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Infimum of a set.

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Let A [tex]\subset[/tex] R and B [tex]\subset[/tex] R be bounded below and define A+B = {x+y| x[tex]\in[/tex] A and y [tex]\in[/tex]B}. Is it true that inf (A+B) = inf(A) + inf(B) ?

    2. Relevant equations



    3. The attempt at a solution

    First I proved that inf(A+B) exist by doing the following.

    [tex]inf(A) \leq x[/tex] and [tex]inf(B) \leq y[/tex] for all x and y so,

    [tex]x- inf(A) + y-inf(B) \geq 0[/tex]

    [tex]x + y \geq inf(A) + inf(B)[/tex]

    So the set A +B is bounded below and therefore has a greatest lower bound; that is, inf(A+B) exist.

    My next step was to show that inf(A+B) = inf(A) + inf(B).

    To do this, I showed did the following:


    Suppose there is a number [tex]\alpha[/tex] which is a lower bound of A+B and [tex]\alpha[/tex] [tex] > inf(A) + inf(B)[/tex] . I tried to derive a contradiction.


    [tex]\alpha[/tex] [tex]> inf(A) + inf(B)[/tex] and since [tex]\alpha[/tex] is a lower bound...
    [tex]\alpha[/tex][tex]\leq x+ y[/tex] for all x and y ( which are elements of A and B, respectively )

    [tex]\alpha = inf(A) + inf(B) + \epsilon[/tex]
    [tex]\alpha = inf(A) + \frac{\epsilon}{2} + inf(B)+\frac{\epsilon}{2} [/tex]

    We know that since inf(A) and inf(B) exist there is an [tex]x_{0}\in A[/tex] and [tex]y_{0}\in B[/tex] such that

    [tex] x_{0} < inf(A) + \frac{\epsilon}{2}[/tex] and

    [tex] y_{0} < inf(B) + \frac{\epsilon}{2}[/tex].

    So this means

    [tex]\alpha > x_{0} + y_{0}[/tex] which is a contradiction since [tex]\alpha[/tex] is a lower bound.

    From this it is clear that
    [tex] inf(A+B) = inf(A) + inf(B)[/tex]


    How is this ? Is it clear enough ?
     
    Last edited: Aug 31, 2010
  2. jcsd
  3. Aug 31, 2010 #2
    Did I do something incorrectly ?
     
  4. Aug 31, 2010 #3

    Office_Shredder

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    Looks good to me
     
  5. Aug 31, 2010 #4
    Great! Thanks.
     
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