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Journal of Applied Mathematics
VolumeΒ 2012, Article IDΒ 847958, 15 pages
Research Article

A Minimum Problem for Finite Sets of Real Numbers with Nonnegative Sum

1Dipartimento di Matematica, UniversitΓ‘ della Calabria, Via Pietro Bucci, Cubo 30B, 87036 Arcavacata di Rende, Italy
2Dipartimento di Matematica, UniversitΓ‘ della Calabria, Via Pietro Bucci, Cubo 30C, 87036 Arcavacata di Rende, Italy

Received 6 February 2012; Accepted 2 March 2012

Academic Editor: YonghongΒ Yao

Copyright Β© 2012 G. Chiaselotti et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


Let 𝑛 and π‘Ÿ be two integers such that 0<π‘Ÿβ‰€π‘›; we denote by 𝛾(𝑛,π‘Ÿ)[πœ‚(𝑛,π‘Ÿ)] the minimum [maximum] number of the nonnegative partial sums of a sum βˆ‘π‘›1=1π‘Žπ‘–β‰₯0, where π‘Ž1,…,π‘Žπ‘› are 𝑛 real numbers arbitrarily chosen in such a way that π‘Ÿ of them are nonnegative and the remaining π‘›βˆ’π‘Ÿ are negative. We study the following two problems: (𝑃1) which are the values of 𝛾(𝑛,π‘Ÿ) and πœ‚(𝑛,π‘Ÿ) for each 𝑛 and π‘Ÿ, 0<π‘Ÿβ‰€π‘›? (𝑃2) if π‘ž is an integer such that 𝛾(𝑛,π‘Ÿ)β‰€π‘žβ‰€πœ‚(𝑛,π‘Ÿ) , can we find 𝑛 real numbers π‘Ž1,…,π‘Žπ‘›, such thatπ‘Ÿ of them are nonnegative and the remaining π‘›βˆ’π‘Ÿ are negative with βˆ‘π‘›1=1π‘Žπ‘–β‰₯0, such that the number of the nonnegative sums formed from these numbers is exactly π‘ž?

1. Introduction

In [1] Manickam and MiklΓ³s raised several interesting extremal combinatorial sum problems, two of which will be described below. Let 𝑛 and π‘Ÿ be two integers such that 0<π‘Ÿβ‰€π‘›; we denote by 𝛾(𝑛,π‘Ÿ)[πœ‚(𝑛,π‘Ÿ)] the minimum [maximum] number of the nonnegative partial sums of a sum βˆ‘π‘›1=1π‘Žπ‘–β‰₯0, when π‘Ž1,…,π‘Žπ‘› are 𝑛 real numbers arbitrarily chosen in such a way that π‘Ÿ of them are nonnegative and the remaining π‘›βˆ’π‘Ÿ are negative. Put 𝐴(𝑛)=min{𝛾(𝑛,π‘Ÿ)∢0<π‘Ÿβ‰€π‘›}. In [1] the authors answered the following question.

(𝑄1) Which is the value of 𝐴(𝑛)?

In [1, Theorem 1] they found that 𝐴(𝑛)=2π‘›βˆ’1. On the other side, from the proof of Theorem 1 of [1] also it follows that 𝛾(𝑛,π‘Ÿ)β‰₯2π‘›βˆ’1 for each π‘Ÿ and that 𝛾(𝑛,1)≀2π‘›βˆ’1; therefore 𝛾(𝑛,1)=2π‘›βˆ’1 (since 2π‘›βˆ’1=𝐴(𝑛)≀𝛾(𝑛,1)≀2π‘›βˆ’1). It is natural to set then the following problem which is a refinement of (𝑄1).

(𝑃1) Which are the values of 𝛾(𝑛,π‘Ÿ)and πœ‚(𝑛,π‘Ÿ) for each 𝑛 and π‘Ÿ with 0<π‘Ÿβ‰€π‘›?

In the first part of this paper we solve the problem (𝑃1), and we prove (see Theorem 4.1) that 𝛾(𝑛,π‘Ÿ)=2π‘›βˆ’1 and πœ‚(𝑛,π‘Ÿ)=2π‘›βˆ’2π‘›βˆ’π‘Ÿ for each positive integer π‘Ÿβ‰€π‘›.

A further question that the authors raised in [1] is the following.

(𝑄2)β€‰β€‰β€œWe do not know what is the range of the possible numbers of the nonnegative partial sums of a nonnegative 𝑛-element sum. The minimum is 2π‘›βˆ’1 as it was proven and the maximum is obviously 2π‘›βˆ’1 but we do not know which are the numbers between them for which we can find reals π‘Ž1,…,π‘Žπ‘› with βˆ‘π‘›1=1π‘Žπ‘–β‰₯0 such that the number of the nonnegative sums formed from these numbers is equal to that number.’’

The following problem is a natural refinement of (Q2).

(𝑃2)  If π‘ž is an integer such that 𝛾(𝑛,π‘Ÿ)β‰€π‘žβ‰€πœ‚(𝑛,π‘Ÿ), can we find 𝑛 real numbers π‘Ž1,…,π‘Žπ‘›, such that π‘Ÿ of them are nonnegative and the remaining π‘›βˆ’π‘Ÿ are negative with βˆ‘π‘›1=1π‘Žπ‘–β‰₯0, such that the number of the nonnegative sums formed from these numbers is exactly π‘ž?

