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Thread: 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 $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - x_{0}| < \delta \implies |f(x)-f(x_{0})| <\epsilon$

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

    Where does this fail in implying continuity, how is this insufficient to showing that the implication:if $\displaystyle \lim_{n \to \infty} x_{n} = x_{0}$, then $\displaystyle \lim_{n \to \infty}f(x_{n}) = f(x_{0})$ 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 $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - x_{0}| < \delta \implies |f(x)-f(x_{0})| <\epsilon$

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

    Where does this fail in implying continuity, how is this insufficient to showing that the implication:if $\displaystyle \lim_{n \to \infty} x_{n} = x_{0}$, then $\displaystyle \lim_{n \to \infty}f(x_{n}) = f(x_{0})$ 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$\displaystyle \epsilon > 0$, there exists $\displaystyle \delta > 0$ such that $\displaystyle |f(x) - f(x_{0})| < \epsilon$ and $\displaystyle |x - x_{0}| \ge \delta$ 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 $\displaystyle \lim_{n \to \infty} f(x_{n}) = f(x_{0})$, then $\displaystyle \lim_{n \to \infty} x_{n} = x_{0}$ but then I'd just have to show that that is not equivalent to if $\displaystyle \lim_{n \to \infty} x_{n} = x_{0}$, then $\displaystyle \lim_{n \to \infty}f(x_{n}) = f(x_{0})$ 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 $\displaystyle f(x) = x+1$ when $\displaystyle x>0$ and $\displaystyle f(0)=0$ on the domain $\displaystyle [0,\infty)$. Kinda stuck on showing if the converse is also not necessarily true.
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