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Difficult analysis problem involve supremum and function concepts

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    f(a) > c > f(b)

    A = { x : b > x > y > a implies f(a) > f(y) }

    let u = sup(A)

    show that f(u) = c

    2. Relevant equations

    I have no idea in particular, save for the definition of the supremum:
    [tex]\forall x \in A x \le u[/tex]
    if [tex]v[/tex] is an upper bound of A, then [tex]u \le v[/tex]

    3. The attempt at a solution

    My intuition led me to attempt a proof by contradiction. If you let f(x*) = c, assume that x* < u to arrive at a contradiction. Then assume that x* > u to arrive at a contradiction. Then to conclude that x* must be u. I don't know how to do this, or even if I can/should be done.
     
  2. jcsd
  3. Nov 3, 2009 #2

    jbunniii

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    There must be some information missing. What is [itex]c[/itex], just some random point between [itex]f(b)[/itex] and [itex]f(a)[/itex]? What is known about [itex]f[/itex]? Is it monotonically decreasing? Is it continuous?
     
  4. Nov 3, 2009 #3
    There is nothing missing from the problem. I guess we can assume that it's continuous. It isn't necessarily a decreasing monotone function. Also, yes, c is any point between f(b) and f(a).
     
  5. Nov 3, 2009 #4

    jbunniii

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    Well, if that's the case then it's not true.

    Let f(x) = -x for all x.

    Let a = 0, b = 1, c = -0.5.

    Then A = (0, 1), u = sup(A) = 1, but f(u) does not equal c.
     
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