Two questions about the extent to which simple arcs and simple closed curves can fill up higher dimensional Euclidean spaces

Question

For each positive integer n, let E(n) be n-dimensional Euclidean space with its standard metric and let p(n) be some fixed point of E(n). The so-called "Osgood Curve" shows that there can exist simple arcs and simple closed curves in E(2) which have positive 2-dimensional Lebesgue measure.

Question (1): Do there exist-for arbitrarily large n-simple arcs or simple closed curves in E(n) which have positive n-dimensional Lebesgue measure?

Question (2): Is it true-for arbitrarily large n-that there exist simple arcs or simple closed curves in E(n) which do not contain p(n) but which intersect each ray in E(n) originating from p(n) in at least one point? This is clearly true when n=2.

Answer 1

Question 1: Yes. Such curves are called Osgood curves. I couldn't find a construction of an Osgood curve in higher dimensions, but some of the constructions of Osgood curves in the plane generalize. For example, Riesz suggested showing that any totally disconnected set is a subset of a simple curve (see the Denjoy-Riesz theorem), and this is easy for a Cartesian product of symmetric Cantor sets, particularly for $n \gt 2$. In $n$ dimensions, you can cut the product into $2^n$ half-sized pieces, order them so that the most negative piece is first and the most positive piece is last, and connect the most positive corner of the $i$th piece with the most negative corner of the $(i+1)$st piece by disjoint curves. Fill in the $2^n$ intervals recursively, extending by continuity. You can take the product of Cantor sets to have positive measure, which forces the curve through them to have positive measure.

Question 2: Yes. Take a non-simple space-filling curve in one lower dimension $f:\mathbb{R} \twoheadrightarrow \mathbb{R}^{n-1}$ and follow it with a surjection $g:\mathbb{R}^{n-1} \twoheadrightarrow S^{n-1}$ where $S^{n-1}$ is the unit $(n-1)$-sphere in $\mathbb{R}^n$. For example, $g$ could be the exponential map on the tangent space on one point. The composition $g \circ f$ is a surjection $\mathbb{R} \twoheadrightarrow S^{n-1}$. Then the image of $t \mapsto e^t g(f(t))$ intersects every ray through the origin, and scaling by $e^t$ makes the curve simple.