Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I do not know an example. Will ask question if in doubt of the proofs provided thank you!!

share|cite|improve this question
1 – user44923 Oct 16 '12 at 21:24
up vote 4 down vote accepted

$$ F(x) = \left\{ \begin{array}{rcl} x,& \mbox{if} & x \in \mathbb{Q}\\ -x , & \mbox{if} & x \notin \mathbb{Q} \\ \end{array} \right. $$

$|x-0| < \varepsilon \Rightarrow |f(x) - f(0)| = |x| <\varepsilon$. If $x_0 > 0$. There is $x \notin \mathbb{Q}$ sufficiently near $x_0$ such that $f(x) = -x$ is sufficiently near $-x_0$. Thus do not sufficiently near $f(x_0) = x_0$.

share|cite|improve this answer
What is the difference between your answer amd tomas'? – Maximiliano Oct 16 '12 at 21:21
@Maximum: It is simpler, and therefore more elegant. – TonyK Oct 16 '12 at 21:23
can you explain how does this show that this function is only continous at 0? plz – Maximiliano Oct 16 '12 at 21:25
I add explanation in the answer. – user29999 Oct 16 '12 at 21:32
AWESOME thank you!! – Maximiliano Oct 16 '12 at 21:34

Let $g(x)=1$ if $x$ is rational, and $0$ if $x$ is irrational. Let $f(x)=xg(x)$.

share|cite|improve this answer

$$ f(x) = \left\{ \begin{array}{rl} x^{2} &\mbox{ if $x$ is rational} \\ -x^{2} &\mbox{ otherwise} \end{array} \right. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.