FormalConjectures/Books/BugeaudDistributionModuloOne/Problem10_4.lean

Bugeaud Collection of Conjectures and Open Questions: Spectrum of Sequence

References:

• [Bug12] Bugeaud, Yann. "Distribution modulo one and Diophantine approximation." Vol. 193. Cambridge University Press, 2012. Chapter 10.
• [Men73] Mendès France, Michel. "Les ensembles de Bésineau." Séminaire Delange-Pisot-Poitou 15.1 (1973): 1-6.
• [Özc26] Özcan, Hikmet Burak. "The spectrum of $(\xi\alpha^n)$ can be uncountable." arXiv:2609.07714 (2026).

Counterexample

Problem 10.4 is false. Özcan [Özc26] proves that for every real $\alpha > 1$ there are $2^{\aleph_0}$ reals $\xi > 0$ whose spectrum contains one and the same uncountable set. The counterexample below is an independent, self-contained instance of this for the base $\alpha = 64$, where the digits of $\xi$ can be prescribed directly.

For a bit sequence $u \in \{0, 1\}^{\mathbb{N}}$ put $$\theta_u = \sum_{i \ge 0} (1 + u_i) \, 8^{-e(i)}, \qquad e(i) = 4^{i+1} + i + 6.$$ The map $u \mapsto \theta_u$ is injective with values in $(0, 1)$.

Split the indices into the blocks $[5 m_k, 6 m_k)$ with $m_k = 20 \cdot 8^k$; these are pairwise disjoint. Block number $k = \langle L, c \rangle$, where $\langle \cdot, \cdot \rangle$ is the Cantor pairing, is reserved for the bit string $c$ of length $L$. On that block the base-$64$ digits of $\xi$ are chosen so that $\{\xi 64^n\}$ shadows $\{n \theta\}$ for every $\theta$ whose first $L$ bits are $c$. This is possible because one base-$64$ digit pins down $\{\xi 64^n\}$ up to $2/64$, and because $k \le 4^{L+1}$ forces the block to sit far to the left of the precision $8^{-e(L)}$ of the truncation of $\theta$, so that $n$ times the truncation error stays below $1/512$ on the block.

Consequently $\{\xi 64^n - n \theta_u\} < 1/10$ for all $n$ in the block attached to the first $L$ bits of $u$, for every $L$. That block is the last sixth of $[0, 6 m_k)$, so at the times $N = 6 m_k - 6$ at most a proportion $5/6$ of the points $\{\xi 64^n - n\theta_u\}$ with $n < N$ lies in $[1/10, 1]$, whereas uniform distribution modulo one would force the proportion $9/10$. Hence every irrational $\theta_u$ lies in the spectrum of $(\xi 64^n)$, and since $\{\theta_u\}$ is uncountable the spectrum cannot be countable.