# A wild embedding of $\mathbb{S}^1$ into $\mathbb{R}^3$

Can one construct an embedding of $$\mathbb{S}^1$$ into $$\mathbb{R}^3$$ so that every orthogonal projection onto a two dimensional plane is a unit disc?

It is easy to construct an embedding of $$\mathbb{S}^1$$ into $$\mathbb{R}^3$$ so that one orthogonal projection is a unit disc: take a Peano-type curve $$\gamma=(\gamma_1,\gamma_2):[0,1]\to\mathbb{R}^2$$ that fills the unit disc and define $$\Gamma(t)=(\gamma_1(t),\gamma_2(t),t):[0,1]\to\mathbb{R}^3$$. Clearly $$\Gamma$$ is one-to-one so it is an embedding and its projection onto the first two coordinates fills the disc. It remains now to "glue" the ends to make it an embedding of $$\mathbb{S}^1$$.

No.

Assume it is possible, that is, there is an embedding $$f\colon\mathbb{S}^1\to \mathbb R^3$$ such that any projection of $$f(\mathbb{S}^1)$$ is a unit disc.

By this answer, the convex hull of the image $$f(\mathbb{S}^1)$$ is a unit ball. Further note that every extreme point lies in the image; that is, $$f(\mathbb{S}^1)\supset\mathbb{S}^2$$ --- a contradiction.

P.S. There are embeddings $$f\colon\mathbb{S}^1\to \mathbb R^3$$ such that any projection of $$f(\mathbb{S}^1)$$ contains a unit disc. Moreover one can assume that the convex hull $$W=\mathop{\rm Conv}f(\mathbb{S}^1)$$ is arbitrary close to the unit ball. (Typically, the set of extreme points of $$W$$ is a Cantor set.) To construct $$f$$, modify a space filling curve by making it injective, but still intersecting all the lines pass thru the unit ball.

• Add the link to this question mathoverflow.net/questions/39127/… – Taras Banakh Dec 4 '18 at 8:08
• @TarasBanakh Thank you for the link. It greatly complements Anton's answer. – Piotr Hajlasz Dec 4 '18 at 12:59
• @PiotrHajlasz You are welcome. – Taras Banakh Dec 4 '18 at 13:25
• This answer is (was) completely impossible to understand if one does not click the link "is a unit ball", I spent 5 minutes puzzled before clicking... I rephrased to emphasize that this is the main point, not just "note that". – YCor Dec 4 '18 at 21:14
• What if we only ask for every projection to contain a disk? i.e. have nonempty interior? – Wojowu Dec 4 '18 at 21:29