# Proof with supremum.

1. Nov 3, 2006

### MathematicalPhysicist

let A,B be nonempty sets of real numbers, prove that:
if A,B are bounded and they are disjoint, then supA doesnt equal supB.

here's my proof:
assume that supA=supB=c
then for every a in A a<=c and for every b in B b<=c.
bacuse A.B are bounded then: for every e>0 there exists x in A such that
c-e<x<=c and there exists y in B such that c-e<y<=c so we have two elements that are both in A and B, but this is a contradiction.
is this proof valid?
i feel that i should show that y=x, but i think bacuse e is as we choose, we have to find elements which are both in A and B.

2. Nov 3, 2006

### quasar987

What about the disjoint sets A=(a,b), B={b} ?
supA = supB=b

3. Nov 3, 2006

### quasar987

This does not follow from boundedness, it follows from the fact that c is the sup. If there was an e such that there is no element of A btw c and c-e, then it would means that c-e is an upper bound of A that is smaller than c. ==><==

4. Nov 3, 2006

### MathematicalPhysicist

thanks, im having troubles of finding counter examples.

5. Nov 3, 2006

### matt grime

It isn't true. Counter examples abound: take any interval like (0,1) and pick two disjoint dense subsets (there are uncountably many disjoint dense subsets so this can be done).

6. Nov 3, 2006

### MathematicalPhysicist

ok thanks.
just to clear on other matters, am i right in saying that the following arent correct:
if A is infinite set and doesnt have a minimum then it's not bounded, the simple counter example is the interval (0,1) it's infinite doesnt have a minimum but it's bounded.
another statement is if A,B are bounded and supA=infB then the intersection has only one element.
i found the next counter example, A=(0,1) B=(1,2) A and B are bounded and supA=infB, but they are disjoint.

7. Nov 3, 2006

### quasar987

Both look good.

8. Nov 14, 2006

### tuananh

"for every e>0 there exists x in A such that
c-e<x<=c and there exists y in B such that c-e<y<=c so we have two elements that are both in A and B, but this is a contradiction."

Wrong conclusion ! Ok, for every e>0, there will be x in A and y in B such that c-e<x<=c and c-e<y<=c, but from this, we can not say anything about A and B contains the same elements.

Note that x and y depend on each e>0 (they "move" when e is changed), so you can not show x = y in anyway

9. Nov 14, 2006

### tuananh

"for every e>0 there exists x in A such that
c-e<x<=c and there exists y in B such that c-e<y<=c so we have two elements that are both in A and B, but this is a contradiction."

Wrong conclusion ! Ok, for every e>0, there will be x in A and y in B such that c-e<x<=c and c-e<y<=c, but from this, we can not say anything about A and B contains the same elements.

Note that x and y depend on each e>0 (they "move" when e is changed), so you can not show x = y in anyway

10. Nov 14, 2006

### CRGreathouse

Both are incorrect, as desired.

For the second, there are two good ways to make this true that I can see:
* If sup A = inf B then the intersection has at most one element.
* If sup A = inf B then the intersection of the closures has exactly one element.

11. Nov 15, 2006

### climber/jumper

what if A is strictly irrational and B is strictly rational, couldn't you use that to form a counter example?

12. Nov 17, 2006

### CRGreathouse

* If sup A = inf B then the intersection has at most one element.

This holds in that case, since the intersection is empty.

* If sup A = inf B then the intersection of the closures has exactly one element.

This also holds. Sup A might not be in A; if it is in A, then inf B is not in B.