In the latter part of this paper (see Theorem 4.5) we give a partial solution to the problem (𝑃2).

To be more precise in the formulation of the problems that we study and to better underline the links with some interesting problems raised in [1], it will be convenient-identify a finite set of real-numbers with an appropriate real-valued function. Let then 𝑛 and π‘Ÿ be two fixed integers such that 0<π‘Ÿβ‰€π‘› and let 𝐼𝑛={1,2,…,𝑛} (we call 𝐼𝑛 the index set). We denote by π‘Š(𝑛,π‘Ÿ), the set of all the functions π‘“βˆΆπΌπ‘›β†’β„ such that βˆ‘π‘₯βˆˆπΌπ‘›π‘“(π‘₯)β‰₯0 and |{π‘₯βˆˆπΌπ‘›βˆΆπ‘“(π‘₯)β‰₯0}|=π‘Ÿ. If π‘“βˆˆπ‘Š(𝑛,π‘Ÿ), we set 𝛼(𝑓)=|{π‘ŒβŠ†πΌπ‘›βˆΆβˆ‘π‘¦βˆˆπ‘Œπ‘“(𝑦)β‰₯0}|. It is easy to observe that 𝛾(𝑛,π‘Ÿ)=min{𝛼(𝑓)βˆΆπ‘“βˆˆπ‘Š(𝑛,π‘Ÿ)} and πœ‚(𝑛,π‘Ÿ)=max{𝛼(𝑓)βˆΆπ‘“βˆˆπ‘Š(𝑛,π‘Ÿ)}. We can reformulate the problem (𝑃2) in an equivalent way using the functions terminology instead of the sets terminology.

(𝑃2)  If π‘ž is an integer such that 𝛾(𝑛,π‘Ÿ)β‰€π‘žβ‰€πœ‚(𝑛,π‘Ÿ), does there exist a function π‘“βˆˆπ‘Š(𝑛,π‘Ÿ) with the property that 𝛼(𝑓)=π‘ž?

To solve the problem (𝑃1) and (partially) (𝑃2), we use some abstract results on a particular class of lattices introduced in [2, 3]. In this paper we substantially continue the research project started in [4], which is the attempt to solve some extremal sum problems as started in [1] and further studied in [3, 5–11].

2. A Partial Order on the Subsets of 𝐼𝑛

Of course if we take two functions 𝑓,π‘”βˆˆπ‘Š(𝑛,π‘Ÿ) such that 𝑓(𝐼𝑛)=𝑔(𝐼𝑛), then 𝛼(𝑓)=𝛼(𝑔). This implies that if we define on π‘Š(𝑛,π‘Ÿ) the equivalence relation π‘“βˆΌπ‘” iff 𝑓(𝐼𝑛)=𝑔(𝐼𝑛) and we denote by [𝑓] the equivalence class of a function π‘“βˆˆπ‘Š(𝑛,π‘Ÿ), then the definition 𝛽([𝑓])=𝛼(𝑓) is well placed. It is also clear that it holds 𝛾(𝑛,π‘Ÿ)=min{𝛽([𝑓])∢[𝑓]βˆˆπ‘Š(𝑛,π‘Ÿ)/∼} and πœ‚(𝑛,π‘Ÿ)=max{𝛽([𝑓])∢[𝑓]βˆˆπ‘Š(𝑛,π‘Ÿ)/∼}. Now, when we take an equivalence class [𝑓]βˆˆπ‘Š(𝑛,π‘Ÿ)/~, there is a unique π‘“βˆ—βˆˆ[𝑓] such thatπ‘“βˆ—(π‘Ÿ)β‰₯β‹―β‰₯π‘“βˆ—(1)β‰₯0>π‘“βˆ—(π‘Ÿ+1)β‰₯β‹―β‰₯π‘“βˆ—(𝑛).(2.1) We can then identify the quotient set π‘Š(𝑛,π‘Ÿ)/~  with the subset of all the functions π‘“βˆ—βˆˆπ‘Š(𝑛,π‘Ÿ) that satisfy the condition (2.1). This simple remark conducts us to rename the indexes of 𝐼𝑛 as follows: Μƒπ‘Ÿ instead of Μƒ1π‘Ÿ,…, instead of 1, 1 instead of π‘Ÿ+1,…,π‘›βˆ’π‘Ÿ instead of 𝑛. Therefore, if we set ̃𝐼(𝑛,π‘Ÿ)={1,…,Μƒπ‘Ÿ,1,…,π‘›βˆ’π‘Ÿ}, we can identify the quotient set π‘Š(𝑛,π‘Ÿ)/βˆΌβ€‰β€‰with the set of all the functions π‘“βˆΆπΌ(𝑛,π‘Ÿ)→ℝ that satisfy the following two conditions:ξ€·Μƒ1𝑓(Μƒπ‘Ÿ)+β‹―+𝑓+𝑓1+β‹―+𝑓̃1ξ€Έξ‚€π‘›βˆ’π‘Ÿβ‰₯0,𝑓(Μƒπ‘Ÿ)β‰₯β‹―β‰₯𝑓β‰₯0>𝑓1β‰₯β‹―β‰₯𝑓.π‘›βˆ’π‘Ÿ(2.2) Now, if a generic function π‘“βˆΆπΌ(𝑛,π‘Ÿ)→ℝ that satisfies (2.2) is given, we are interested to find all the subsets π‘ŒβŠ†πΌ(𝑛,π‘Ÿ) such that βˆ‘π‘¦βˆˆπ‘Œπ‘“(𝑦)β‰₯0. This goal becomes then easier if we can have an appropriate partial orderβŠ‘ on the power set 𝒫(𝐼(𝑛,π‘Ÿ)) β€œcompatible” with the total order of the partial sums inducted by 𝑓, that is, a partial order βŠ‘ that satisfies the following monotonicity property: if π‘Œ,π‘βˆˆπ’«(𝐼(𝑛,π‘Ÿ)), then βˆ‘π‘§βˆˆπ‘βˆ‘π‘“(π‘₯)β‰€π‘¦βˆˆπ‘Œπ‘“(𝑦) whenever π‘βŠ‘π‘Œ. To have such a partial order βŠ‘on 𝒫(𝐼(𝑛,π‘Ÿ)) that has the monotonicity property we must introduce a new formal symbol that we denote by0Β§. We add this new symbol to the index set 𝐼(𝑛,π‘Ÿ), so we set 𝐴(𝑛,π‘Ÿ)=𝐼(𝑛,π‘Ÿ)βˆͺ{0Β§}. We introduce on 𝐴(𝑛,π‘Ÿ) the following total order:π‘›βˆ’π‘Ÿβ‰Ίβ‹―β‰Ί2β‰Ί1β‰Ί0Β§β‰ΊΜƒΜƒ1β‰Ί2β‰Ίβ‹―β‰ΊΜƒπ‘Ÿ.(2.3)If 𝑖,π‘—βˆˆπ΄(𝑛,π‘Ÿ), then we write: 𝑖βͺ―𝑗 for 𝑖=𝑗 or 𝑖≺𝑗. We denote by 𝑆(𝑛,π‘Ÿ) the set of all the formal expressions 𝑖1β‹―π‘–π‘Ÿβˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿ (hereafter called strings) that satisfy the following properties:(i)𝑖1,…,π‘–π‘ŸΜƒβˆˆ{1,…,Μƒπ‘Ÿ,0Β§}, (ii)𝑗1,…,π‘—π‘›βˆ’π‘Ÿβˆˆ{1,…,π‘›βˆ’π‘Ÿ,0Β§}, (iii)𝑖1β‰½β‹―β‰½π‘–π‘Ÿβ‰½0§≽𝑗1β‰½β‹―β‰½π‘—π‘›βˆ’π‘Ÿ, (iv)the unique element which can be repeated is 0Β§.

