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Setting

For an integer aa and a positive integer nn, a(n)=a+nZa(n)=a+n\mathbb Z. The system is

A={as(ns)}s=1kA=\{a_s(n_s)\}_{s=1}^k

with positive moduli n1,…,nkn_1,\ldots,n_k (the paper's (1), preprint p. 1). For x∈Zx\in\mathbb Z let S(x)={1≤s≤k:x≡as(modns)}S(x)=\{1\le s\le k:x\equiv a_s\pmod{n_s}\}; the covering multiplicity is m(A)=inf⁡x∈Z∣S(x)∣m(A)=\inf_{x\in\mathbb Z}\lvert S(x)\rvert (the paper's (2)), and AA is an mm-cover when m(A)≥mm(A)\ge m. A minimal mm-cover is an mm-cover none of whose proper subsystems is one, and an exact mm-cover has ∣S(x)∣=m\lvert S(x)\rvert=m for every xx (pp. 1--2). [x][x] and {x}\{x\} are the integral and fractional parts of a real xx. For J⊆{1,…,k}J\subseteq\{1,\ldots,k\}, J−={1,…,k}∖JJ^-=\{1,\ldots,k\}\setminus J, and N(J)N(J) is the least common multiple of the nsn_s with s∈Js\in J.

The paper also records (its (3), p. 1, with a pointer to earlier work) that ∑s=1k1/ns≥m(A)\sum_{s=1}^k1/n_s\ge m(A), with equality if and only if AA covers each integer exactly mm times for some positive integer mm.

Statement

Theorem 1 (preprint p. 2). Let AA be a system as above and J⊆{1,…,k}J\subseteq\{1,\ldots,k\}.

(i) For all integers m1,…,mkm_1,\ldots,m_k,

∣{I⊆{1,…,k}: I≠J, {∑s∈Imsns}={∑s∈Jmsns}}∣ ≥ m(A).\Bigl\lvert\Bigl\{I\subseteq\{1,\ldots,k\}:\ I\ne J,\ \Bigl\{\sum_{s\in I}\frac{m_s}{n_s}\Bigr\} =\Bigl\{\sum_{s\in J}\frac{m_s}{n_s}\Bigr\}\Bigr\}\Bigr\rvert\ \ge\ m(A).

(ii) Suppose ∅≠J⊆S(x)\emptyset\ne J\subseteq S(x) for some x∈Zx\in\mathbb Z with ∣S(x)∣=m(A)\lvert S(x)\rvert=m(A), and for each s∈J−s\in J^- let msm_s be a positive integer prime to nsn_s. Then there is α∈[0,1)\alpha\in[0,1) such that for every integer rr with 0≤r<N(J)0\le r<N(J) some I⊆J−I\subseteq J^- satisfies

[∑s∈Imsns]≥m(A)−∣J∣and{∑s∈Imsns}=α+rN(J).\Bigl[\sum_{s\in I}\frac{m_s}{n_s}\Bigr]\ge m(A)-\lvert J\rvert \qquad\text{and}\qquad \Bigl\{\sum_{s\in I}\frac{m_s}{n_s}\Bigr\}=\frac{\alpha+r}{N(J)}.

The paper writes (ii) as an inclusion of sets (its (5)): the set of fractional parts {∑s∈Ims/ns}\{\sum_{s\in I}m_s/n_s\} over I⊆J−I\subseteq J^- with [∑s∈Ims/ns]≥m(A)−∣J∣[\sum_{s\in I}m_s/n_s]\ge m(A)-\lvert J\rvert contains {a/N(J):0≤a<N(J), {a}=α}\{a/N(J):0\le a<N(J),\ \{a\}=\alpha\}, where aa ranges over reals; those aa are exactly α+r\alpha+r with r=0,…,N(J)−1r=0,\ldots,N(J)-1.

Consequences stated on p. 2. The paper notes, as properties of an mm-cover AA that follow from Theorem 1 (taking every ms=1m_s=1):

  • (a) for each J⊆{1,…,k}J\subseteq\{1,\ldots,k\} there are at least mm subsets I≠JI\ne J with ∑s∈I1/ns−∑s∈J1/ns∈Z\sum_{s\in I}1/n_s-\sum_{s\in J}1/n_s\in\mathbb Z;
  • (b) if AA is a minimal mm-cover, then for each t=1,…,kt=1,\ldots,k there is αt∈[0,1)\alpha_t\in[0,1) such that for every r=0,1,…,nt−1r=0,1,\ldots,n_t-1 some I⊆{1,…,k}∖{t}I\subseteq\{1,\ldots,k\}\setminus\{t\} has [∑s∈I1/ns]≥m−1[\sum_{s\in I}1/n_s]\ge m-1 and {∑s∈I1/ns}=(αt+r)/nt\{\sum_{s\in I}1/n_s\}=(\alpha_t+r)/n_t.

