1
$\begingroup$

Claim: If a function $\mathbb{R}^n \rightarrow \mathbb{R}^m$ is continuous at $\vec a \in \mathbb{R}^n$, it is continuous in some open ball around $\vec a$.

Is this claim false? In other words, is it possible for a function to be continuous at a single point $\vec a$ only, but not in the points around $\vec a$?

$\endgroup$
  • 3
    $\begingroup$ You asked two questions: «is this claim correct?» and (essentially) «is this claim false?», which have opposite answers! Now people write answers which start with «Yes., blah» and it is all unnecessarily confusing $\endgroup$ – Mariano Suárez-Álvarez Jul 30 '15 at 6:08
  • $\begingroup$ Edited to avoid this confusion. $\endgroup$ – FreshAir Jul 30 '15 at 6:11
  • $\begingroup$ Your question title is still asking the opposite question as your last paragraph. $\endgroup$ – Ilmari Karonen Jul 30 '15 at 7:20
  • $\begingroup$ I'm thinking of changing it to "Existence of a Case for Continuity not over an Entire Open Set". Would you have a better suggestion? $\endgroup$ – FreshAir Jul 30 '15 at 9:43
11
$\begingroup$

$$f(x) = \begin{cases} x &\mbox{if } x \in \mathbb Q \\ 0 & \mbox{if } x \in \mathbb R \setminus \mathbb Q \end{cases}$$ is continuous at $x = 0$ and discontinuous elsewhere (sequence definition of continuity helps here).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You beat me by just a moment with exactly this example. NB that since $|f(x)| \leq |x|$, one can just as well show continuity at $x = 0$ using the $\delta$-$\epsilon$ definition, in particular taking $\epsilon = \delta$. $\endgroup$ – Travis Willse Jul 30 '15 at 6:09
  • $\begingroup$ I was about to make an edit that proves continuity at $x = 0$ using $\delta - \varepsilon$ but decided not to. $\endgroup$ – user217285 Jul 30 '15 at 6:10
3
$\begingroup$

Yes, consider the function $f\colon \mathbb R \to \mathbb R$ given by $f(x)=x$ if $x\in \mathbb Q$ and $f(x)=-x$ otherwise. You can even improve this example to obtain a function that is differentiable at a point but no continuous at any other point.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ yes Amitai, something like that :) $\endgroup$ – Ittay Weiss Jul 30 '15 at 6:19
  • $\begingroup$ Would be cool if you could show the second example too! $\endgroup$ – user541686 Jul 30 '15 at 6:40
  • 1
    $\begingroup$ @Mehrdad: I believe $g(x) = x^2$ for $x \in \mathbb Q$, $g(x) = -x^2$ otherwise, should qualify. $\endgroup$ – Ilmari Karonen Jul 30 '15 at 7:27

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.