In the sequel we often use the lowercase letters 𝑒,𝑀,𝑧,… to denote a generic string in 𝑆(𝑛,π‘Ÿ). Moreover to make smoother reading, in the numerical examples the formal symbols which appear in a string will be written without ~β€‰β€‰βˆ’ and Β§; in such way the vertical bar ∣ will indicate that the symbols on the left of | are in {Μƒ1,…,Μƒπ‘Ÿ,0Β§} and the symbols on the right of ∣ are elements in {0Β§,1,…,π‘›βˆ’π‘Ÿ}. For example, if 𝑛=3 and π‘Ÿ=2, then ̃̃𝐴(3,2)={2≻1≻0§≻1} and 𝑆(3,2)={21|0,21|1,10|0,20|0,10|1,20|1,00|1,00|0}.

Note that there is a natural bijective setcorrespondence βˆ—βˆΆπ‘€βˆˆπ‘†(𝑛,π‘Ÿ)β†¦π‘€βˆ—βˆˆπ’«(𝐼(𝑛,π‘Ÿ))between 𝑆(𝑛,π‘Ÿ) and 𝒫(𝐼(𝑛,π‘Ÿ)) defined as follows: if 𝑀=𝑖1β‹―π‘–π‘Ÿβˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆˆπ‘†(𝑛,π‘Ÿ), then π‘€βˆ— is the subset of 𝐼(𝑛,π‘Ÿ) made with the elements π‘–π‘˜ and 𝑗𝑙 such that π‘–π‘˜β‰ 0Β§ and 𝑗𝑙≠0Β§. For example, if 𝑀=4310∣013βˆˆπ‘†(7,4), then π‘€βˆ—ΜƒΜƒΜƒ={1,3,4,1,3}. In particular, if 𝑀=0β‹―0∣0β‹―0, then π‘€βˆ—=βˆ….

Now, if 𝑣=𝑖1β‹―π‘–π‘Ÿβˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿ and 𝑀=π‘–ξ…ž1β‹―π‘–ξ…žπ‘Ÿβˆ£π‘—ξ…ž1β‹―π‘—ξ…žπ‘›βˆ’π‘Ÿ are two strings in 𝑆(𝑛,π‘Ÿ), we define: π‘£βŠ‘π‘€ iff 𝑖1βͺ―π‘–ξ…ž1,…,π‘–π‘Ÿβͺ―π‘–ξ…žπ‘Ÿ,𝑗1βͺ―π‘—ξ…žπ‘Ÿ,…,π‘—π‘›βˆ’π‘Ÿβͺ―π‘—ξ…žπ‘›βˆ’π‘Ÿ.

