Cfrgtkky

I was working with an exercise of general topology and I had a question: are there an homeomorphism $f:mathbb{R}to(0,1)$ such that $f(x)inmathbb{Q}$ if and only if $xinmathbb{Q}$?, i.e., the homeomorphism maps the rationals to rationals and therefore the irrationals to irrationals.

My intuition says that the answer is yes but I can't find an example. The closer example was $g:mathbb{R}to(0,1)$ defined by $g(x)=dfrac{1}{1+2^{-x}}$. But I think that doesn't works.

$f(x)=(frac x {1+|x|}+1)/2$ is such a map. Its inverse is $y to frac {2y-1} {1-|2y-1|}$.

Yes. By a well-known result, any two countable totally-ordered sets
with no largest and smallest elements are order-isomorphic. Therefore
there is an order-isomorphism $F:Bbb QtoBbb Qcap(0,1)$. Now $Bbb R$
and $(0,1)$ can be constructed by Dedekind cuts on $Bbb Q$ and
$Bbb Qcap(0,1)$ respectively. Thus $F$ extends to an order-isomorphism
$f:Bbb Rto(0,1)$. As both $Bbb R$ and $(0,1)$ have the topologies
induced by their ordering, $f$ is a homeomorphism.

You can construct an example by hand, choosing a rational at each integer (say) and imposing piecewise linearity.

For example, let $fcolonmathbb{R}to(0,1)$ be
$$
f(x)=
begin{cases}
1-dfrac1{2(x+1)} & text{if }xinmathbb{N}\
dfrac1{2(-1-x)} & text{if }-xinmathbb{N}\
frac12 & text{if }n=0\
text{linear} & text{otherwise}.
end{cases}
$$