Let m, n, S1, S2, …, Sn, be non-negative integers with 0≤m≤n. Assume μ(S1, S2, …, Sn)={(a1, a2, …, an)|0≤ai≤Si for each i} is a poser, Where (a1, a2, …, an)(b1, b2, …, bn) if and only if aibi for all i. A subset of μ(s1, s2, …, Sn) is called a two-part Sperner family in μ(s1, s2, …, sn) if for any a=(a1, a2, …, an), b=(b1, b2, …, bn) ∈μ(s1, s2, …, sn), (i) ai=bi(1≤i≤m) and ai≤bi(m+1≤i≤n) imply ai=bi for all i, and (ⅱ) ai≤bi(1≤i≤m) and ai=bi(m+1≤i≤n) imply ai=bi for all i.In this paper, we prove that if is a two-part Sperner family in μ(s1, s2,…, sn), then
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机译:设M,N,S 1 sub>,s 2 sub>,...,s n sub>,是0≤m≤n的非负整数。假设μ(S 1 sub>,s 2 sub>,...,s n sub>)= {(a 1 sub>,a <子> 2 sub>,..., n sub>)|0≤a i sub>≤s i sub>对于每个i}是一个poser,其中(a 1 sub>, 2 sub>,..., n sub>)<(b 1 sub>,b 2 sub>,...,b n sub>)如果只有a i sub> i sub>,我都是。 μ(S 1 sub>,s 2 sub>,...,s n sub>)称为μ的双部分尖垂系列(s <子> 1 sub>,s 2 sub>,...,s n sub>)如果任何a =(a 1 sub>,a 2 sub>,...,a),b =(b 1 sub>,b 2 sub>,...,b n sub>)∈μ(s 1 sub>,s 2 sub>,...,s n sub>),(i)a i sub> = b i sub>(1≤i≤m)和 i sub>≤b i sub>(m +1≤i≤n)意味着 i sub> = B i sub>对于所有I,(Ⅱ) i sub>≤b i sub>(1≤i≤m)和 i sub> = b i sub>(m +1≤i≤n)意味着所有I.本文的 i sub> = b i sub> ,我们证明,如果是μ的双部分斜氏族系列(s 1 sub>,s 2 sub>,...,s n sub>)那么
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