It is easily seen that (1)(𝑆(𝑛,π‘Ÿ),βŠ‘) is a finite distributive (hence also graded) lattice with minimum element 0β‹―0∣12β‹―(π‘›βˆ’π‘Ÿ) and maximum element π‘Ÿ(π‘Ÿβˆ’1)β‹―21∣0β‹―0; (2)(𝑆(𝑛,π‘Ÿ),βŠ‘)has the following unary complementary operation 𝑐: 𝑝1β‹―π‘π‘˜0β‹―0∣0β‹―0π‘ž1β‹―π‘žπ‘™ξ€Έπ‘=π‘ξ…ž1β‹―π‘ξ…žπ‘Ÿβˆ’π‘˜0β‹―0∣0β‹―0π‘žξ…ž1β‹―π‘žξ…žπ‘›βˆ’π‘Ÿβˆ’π‘™,(2.4)where {π‘ξ…ž1,…,π‘ξ…žπ‘Ÿβˆ’π‘˜} is the usual complement of {𝑝1,…,π‘π‘˜} in {Μƒ1,…,Μƒπ‘Ÿ} and {π‘žξ…ž1,…,π‘žξ…žπ‘›βˆ’π‘Ÿβˆ’π‘™} is the usual complement of {π‘ž1,…,π‘žπ‘™} in {1,…,π‘›βˆ’π‘Ÿ} (e.g., in 𝑆(7,4), we have that (4310∣001)𝑐=2000∣023).

Since we have the formal necessity to consider functions 𝑓 defined on the extended set 𝐴(𝑛,π‘Ÿ) instead of on the indexes set 𝐼(𝑛,π‘Ÿ), then we will put 𝑓(0Β§)=0. Precisely we can identify the quotient set π‘Š(𝑛,π‘Ÿ)/∼ with the set π‘ŠπΉ(𝑛,π‘Ÿ), defined by ξ€½ξ€·Μƒ1ξ€Έξ€·0π‘ŠπΉ(𝑛,π‘Ÿ)=π‘“βˆΆπ΄(𝑛,π‘Ÿ)β†’β„βˆΆπ‘“(Μƒπ‘Ÿ)β‰₯β‹―β‰₯𝑓β‰₯𝑓§=0>𝑓1β‰₯β‹―β‰₯𝑓̃1ξ€Έξ‚€π‘›βˆ’π‘Ÿ,𝑓(Μƒπ‘Ÿ)+β‹―+𝑓+𝑓1+β‹―+𝑓.π‘›βˆ’π‘Ÿβ‰₯0(2.5)

We call an element of π‘ŠπΉ(𝑛,π‘Ÿ) a (𝑛,π‘Ÿ)-weight function, and if π‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ), we will continue to set βˆ‘π›Ό(𝑓)∢=|{π‘ŒβŠ†πΌ(𝑛,π‘Ÿ)βˆΆπ‘¦βˆˆπ‘Œπ‘“(𝑦)β‰₯0}|. Therefore, with these last notations we have that 𝛾(𝑛,π‘Ÿ)=min{𝛼(𝑓)βˆΆπ‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ)}, πœ‚(𝑛,π‘Ÿ)=max{𝛼(𝑓)βˆΆπ‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ)} and the question (𝑃2) becomes equivalent to the following.

(𝑃2)  If π‘ž is an integer such that 𝛾(𝑛,π‘Ÿ)β‰€π‘žβ‰€πœ‚(𝑛,π‘Ÿ), does there exist a function π‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ)with the property that 𝛼(𝑓)=π‘ž?

3. Boolean Maps Induct by Weight Functions

We denote by 𝟐 the boolean lattice composed of a chain with 2 elements that we denote 𝑁 (the minimum element) and 𝑃 (the maximum element). A Boolean map (briefly BM) on 𝑆(𝑛,π‘Ÿ) is a map 𝐴∢dom(𝐴)βŠ†π‘†(𝑛,π‘Ÿ)β†’πŸ; in particular if dom(𝐴)=𝑆(𝑛,π‘Ÿ),we also say that 𝐴 is a Boolean total map (briefly BTM) on 𝑆(𝑛,π‘Ÿ). If 𝐴 is BM on 𝑆(𝑛,π‘Ÿ), we set 𝑆+𝐴(𝑛,π‘Ÿ)={π‘€βˆˆdom(𝐴)∢𝐴(𝑀)=𝑃}.

If π‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ), the sum function Ξ£π‘“βˆΆπ‘†(𝑛,π‘Ÿ)→ℝ induced by 𝑓 on 𝑆(𝑛,π‘Ÿ) is the function that associates to 𝑀=π‘–π‘Ÿβ‹―π‘–1βˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆˆπ‘†(𝑛,π‘Ÿ) the real number Σ𝑓(𝑀)=𝑓(𝑖1)+β‹―+𝑓(π‘–π‘Ÿ)+𝑓(𝑗1)+β‹―+𝑓(π‘—π‘›βˆ’π‘Ÿ), and therefore we can associate to π‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ) the map π΄π‘“βˆΆπ‘†(𝑛,π‘Ÿ)β†’πŸ setting:π΄π‘“βŽ§βŽͺ⎨βŽͺ⎩(𝑀)=𝑃ifΣ𝑓(𝑀)β‰₯0,𝑀≠0β‹―0∣0β‹―0,𝑁if𝑀=0β‹―0∣0β‹―0,𝑁ifΣ𝑓(𝑀)<0.(3.1)

Let us note that |𝑆+𝐴𝑓(𝑛,π‘Ÿ)|=|{π‘€βˆˆπ‘†(𝑛,π‘Ÿ)βˆΆπ΄π‘“(𝑀)=𝑃}|=𝛼(𝑓), and so 𝛾(𝑛,π‘Ÿ)=min{|𝑆+𝐴𝑓(𝑛,π‘Ÿ)|βˆΆπ‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ)},πœ‚(𝑛,π‘Ÿ)=max{|𝑆+𝐴𝑓(𝑛,π‘Ÿ)|βˆΆπ‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ)}.

