Is there, e.g. in $\mathbb R^4$ a simple curve that does not contain the origin and intersects every subspace of dimension 2?

Sorry if the question is too easy, but I just cannot figure it out. In three dimensions such a curve exists, but I cannot imagine four dimensions. Is it possible to somehow lift-up the Peano-curve? What about higher dimensions and higher dimensional subspaces?

| cite | improve this question | | | | |
  • 1
    $\begingroup$ It is equivalent to ask for a curve in $\Bbb R^3$ that intersects every vector line. $\endgroup$ – Daniele Zuddas Aug 25 '12 at 13:03
  • $\begingroup$ A Peano-curve type construction does intersect every line through the origin, so I suppose then the answer to my original question is yes. But what about five dimensions then? $\endgroup$ – domotorp Aug 25 '12 at 13:18
  • $\begingroup$ Hm, even this works in any dimension, oh well... $\endgroup$ – domotorp Aug 25 '12 at 13:26

Consider closed space filling curve $\theta:\mathbb S^1\to\mathbb S^3$. You can choose $\theta$ so that $\theta^{-1}(x)$ is finite for any $x\in\mathbb S^3$ and $|\theta^{-1}(x)|=1$ for all but countable set in $\mathbb S^3$.

Then it is easy to find a function $\rho:\mathbb{S}^1\to\mathbb R_+$ so that the curve $\gamma(x)=(\theta(x),\rho(x)$ in polar coordinates is the curve you are looking for.

| cite | improve this answer | | | | |
  • $\begingroup$ I see, and this of course works for every dimension. Thank you! $\endgroup$ – domotorp Aug 25 '12 at 13:25

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.