2
$\begingroup$

I found an old exam and one task is: What does "$f$ is continuous in $x_{0}$" mean, with reference to $\varepsilon-\delta$ definition.

$f$ is a function and $x_{0}$ is the position where we check if the function is continuous...

Would it be enough if I just wrote:

Let $f: A \rightarrow \mathbb{R}$

$f$ is continuous in $x_{0}$ $\Leftrightarrow$ for all $\varepsilon>0 $ there exists a $\delta>0$ so that for all $x \in A:$ $|x-x_{0}|<\delta: |f(x)-f(x_{0})|< \varepsilon$


Is it correct like that? Or would you do it completely different than me? Please let me know, it's very important!

$\endgroup$
2
$\begingroup$

Your definition is correct if $A$ is a subset of $\mathbb R$. Otherwise, $|x-x_0|$ is not defined.

Also, it would be nice (but it's not really a mistake if you don't) to mention somewhere that $x_0\in A$.

$\endgroup$
1
$\begingroup$

Your answer is wrong due to the use of colons. A colon means "such that", which means the last colon is wrong. A correct definition would be:

for all $\varepsilon>0 $ there exists a $\delta>$ so that for all $x \in A:$ $|x-x_{0}|<\delta \Rightarrow |f(x)-f(x_{0})|< \varepsilon$

$\endgroup$
  • $\begingroup$ Oh thank you! Didn't know how to translate German word "sodass" correctly into English... :-) $\endgroup$ – cnmesr Sep 29 '16 at 16:49
  • $\begingroup$ @cnmesr Ja, ja. Deutsch ist so eine schwere Sprache, dass sogar die Übersetzung schwieriger ist als bei den meisten anderen Sprachen. $\endgroup$ – callculus Apr 25 '18 at 14:50

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.