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.
2026-03-28 22:35:44.1774737344
Proof of fractal dimension of Thomae's function
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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.