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(related, but not a duplicate: curve which crosses itself at every point )

When reading the comments to the question above, it has been pointed out that if by "cross" we mean that for every $\alpha\in [0,1]$ there is $\beta\neq\alpha$ such that $\gamma(\alpha)=\gamma(\beta)$, then there are simple examples of such curves. However, all the examples I can think of have the following property:

There are two disjoint (non-degenerate) intervals $I_1,I_2\subseteq [0,1]$ such that $\gamma(I_1)=\gamma(I_2)$.

(intuitively, this means that the curve goes over some segment twice in the same way or reversed).

My question is, does this always have to happen? To be precise:

Does there exist a continuous curve $\gamma:[0,1]\rightarrow\Bbb R^2$ such that for every $\alpha\in [0,1]$ there is $[0,1]\ni\beta\neq\alpha$ such that $\gamma(\alpha)=\gamma(\beta)$, but there are no two disjoint intervals $I_1,I_2\subseteq [0,1]$ we have that $\gamma(I_1)=\gamma(I_2)$?

I believe the answer is yes, and that this is achieved by some space-filling curve, but I am not sure.

Thanks in advance.

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Wojowu
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  • Well... if you don't care for continuity then I'm sure that a minor modification of the Conway Base 13 function could do the trick. :-) – Asaf Karagila Jun 18 '15 at 18:32
  • @AsafKaragila Note that I have specified that I want a Jordan curve, which by definition is continuous. But your idea is also an interesting one :D – Wojowu Jun 18 '15 at 18:33
  • You're essentially saying you want the curve to really "cross" itself (or become tangent to itself) rather than "lie along" itself for some interval. Surely this can only happen countably many times? – MPW Jun 18 '15 at 18:35
  • @MPW This might be true, but it's not immediately clear to me why so. – Wojowu Jun 18 '15 at 18:39
  • @MPW According to answer to this question http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function this can happen uncountably many times without the curve lying long itself: take a curve to be interval from 0 to 1 connected to the graph of function having Cantor set as its root set. – Wojowu Jun 18 '15 at 18:41
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    I don't think you mean "Jordan curve". By definition a Jordan curve doesn't intersect itself. Presumably you just want a continuous curve? – Jim Belk Jun 18 '15 at 19:01
  • @JimBelk Woops, that sounds about right! I have messed up the definition of Jordan curve in my head. I'll edit right away. – Wojowu Jun 18 '15 at 19:02
  • My example of a curve which takes each one of its values continuum number of times is the projection of a space filling curve. So the image is [0,1] (This doesn't satisfy the disjoint interval property though). Also, I don't think that the usual space filling curves have this property – hot_queen Jun 18 '15 at 19:30
  • @hot_queen What is your example of the curve? – Wojowu Jun 18 '15 at 19:32
  • $t \mapsto x(t)$ where $t \mapsto (x(t), y(t))$ is a continuous surjection from $[0, 1]$ to $[0, 1]^2$. – hot_queen Jun 18 '15 at 19:34
  • @Wojowu What examples did you have in mind? Maybe I can use them. – hot_queen Jun 18 '15 at 19:50
  • @hot_queen The simplest one would be $f(x)=(\cos 2x,\sin 2x)$ (a curve going around a circle twice). A trivial one would be $f(x)=const$. Basically all my examples were just a curve going one way, and then returning the same track, or going same track again. – Wojowu Jun 18 '15 at 19:53

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Edit: As Rahul points out, this curve actually fails to satisfy the given criterion, since the images of $[0,1)$ and $(1,2]$ are the same. It's not clear how this can be fixed.

Original Post: The answer is yes, and the easy example is to concatenate two space filling curves with different local structures. For example, let $h\colon [0,1]\to [0,1]^2$ be a Hilbert curve and $p\colon [1,2]\to[0,1]^2$ a Peano curve such that $h(1) = p(1)$. Then the union $h\cup p\colon [0,2]\to[0,1]^2$ has the desired property.

Note: Technically this curve fails to cross itself at the point $h(1)=p(1)$, but this can be solved by, say, appending a line segment to the end of the curve that goes from $p(2)$ back to $h(1)=p(1)$.

Jim Belk
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  • Really nice idea! I am going to postpone the acceptance of your answer, because I am interested in some examples other people might come up with. – Wojowu Jun 18 '15 at 19:09
  • By the way, do you know of any simple argument on why both Hilbert or Peano curves have the property of not having fragments which "lie along" the curve? It seems intuitively clear, but I'm interested in something a bit more formal. – Wojowu Jun 18 '15 at 19:10
  • Doesn't this still fail to satisfy the criterion? You have $\gamma = h\cup p : [0,2] \to [0,1]^2$, and if you take $I_1=[0,1]$, $I_2=[1,2]$ you have $\gamma(I_1)=\gamma(I_2)$. –  Jun 18 '15 at 19:17
  • @Rahul Good point! I'll leave the answer up since it may be helpful to others. – Jim Belk Jun 18 '15 at 19:25
  • @Rahul It turns out that my restriction is actually stronger than just "one part of a curve lies along the other", because for a space filling curves curve can't move around the filled region. Nice observation. – Wojowu Jun 18 '15 at 19:30
  • Actually, I think this can be fixed by doing the first half of a space-filling curve in reverse, and then a second space-filling curve, and then the second half of the original space filling curve. However, one would need to use two incompatible space-filling curves that have the same begin and end points, and one would need to be careful to check that no two disjoint intervals have the same images. – Jim Belk Jun 18 '15 at 19:36