# Is this parametric curve space-filling? Why or why not?

Really, the curve in question is the polar plot $r = cos( K * \theta)$, where $K$ is any irrational number (I use $\pi$), but the transformation to a parametric one on $x$ and $y$ with domain $t$ is an unsurprising one.

It would appear that the curve is confined to the unit circle, and also that it never repeats -- that it is aperiodic.

Given these three things

• The domain of the function is any real number
• The range of the function is confined to a finite space
• The function is aperiodic

Does it mean that the unit circle is completely filled for this parametric function from negative infinity to infinity?

Can we say that for any given point in the unit circle, there is a number for $t$ where the curve intersects it?

I want to say no. I really do. My intuition says so. But why?

Are there any other curves with the three bullet pointed conditions above that can be shown to be more clearly non-space-filling?

What is the appropriate mathematical term for the way thi curve acts on the unit circle?

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When you say "it never overlaps itself/repeats", you should note that it intersects itself at a countably infinite number of points, especially at the origin where it does so at a countably infinite number of times. – Henry Mar 2 '11 at 11:12
@Henry - Ah thanks, I forgot to fix that. I did at my bullet point description but not at the normal text one. That is true, but I would figure that the non-intersecting points are also countably infinitely many. – Justin L. Mar 2 '11 at 18:28
No smooth curve is space filling. It can only fill a set which is both of zero measure (en.wikipedia.org/wiki/Null_set#Lebesgue_measure) and meagre (en.wikipedia.org/wiki/Meagre_set). – George Lowther Mar 2 '11 at 20:40

Consider the intersection of the curve with the ray $\theta = 0$. These are points at a distance $r = \cos(2\pi n K)$ from the origin, for all $n \in \mathbb Z$. Note that there are only countably many of them.
Dense means it gets arbitrarily close to all points, but does not require that the set is countable. In particular, $\mathbb{R}$ is dense in $\mathbb{R}$ – Ross Millikan Mar 2 '11 at 14:54