8
$\begingroup$

Let $f\colon \mathbb{R}^n \to \mathbb{R}^{n-1}$ be a continuous function and $K\subset \mathbb{R}^n$ a subset of positive Lebesgue measure. Is it possible that $f$ is one-to-one on $K$?

If $K$ contains a (nonempty) open set this is impossible because of the invariance of domain theorem. But can we say anything for arbitrary (measurable) sets?

$\endgroup$
  • $\begingroup$ This smells like Baire's category theorem and absolute continuity to me, but that's just a first impression $\endgroup$ – Ben Grossmann Nov 16 '15 at 14:33
7
$\begingroup$

Let $n = 2$, $C \subseteq \mathbf R$ a fat cantor set, $f \colon C \times C \to C$ a homeomorphism. As $C \times C \subseteq \mathbf R^2$ is closed, there is a continuous extension $F \colon \mathbf R^2 \to [0,1]$ by the Tietze extension theorem. Now $\lambda(C \times C) > 0$ and $F|_{C \times C} = f$ is one-to-one.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Maybe it's worth mentioning that every two Cantor spaces are homeomorphic, with some reference. $\endgroup$ – Silvia Ghinassi Nov 16 '15 at 14:49
  • $\begingroup$ Ok, so this reduces the problem to finding a homeomorphism from $C\times C \to C$. Is this well known? I think the reference to homeomorphic cantor spaces won't suffice for me, as I don't know why $C\times C$ is a Cantor space... $\endgroup$ – KoliG Nov 16 '15 at 14:58
  • 2
    $\begingroup$ $C$ is known to be homeomorphic to $2^{\mathbf N}$ (the countable infinite product of the space $2 = \{0,1\}$ [with the discrete topology] with itself), hence $C^2 \cong 2^{\mathbf N} \times 2^{\mathbf N}$, but this is homeomorphic to $2^{\mathbf N}$ by mapping the first factor to the even, the second to the odd coordinates. $\endgroup$ – martini Nov 16 '15 at 15:02
  • $\begingroup$ Thanks! So continuous is obviously not enough to keep dimension with a one-to-one mapping. I will ask a new question for more regular mappings. There's just one typo in your answer, n=2 ;) $\endgroup$ – KoliG Nov 16 '15 at 15:06

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.