Vtyjkyuk

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

Result:

Let $J=mathbbR setminus mathbbQ$ denote the set of irrational numbers. There is no continuous map $f: mathbbR rightarrow mathbbR$ such that $f(mathbbQ) subseteq J$ and $f(J) subseteq mathbbQ$.

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

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

If $f(mathbbQ)subseteq mathbb Rsetminusmathbb Q$ and $f(mathbb Rsetminus mathbb Q)subseteqmathbb Q$, 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 $left[minf(0),f(sqrt 2),maxf(0),f(sqrt 2)right]$. 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 $mathbb Qcup f(mathbb Q)$, whereas the intermediate value theorem and uncountability of $mathbb R$ imply that a nonconstant continuous function $f:mathbb Rtomathbb R$ has uncountable range.

Suppose there is such a mapping $f$. Consider $g:[0,1]to mathbbR$ defined by $$g(x)=f(x)-x.$$ Suppose that $g(x)in mathbbQ$ for some $xin [0,1]$. Then:

• if $xin J$, then $g(x)-f(x)in mathbbQ$, i.e. $xin mathbbQ$.


• if $xin mathbbQ$, then $g(x)+xin mathbbQ$, i.e. $f(x)in
  mathbbQ$, i.e. $xin J$

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 $zin[0,1]$ 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 $cin J$. Particularly, $f(c)=2c$ which, as Jonas pointed, is contradictory to the hypothesis. Therefore $f$ can not exist.

Another simple proof:

Because $f$ is continuous and by connectedness, $f([0,1])=[a,b]$ for some $a<b$. Now define $$g : x mapsto frac1p left(f(x)-q right), textwhere p,q in mathbbQ.$$

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

Notice that $g([0,1])= left[ fraca-qp, fracb-qp right]$. 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 $x_0 in mathbbQ$ iff $x_0 = g(x_0) notin mathbbQ$, a contradiction.