FormalConjectures/ErdosProblems/464.lean
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1 declarations, 1 contain sorry, 0 declared as axioms
Erdős Problem 464
References:
- erdosproblems.com/464
- [Er75i] Erdős, P., Répartition modulo $1$. (1975), iv+258.
- [ErGr80] Erdős, P. and Graham, R., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathématique (1980).
- [Er82e] Erdős, Paul, Some of my favourite problems which recently have been solved. (1982), 59--79.
- [AkMo04] Akhunzhanov, R. K. and Moshchevitin, N. G., On the chromatic number of a distance graph associated with a lacunary sequence. Dokl. Akad. Nauk (2004), 295-296.
- [Du06] Dubickas, Artūras, On the fractional parts of lacunary sequences. Math. Scand. (2006), 136-146.
- [Ka01] Katznelson, Y., Chromatic numbers of Cayley graphs on $\mathbb{Z}$ and recurrence. Combinatorica (2001), 211-219.
- [PeSc10] Peres, Yuval and Schlag, Wilhelm, Two Erdős problems on lacunary sequences: chromatic number and Diophantine approximation. Bull. Lond. Math. Soc. (2010), 295-300.
- [Po79b] Pollington, A. D., On the density of sequence $\{n_k\xi\}$. Illinois J. Math. (1979), 511-515.
- [dM80] de Mathan, B., Numbers contravening a condition in density modulo $1$. Acta Math. Acad. Sci. Hungar. (1980), 237-241 (1981).
Imports (1)
- theorem
Erdos464.erdos_464Let $A=\{n_10$ with $n_{k+1}\geq (1+\epsilon)n_k$ for all $k$) sorry line 84 sourceSource documentation
Let $A=\{n_1<n_2<\cdots\}\subset \mathbb{N}$ be a lacunary sequence (so there exists some $\epsilon>0$ with $n_{k+1}\geq (1+\epsilon)n_k$ for all $k$). Must there exist an irrational $\theta$ such that $$\{ \|\theta n_k\| : k\geq 1\}$$ is not dense in $[0,1]$ (where $\| x\|$ is the distance to the nearest integer)?
Solved independently by de Mathan [dM80] and Pollington [Po79b], who showed that, given any such $A$, there exists such a $\theta$, with $$\inf_{k\geq 1}\| \theta n_k\| \gg \frac{\epsilon^4}{\log(1/\epsilon)}.$$ This bound was improved by Katznelson [Ka01], Akhunzhanov and Moshchevitin [AkMo04], and Dubickas [Du06], before Peres and Schlag [PeSc10] improved it to $$\inf_{k\geq 1}\| \theta n_k\| \gg \frac{\epsilon}{\log(1/\epsilon)},$$ and note that the best bound possible here would be $\gg \epsilon$.
This problem has consequences for 894.
The conclusion "$\{\|\theta n_k\|\}$ is not dense in $[0,1]$" is formalized as the sequence $(\theta n_k)$ not being dense modulo one; see the formalization notes above.
L4DEL explanation
Let $A=\{n_10$ with $n_{k+1}\geq (1+\epsilon)n_k$ for all $k$). Must there exist an irrational $\theta$ such that $$\{ \|\theta n_k\| : k\geq 1\}$$ is not dense in $[0,1]$ (where $\| x\|$ is the distance to the nearest integer)? Solved independently by de Mathan [dM80] and Pollington [Po79b], who showed that, given any such $A$, there exists such a $\theta$, with $$\inf_{k\geq 1}\| \theta n_k\| \gg \frac{\epsilon^4}{\log(1/\epsilon)}.$$ This bound was improved by Katznelson [Ka01], Akhunzhanov and Moshchevitin [AkMo04], and Dubickas [Du06], before Peres and Schlag [PeSc10] improved it to $$\inf_{k\geq 1}\| \theta n_k\| \gg \frac{\epsilon}{\log(1/\epsilon)},$$ and note that the best bound possible here would be $\gg \epsilon$. This problem has consequences for 894. The conclusion "$\{\|\theta n_k\|\}$ is not dense in $[0,1]$" is formalized as the sequence $(\theta n_k)$ not being dense modulo one; see the formalization notes above.
Sources and attribution: Pinned Lean source (1646ca16afd6)
depends on sorry what this means
Statement uses:
Iff,True,Nat,StrictMono,LT.lt,OfNat.ofNat, IsLacunary,Exists,Real,And,Irrational,Not,Dense,HasQuotient.Quotient,AddSubgroup,AddSubgroup.zmultiples,Set.range,HMul.hMulProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous34 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_464 : answer(True) ↔ ∀ n : ℕ → ℕ, StrictMono n → (∀ k, 0 < n k) → IsLacunary n → ∃ θ : ℝ, Irrational θ ∧ ¬ Dense (Set.range fun k => (↑(θ * n k) : AddCircle (1 : ℝ))) := by …Mentions: IsLacunary