# Do there exist space-filling curves that fill the whole plane? If so, can they be visualized?

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)?

• A curve $\gamma$ is continuous. $[0,1]$ is compact. Hence $\gamma([0,1])$ is ...? – Daniel Fischer Oct 9 '14 at 14:18
• 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. – Arthur Oct 9 '14 at 14:21

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:
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.