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There are numbers c, d, with f(a) < f(x) < f(b) for x in (c,d)

  1. Jun 6, 2013 #1
    1. The problem statement, all variables and given/known data
    If ##f## is continuous on ##[a,b]## and ##f(a) < f(b)##. Prove that there are numbers ##c, d## with ##a \le c < d \le b## such that ##f(c) = f(a)## and ##f(d) = f(b)## and if ##x \in (c,d)## then ##f(a) < f(x) < f(b)##.


    2. Relevant equations



    3. The attempt at a solution

    This is what I tried.
    By considering the set
    ##A = \left\{ x : a \le x < b \land f(x) = f(a) \right\} ##, which is non-empty and bounded above, so it has a least upper bound ##\alpha##, then I showed that ##f(\alpha) = f(a)##. And by considering the set ##B = \left\{ x : \alpha < x \le b \land f(x) = f(b) \right\} ##, which is nonempty and bounded below, so it has a greatest lower bound ##\beta##. I showed ##f(\beta) = f(b)##. And finally showed if ##x \in (\alpha, \beta)## then ##f(a) < f(x) < f(b)##. So letting ##c = \alpha## and ##d = \beta##.
     
  2. jcsd
  3. Jun 6, 2013 #2

    LCKurtz

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    But might it not be that ##\beta < \alpha##?
     
  4. Jun 6, 2013 #3
    Well ##B## is bounded below by ##\alpha##, so ##\alpha## is a lower bound of B, and since ##\beta## is the greatest lower bound of ##B##, ##\alpha \le \beta##
     
  5. Jun 6, 2013 #4

    LCKurtz

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    Yes. Sorry, I looked at your definitions of A and B and didn't notice you had ##\alpha## instead of ##a## for the lower limit in B. So I think your argument looks good.
     
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