# Proof of fractal dimension of Thomae's function

Thomae's function is defined to be $0$ if x is irrational. Its defined to be $1 \over q$ where $x={p \over q}$ in lowest terms and $q \gt 0$. Its measure is $0$ since the set of rational numbers is countable. However, it seems that this function might be a fractal and/or have fractal dimension. If it does, how does one find this fractal dimension? A proof or reference would be appreciated.

• The measure of what is zero? Are you asking about the fractal dimension of the graph of the function? – user225318 Mar 31 '15 at 18:31
• @user225318 That was assumed, input an x, get out a y. And yes, I want to find the fractal dimension of the function. – Zach466920 Mar 31 '15 at 19:09

## 1 Answer

Denote the graph of Thomae's function by $G$. The Hausdorff dimension and the topological dimension of $G$ are both $1$. This is because $G$ is the set of irrational numbers in the unit interval (well known to have full Lebesgue measure) together with countably many isolated points. Thus, $G$ is not a fractal in the strict sense of Mandelbrot. On the other hand, it is certainly an interesting example in real analysis that is not so far removed the ideas of fractal geometry.

• So does that mean the Thomae's function isn't a fractal? – Zach466920 Mar 31 '15 at 23:00
• @Zach466920 Strictly speaking no. I did edit my answer a bit to address this question. – Mark McClure Apr 1 '15 at 0:33
• Thanks for clearing that up. – Zach466920 Apr 1 '15 at 14:05