Our goal is now to underline that some properties of such maps simplify the study of our problems. It is easy to observe that the map 𝐴𝑓 has the following properties.(i)𝐴𝑓 is orderpreserving.(ii)If π‘€βˆˆπ‘†(𝑛,π‘Ÿ) is such that 𝐴𝑓(𝑀)=𝑁, then 𝐴𝑓(𝑀𝑐)=𝑃,(iii)𝐴𝑓(10β‹―0∣0β‹―0)=𝑃, 𝐴(0β‹―0∣0β‹―0)=𝑁, and 𝐴𝑓(π‘Ÿβ‹―21∣12β‹―(π‘›βˆ’π‘Ÿ))=𝑃.

Example 3.1. Let 𝑓 be the following (5,3)-weight function: Μƒ3Μƒ2Μƒ1π‘“βˆΆ12↓↓↓↓↓110.9βˆ’0.8βˆ’2.1(3.2) We represent the map 𝐴𝑓 defined on 𝑆(5,3) by using the Hasse diagram of this lattice, on which we color green the nodes where the map 𝐴𝑓 assumes value 𝑃 and red the nodes where it assumes value 𝑁 (See Figure 1).
Note that if we have a generic Boolean total map π΄βˆΆπ‘†(𝑛,π‘Ÿ)β†’πŸ which has the properties (i), (ii), and (iii) of 𝐴𝑓, that is, the following:(BM1)𝐴 is orderpreserving,(BM2)if π‘€βˆˆπ‘†(𝑛,π‘Ÿ) is such that 𝐴(𝑀)=𝑁, then 𝐴(𝑀𝑐)=𝑃,(BM3)𝐴(10β‹―0∣0β‹―0)=𝑃, 𝐴(0β‹―0∣0β‹―0)=𝑁, and𝐴(π‘Ÿβ‹―21∣12β‹―(π‘›βˆ’π‘Ÿ))=𝑃,in general there does not exist a function π‘“βˆˆπ‘ŠπΉ(𝑛,π‘Ÿ) such that 𝐴𝑓=𝐴 (see [3] for a counterexample).
We denote by 𝒲+(𝑆(𝑛,π‘Ÿ),𝟐) the set of all the maps π΄βˆΆπ‘†(𝑛,π‘Ÿ)β†’πŸ which satisfy (BM1) and (BM2) and by 𝒲+(𝑛,π‘Ÿ) the subset of all the maps in 𝒲+(𝑆(𝑛,π‘Ÿ),𝟐) which satisfy also (BM3). We also set π›Ύβˆ—(𝑛,π‘Ÿ)∢=min{|𝑆+𝐴(𝑛,π‘Ÿ)βˆΆπ΄βˆˆπ’²+(𝑛,π‘Ÿ)} and πœ‚βˆ—(𝑛,π‘Ÿ)∢=max{|𝑆+𝐴(𝑛,π‘Ÿ)|βˆΆπ΄βˆˆπ’²+(𝑛,π‘Ÿ)}. Let us observe that π›Ύβˆ—(𝑛,π‘Ÿ)≀𝛾(𝑛,π‘Ÿ)β‰€πœ‚(𝑛,π‘Ÿ)β‰€πœ‚βˆ—(𝑛,π‘Ÿ). A natural question raises at this point.
(𝑄):if π‘ž is an integer such that π›Ύβˆ—(𝑛,π‘Ÿ)β‰€π‘žβ‰€πœ‚βˆ—(𝑛,π‘Ÿ), does there exist a map π΄βˆˆπ’²+(𝑛,π‘Ÿ) with the property that |𝑆+𝐴(𝑛,π‘Ÿ)|=π‘ž?
The question (𝑄) is the analogue of (𝑃2) expressed in terms of Boolean total maps on 𝑆(𝑛,π‘Ÿ) instead of (𝑛,π‘Ÿ)-weight functions, and if we are able to respond to (𝑄), we provide also a partial answer to (𝑃2). In Section 4 we give an affirmative answer to the question (𝑄), and also we give a constructive method to build the map 𝐴.

