# 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 – Mariano Suárez-Álvarez Jul 30 '15 at 6:08
• Edited to avoid this confusion. – FreshAir Jul 30 '15 at 6:11
• Your question title is still asking the opposite question as your last paragraph. – Ilmari Karonen Jul 30 '15 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? – FreshAir Jul 30 '15 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 $\epsilon = \delta$. – Travis Willse Jul 30 '15 at 6:09
• I was about to make an edit that proves continuity at $x = 0$ using $\delta - \varepsilon$ but decided not to. – user217285 Jul 30 '15 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. – Ilmari Karonen Jul 30 '15 at 7:27