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Proof of Uniform Continuity

  1. Nov 24, 2008 #1
    From my textbook, this is the proof given for a theorem stating that any function continuous in a closed interval is automatically uniformly continuous in that interval.

    Proof: "If f were not uniformly continuous in [a, b] there would exist a fixed [tex]\epsilon > 0[/tex] and points x, z in [a, b] arbitrarily close to each other for which [tex]|f(x) - f(z)| \ge \epsilon[/tex]. It would then be possible for every n to choose points [tex]x_n, z_n [/tex] in [a, b] for which [tex]|f(x_n) - f(z_n)| \ge \epsilon[/tex] and [tex]|x_n - z_n| < 1/n[/tex]. Since the [tex]x_n[/tex] form a bounded sequence of numbers we could find a subsequence converging to a point c of the interval (using the compactness of closed intervals). The corresponding values [tex]z_n[/tex] would then also converge to c: since f is continuous at c, we would find that [tex]c = lim f(x_n) = lim f(z_n) [/tex] for n tending to infinity in the subsequence, which is impossible if [tex]|f(x_n) - f(z_n)| \ge \epsilon[/tex] for all n."

    I understand everything except one thing: Why is c equal to the limit of [tex]f(x_n)[/tex]? If you cut out that part and just say that since [tex]x_n[/tex] and [tex]z_n[/tex] both converge to c, then [tex]lim f(x_n) = lim f(z_n)[/tex] and there we have our contradiction.
    Last edited: Nov 24, 2008
  2. jcsd
  3. Nov 24, 2008 #2
    Shouldn't it be

    \lim_{n\to\infty} f(x_n) = f(c) = \lim_{n\to\infty} f(z_n)?

    Looks like a mistake where c and f(c) are confused.
  4. Nov 24, 2008 #3
    That's EXACTLY what I was thinking. Glad to see I wasn't the only one thinking that :)
  5. Nov 24, 2008 #4


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    given any open cover of a closed bounded interval, there is a positive number e such that every interval of diameter e is entirely contained in one of the open intervals. qed.
  6. Nov 28, 2008 #5
    Hopefully that interval is bounded too...otherwise things could get nasty.
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