Figure 1

4. Main Results

In the sequel of this paper we adopt the classical terminology and notations usually used in the context of the partially ordered sets (see [12–14] for the general aspects on this subject). If π‘βŠ†π‘†(𝑛,π‘Ÿ), we will set ↓𝑍={π‘₯βˆˆπ‘†(𝑛,π‘Ÿ)βˆΆβˆƒπ‘§βˆˆπ‘suchthatπ‘₯βŠ‘π‘§},↑𝑍={π‘₯βˆˆπ‘†(𝑛,π‘Ÿ)βˆΆβˆƒπ‘§βˆˆπ‘suchthatπ‘§βŠ‘π‘₯}. In particular, if π‘§βˆˆπ‘†(𝑛,π‘Ÿ), we will set ↓𝑧=↓{𝑧}={π‘₯βˆˆπ‘†(𝑛,π‘Ÿ)βˆΆπ‘§βŠ’π‘₯}, ↑𝑧=↑{𝑧}={π‘₯βˆˆπ‘†(𝑛,π‘Ÿ)βˆΆπ‘§βŠ‘π‘₯}. 𝑍 is called a downset of 𝑆(𝑛,π‘Ÿ) if for π‘§βˆˆπ‘ and π‘₯βˆˆπ‘†(𝑛,π‘Ÿ) with π‘§βŠ’π‘₯, then π‘₯βˆˆπ‘. 𝑍 is called an upset of 𝑆(𝑛,π‘Ÿ) if for π‘§βˆˆπ‘ and π‘₯βˆˆπ‘†(𝑛,π‘Ÿ) with π‘§βŠ‘π‘₯, then π‘₯βˆˆπ‘. ↓𝑍 is the smallest down-set of 𝑆(𝑛,π‘Ÿ) which contains 𝑍, and 𝑍 is a downset of 𝑆(𝑛,π‘Ÿ) if and only if 𝑍=↓𝑍. Similarly, ↑𝑍 is the smallest up-set of 𝑆(𝑛,π‘Ÿ) which contains 𝑍, and 𝑍 is an upset in 𝑆(𝑛,π‘Ÿ) if and only if 𝑍=↑𝑍.

Theorem 4.1. If 𝑛 and π‘Ÿ are two integers such that 0<π‘Ÿ<𝑛, then 𝛾(𝑛,π‘Ÿ)=π›Ύβˆ—(𝑛,π‘Ÿ)=2π‘›βˆ’1 and πœ‚(𝑛,π‘Ÿ)=πœ‚βˆ—(𝑛,π‘Ÿ)=2π‘›βˆ’2π‘›βˆ’π‘Ÿ.

