50
$\begingroup$

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

Result:

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 $f(J) \subseteq \mathbb{Q}$.

http://planetmath.org/encyclopedia/ThereIsNoContinuousFunctionThatSwitchesTheRationalNumbersWithTheIrrationalNumbers.html

$\endgroup$
6
  • 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. $\endgroup$ Aug 4, 2011 at 18:18
  • $\begingroup$ @Ross Millikan: just edited it. $\endgroup$
    – user10
    Aug 4, 2011 at 18:22
  • 1
    $\begingroup$ @user10 actually the link is down. $\endgroup$ Feb 21, 2014 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. $\endgroup$ Jul 7, 2016 at 14:42
  • $\begingroup$ Linked. $\endgroup$ Sep 23, 2017 at 1:05

4 Answers 4

72
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ thank you, beautiful argument. $\endgroup$
    – user10
    Aug 4, 2011 at 18:31
  • 5
    $\begingroup$ More elementary than a proof using Baire Category! $\endgroup$
    – GEdgar
    Aug 4, 2011 at 18:52
  • 5
    $\begingroup$ I like the second paragraph version $\endgroup$ Dec 29, 2013 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 $\endgroup$ Mar 22, 2017 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. $\endgroup$ Mar 23, 2017 at 1:35
21
$\begingroup$

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. $f(x)\in \mathbb{Q}$, but $f(x)\in J$.

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, $f(c)=2c\not\in \mathbb Q$ which, as Jonas pointed, is contradictory to the hypothesis. Therefore such an $f$ can not exist.

$\endgroup$
6
  • 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 $\mathbb R$. $\endgroup$ Aug 4, 2011 at 19:43
  • $\begingroup$ yes, that's true and thank you @Jonas, I will correct for the sake of simplicity. $\endgroup$
    – leo
    Aug 4, 2011 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. $\endgroup$ Aug 4, 2011 at 20:06
  • $\begingroup$ Yes, I had some doubts after I wrote my comment. $\endgroup$
    – leo
    Aug 4, 2011 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, 2016 at 9:24
5
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ This uses compactness as well as connectedness of $[0,1]$ to conclude that $f([0,1])$ is a closed, bounded interval. $\endgroup$ Jul 23, 2013 at 15:10
  • $\begingroup$ Very nice proof. $\endgroup$
    – leo
    Dec 30, 2013 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, 2015 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. $\endgroup$
    – user170039
    Aug 13, 2015 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, 2015 at 6:46
2
$\begingroup$

Suppose such a continuous function exists. $f(\Bbb R)$ is connected, and hence uncountable. Then, $f(\Bbb Q)$ is countable because $\Bbb Q$ is so. Moreover, $f(\Bbb R\setminus\Bbb Q)$ is countable since it is contained in $\Bbb Q$. This is a contradiction.

$\endgroup$
2
  • $\begingroup$ This is just the last paragraph of Jonas Meyer's answer above, rephrased. $\endgroup$ Mar 5, 2021 at 4:07
  • $\begingroup$ Oh, you're right! I just wanted to see an answer that explicitly mentioned connectedness. $\endgroup$ Mar 5, 2021 at 4:19

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.