Is there a way to prove the following result using connectedness?


Let $J=\mathbb{R} \setminus \mathbb{Q}$ denote the set of irrational numbers. There is no continuous map $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(\mathbb{Q}) \subseteq J$ and $f(J) \subseteq \mathbb{Q}$.


  • 1
    $\begingroup$ Please make your post self-contained by incorporating the statement. It takes forever for planetmath to render for me. Maintaining the link is fine, so one can see how it is proved there. $\endgroup$ – Ross Millikan Aug 4 '11 at 18:18
  • $\begingroup$ @Ross Millikan: just edited it. $\endgroup$ – user10 Aug 4 '11 at 18:22
  • 1
    $\begingroup$ @user10 actually the link is down. $\endgroup$ – Gabriel Romon Feb 21 '14 at 23:56
  • $\begingroup$ I searched planetmath to find the original linked argument; I think it is here and here If it moves again but stays on the site, a Google search such as site:planetmath.org switch rational irrational baire will find it. $\endgroup$ – Jonas Meyer Jul 7 '16 at 14:42
  • $\begingroup$ Linked. $\endgroup$ – Alex Ravsky Sep 23 '17 at 1:05

Here's a way to use connectedness, really amounting to using the intermediate value theorem.

If $f(\mathbb{Q})\subseteq \mathbb R\setminus\mathbb Q$ and $f(\mathbb R\setminus \mathbb Q)\subseteq\mathbb Q$, then $f(0)\neq f(\sqrt 2)$. Because intervals are connected in $\mathbb R$ and $f$ is continuous, $f[0,\sqrt 2]$ is connected. Because connected subsets of $\mathbb R$ are intervals, $f[0,\sqrt 2]$ contains the interval $\left[\min\{f(0),f(\sqrt 2)\},\max\{f(0),f(\sqrt 2)\}\right]$. The set of irrational numbers in this interval is uncountable, yet contained in the countable set $f(\mathbb Q)$, a contradiction.

A slightly briefer outline: The hypothesis implies that $f$ is nonconstant with range contained in the countable set $\mathbb Q\cup f(\mathbb Q)$, whereas the intermediate value theorem and uncountability of $\mathbb R$ imply that a nonconstant continuous function $f:\mathbb R\to\mathbb R$ has uncountable range.

  • $\begingroup$ thank you, beautiful argument. $\endgroup$ – user10 Aug 4 '11 at 18:31
  • 5
    $\begingroup$ More elementary than a proof using Baire Category! $\endgroup$ – GEdgar Aug 4 '11 at 18:52
  • 4
    $\begingroup$ I like the second paragraph version $\endgroup$ – Hagen von Eitzen Dec 29 '13 at 23:13
  • $\begingroup$ @Jonas Meyer How to prove rigoruosly that $f(\mathbb{Q})$ is countable...a hint would do...I have to do this as a homework question ...i have understood the general idea but unable to express rigoruously the above cardinality argument..thanks $\endgroup$ – spaceman_spiff Mar 22 '17 at 9:23
  • 2
    $\begingroup$ @spaceman_spiff: If $A=\{a_1,a_2,a_3,\ldots\}$ is a countable set and $f$ is a function defined on $A$, then $f(A)=\{f(a_1),f(a_2),f(a_3),\ldots\}$ is a countable set. $\endgroup$ – Jonas Meyer Mar 23 '17 at 1:35

Suppose there is such a mapping $f$. Consider $g:[0,1]\to \mathbb{R}$ defined by $$g(x)=f(x)-x.$$ Suppose that $g(x)\in \mathbb{Q}$ for some $x\in [0,1]$. Then:

