# Image of Hilbert's Space filling Curve

I am trying to find the image of $$[0,1/2)$$ in Hilbert's Space Filling Curve.

What I thought is $$[0,1/2]\times$$ $$[0,1]$$ \ $$[1/4,1/2]\times[1/2,3/4]$$. Is my understanding correct?

The sets considered are as usual. From $$[0,1]$$ to $$[0,1]^2$$.

• Why isn't it $[0,1/2]\times[0,1]\setminus\{(1/2,1/2)\}?$ – saulspatz Jul 14 at 3:03

What I thought is ...

From the definition of Hilbert Curve;
for each parameter $$t \in I := [0,1]$$ a sequence of nested intervals $$I \supset [a_1, b_1] \supset \dots \supset [a_n, b_n] \supset \dots$$
exists, such that each interval is obtained by splitting its predecessor into four parts of equal size.
Any such sequence of intervals can be mapped one by one to a sequence of nested 2D interval. These nested intervals will converge to a uniquely defined point $$h(t) \in Q:= [0,1]\times[0,1]$$.

The point to point mapping is determined by the iteration of the curve.

• Iteration 0 :

$$[0,1] \longmapsto [0,1]\times[0,1]$$

• Iteration 1 :

$$[0,\frac{1}{4}] \longmapsto [0,\frac{1}{2}]\times[0,\frac{1}{2}]\\ [\frac{1}{4},\frac{1}{2}] \longmapsto [0,\frac{1}{2}]\times[\frac{1}{2},1]\\ [\frac{1}{2},\frac{3}{4}] \longmapsto [\frac{1}{2},1]\times[0,\frac{1}{2}] \\ [\frac{3}{4},1] \longmapsto [\frac{1}{2},1]\times[\frac{1}{2},1] \\$$

and so on. The following image shows the said iterations of the hilbert curve (src Why does the Hilbert curve fill the whole square?)