MHB Prove that A and B are disjoint

[alexmahone]

How do I prove that 2 sets A and B are disjoint?

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

 

[Plato2]

How do I prove that 2 sets A and B are disjoint?
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.
That said, you might suppose that \exists x\in A\cap B.
Prove that leads to a contradiction.

 

[alexmahone]

As stated, the question is too vague to give a clear answer.
That said, you might suppose that \exists x\in A\cap B.
Prove that leads to a contradiction.

Thanks for your answer, but why is the question too vague?

 

[Plato2]

Thanks for your answer, but why is the question too vague?

Because you told us nothing about $A\text{ or }B$.

 

[alexmahone]

Because you told us nothing about $A\text{ or }B$.

A and B are 2 arbitrary sets.

 

[Plato2]

A and B are 2 arbitrary sets.

\left( {\forall x \in \mathcal{ U}} \right)\left[ x \notin A\cap B \right].

 

[Deveno]

if for ALL x in A, x is not in B, that would be sufficient. why? because that asserts that A is a subset of A - B = A - A∩B. since A - B is automatically a subset of A, we get immediately that A = A - B = A - A∩B, hence A∩B = Ø (if y was in A∩B, then y would not be in A-B, a contradiction since A = A-B).

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).

 

[HallsofIvy]

It is sufficient to prove "if x is in A then it is NOT in B" or "if x is in B then it is NOT in A". You do not have to prove both.

 

[Dustinsfl]

How do I prove that 2 sets A and B are disjoint?

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

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.