  • if $x\in J$, then $g(x)-f(x)\in \mathbb{Q}$, i.e. $x\in \mathbb{Q}$.
  • if $x\in \mathbb{Q}$, then $g(x)+x\in \mathbb{Q}$, i.e. $f(x)\in \mathbb{Q}$, i.e. $x\in J$

both produce contradictions. Thus $g([0,1])\subseteq J$. Since $f$ is continuous, $g$ is continuous, and then $g([0,1])=[\min g,\max g]$. If $g$ is not constant then there exists $r$ a rational in $[\min g,\max g]$. By the intermediate value theorem, there exists $z\in[0,1]$ such that $g(z)=r$, but this is impossible because $g([0,1])\subseteq J$. Therefore, $g$ must be constant and then $$f(x)=c+x$$ with $c\in J$. Particularly, $f(c)=2c$ which, as Jonas pointed, is contradictory to the hypothesis. Therefore $f$ can not exist.

  • 1
    $\begingroup$ A quicker way to finish once you have $f(x)=c+x$ is to note that $f(c)=2c$ is irrational, contradicting the hypothesis. +1: This is a nice alternative that can handle a more general situation. Note that cardinality need not be considered, and in fact $\mathbb Q$ can be replaced by any subgroup of $\mathbb R$. $\endgroup$ – Jonas Meyer Aug 4 '11 at 19:43
  • $\begingroup$ yes, that's true and thank you @Jonas, I will correct for the sake of simplicity. $\endgroup$ – leo Aug 4 '11 at 19:48
  • 1
    $\begingroup$ To clarify my previous comment, I should have said that this argument applies verbatim if $\mathbb Q$ is replaced by any dense subgroup of $\mathbb R$ containing $2$. The last restriction could be handled by rescaling if necessary. $\endgroup$ – Jonas Meyer Aug 4 '11 at 20:06
  • $\begingroup$ Yes, I had some doubts after I wrote my comment. $\endgroup$ – leo Aug 4 '11 at 20:39
  • $\begingroup$ Hey Leo, how do you conclude that $g([0,1]) \subset $\mathbb{I}$ from the two contradictions above? $\endgroup$ – Kamil Jul 7 '16 at 9:24

Another simple proof:

Because $f$ is continuous and by connectedness, $f([0,1])=[a,b]$ for some $a<b$. Now define $$g : x \mapsto \frac{1}{p} \left(f(x)-q \right), \ \text{where} \ p,q \in \mathbb{Q}.$$

In particular, $g(x)$ is rational iff $f(x)$ is rational, i.e. $g$ has the same property that $f$.

Notice that $g([0,1])= \left[ \frac{a-q}{p}, \frac{b-q}{p} \right]$. Therefore, if $b-1 \leq q \leq a$ and $p \geq b-q$ then $g : [0,1] \to [0,1]$ and classically $g$ has a fixed point $x_0 \in [0,1]$.

Finally we deduce that $x_0 \in \mathbb{Q}$ iff $x_0 = g(x_0) \notin \mathbb{Q}$, a contradiction.

  • 1
    $\begingroup$ This uses compactness as well as connectedness of $[0,1]$ to conclude that $f([0,1])$ is a closed, bounded interval. $\endgroup$ – Jonas Meyer Jul 23 '13 at 15:10
  • $\begingroup$ Very nice proof. $\endgroup$ – leo Dec 30 '13 at 1:51
  • $\begingroup$ If $f([0,1])=[-1,1]$, then you are suggesting to consider $g(x)=f(x)$ in order to find a function $[0,1] \to [0,1]$, so it doesn't seem to work. $\endgroup$ – Seirios Aug 12 '15 at 5:49
  • $\begingroup$ Yes, you are right. Define $g(x)=\left\lvert\dfrac{f(x)}{\left\lceil \max(\left\lvert a \right\rvert,\left\lvert b\right\rvert)\right\rceil}\right\rvert$. I think that it will work. $\endgroup$ – user 170039 Aug 13 '15 at 13:48
  • $\begingroup$ I think so. But now, your solution does not seem to be really simpler than the original. $\endgroup$ – Seirios Aug 14 '15 at 6:46

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.