Therefore we can pick x in the intersection.

And *that * we can show is imposible.

$x \in (A-B) \cap (B-A)$ so $x \in A-B $ and $x\in B-A $. So $x\in A$, $x\not \in B$, $x \in B $, and $x \not \in A$. Which is contradictory.

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