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 $\vec a$?

  • 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 $\endgroup$ Jul 30, 2015 at 6:08
  • $\begingroup$ Edited to avoid this confusion. $\endgroup$
    – FreshAir
    Jul 30, 2015 at 6:11
  • $\begingroup$ Your question title is still asking the opposite question as your last paragraph. $\endgroup$ Jul 30, 2015 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, 2015 at 9:43

2 Answers 2


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

  • $\begingroup$ 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 $\epsilon = \delta$. $\endgroup$ Jul 30, 2015 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, 2015 at 6:10

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.

  • $\begingroup$ yes Amitai, something like that :) $\endgroup$ Jul 30, 2015 at 6:19
  • $\begingroup$ Would be cool if you could show the second example too! $\endgroup$
    – user541686
    Jul 30, 2015 at 6:40
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    $\begingroup$ @Mehrdad: I believe $g(x) = x^2$ for $x \in \mathbb Q$, $g(x) = -x^2$ otherwise, should qualify. $\endgroup$ Jul 30, 2015 at 7:27

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