Does there exists an injective function from $\mathbb R^2$ to $\mathbb R$ such that image of every open set is open ? Thank you.


closed as off-topic by user26857, Watson, Daniel W. Farlow, choco_addicted, Chill2Macht Jun 13 '16 at 14:58

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  • $\begingroup$ The image of an open set, or of any open set? $\endgroup$ – Alex M. Jun 13 '16 at 14:51

There does not exist such a function even if we do not assume continuity.

$f(\mathbb R^2)\subset \mathbb R$ must be an open set, and we have a bijective continuous function $f^{-1}:f(\mathbb R^2)\to \mathbb R^2$. This map is a homeomorphism onto its image when restricted to each closed subinterval of $f(\mathbb R^2)$, and a subspace of $\mathbb R^2$ homeomorphic to a closed interval in $\mathbb R$ has empty interior. Since $f(\mathbb R^2)$ is a countable union of such closed intervals, so is $\mathbb R^2$. But this contradicts the Baire category theorem.

  • $\begingroup$ This is false that a topological arc in the plane has zero 2-dimensional Lebesgue measure, see "Osgood curves", mathoverflow.net/questions/228819/…. But you can use a Baire category argument. $\endgroup$ – Moishe Kohan May 10 '16 at 18:32
  • $\begingroup$ @studiosus fixed, thanks. $\endgroup$ – Matt Samuel May 10 '16 at 20:50

I assume by map you mean 'continuous function.' If so, there does not exist such a function.

Assume for contradiction such a map $f:\mathbb{R}^2 \to \mathbb{R}$ exists. Then its image must be a connected set in $\mathbb{R}$. Now, because $f$ is injective by hypothesis, the pre-image of any point in the image of $f$ is a unique point in the domain. But if I remove a point in the image, the image ceases to be connected; however $\mathbb{R}^2$ less one point remains connected.

In that spirit, let $a\in f(\mathbb{R}^2)$. Then using $f$ I can define the continuous injective map $g: \mathbb{R}^2 \setminus f^{-1}(a) \to \mathbb{R}$ via $(x,y) \mapsto f(x,y)$. By construction, $g$ is continuous and injective since $f$ is. But $g$ cannot exist: it continuously maps a connected set into a disconnected one. Contradiction.

  • $\begingroup$ A much more fun question! In that case, Matt's answer looks like what you're after. $\endgroup$ – Pete Caradonna May 10 '16 at 9:46

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