# Continuity Must Hold in an Entire Open Set?

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 $\vec a$?

• 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 Jul 30, 2015 at 6:08
• Edited to avoid this confusion. Jul 30, 2015 at 6:11
• Your question title is still asking the opposite question as your last paragraph. Jul 30, 2015 at 7:20
• 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? Jul 30, 2015 at 9:43

$$f(x) = \begin{cases} x &\mbox{if } x \in \mathbb Q \\ 0 & \mbox{if } x \in \mathbb R \setminus \mathbb Q \end{cases}$$ is continuous at $x = 0$ and discontinuous elsewhere (sequence definition of continuity helps here).
• 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 $\epsilon = \delta$. Jul 30, 2015 at 6:09
• I was about to make an edit that proves continuity at $x = 0$ using $\delta - \varepsilon$ but decided not to. 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.
• @Mehrdad: I believe $g(x) = x^2$ for $x \in \mathbb Q$, $g(x) = -x^2$ otherwise, should qualify. Jul 30, 2015 at 7:27