2
$\begingroup$

I have been told that a function $f$ is continuous at $c$ if for every $\epsilon >0 $ there exists a $ \delta >0 $ such that:

$|x-c|<\delta \Rightarrow |f(x)-f(c)|< \epsilon$

This may be a stupid question, but is this the same as saying $f$ is continuous at $c$ if for every $\delta >0 $ there exists an $ \epsilon >0 $ such that:

$|f(x)-f(c)|< \epsilon \Rightarrow |x-c|<\delta$ ?

Thank you in advance.

$\endgroup$
  • 1
    $\begingroup$ It's not $\iff$ but only $\implies$. $\endgroup$ – Xam Nov 22 '16 at 21:49
2
$\begingroup$

No, it is not the same.

  • $$\boldsymbol{f(x) = \sin x}$$ is an example of a function that is continuous, but does not satisfy your alternate condition anywhere. Why? Consider for example $c = 0$. Pick any $\delta > 0$ and there must be an $\epsilon > 0$ such that whenever $|\sin x| < \epsilon$, $|x| < \delta$. But taking $x = k \pi$, since $|\sin (k \pi)| = 0 < \epsilon$, that means that $|k \pi| < \delta$ for all $k$, which never happens.

  • $$\boldsymbol{f(x) = \begin{cases} x + 1 &\text{if } x > 0 \\ 0 &\text{if } x = 0 \\ x - 1 &\text{if } x < 0 \end{cases}}$$ is an example of a function that is not continuous (at $0$), but satisfies your alternate condition everywhere.

$\endgroup$
  • $\begingroup$ thank you, could you help me furthermore? Is the definition the same if you say "for every $\delta > 0$ there exists $\epsilon > 0$"? (instead of for every epsilon there exists a delta). $\endgroup$ – Desmoz Nov 22 '16 at 23:29
  • $\begingroup$ @ConnorGaughan No, that would not be the same either. That condition is the same as the function being bounded. If it is bounded, then for any $\delta$ we can just take $\epsilon$ to be bigger than the bound on the function. $\endgroup$ – 6005 Nov 23 '16 at 1:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.