A proof of Poincaré-Hurwitz Theorem in English?

I was reading this paper by Shane Chern: https://arxiv.org/abs/1602.02844 and I found the following theorem:

Let $$E$$ be a nonsingular cubic curve in $$\mathbb{P}^2$$ which is defined over $$\mathbb{Q}$$. If the set $$E\left (\mathbb{Q}\right )$$ is infinite, then every open subset of $$\mathbb{P}^2\left (\mathbb{R}\right )$$ which contains one point of $$E\left (\mathbb{Q}\right )$$ must contain infinitely many points of $$E\left (\mathbb{Q}\right )$$

I could not find any proof of this which is written in a language I can understand (I only understand English and Spanish). The reference that Shane Chern provides is an article by Poincaré, which is written in French, and an article by Hurwitz, which is written in German.

Reading another article which also makes use of this result ( http://archive.ymsc.tsinghua.edu.cn/pacm_download/21/129-2012A_DIOPHANTINE_PROBLEM_FROM_MATHEMATICAL_PHYSICS.pdf ) I found another source: "T. Skolem, Diophantische gleichungen, Ergebnisse der Mathematik und ihrer Grenzgebiete, New Chelsea Publishing Company, 1950, p. 78", which provides the following proof:

but I do not understand German.

So, do you know where to find a proof in English? And, if you know or attain a proof, could you please share it? Thank you in advance!

This statement follows by combining the following two claims:

1. As a Lie group, we must have $$E(\mathbf R) \cong \begin{cases} \mathbf R/\mathbf Z \\ \{\pm 1 \} \times \mathbf R/\mathbf Z \end{cases}$$ depending on whether the real locus has one or two connected components.

2. An infinite subgroup of $$\mathbf R/\mathbf Z$$ is everywhere dense.

I hope you can take it from here - let me know if you need a further hint!

I decided to say a bit more. I will give a slightly different argument than the one outlined above, adapted to proving only the specific conclusion you wanted.

Lemma. Let $$G$$ be a compact Lie group and $$\Gamma$$ an infinite subgroup of $$G$$. Every neighborhood of $$0$$ in $$G$$ contains infinitely many elements of $$\Gamma$$. Proof. Put a translation invariant metric on $$G$$. For every $$\epsilon >0$$ cover $$G$$ by finitely many balls of radius $$\epsilon/2$$; at least one such ball $$U$$ contains infinitely many elements of $$\Gamma$$ by the pigeonhole principle. If $$\gamma \in U \cap \Gamma$$ then $$\gamma^{-1}U$$ is contained in the ball of radius $$\epsilon$$ around $$0$$ and contains infinitely many elements of $$\Gamma$$. QED

Now to prove the result let $$E$$ be an elliptic curve. Then $$E(\mathbf R)$$ is a compact Lie group (it is a closed subgroup of $$E(\mathbf C) \cong \mathbf C/\Lambda$$). If $$E(\mathbf Q)$$ is infinite then by the lemma every neighborhood of $$0$$ contains infinitely many elements of $$E(\mathbf Q)$$. By translation using the group structure it follows that every neighborhood of every point of $$E(\mathbf Q)$$ contains infinitely many elements of $$E(\mathbf Q)$$.

• Yes, that's exactly the idea I had. I elaborated the answer a little. Commented Apr 2, 2020 at 8:23