Does there exist a continuous onto function from $\mathbb{R}-\mathbb{Q}$ to $\mathbb{Q}$? (where domain is all irrational numbers)

I found many answers for contradicting the fact that there doesnt exist a continuous function which maps rationals to irrationals and vice versa.

But proving that thing was easier since our domain of definition of function was a connected set, we could use that connectedness or we could use the fact that rationals are countable and irrationals are uncountable.

But in this case those properties are not useful. I somehow think that baire category theorem might be useful but I am not good at using it.

  • $\begingroup$ Try $f(x)=1$... $\endgroup$ – David C. Ullrich Sep 13 '15 at 18:07
  • 1
    $\begingroup$ you missed onto. $\endgroup$ – Landon Carter Sep 13 '15 at 18:07
  • 1
    $\begingroup$ @DavidC.Ullrich - OP asked for an onto function. $\endgroup$ – Sam Cappleman-Lynes Sep 13 '15 at 18:07

Yes. Say $E_n$ is the set of irrationals in the interval $(n,n+1)$. Say $(q_n)$ is an enumeration of $\Bbb Q$. Define $f(x)=q_n$ for $x\in E_n$.

  • $\begingroup$ This is amazing!! How could you create such a function? I mean how did you come up with this? $\endgroup$ – Landon Carter Sep 13 '15 at 18:14
  • 1
    $\begingroup$ Can you give more explanation , what is enumeration of rational ? $\endgroup$ – Shubham Ugare Sep 13 '15 at 18:21
  • $\begingroup$ I guess this construction can also be used to show that, if $X$ is any topological space which can be partitioned into a countably infinite collection of open sets (which are then automatically closed too) and $Y$ is any countable topological space, then there exists a continuous surjection $X \to Y$. $\endgroup$ – Mike F Sep 13 '15 at 18:25
  • $\begingroup$ @ShubhamUgare The rationals are countable. An enumeration of the rationals is just a sequence containing each rational exactly once. (Here just for convenience we took that "sequence" to be indexed by integers instead of natural numbers.) $\endgroup$ – David C. Ullrich Sep 13 '15 at 18:27
  • $\begingroup$ @LandonCarter I have a hard time saying how I came up with that, sorry. Seemed like an obvious thing... $\endgroup$ – David C. Ullrich Sep 13 '15 at 18:30

Yes, there exists such a function.

Biject $\mathbb{Q}$ with $\mathbb{Z}$ to get $\mathbb{Q} = \{q_n \mid n \in \mathbb{Z}\}$, and let $\mathbb{I}$ be the set of irrational numbers.

Define $I_n$ for $n \in \mathbb{Z}$ as $\mathbb{I} \cap (n, n+1)$. Then define $f(x) = q_n$ for all $x \in I_n$.

This is continuous, since for any irrational number in $I_n$, there is a small neighbourhood of it which is contained entirely within $I_n$ (because the "endpoints" of $I_n$ were chosen to be rational)

  • $\begingroup$ Did you notice that that's exactly the same as the example I gave? $\endgroup$ – David C. Ullrich Sep 13 '15 at 18:26
  • 2
    $\begingroup$ Yes - but we answered at almost the same time. $\endgroup$ – Sam Cappleman-Lynes Sep 13 '15 at 18:27

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.