MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
    
It could be the origin or any other point of E(n) which is fixed in advance. – Garabed Gulbenkian Jan 19 at 21:09
up vote 1 down vote accepted

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.

share|cite|improve this answer
    
Many thanks for these very complete responses. I could never have found the proofs on my own. – Garabed Gulbenkian Jan 20 at 21:00
    
I must apologize. I just discovered that my question (1) is a duplicate of MATHOVERFLOW.NET question No. 129867 which I asked nearly 3 years ago and which I had completely forgotten about. I received an affirmative answer at the time, but this time I am getting an added bonus-a demonstration of how to construct the arcs that have the property I was looking for. – Garabed Gulbenkian Jan 24 at 16:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.