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Math Help - Epsilon-delta definition of a continuous function

  1. #1
    Senior Member Pinkk's Avatar
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    Epsilon-delta definition of a continuous function

    The definition for a function being continuous at a point is obviously well known: Given \epsilon > 0 there exists \delta > 0 such that |x - x_{0}| < \delta \implies |f(x)-f(x_{0})| <\epsilon

    But assume we use the following "definition"; Given \delta > 0 there exists \epsilon >  0 such that |f(x)-f(x_{0})| <\epsilon \implies |x - x_{0}| < \delta.

    Where does this fail in implying continuity, how is this insufficient to showing that the implication:if \lim_{n \to \infty} x_{n} = x_{0}, then \lim_{n \to \infty}f(x_{n}) = f(x_{0}) holds (which is the sequential definition of continuity)?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    The definition for a function being continuous at a point is obviously well known: Given \epsilon > 0 there exists \delta > 0 such that |x - x_{0}| < \delta \implies |f(x)-f(x_{0})| <\epsilon

    But assume we use the following "definition"; Given \delta > 0 there exists \epsilon >  0 such that |f(x)-f(x_{0})| <\epsilon \implies |x - x_{0}| < \delta.

    Where does this fail in implying continuity, how is this insufficient to showing that the implication:if \lim_{n \to \infty} x_{n} = x_{0}, then \lim_{n \to \infty}f(x_{n}) = f(x_{0}) holds (which is the sequential definition of continuity)?
    What have you done?
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  3. #3
    Senior Member Pinkk's Avatar
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    Well, the only method I could see working is that if that definition worked, then it's negation must always be false, that: given \epsilon > 0, there exists \delta > 0 such that |f(x) - f(x_{0})| < \epsilon and |x - x_{0}| \ge \delta must always be false. If I can find a case where the negation is true, then obviously the negation is not always false and therefore the new definition cannot imply continuity. But that is as far as I have gotten so far.

    Or I guess the new definition just shows that if \lim_{n \to \infty} f(x_{n}) = f(x_{0}), then \lim_{n \to \infty} x_{n} = x_{0} but then I'd just have to show that that is not equivalent to if \lim_{n \to \infty} x_{n} = x_{0}, then \lim_{n \to \infty}f(x_{n}) = f(x_{0}) but I do not know how to go about that exactly.
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  4. #4
    Senior Member Pinkk's Avatar
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    Okay, so I showed that the new definition does not imply the sequential definition of continuity by using the example of f(x) = x+1 when x>0 and f(0)=0 on the domain [0,\infty). Kinda stuck on showing if the converse is also not necessarily true.
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