Proof. Assume that 0<π‘Ÿ<𝑛. We denote by 𝑆1(𝑛,π‘Ÿ) the sublattice of 𝑆(𝑛,π‘Ÿ) of all the strings 𝑀 of the form 𝑀=𝑖1β‹―π‘–π‘Ÿβˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1(π‘›βˆ’π‘Ÿ), with 𝑗1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1∈{0Β§,1,…,π‘›βˆ’π‘Ÿβˆ’1} and by 𝑆2(𝑛,π‘Ÿ) the sublattice of 𝑆(𝑛,π‘Ÿ) of all the strings 𝑀 of the form 𝑀=𝑖1β‹―π‘–π‘Ÿβˆ£0𝑗2β‹―π‘—π‘›βˆ’π‘Ÿ, with 𝑗2β‹―π‘—π‘›βˆ’π‘Ÿβˆˆ{0Β§,1,…,π‘›βˆ’π‘Ÿβˆ’1}. It is clear that 𝑆(𝑛,π‘Ÿ)=𝑆1̇⋃𝑆(𝑛,π‘Ÿ)2(𝑛,π‘Ÿ)and 𝑆1(𝑛,π‘Ÿ)≅𝑆2(𝑛,π‘Ÿ)≅𝑆(π‘›βˆ’1,π‘Ÿ). We consider now the following further sublattices of 𝑆(𝑛,π‘Ÿ): 𝑆+1ξ€½(𝑛,π‘Ÿ)∢=π‘€βˆˆπ‘†1(𝑛,π‘Ÿ)βˆΆπ‘€=π‘Ÿ(π‘Ÿβˆ’1)β‹―21βˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1ξ€Ύ,𝑆(π‘›βˆ’π‘Ÿ)Β±1ξ€½(𝑛,π‘Ÿ)∢=π‘€βˆˆπ‘†1(𝑛,π‘Ÿ)βˆΆπ‘€=𝑖1β‹―π‘–π‘Ÿβˆ’10βˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1(π‘›βˆ’π‘Ÿ),𝑖1≻0Β§ξ€Ύ,π‘†βˆ’1ξ€½(𝑛,π‘Ÿ)∢=π‘€βˆˆπ‘†1(𝑛,π‘Ÿ)βˆΆπ‘€=0β‹―0βˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1ξ€Ύ,𝑆(π‘›βˆ’π‘Ÿ)+2ξ€½(𝑛,π‘Ÿ)∢=π‘€βˆˆπ‘†2(𝑛,π‘Ÿ)βˆΆπ‘€=π‘Ÿ(π‘Ÿβˆ’1)β‹―21∣0𝑗2β‹―π‘—π‘›βˆ’π‘Ÿξ€Ύ,𝑆±2(𝑛,π‘Ÿ)∢=π‘€βˆˆπ‘†2(𝑛,π‘Ÿ)βˆΆπ‘€=𝑖1β‹―π‘–π‘Ÿβˆ’10∣0𝑗2β‹―π‘—π‘›βˆ’π‘Ÿ,𝑖1≻0Β§ξ€Ύ,π‘†βˆ’2ξ€½(𝑛,π‘Ÿ)∢=π‘€βˆˆπ‘†2(𝑛,π‘Ÿ)βˆΆπ‘€=0β‹―0∣0𝑗2β‹―π‘—π‘›βˆ’π‘Ÿξ€Ύ.(4.1)It occurs immediately that 𝑆𝑖(𝑛,π‘Ÿ)=𝑆+𝑖̇⋃𝑆(𝑛,π‘Ÿ)±𝑖̇⋃𝑆(𝑛,π‘Ÿ)βˆ’π‘–(𝑛,π‘Ÿ), for 𝑖=1,2 and 𝑆±𝑖(𝑛,π‘Ÿ) is a distributive sublattice of 𝑆𝑖(𝑛,π‘Ÿ) with 2π‘›βˆ’1βˆ’2β‹…2π‘›βˆ’π‘Ÿβˆ’1=2π‘›βˆ’1βˆ’2π‘›βˆ’π‘Ÿ elements, for 𝑖=1,2.
Now we consider the following (𝑛,π‘Ÿ)-weight function π‘“βˆΆπ΄(𝑛,π‘Ÿ)→ℝ:Μƒπ‘“βˆΆΜƒπ‘Ÿβ‹―10Β§1β‹―(π‘›βˆ’π‘Ÿβˆ’1)(π‘›βˆ’π‘Ÿ)↓↓↓↓↓↓↓↓(π‘›βˆ’π‘Ÿ)β‹―(π‘›βˆ’π‘Ÿ)0βˆ’1β‹―βˆ’1(π‘›βˆ’π‘Ÿ)(1βˆ’π‘Ÿ)βˆ’1(4.2) Then it follows that Ξ£π‘“βˆΆπ‘†(𝑛,π‘Ÿ)→ℝ is such that Σ𝑓(𝑀)β‰₯0ifπ‘€βˆˆπ‘†Β±2(𝑛,π‘Ÿ),<0ifπ‘€βˆˆπ‘†Β±1(𝑛,π‘Ÿ).(4.3) It means that the boolean map π΄π‘“βˆˆπ’²+(𝑛,π‘Ÿ) is such that 𝐴𝑓(𝑀)𝑃ifπ‘€βˆˆπ‘†Β±2(𝑛,π‘Ÿ),𝑁ifπ‘€βˆˆπ‘†Β±1(𝑛,π‘Ÿ).(4.4) This shows that |𝑆+𝐴𝑓(𝑛,π‘Ÿ)|=|𝑆+1̇⋃𝑆(𝑛,π‘Ÿ)+2̇⋃𝑆(𝑛,π‘Ÿ)Β±2(𝑛,π‘Ÿ)|=2π‘›βˆ’π‘Ÿβˆ’1+2π‘›βˆ’π‘Ÿβˆ’1+2π‘›βˆ’1βˆ’2β‹…2π‘›βˆ’π‘Ÿβˆ’1=2π‘›βˆ’1.
In [1, Theorem 1] it has been proved that 𝛾(𝑛,1)=2π‘›βˆ’1 and 𝛾(𝑛,π‘Ÿ)β‰₯2π‘›βˆ’1. Since 𝛾(𝑛,π‘Ÿ)β‰₯π›Ύβˆ—(𝑛,π‘Ÿ), using a technique similar to that used in the proof Theorem 1 of [1], it easily follows that π›Ύβˆ—(𝑛,π‘Ÿ)β‰₯2π‘›βˆ’1. As shown above, it results that |𝑆+𝐴𝑓(𝑛,π‘Ÿ)|=2π‘›βˆ’1. Hence 2π‘›βˆ’1β‰₯𝛾(𝑛,π‘Ÿ)β‰₯π›Ύβˆ—(𝑛,π‘Ÿ)β‰₯2π‘›βˆ’1, that is, 𝛾(𝑛,π‘Ÿ)=π›Ύβˆ—(𝑛,π‘Ÿ)=2π‘›βˆ’1. This proves; the first part of theorem; it remains to prove the latter part.
We consider now the following (𝑛,π‘Ÿ)-positive weight function π‘”βˆΆπ΄(𝑛,π‘Ÿ)→ℝ: ΜƒΜƒπ‘Ÿβ‹―10Β§1β‹―(π‘›βˆ’π‘Ÿβˆ’1)(π‘›βˆ’π‘Ÿ)π‘”βˆΆβ†“β†“β†“β†“β†“β†“β†“β†“1β‹―10βˆ’1β‹―π‘›βˆ’π‘Ÿβˆ’1π‘›βˆ’π‘Ÿβˆ’1π‘›βˆ’π‘Ÿ(4.5) It results then that the sum Ξ£π‘”βˆΆπ‘†(𝑛,π‘Ÿ)→ℝ is such that Σ𝑔(𝑀)β‰₯0 if π‘€βˆˆπ‘†Β±1̇⋃𝑆(𝑛,π‘Ÿ)Β±2(𝑛,π‘Ÿ); that is, the boolean map π΄π‘”βˆˆπ’²+(𝑛,π‘Ÿ) is such that 𝐴𝑔(𝑀)=𝑃if π‘€βˆˆπ‘†Β±1̇⋃𝑆(𝑛,π‘Ÿ)Β±2(𝑛,π‘Ÿ). This shows that |𝑆+𝐴𝑔(𝑛,π‘Ÿ)|=|𝑆±1̇⋃𝑆(𝑛,π‘Ÿ)Β±2̇⋃𝑆(𝑛,π‘Ÿ)+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ)|=(2π‘›βˆ’1βˆ’2π‘›βˆ’π‘Ÿ)+(2π‘›βˆ’1βˆ’2π‘›βˆ’π‘Ÿ)+2π‘›βˆ’π‘Ÿβˆ’1+2π‘›βˆ’π‘Ÿβˆ’1 = 2π‘›βˆ’2π‘›βˆ’π‘Ÿ.
On the other hand, it is clear that for any π΄βˆˆπ’²+(𝑛,π‘Ÿ) it results 𝐴(𝑀)=𝑃 for each π‘€βˆˆπ‘†+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ)and 𝐴(𝑀)=𝑁 for each π‘€βˆˆπ‘†βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ).The number (2π‘›βˆ’2π‘›βˆ’π‘Ÿ) is the biggest number of values 𝑃 that a boolean map π΄βˆˆπ’²+(𝑛,π‘Ÿ) can assume. Hence, since πœ‚(𝑛,π‘Ÿ)β‰€πœ‚βˆ—(𝑛,π‘Ÿ), we have 2π‘›βˆ’2π‘›βˆ’π‘Ÿ=|𝑆+𝐴𝑔(𝑛,π‘Ÿ)|β‰€πœ‚(𝑛,π‘Ÿ)β‰€πœ‚βˆ—(𝑛,π‘Ÿ)≀2π‘›βˆ’2π‘›βˆ’π‘Ÿ; that is be; πœ‚(𝑛,π‘Ÿ)=πœ‚βˆ—(𝑛,π‘Ÿ)=2π‘›βˆ’2π‘›βˆ’π‘Ÿ. This concludes the proof of Theorem 4.1.