Property (b) is (ii) with J={t}J=\{t\} (a reading of this page): since A∖{at(nt)}A\setminus\{a_t(n_t)\} is not an mm-cover, some x∈at(nt)x\in a_t(n_t) lies in exactly mm classes of AA, so m(A)=mm(A)=m and {t}⊆S(x)\{t\}\subseteq S(x) with ∣S(x)∣=m(A)\lvert S(x)\rvert=m(A). The paper's Corollary 4 (p. 4) applies (ii) with J={t}J=\{t\} in the same way, for positive msm_s prime to nsn_s, and states the result with differences of two subset sums. The paper also notes (p. 2) that the case J=∅J=\emptyset of (i) for a 1-cover gives a nonempty II with ∑s∈I1/ns∈Z\sum_{s\in I}1/n_s\in\mathbb Z, which it attributes to M. Z. Zhang.

Source. Zhi-Wei Sun, On covering multiplicity, Proc. Amer. Math. Soc. 127 (1999), no. 5, 1293--1300, doi:10.1090/S0002-9939-99-04817-0, read in the author's preprint identified on the source card, whose pages are numbered 1 to 9: the theorem on p. 2, the proof of (i) on pp. 5--7 and of (ii) on pp. 7--8.

Read depth. Claims checked: the definitions, the statement and the consequences (a) and (b) were read clause by clause on the page images. The proof was read but not checked step by step; nothing here is independently reviewed.

Proof pointer

Section 2 (pp. 5--8). Both parts rest on a characterization of mm-covers that the paper quotes from the author's earlier work (Proposition 1, p. 5): a system of real arithmetic sequences αs+βsZ\alpha_s+\beta_s\mathbb Z is an mm-cover exactly when certain signed sums of binomial coefficients times exponentials, taken over the subsets II with a given fractional part {∑s∈I1/βs}\{\sum_{s\in I}1/\beta_s\}, vanish. Lemma 1 (p. 5) says AA is an mm-cover if and only if every subsystem obtained by deleting m−nm-n classes is an nn-cover. For (i) (pp. 5--7), applied to the rescaled classes as+(ns/ms)Za_s+(n_s/m_s)\mathbb Z, these give, case by case on whether JJ or J−J^- has at least mm elements and then on the classes of modulus 1, m+1m+1 distinct subsets with the same fractional part. For (ii) (pp. 7--8), Lemma 2 (p. 7) shows that a weighted sum C(a)C(a) over the subsets with fractional part a/N(J)a/N(J) depends only on {a}\{a\}; the hypothesis that xx lies in exactly m(A)−∣J∣m(A)-\lvert J\rvert classes of J−J^- and the coprimality of msm_s and nsn_s make the rescaled system fail to be an (m(A)−∣J∣+1)(m(A)-\lvert J\rvert+1)-cover, so Proposition 1 yields a θ\theta with C(N(J)θ)≠0C(N(J)\theta)\ne0, and α={N(J)θ}\alpha=\{N(J)\theta\} works.

Bears on

  • Problem 1189: the problem counts irreducible covering sets, sets of distinct moduli 1<n1<⋯<nk1<n_1<\cdots<n_k that admit a covering choice of residues while no proper subset does. For such a set, any covering choice of residues is a minimal 1-cover, so consequence (b) applies to it with m=1m=1, where the condition [∑s∈I1/ns]≥0[\sum_{s\in I}1/n_s]\ge0 is empty. The ntn_t values (αt+r)/nt(\alpha_t+r)/n_t are then fractional parts of subset sums over distinct sets I⊆{1,…,k}∖{t}I\subseteq\{1,\ldots,k\}\setminus\{t\}, which gives nt≤2k−1n_t\le2^{k-1} for every tt (an observation of this page; the paper draws no conclusion about irreducible covering sets). This is a necessary condition on the moduli, not a count, and the theorem does not answer the problem.