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Space filling curves: two and a half questions

  1. Oct 10, 2008 #1
    I've been reading a bit on these, not in a rigorous way yet, and it's an enjoyable read. But now I've a few questions.

    As I understand it, they allow for a continuous index set in [tex]\Re[/tex] to completely cover a higher dimensional [tex]\Re[/tex][tex]^{n}[/tex]. Everywhere continuous, but nowhere differentiable, so it can't be used to represent the codomain in any way. That's kind of what non-holomorphic means, yes? That's sort of disappointing, but it also sort of makes sense.

    There can be many space filling curves for the same space - infinitely many, right? I think there should be uncountably many, but I'm not sure about that, which is question 1. If so, I'd expect it could be possible for at least one such curve to be differentiable somewhere, even if only in a tiny little corner. Is this possible? That's question 2. Is there any loophole in "nowhere differentiable" that might work out as "somewhere, but not elsewhere." If the answer to question 2 is yes, which I'm not expecting, then question 3 is if that tiny section might ever possibly be smooth.

    Thanks much.
  2. jcsd
  3. Oct 10, 2008 #2
    I don't know much about holomorphic stuff but it should be quite obvious that there are infinitely many (uncountably) space filling curves.. For instance, you can just re-parametrize the curve and compose different space filing curves in various ways (one to fill the lower left corner, the other the top left corner... etc). That'll only score you countably many curves though. But you can mix things up so that the curve at 1 is at a different position each time, then take the limit of certain curve compositions, that'll give you uncountably many.

    Of course differentiability is possible since you can just put two curves on top of each other and whatever that differentiable piece doesn't cover, it'll be covered by the other piece. Further, anything curve higher or equal to C1 will fail to cover space, simply due to inverse function theorem.
    Last edited: Oct 10, 2008
  4. Oct 12, 2008 #3
    Thanks for the reply, Tim (or is it Lou?) I agree it is pretty obvious. But it brings up other questions. Setting the curve at 1 means fixing the value of the end point of the indexing interval, if I understand you.The more I think about this "last" point the more uncomfortable I get. It imposes a structure of firstness and lastness on the higher dimensional space that doesn't exist. This can be avoided by making the space filling curve "close upon itself" by having the last point at connect to the first point - something we can do with an equivalence class, right? Seen that way, changing the value at 1 might not give us a different curve at all. We might be on exactly the same curve, only further down along the path.

    I don't understand this. Perhaps I don't know what you mean by "cover." It almost sounds like your saying you can sum over non-differentiable functions to get one that is differentiable, which I don't think is true. So I'm confused. And doesn't the inverse function theorem require continuous differentiability, so it might not apply.

    I learned today that there are terms "nowhere" differentiable, almost "nowhere", and "almost everywhere." That was the thought I was trying to get at.
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