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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?

flag	asked Aug 25 at 11:37 domotorp 4,068●7●25

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It is equivalent to ask for a curve in $\Bbb R^3$ that intersects every vector line. – Daniele Zuddas Aug 25 at 13:03

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? – domotorp Aug 25 at 13:18

Hm, even this works in any dimension, oh well... – domotorp Aug 25 at 13:26

Anton Petrunin · Accepted Answer · 2012-08-25 13:17:01Z UTC

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.

Can a simple curve intersect every subspace of dim 2 and avoid the origin?

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