1
$\begingroup$

Just curious, the definition of a limit is:

For every $\epsilon\gt0$, there exists a $\delta\gt0$, such that for every $x$, the expression $0\lt|x-c|\lt\delta$ implies $|f(x)-L|\lt\epsilon$.

Is there a reason why this definition uses an open ball around L and an open, punctured disc around c instead of closed discs/balls? To be more precise, would it be incorrect to say:

For every $\epsilon\gt0$, there exists a $\delta\gt0$, such that for every $x$, the expression $0\lt|x-c|\le\delta$ implies $|f(x)-L|\le\epsilon$.

as the definition of a limit?

$\endgroup$
  • 1
    $\begingroup$ they are equivalent. $\endgroup$ – yoyo Oct 11 '14 at 1:06
  • $\begingroup$ @user1346994 If you had equality, couldn't you just subtract a really small number from $\delta$ and $\epsilon$ you found? If your answer is yes (and it should be since $\mathbb{R}$ is complete), then the two definitions are equivalent. $\endgroup$ – epsilon-delta Oct 11 '14 at 1:07
  • 1
    $\begingroup$ @epsilon-delta this makes sense. Is there a reason it's always phrased with the open ball/disc? I was thinking maybe it had something to do with topological spaces and neighborhoods. $\endgroup$ – Justin Oct 11 '14 at 1:24

Your Answer

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

Browse other questions tagged or ask your own question.