7
$\begingroup$

I know that there are space-filling curves from $[0, 1]$ to the unit square, and this question addresses curves transforming the real line into the entire plane. But what about transforming the unit interval into the whole plane?

And if such curves exist, can their construction somehow be depicted, perhaps by showing a few examples of a sequence of curve that converges to the space-filling curve (as is often done for curves mapping the unit interval to the unit square, see below)?

Construction of Hilbert Curve

Image from http://en.wikipedia.org/wiki/Space-filling_curve

$\endgroup$
  • 4
    $\begingroup$ A curve $\gamma$ is continuous. $[0,1]$ is compact. Hence $\gamma([0,1])$ is ...? $\endgroup$ – Daniel Fischer Oct 9 '14 at 14:18
  • $\begingroup$ Going further with what Daniel Fischer said, seeing as there are continuous transformations from both $(0, 1)$ and $[0, 1)$ onto the real line, those can be used as a parametrisation to fill the plane. $\endgroup$ – Arthur Oct 9 '14 at 14:21
5
$\begingroup$

Update: Here is a neater construction!

enter image description here

As Daniel Fischer has pointed out this cannot be done with $[0,1]$ as parameter interval. Therefore we shall map the half-open interval $[0,1[\ $ continuously onto ${\mathbb R}^2$. Use the redraw rule shown in the above figure to create a Peano curve $$\gamma_*: \quad\bigl[0,1]\to R:=[0,\sqrt{3}]\times[0,1]\ .$$ (Sketch of proof: Let $t\mapsto\phi_0(t)\in R$ be the parametrization of the diagonal in the lefthand figure above. The redraw rule, applied recursively in nested triadic subintervals of $[0,1]$, produces a sequence $(\phi_n)_{n\geq0}$ of piecewise linear maps $\phi_n:\>[0,1]\to R$. It is easy to check that $$|\phi_{n+1}(t)-\phi_n(t)|\leq 3^{-n/2}\qquad(0\leq t\leq1)\ ,$$ and this implies that the $\phi_n$ converge uniformly to a continuous map $\gamma_*:\>[0,1]\to R$. Since $\gamma_*(I)$ is closed and dense in $R$ it has to be all of $R$.)

Stretch this $\gamma_*$ horizontally by a factor $\sqrt{2\over3}$, so that we now have a Peano curve $$\gamma_0:\quad\bigl[0,1]\to[0,\sqrt{2}]\times[0,1]\ ,$$ beginning at $(0,0)$ and ending at $(\sqrt{2},1)$. This $\gamma_0$ is now concatenated with a copy of itself, then with ever larger copies scaled by factors $2^{n/2}$, in a spiraling way, see the following figure which shows only the "diagonal" of each copy:

enter image description here

For the concatenation we allot the time interval $\bigl[0,{1\over2}\bigr]$ for $\gamma_0$, then the subsequent time intervals $$\bigl[1-2^{-n},1-2^{-(n+1)}\bigr]\quad(n\geq1)$$ for the subsequent rectangle filling curves.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Cool. I don't recall what it takes to prove that the limit of such a sequence of functions exists and is surjective but this seems intuitively to do it. $\endgroup$ – kuzzooroo Oct 10 '14 at 14:01
  • $\begingroup$ @kuzzooroo: See my edit, with sketch of proof. $\endgroup$ – Christian Blatter Oct 10 '14 at 14:49
1
$\begingroup$

We can also use Hilbert type curves to construct a line which fills into the entire plane. I propose a kind of '''Hilbert spiral'''. The basic idea is quite simple - besides the standard downward recursive construction of Hilbert-type curves, we add an upward process to make a spiral.

expanding Hilbert curve to the the entire plane

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.