Space-filling curve

By | October 8, 2017

Theorem. There exists a continuous curve in \mathbb{R}^{2} that passes through every point of the unit square \left[0,1\right]\times\left[0,1\right].

Proof. Define \phi:\left[0,2\right]\rightarrow\mathbb{R} by

    \[ \phi\left(t\right)=\begin{cases} 0 & \text{if }0\le t\le\frac{1}{3},\quad\text{or if }\frac{5}{3}\le t\le2,\\ 3t-1 & \text{if }\frac{1}{3}\le t\le\frac{2}{3}\\ 1 & \text{if }\frac{2}{3}\le t\le\frac{4}{3}\\ -3t+5 & \text{if }\frac{4}{3}\le t\le\frac{5}{3}. \end{cases} \]

Extend \phi to all of \mathbb{R} by making \phi periodic with period 2.


    \[ f_{1}\left(t\right)=\sum_{n=1}^{\infty}\frac{\phi\left(3^{2n-2}t\right)}{2^{n}},\quad f_{2}\left(t\right)=\sum_{n=1}^{\infty}\frac{\phi\left(3^{2n-1}t\right)}{2^{n}}. \]

By the Weierstrass M-test, both f_{1} and f_{2} converges uniformly on \mathbb{R}. Moreover, it is continuous on \mathbb{R}. Now define f=\left(f_{1},f_{2}\right) and let \Gamma denote the image of the unit interval \left[0,1\right] under f. We show \Gamma=\left[0,1\right]\times\left[0,1\right].

Observe 0\le f_{1}\left(t\right)\le1 and 0\le f_{2}\left(t\right)\le1. Hence \Gamma\subset\left[0,1\right]\times\left[0,1\right]. Let \left(a,b\right)\in\left[0,1\right]\times\left[0,1\right]. Write

    \[ a=\sum_{n=1}^{\infty}\frac{a_{n}}{2^{n}},\quad b=\sum_{n=1}^{\infty}\frac{b_{n}}{2^{n}} \]

with a_{n},b_{n}\in\left\{ 0,1\right\}. Now let

    \[ c=2\sum_{n=1}^{\infty}\frac{c_{n}}{3^{n}},\quad\text{where }c_{2n-1}=a_{n},\quad c_{2n}=b_{n}. \]

Then 0\le c\le1 since 2\sum_{n=1}^{\infty}\frac{1}{3^{n}}=1. Now we show f_{1}\left(c\right)=a and f_{2}\left(c\right)=b. We show \phi\left(3^{k}c\right)=c_{k+1} for each k=0,1,2,… If we can show this, then we have \phi\left(3^{2n-2}c\right)=c_{2n-1}, \phi\left(3^{2n-1}c\right)=c_{2n}=b_{n} and this gives f_{1}\left(c\right)=a, f_{2}\left(c\right)=b.

Now write

    \begin{align*} 3^{k}c & =2\sum_{n=1}^{\infty}\frac{c_{n}}{3^{n-k}}+2\sum_{n=k+1}^{\infty}\frac{c_{n}}{3^{n-k}}\\ & =\text{even integer}+d_{k}, \end{align*}


    \[ d_{k}=2\sum_{n=1}^{\infty}\frac{c_{n+k}}{3^{n}}. \]

Since \phi has period 2, we have

    \[ \phi\left(3^{k}c\right)=\phi\left(d_{k}\right). \]

If c_{k+1}=0, then 0\le d_{k}\le2\sum_{n=2}^{\infty}3^{-n}=\frac{1}{3} and hence \phi\left(d_{k}\right)=0. So \phi\left(3^{k}c\right)=c_{k+1} in this case. If c_{k+1}=1, then \frac{2}{3}\le d_{k}\le1 and hence \phi\left(d_{k}\right)=1. Therefore, \phi\left(3^{k}c\right)=c_{k+1}. This proves f_{1}\left(c\right)=a, f_{2}\left(c\right)=b. So \Gamma=\left[0,1\right]\times\left[0,1\right].

Meaning of this theorem. Observe that f:\mathbb{R}\rightarrow\left[0,1\right]\times\left[0,1\right] is continuous. \mathbb{R} has dimension 1 and \left[0,1\right]\times\left[0,1\right] has dimension 2. By the above theorem, the continuity does not guarantee the dimension of spaces. The above curve we constructed is nowhere differentiable. This is proved by Alsina.

It seems that this kind of curve is not useful, but it has quite a lot of applications. One can use this kind of fact to probability theory, topology, etc. It has an application to industry, e.g. Google map.


  1.  J. Alsina,  The Peano curve of Schoenberg is nowhere differentiable, Journal of Approximation theory, Vol. 33 (1), 28– 42.
  2. T. Apostol, Mathematical Analysis
  3. C. S. Perone, Google’s S2, geometry on the sphere, cells and Hilbert curve 

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