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#### Alexmahone

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- Jan 26, 2012

- 268

Do I let $\displaystyle x\in A$ and prove that $\displaystyle x\notin B$? Do I also have to prove the converse?

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- Jan 26, 2012

- 268

Do I let $\displaystyle x\in A$ and prove that $\displaystyle x\notin B$? Do I also have to prove the converse?

As stated, the question is too vague to give a clear answer.

Do I let $\displaystyle x\in A$ and prove that $\displaystyle x\notin B$? Do I also have to prove the converse?

That said, you might suppose that [tex]\exists x\in A\cap B[/tex].

Prove that leads to a contradiction.

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- Jan 26, 2012

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Thanks for your answer, but why is the question too vague?As stated, the question is too vague to give a clear answer.

That said, you might suppose that [tex]\exists x\in A\cap B[/tex].

Prove that leads to a contradiction.

Because you told us nothing about $A\text{ or }B$.Thanks for your answer, but why is the question too vague?

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- Jan 26, 2012

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A and B are 2 arbitrary sets.Because you told us nothing about $A\text{ or }B$.

[tex]\left( {\forall x \in \mathcal{ U}} \right)\left[ x \notin A\cap B \right][/tex].A and B are 2 arbitrary sets.

- Feb 15, 2012

- 1,967

however, you have to be careful. just picking "some" x in A, and showing it is not in B, just shows x is in A-B, which can happen when A and B are not disjoint. the choice of x has to be completely arbitrary (a "for all" choice, not a "there exists" choice).

for example, let's use "your method" to prove (0,1) and (3,4) (these are real intervals) are disjoint. let x be any real number in (0,1). then 0 < x < 1. since 1 < 3, by the transitivity of < we have x < 3. hence (x > 3)&(x < 4) is false (trichotomy property: exactly one of x < 3, x = 3 or x > 3 can be true...note this is a specific property of real numbers, we are using the fact that R is an ordered field, here), that is: 3 < x < 4 is false, so x is not in (3,4). since x is arbitrary, we conclude (0,1) and (3,4) are disjoint.

the following is a "bad proof": let x be in (0,1). then x is not in (3,4), so (0,1) and (3,4) are disjoint. why is this bad? because x might be 1/2, and all we have shown is 1/2 is not in (3,4).

it's a subtle difference, and Plato's posts are meant to underscore the important part: disjoint sets don't "overlap". put another way: quantification matters (logically speaking).

- Jan 29, 2012

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In Munkres (Topology), there is a proof about showing sets are disjoint which is probably along the lines you are looking for. However, I can't remember off hand but it may not be too set theoretic. Chapter 2 or 3 I believe. It will be on a left hand side page at the bottom continuing to right page on the top.

Do I let $\displaystyle x\in A$ and prove that $\displaystyle x\notin B$? Do I also have to prove the converse?