On irrationals with Lagrange value exactly 3
Zhe Cao, Harold Erazo, Carlos Gustavo Moreira
TL;DR
The paper resolves a gap in the understanding of the Lagrange spectrum at the critical threshold 3 by proving that X_3 is uncountable and, more strongly, that for every n∈N∪{∞}, the sets X_3(n) and the projected Y_3(n) are uncountable. The authors deploy Bombieri's renormalization of bi-infinite words, analyze admissible word structures, and construct uncountably many limit words with Lagrange value exactly 3 by controlling cuts and their values. They establish a continuum of elements with k(x)=3 and precisely n representations of near-3-approximation, including infinite (n=∞) and finite (n≥0) cases, by carefully gluing finite blocks to tail sequences. The work integrates renormalization, Sturmian-like dynamics, and symbolic dynamics to extend classical Markov–Hurwitz insights and to illuminate the fine structure around the 3-accumulation point in the Lagrange/Markov landscapes, with implications for the generalized spectra and their projections.
Abstract
For $c>0$, let $X_c$ denote the set of $x\in\mathbb{R}\backslash\mathbb{Q}$ such that $\left| x-\frac{p}{q} \right|<\frac{1}{cq^2}$ has only finitely many rational solutions $\frac{p}{q}$. It is a classical fact, known since the 1950s, that $X_c$ is uncountable for $c>3$ and countable for $c<3$. However, the cardinality of $X_3$ does not appear to be present in the literature. We prove that $X_3$ is uncountable. More generally, we show that for any $n\in\mathbb{N}\cup\{\infty\}$, the set of $x\in\mathbb{R}\backslash\mathbb{Q}$ with Lagrange value exactly $3$ and such that $\left| x-\frac{p}{q} \right|<\frac{1}{3q^2}$ has exactly $n$ rational solutions $\frac{p}{q}$ is also uncountable.