# Epsilon-delta definition of a continuous function

• March 15th 2010, 04:08 PM
Pinkk
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)?
• March 15th 2010, 04:13 PM
Drexel28
Quote:

Originally Posted by Pinkk
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?
• March 15th 2010, 04:17 PM
Pinkk
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.
• March 15th 2010, 05:20 PM
Pinkk
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.