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Prove that A and B are disjoint

Alexmahone

Active member
Jan 26, 2012
268
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?
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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 [tex]\exists x\in A\cap B[/tex].
Prove that leads to a contradiction.
 

Alexmahone

Active member
Jan 26, 2012
268
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.
Thanks for your answer, but why is the question too vague?
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196

Alexmahone

Active member
Jan 26, 2012
268

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.