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Homework Help: Inequality of Supremums

  1. Sep 17, 2005 #1
    Theorem:

    For every non empty set E of real numbers that is bounded above there exists a unique real number sup(E) such that

    1. sup(E) is an upper bound for E.

    2. if y is an upper bound for E then y [itex]\geq[/itex] sup(E).


    Prove:

    [itex]sup(A\cap B)\leq sup(A)[/itex]

    I can show a special case of this,

    if [itex]A\cap B=\emptyset [/itex], then [itex]sup(A\cap B)\leq sup(A)[/itex].

    Nothing is less than something, right?

    Now here's my problem...
    Beyond the trivial case, all I have been able to do is draw pictures of sets on a number line. The pictures make the inequality really obvious, but I don't think that pictorial intuition counts as a real proof.

    Could anyone give me a pointer on how to set up a real proof?

    thanks!
     
    Last edited: Sep 17, 2005
  2. jcsd
  3. Sep 17, 2005 #2
    Ok break that inequality into two parts
    first prove the equal to part. How would you do that . Assume the su pof A and B are something and that they are equal. That way you can prove the equality part
    Then assume that the sup of A is greater than the Sup B. Now what is sup (a inter B) ? What is it less than?
     
  4. Sep 17, 2005 #3
    Ok, this is what I have for the equality part if I'm understanding you right.

    let [tex]sup(A)=sup(B)[/tex]

    then [tex]sup(A\cap B)=sup(A)[/tex]


    And the inequality would look like this?

    let [tex]sup(B)\leq sup(A)[/tex]

    then [tex]sup(A\cap B)\leq sup(A)[/tex]


    Is that right?
    It seems too simple...
     
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