1
$\begingroup$

Can any one help me to answer this question:

Assuming $$f(x)=\begin{cases} x &\text{if }x\in \mathbb{Q} \\ 0 &\text{if }x\in \mathbb{R}\setminus\mathbb{Q} \end{cases}$$

show that $f$ is continuous only at $x=0$?

Notice: use this theorem

Let $f:D\rightarrow \mathbb{R}$ and let $c\in D$. Then $f$ is continous at $c$ if and only if, whenever $X_n$ is a sequence in $D$ that converges to $c$, then $f(X_n)$ converges to $f(c)$

I hope someone can answer this question

Thanks

$\endgroup$
1
  • 1
    $\begingroup$ Which is it that you have problems with - to show that $f$ is continuous at $0$ or to show that is it discontinuous everywhere else? $\endgroup$ – Karolis Juodelė Apr 8 '13 at 7:34
1
$\begingroup$

Hints: To show that $f$ is continuous at $0$, use the easy result that $\lim_{x\to x_0} f(x)=0 \iff \lim_{x\to x_0}|f(x)|=0$ (the proof of this result is almost a tautology, and involves only unpacking the definitions.

To show that $f$ is discontinuous at any point $p\ne 0$, choose two sequence that converge to $p$, one consisting of rationals only and one consisting of irrationals only (you will need to justify the existence of such sequences). Then use the second result you quote.

$\endgroup$
1
1
$\begingroup$

Notice that $|f(x)| \le |x|$, hence if $x_n \to 0$, then we have $f(x_n) \to 0$. Hence $f$ is continuous at $x=0$.

If $x \ne 0$, then let $q_n$ be a sequence of rationals converging to $x$, and let $\alpha_n$ be a sequence of irrationals converging to $x$. Then we have $f(q_n) = q_n \to x$ and $f(\alpha_n) = 0 \to 0$. Since $x \ne 0$, $f$ cannot be continuous at $x$ (since for any sequence $x_n \to x$ we would have $f(x_n) \to f(x)$).

$\endgroup$

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.