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I have been reading several articles on space-filling curves recently. Most of the articles do mention(and it is obvious from the images) that the map $H:[0,1]\rightarrow [0,1]\times [0,1]$ passes through each point of the unit square. How does one show this formally?


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Which map $H$ ? – lhf Sep 26 '11 at 14:52
@Kuku: If it doesn't pass through every point then it's not a space filling curve, is it? Generically, what we usually do is show that the image is dense in $[0, 1] \times [0, 1]$; but $[0, 1]$ is compact and $[0, 1] \times [0, 1]$ is Hausdorff, so the image is also closed, and hence, must be everything. – Zhen Lin Sep 26 '11 at 15:07
I guess, you meant that trajectory of any surjection $H:[0,1]\to [0,1]\times [0,1]$ passes through each point of the unit square. – Ilya Sep 26 '11 at 15:42
@ZhenLin: I know what compact and dense mean, but I'm not familiar with the term Hausdorff. I'd appreciate it if you could elaborate. Thanks. – Kuku Sep 26 '11 at 16:04
@Kuku: A Hausdorff topological space is one which satisfies a certain separation axiom: if you have two distinct points, then there are disjoint open neighbourhoods about each point. In a Hausdorff topological space, compact sets are closed. – Zhen Lin Sep 26 '11 at 16:20
up vote 3 down vote accepted

There are many such constructions. Here is a sketch of one of them:

${\bf Lemma.}$ If $$\phi:\quad [t_0,t_1]\to[a,a+h]\times[b,b+h], \quad t\mapsto \Bigl(a+{t-t_0\over t_1-t_0}h,b+{t-t_0\over t_1-t_0}h\Bigr)\qquad(t_0\leq t\leq t_1)$$ is a parametric representation of the diagonal of the square $Q:=[a_0,a_0+h]\times[b_0,b_0+h]$ then one can find a piecewise linear function $$\phi':\quad [t_0,t_1]\to[a,a+h]\times[b,b+h]\qquad(t_0\leq t\leq t_1)$$ such that $\phi'(t_0)=(a,b)$ and $\phi'(t_1)=(a+h,b+h)$ and such that $\phi'$ traces nine diagonals of the nine equal squares $Q'$ partitioning $Q$ in turn. (Replace $h$ by $-h$ where necessary.) It follows that $$|\phi'(t)-\phi(t)|\leq \sqrt{2}|h|\qquad(t_0\leq t\leq t_1)\ .$$ Starting with $$\phi_0:\quad[0,1]\to Q_0:=[0,1]^2,\quad \quad t \mapsto (t,t)\qquad(0\leq t\leq 1)$$ one constructs a sequence of maps $\phi_n:\quad[0,1]\to Q_0$ as follows: To obtain $\phi_{n+1}$ from $\phi_n$, each diagonal segment of $\phi_n$ is replaced by nine diagonal segments in the manner described above.

Using the estimate $(*)$ one proves that the $\phi_n$ converge uniformly to a limit function $\phi:[0,1]\to Q_0$. As each $\phi_n$ visits all points of the form $(j/3^n,k/3^n)$ in $Q_0$, it is easy to see that $\phi\bigl([0,1]\bigr)$ is dense in $Q_0$, and as $\phi\bigl([0,1]\bigr)$ is compact it has to be all of $Q_0$.

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