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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?

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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 '12 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 '12 at 13:18
    
Hm, even this works in any dimension, oh well... –  domotorp Aug 25 '12 at 13:26
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1 Answer

up vote 8 down vote accepted

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.

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I see, and this of course works for every dimension. Thank you! –  domotorp Aug 25 '12 at 13:25
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