# No continuous function switches $\mathbb{Q}$ and the irrationals

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

• 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. – Ross Millikan Aug 4 '11 at 18:18
• @Ross Millikan: just edited it. – user10 Aug 4 '11 at 18:22
• @user10 actually the link is down. – Gabriel Romon Feb 21 '14 at 23:56
• 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. – Jonas Meyer Jul 7 '16 at 14:42
• – 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.

• thank you, beautiful argument. – user10 Aug 4 '11 at 18:31
• More elementary than a proof using Baire Category! – GEdgar Aug 4 '11 at 18:52
• I like the second paragraph version – Hagen von Eitzen Dec 29 '13 at 23:13
• @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 – spaceman_spiff Mar 22 '17 at 9:23
• @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. – 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.

• 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 $\mathbb R$. – Jonas Meyer Aug 4 '11 at 19:43
• yes, that's true and thank you @Jonas, I will correct for the sake of simplicity. – leo Aug 4 '11 at 19:48
• 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. – Jonas Meyer Aug 4 '11 at 20:06
• Yes, I had some doubts after I wrote my comment. – leo Aug 4 '11 at 20:39
• Hey Leo, how do you conclude that $g([0,1]) \subset$\mathbb{I}$from the two contradictions above? – Kamil Jul 7 '16 at 9:24 Another simple proof: Because$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. • This uses compactness as well as connectedness of$[0,1]$to conclude that$f([0,1])$is a closed, bounded interval. – Jonas Meyer Jul 23 '13 at 15:10 • Very nice proof. – leo Dec 30 '13 at 1:51 • 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. – Seirios Aug 12 '15 at 5:49 • 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. – user 170039 Aug 13 '15 at 13:48
• I think so. But now, your solution does not seem to be really simpler than the original. – Seirios Aug 14 '15 at 6:46