4 ms·
Yes, I've got that part - but that only gets me as far as R * I ^ (d - 1), not R ^ d. (d being the dimensionality of the curve.) As I think about this there's
by yarg 3y ago
Yes, I've got that part - but that only gets me as far as R * I ^ (d - 1), not R ^ d.
(d being the dimensionality of the curve.)
As I think about this there's an unfilled infinitesimal R ^ (d - 1) space orthogonal to every point on the curve.
I don't see how it is that that does not manifest as discontinuities in the filled space.
- Skeime 3y agoYou can get from R^d to I^d (with I the open unit interval) easily by applying x -> (arctan(x) + pi/2)/pi and y -> tan(pi * y - pi/2) to each coordinate. There are also ways to go from R to the closed unit interval. You can then combine this with space-filling curves for I^d.
- yarg 3y agoYou can even go from R to I^d. The problem that I'm having is in getting from R to R^d, which I still don't think is possible - look at the comment where I mention "surjections" and "bijections".
- Skeime 3y agoBy applying the map from I^d to R^d (using the tan-arctan-construction above in each coordinate) after the map from R to R^d, you get a map from R to R^d.