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

Question

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 $\mathbb{S}^1$.

Answer 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.