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Proof with supremum.

  1. Nov 3, 2006 #1

    MathematicalPhysicist

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    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. jcsd
  3. Nov 3, 2006 #2

    quasar987

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    What about the disjoint sets A=(a,b), B={b} ?
    supA = supB=b
     
  4. Nov 3, 2006 #3

    quasar987

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    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. ==><==
     
  5. Nov 3, 2006 #4

    MathematicalPhysicist

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    thanks, im having troubles of finding counter examples.
     
  6. Nov 3, 2006 #5

    matt grime

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    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).
     
  7. Nov 3, 2006 #6

    MathematicalPhysicist

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    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.
     
  8. Nov 3, 2006 #7

    quasar987

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    Both look good.
     
  9. Nov 14, 2006 #8
    "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 #9
    "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
     
  11. Nov 14, 2006 #10

    CRGreathouse

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    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.
     
  12. Nov 15, 2006 #11
    what if A is strictly irrational and B is strictly rational, couldn't you use that to form a counter example?
     
  13. Nov 17, 2006 #12

    CRGreathouse

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    * 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.
     
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