# Find the proof with supremum for bounded and disjoint sets

• MathematicalPhysicist
In summary: But then the intersection of the closures of A and B is {sup A = inf B}.In summary, the statement that if A and B are bounded and disjoint, then their suprema are not equal is valid. The proof provided shows that for every epsilon, there are elements in A and B that are within epsilon of their respective suprema, but this does not mean that these elements are equal. Additionally, the statements about A and B not having a minimum and the intersection of their closures having only one element are incorrect and can be easily disproved with counterexamples.
MathematicalPhysicist
Gold Member
let A,B be nonempty sets of real numbers, prove that:
if A,B are bounded and they are disjoint, then supA doesn't 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.

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

loop quantum gravity said:
let A,B be nonempty sets of real numbers, prove that:
if A,B are bounded and they are disjoint, then supA doesn't 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

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. ==><==

thanks, I am having troubles of finding counter examples.

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).

ok thanks.
just to clear on other matters, am i right in saying that the following arent correct:
if A is infinite set and doesn't have a minimum then it's not bounded, the simple counter example is the interval (0,1) it's infinite doesn't 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.

Both look good.

"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

"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

loop quantum gravity said:
ok thanks.
just to clear on other matters, am i right in saying that the following arent correct:
if A is infinite set and doesn't have a minimum then it's not bounded, the simple counter example is the interval (0,1) it's infinite doesn't 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.

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.

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

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

* 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.

## 1. What is the definition of supremum?

The supremum of a set is the least upper bound, or the smallest number that is greater than or equal to all numbers in the set.

## 2. How is supremum used in mathematical proofs?

Supremum is often used in proofs to establish the existence of a maximum or minimum value in a set, or to show that a particular value is the "best" or "largest" possible value for a given property.

## 3. Can a set have more than one supremum?

Yes, a set can have more than one supremum if it is unbounded or if there are multiple elements that are equal to the supremum value.

## 4. What is the difference between supremum and maximum?

While supremum is the smallest number that is greater than or equal to all numbers in a set, maximum is the largest number in a finite set. In other words, supremum can exist in infinite sets, while maximum only exists in finite sets.

## 5. How is the supremum used to prove the convergence of a sequence?

In order to prove the convergence of a sequence, we use the definition of supremum to show that the sequence approaches a particular limit, as the supremum serves as a boundary for the values of the sequence.

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