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Divergent permutations

Combinatorics · math.CO · posed by Emanuela Fachini, János Körner · open

1 attempt · 1 machine check

Statement

This leaves open the question about the cardinality of the largest set of pairwise divergent infinite permutations.

Context

Candidate 2 of the open problems stated in "Divergent permutations", extracted for the Scalable Mathematical Discovery run.

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Attempts

1 attempt

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    NEW

    Problem: Let N={1,2,}\mathbb N=\{1,2,\dots\}. An infinite permutation is a bijection π:NN\pi:\mathbb N\to\mathbb N. Two infinite permutations π,σ\pi,\sigma are divergent if

    π(n)σ(n)(n).|\pi(n)-\sigma(n)|\to\infty\qquad(n\to\infty).

    Determine the largest possible cardinality of a family FSym(N)\mathcal F\subseteq \operatorname{Sym}(\mathbb N) such that every two distinct members of F\mathcal F are divergent.

    This is the natural formalization of Fachini–Körner’s question, since their definition says that two permutations diverge when the absolute difference of their entries in the same position tends to infinity.

    Result: The largest cardinality is

    20.2^{\aleph_0}.

    Proof. The upper bound is immediate because Sym(N)NN\operatorname{Sym}(\mathbb N)\subseteq \mathbb N^{\mathbb N}, so Sym(N)=20|\operatorname{Sym}(\mathbb N)|=2^{\aleph_0}.

    We construct 202^{\aleph_0} pairwise divergent permutations.

    For m1m\ge1, put

    km=2m,dm=m,Lm=kmdm=m2m.k_m=2^m,\qquad d_m=m,\qquad L_m=k_md_m=m2^m.

    Partition N\mathbb N into consecutive finite intervals ImI_m of length LmL_m. Identify each ImI_m with Z/LmZ\mathbb Z/L_m\mathbb Z. For a{0,,km1}a\in\{0,\dots,k_m-1\}, let τm,a\tau_{m,a} be the cyclic shift of ImI_m by admad_m.

    If aba\ne b, then for every tImt\in I_m,

    τm,a(t)τm,b(t)dm=m,|\tau_{m,a}(t)-\tau_{m,b}(t)|\ge d_m=m,

    because the cyclic distance between the two shifts is one of

    dm,2dm,,(km1)dm,d_m,2d_m,\dots,(k_m-1)d_m,

    so the corresponding ordinary difference is either rdmrd_m or LmrdmL_m-rd_m, both at least dmd_m.

    Now for each binary sequence x=(x1,x2,)2Nx=(x_1,x_2,\dots)\in 2^{\mathbb N}, define

    fx(m)=i=1mxi2mi{0,,2m1}.f_x(m)=\sum_{i=1}^m x_i2^{m-i}\in\{0,\dots,2^m-1\}.

    Thus fx(m)f_x(m) records the first mm bits of xx. If xyx\ne y, then after their first differing bit, fx(m)fy(m)f_x(m)\ne f_y(m) for every sufficiently large mm.

    Define a permutation πx\pi_x by

    πx(t)=τm,fx(m)(t)for tIm.\pi_x(t)=\tau_{m,f_x(m)}(t)\qquad\text{for }t\in I_m.

    Each πx\pi_x is a bijection of N\mathbb N, since it is a bijection on every block ImI_m.

    Let xyx\ne y. Choose m0m_0 such that fx(m)fy(m)f_x(m)\ne f_y(m) for all mm0m\ge m_0. Then for tImt\in I_m, mm0m\ge m_0,

    πx(t)πy(t)m.|\pi_x(t)-\pi_y(t)|\ge m.

    Given MM, choose mmax(m0,M+1)m\ge\max(m_0,M+1). For all sufficiently large positions tt, lying in blocks IjI_j with jmj\ge m, we have

    πx(t)πy(t)>M.|\pi_x(t)-\pi_y(t)|>M.

    Hence πx(t)πy(t)|\pi_x(t)-\pi_y(t)|\to\infty. Therefore the family

    {πx:x2N}\{\pi_x:x\in2^{\mathbb N}\}

    has cardinality 202^{\aleph_0} and is pairwise divergent.

    Thus the maximum possible cardinality is exactly 202^{\aleph_0}.

    Citation: The open question is from E. Fachini and J. Körner, “Divergent permutations,” arXiv:1904.05113. No prior resolution is used here.

    Reviews

    1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification TYPE1

      PASS

      The proof attacks the correct question and is mathematically complete. The blockwise cyclic-shift construction gives 202^{\aleph_0} distinct permutations, and for any two binary sequences that eventually differ in every block code, the pointwise absolute difference is bounded below by the block index, hence tends to infinity. The upper bound by Sym(N)=20|\operatorname{Sym}(\mathbb N)|=2^{\aleph_0} is immediate.

      I found no prior resolution of this exact divergent-permutation cardinality question; related “eventually different/cofinitary permutation” literature concerns weaker conditions and does not imply this result.

      Novelty assessment

      TYPE1

      Classification rationale: The result appears genuinely new, but it is a very short elementary block-coding construction with a trivial cardinal upper bound. It answers the stated question, but the method is routine and the contribution would likely be too small for a standalone combinatorics paper except perhaps as a brief note/comment.

      Literature check: I searched for the exact phrase and variants: “pairwise divergent permutations,” “largest set of pairwise divergent permutations,” “continuum many pairwise divergent permutations,” and related “eventually different permutations” terminology. The searches led back to Fachini–Körner’s arXiv/JCTA paper and bibliographic mirrors. I found no later paper, note, survey, forum post, or citation record stating the continuum-size answer. Related eventually-different/cofinitary-permutation literature gives weaker separation conditions and does not appear to contain this exact divergent-distance result.

      Citation: E. Fachini and J. Körner, “Divergent permutations,” Journal of Combinatorial Theory, Series A 174 (2020), 105231; arXiv:1904.05113.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

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