3
$\begingroup$

Does there exist continous function $f(x)$ such that $$f(x)=\begin{cases} \frac{m}{n} & \text{if } x \text{ is irrational,} \\ \text{irrational} & \text{if } x \text{ is rational} \end{cases}$$ I think it's impossible, as definition of that function is similar to Dirichlet Function or Thomae's function. And these functions are always discontinous somewhere. Please help, I don't know what to start with. I'm first year undergraduate

$\endgroup$
  • $\begingroup$ So you want a continuous function which sends rationals to irrationals and vice versa? $\endgroup$ – Hanul Jeon Jan 30 '17 at 14:55
  • 2
    $\begingroup$ What do you denote with $m/n$ ? $\endgroup$ – Yves Daoust Jan 30 '17 at 14:57
  • $\begingroup$ yes, I do. Sorry for my formulas. I started to learn LaTEX recently $\endgroup$ – ioleg19029700 Jan 30 '17 at 15:00
  • $\begingroup$ $\frac{m}{n}$ is rational number $\endgroup$ – ioleg19029700 Jan 30 '17 at 15:01
  • $\begingroup$ Note the proper use of \text{} in MathJax, as seen in my edit. $\endgroup$ – Michael Hardy Jan 30 '17 at 16:34
3
$\begingroup$

No, there is no such function.

FIrst, $f$ cannot be constant, so by the intermediate value theorem, there exists an interval $[a,b]$ such that $[a,b] \subset f(\Bbb R)$ ($a < b$)

So we have $[a,b] \cap (\Bbb{R}-\Bbb{Q}) \subset f( \Bbb{Q} )$

But $\Bbb{Q}$ is countable, so $f( \Bbb{Q} )$ is countable (or finite) too. In the other hand, $[a,b] \cap (\Bbb{R}-\Bbb{Q})$ is uncountable : contradiction

$\endgroup$
0
$\begingroup$

Let $a$ be a rational number and take a sequence $x_n\rightarrow a$ with each $x_n$ irrational. By definition $\lim_{n\rightarrow \infty}f(x_n)=f(a)$. But then $\lim_{n\rightarrow \infty}f(x_n)=\frac{m}{n}=f(a)$, which is impossible.

$\endgroup$

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.