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Consider a continuous surjection $f:[0,1]\to [0,1]\times[0,1]$. Is $$\{x:\exists(t_1\not=t_2) f(t_1)=f(t_2)=x\}=[0,1]\times[0,1]?$$

Michal
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1 Answers1

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No. Consider for instance the Hilbert space-filling curve, on which Brian Hayes wrote a nice popular article recently (HTML, PDF).

This mapping from $[0, 1]$ to $[0, 1] \times [0, 1]$ can be viewed as a mapping from digit sequences $0.d_1d_2d_3\dots$ in base-$4$ (where $0 \le d_i < 4$) to $[0, 1] \times [0, 1]$. The first digit tells us which quadrant a point lies in, the second digit tells us which sub-quadrant of that it lies in, and so on. (See the second image on page 4.) Conversely, given a point in $[0, 1] \times [0, 1]$, we can invert the curve -- find all $t$ in $[0, 1]$ that are mapped to this point -- by noting down which quadrant it lies in, then which sub-quadrant of that it lies in, and so on, recursively.

The only points that are the image of multiple $t$ are those that, at some granularity, lie in multiple quadrants. These are precisely those points $(a, b)$ such that either $a$ or $b$ can be written as $\dfrac{m}{2^k}$ for some integers $k \ge 1$ and $1 \le m < 2^k$. Every other point is the image of exactly one $t$.

So for the Hilbert curve at least, the set of points at which the curve is "self-intersecting" in your sense is, far from being every point, actually of measure $0$.

ShreevatsaR
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  • I'm not able to rigorously show that the set of points $(a, b)$ with the restriction that "either $a$ or $b$ is a binary rational" is a measure-$0$ set... is this even true? – ShreevatsaR Jun 10 '13 at 12:26
  • Isn't it a countable sum of measure-$0$ set? Thus measure-$0$ set? I mean $\bigcup_{p,k} {(x,\frac{p}{2^k}):x\in [0,1]} \cup {(\frac{p}{2^k},x):x\in [0,1]}$ – Michal Jun 10 '13 at 12:58
  • I think he wants measure $0$ in $[0,1]$ :) – Ted Shifrin Jun 10 '13 at 13:10
  • @Michal: Oh yes, that's right: it's the union of two sets which have measure $0$ in $[0,1]\times[0,1]$. So there. :) – ShreevatsaR Jun 10 '13 at 13:25
  • So I have other 2 questions, I wonder if to put it in questions here 1. Is there a theorem that self-intersection must be dense? Is there any arc of points that are not self-interesections? – Michal Jun 10 '13 at 13:33
  • Ok, self intersection must be dense. I think it would violate Baire theorem if there will be no self-intersections on some open subset. So still it's iteresting for me if that can be an arc of not self-intersection points. – Michal Jun 10 '13 at 13:44
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    @Michal: As an aside, a trivial observation: the set of points of non self-intersection can be made empty. Just define a new function $g$ which goes at double the speed, and then doubles back. In other words, define $g(t) = f(2t)$ if $t \le 1/2$, and $g(t) = f(2-2t)$ otherwise. :) (Or if $f(1/2)$ is non-self-intersecting here, you can go even faster and cover the whole of $[0, 1]\times[0,1]$ more times.) – ShreevatsaR Jun 10 '13 at 13:50
  • @user14111: That's beautiful! Thanks... – ShreevatsaR Jun 10 '13 at 16:09