Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Infinite curve with two endpoints?

  1. Jul 8, 2010 #1
    I'm wondering if such a thing can exist. It seems a non intersecting smooth or Peano type curve can be defined over a finite surface area that passes once through (or terminates at) every point on the surface, including its two end points.

    EDIT: I don't think this is the same question as the number of points on a line segment. I asking about a curve of infinite length.
     
    Last edited: Jul 8, 2010
  2. jcsd
  3. Jul 9, 2010 #2
    If you mean a curve with infinite length, yes. Consider fractals.
     
  4. Jul 9, 2010 #3
    Yes, such as the Peano curve I mentioned. Is it true for any smooth non intersecting curve as well? I'm thinking that an everywhere smooth non intersecting curve cannot completely pave a plane, but could still be infinite with two endpoints as weird as that sounds.

    EDIT: I partially rescind that. A smooth non intersecting curve can pave a plane with smooth boundaries, but not other kinds of boundaries which contain singular points. I'm thinking an infinite non intersecting smooth curve must pass through (or terminate at) every point in a plane once and only once.
     
    Last edited: Jul 9, 2010
  5. Jul 13, 2010 #4
    I wonder if we could keep the Peano curve smooth throughout its construction ; so that the sequence of smooth curves is smooth in the limit. We have a curve which is smooth almost everywhere already.
     
  6. Jul 13, 2010 #5
    I don't think so, but I'm not that deep into topology. Doesn't the Peano curve have singular points by definition such that line segments are not eliminated at the limit? That is, the Peano curve cannot be smoothed?
     
    Last edited: Jul 13, 2010
  7. Jul 14, 2010 #6
    Suppose that we 'pare off' the corners at each stage of the construction of the Peano curve , so that the resulting curve is sufficiently smooth. The sequence of such curves ought to converge to a differentiable,space-filling function (under some restrictions, of course). Its image won't be a square, but a compact subset of the square.
     
  8. Jul 14, 2010 #7
    I'm not sure what a Peano curve is, but I know a simple case where you can have this to be true. If you take a toroidal curve with an irrational rotational transform (the ratio of poloidal period covered in one full toroidal period), then as the number of transits goes to infinity, this curve will fill an entire toroidal surface, and it will never intersect its tail.
     
  9. Oct 4, 2011 #8

    lavinia

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    This seems wrong. A line on the torus can only intersect a circle in countably many points.

    If the torus is viewed as a square with opposite edges identified then a line drawn at an irrational angle to its lower edge can only intersect that edge in countably many points. Otherwise the line would be uncountably long i.e. could be divided into uncountably many disjoint intervals of equal length.
     
    Last edited: Oct 4, 2011
  10. Oct 4, 2011 #9

    lavinia

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The sample paths of a continuous Brownian motion are almost surely continuous and have infinite variation on any interval.
     
  11. Oct 4, 2011 #10

    lavinia

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I do not think a space filling curve can be 1 to 1. Not sure of the proof.
     
  12. Oct 4, 2011 #11
    A 1-1 space-filling curve would be a continuous bijection between compact and Hausdorff...
     
  13. Oct 5, 2011 #12

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    A smooth curve has finite length -- the integrand defining its length is a continuous function of the parameter, and thus bounded.

    Any subset of the plane of finite length cannot have positive area; its image cannot contain any non-empty open set.
     
  14. Oct 5, 2011 #13
    Yes, what I meant is that if you had a continuous bijection into the image, you would get an embedding which would send open sets in R into relatively-open subsets of the square, violating invariance of domain.
     
  15. Oct 5, 2011 #14

    lavinia

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I don't see why this is true. Can you explain?
     
  16. Oct 5, 2011 #15
    Well, the general result is that a continuous bijection between a compact space and a Hausdorff space is a homeomorphism. Then, if the unit interval I could be mapped bijectively (and, of course, continuously), into the unit square, that would mean that there was an embedding of I into the unit square, so that open subsets of the unit square are mapped to relatively open subsets of the unit square.
     
  17. Oct 5, 2011 #16

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is the key part of your argument, I think. You're being somewhat misleading by emphasizing every detail except this one.
     
  18. Oct 5, 2011 #17
    How so? I don't get your point; this is a standard result from point set topology.

    The idea is that a closed subset of a compact space is Hausdorff , and that a continuous

    function sends compact to compact. Then a compact subset of Hausdorff is closed, so that the

    inverse function takes closed sets to closed sets , and so the function is continuous with a continuous inverse.
     
    Last edited: Oct 5, 2011
  19. Oct 5, 2011 #18

    lavinia

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Right if the mapping is of a closed interval. If the interval is open this doesn't work.
     
  20. Oct 5, 2011 #19
    Right, but isn't the Peano curve an image of the closed interval?
     
  21. Oct 5, 2011 #20
    I think we can discount the case of f being smooth by using Sard's theorem:

    The Jacobian J of a map from R1 into R2 is a 1x2-matrix, so that every single value in the domain is a critical value of J. Then the set f(I) (whether I is--I am?-- open or closed) has measure zero in R2. Tho I don't know if asking for smoothness is overkill.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook