# Rational points on an analytic curve

Let $$\Gamma$$ be a transcendental analytic curve in $$\mathbb{R}^2$$. I am interested in the topology of its rational points $$\Gamma(\mathbb{Q}):=\Gamma\cap\mathbb{Q}^2$$.

We know by Pila-Wilkie that if $$\Gamma$$ is compact then it has few rational points.

Is it true that $$\Gamma(\mathbb{Q})$$ is locally finite? Is it possible for $$\Gamma(\mathbb{Q})$$ to be dense in some open subset of $$\Gamma$$?

Any example/counterexample will be appreciated.

• Commented Aug 4 at 20:22

Note that the results of (Bombieri-)Pila-Wilkie only tell you that there are "few" rational points when you count them by looking at their height. If you ignore this aspect, very little can be said: in particular, one can construct analytic transcendental functions $$f : [0,1] \to [0,1]$$ which are bijective on the rational numbers in this interval (hence, in particular, the graph $$\Gamma$$ of $$f$$ has "many" rational points), see https://ems.press/journals/rlm/articles/14595. See also related work by Marques and Moreira on several problems raised by Mahler (e.g., https://link.springer.com/article/10.1007/s00208-016-1485-z, where they construct entire analytic functions inducing bijections of the set of algebraic numbers).

• Interesting question and accurate answer! Some thoughts after reading: The construction mainly uses approximation by polynomials. Since the rational points are countable, we can find a polynomial that vanishes at the first n points, while maintaining good control over its growth.
– zxx
Commented Aug 4 at 15:32
• can these examples be made into a closed curve in $\mathbb R^2$? Commented Aug 4 at 16:03
• Is the curve $x^2+(f(y))^2=1$ an example? $f$ is the function described in the answer.
– zxx
Commented Aug 4 at 18:13
• I'm not sure if the previous suggestion works, but a slight variant should: let $A=[-1,1] \cap \mathbb{Q}$ and $B=\{ \pm \sqrt{1-r^2} : r \in A\}$. Using the same method (or applying an older result of Franklin, "Analytic transformations of everywhere dense point sets") it should be possible to construct $g : [-1,1] \to [-1,1]$ that gives a bijection from $A$ to $B$ (both are countable and dense, which are essentially the only properties used). For this $g$, $x^2 + g(y)^2 = 1$ should work (to ensure smoothness, we should also impose $g'(-1) \neq 0, g'(1) \neq 0$). Commented Aug 4 at 18:32
• If you notice that rational points on the curve $x^2 + y^2 = 1$ are dense, then you can see that rational points on the curve $x^2 + (f(y))^2 = 1$ are also dense.
– zxx
Commented Aug 4 at 20:02