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Math Help - uniform continuity

  1. #1
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    uniform continuity

    Show that the function  f: (0,2) \to \bold{R} defined by  f(x) := 1/x is not uniformly continuous. I want to show this using the following corollary: Let  X be a subset of  \bold{R} , let  f: \bold{X} \to \bold{R} be a uniformly continuous function, and let  x_0 be an adherent point of  X . Then  \lim_{x \to x_{0}: x \in \bold{X}} f(x) exists and is a real number.

    So  0 is an adherent point of  X = (0,2) , because  [0- \varepsilon, 0+ \varepsilon] contains a point of  X for every  \varepsilon > 0 (its probably assumed that  0 < \varepsilon < 1 ). And  \lim_{x \to 0; x \in \bold{X}} f(x) does not exist, which implies that  f(x) is not uniformly continuous. I used the contrapositive of the corollary.

    Is this correct?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by particlejohn View Post
    Show that the function  f: (0,2) \to \bold{R} defined by  f(x) := 1/x is not uniformly continuous. I want to show this using the following corollary: Let  X be a subset of  \bold{R} , let  f: \bold{X} \to \bold{R} be a uniformly continuous function, and let  x_0 be an adherent point of  X . Then  \lim_{x \to x_{0}: x \in \bold{X}} f(x) exists and is a real number.

    So  0 is an adherent point of  X = (0,2) , because  [0- \varepsilon, 0+ \varepsilon] contains a point of  X for every  \varepsilon > 0 (its probably assumed that  0 < \varepsilon < 1 ). And  \lim_{x \to 0; x \in \bold{X}} f(x) does not exist, which implies that  f(x) is not uniformly continuous. I used the contrapositive of the corollary.

    Is this correct?
    Yes, but you need more carefull wording, something like:

    Assume f(x) is uniformly continuous, then by the corrolarly

     \lim_{x \to 0: x \in \bold{X}} f(x)=c; \ \ c \in \mathbb{R},\ \ \ \ \ ...(1)

    that is the limit exists and is a real number as  0 is an adherent point of  X = (0,2) .

    But

     \lim_{x \to 0: x \in \bold{X}} f(x) =\infty

    which contradicts (1), hence the assumption that f(x) is uniformly continuous is false and f(x) is not uniformly continuous.

    RonL
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