I can see space filling curves through I^I, but I can't see them covering R^I.
How does a space-filling curve cover a plane without being filled with infinitesimal discontinuities?
(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.
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".
It is true that there is no continuous bijection R -> R^2. It is also true that this has nothing to with cardinality, which is only about the existence of bijections (and indeed is defined for arbitrary sets with no topological structure for which continuity makes no sense)
> Represent each real number x∈[0,1) as a sequence of DIGITS, where each DIGIT is either in {0,…,8} or is of the form 10∗(10k−1)+i with k≥1 and i∈{0,…,8} (i.e., in {90,…,98;990,…,998;9990,…,9998;…}.
He's iterating to infinity, that sure as shit sounds countable to me.
I think his iterator is generating infinitesimals of rational and not continuous nature.
If there are holes in the map does it really cover the whole space?
As a simpler example: The map that multiplies each rational number by -1 and keeps irrational numbers as they are is bijective (it’s its own inverse), however it is nowhere continuous.
Aleph_0^2 = Aleph_0
Aleph_1^2 = Aleph_1