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

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?

• 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

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.