To better visualize the previous result, we give a numerical example on a specific Hasse diagram. Let 𝑛=6 and π‘Ÿ=2 and let 𝑓 be as given in previous theorem; that is Μƒ2Μƒπ‘“βˆΆ10Β§1234↓↓↓↓↓↓↓440βˆ’1βˆ’1βˆ’1βˆ’5(4.6) In Figure 2 we have shown the Hasse diagram of the lattice 𝑆(6,2), where 𝑆+1(𝑛,π‘Ÿ) is black, 𝑆±1(𝑛,π‘Ÿ) is violet, π‘†βˆ’1(𝑛,π‘Ÿ) is red, 𝑆+2(𝑛,π‘Ÿ) is blue, 𝑆±2(𝑛,π‘Ÿ) is brown, π‘†βˆ’2(𝑛,π‘Ÿ) is green. Therefore 𝐴𝑓 assume the following values.(i)The blue, black, and brown nodes correspond to values 𝑃 of 𝐴𝑓,(ii)The violet, red, and green nodes correspond to values 𝑁 of 𝐴𝑓.

Figure 2

First to give the proof of Theorem 4.5 we need to introduce some useful results and the concept of basis in 𝑆(𝑛,π‘Ÿ). In the following first lemma we show some properties of the sublattices of 𝑆(𝑛,π‘Ÿ).

Lemma 4.2. Here the following properties hold, where πœƒ=00β‹―0∣0β‹―0 and Θ=π‘Ÿβ‹―21∣12β‹―(π‘›βˆ’π‘Ÿ). (i)β†‘Ξ˜=𝑆+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ).(ii)β†“πœƒ=π‘†βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ).(iii)↑𝑆±2(𝑛,π‘Ÿ)βŠ†π‘†Β±2̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ).(iv)↓𝑆±1(𝑛,π‘Ÿ)βŠ†π‘†Β±1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’1(𝑛,π‘Ÿ).(v)((𝑆±1(𝑛,π‘Ÿ))𝑐=𝑆±2(𝑛,π‘Ÿ).

Proof. (i) If π‘€βˆˆ(β†‘Ξ˜), then Ξ˜βŠ‘π‘€; that is, it has the form 𝑀=π‘Ÿβ‹―1βˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿ, where 𝑗1β‹―π‘—π‘›βˆ’π‘Ÿβˆˆ{0Β§,1,…,π‘›βˆ’π‘Ÿ}; therefore π‘€βˆˆπ‘†+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ). If π‘€βˆˆπ‘†+1(𝑛,π‘Ÿ), it has the form 𝑀=π‘Ÿβ‹―1βˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1(π‘›βˆ’π‘Ÿ), where 𝑗1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1∈{0Β§,1,…,π‘›βˆ’π‘Ÿβˆ’1}; if π‘€βˆˆπ‘†+2(𝑛,π‘Ÿ), it has the form 𝑀=π‘Ÿβ‹―1∣0𝑗2β‹―π‘—π‘›βˆ’π‘Ÿ, where 𝑗2β‹―π‘—π‘›βˆ’π‘Ÿβˆˆ{0Β§,1,…,π‘›βˆ’π‘Ÿβˆ’1}. In both cases it results that Ξ˜βŠ‘π‘€, that is, π‘€βˆˆ(β†‘Ξ˜).(ii)It is analogue to (i).(iii)The minimum of the sublattice 𝑆±2(𝑛,π‘Ÿ)is 𝛼=10β‹―0∣01β‹―(π‘›βˆ’π‘Ÿβˆ’1); since ↑𝑆±2(𝑛,π‘Ÿ)βŠ†β†‘π›Ό, it is sufficient to show that ↑𝛼=𝑆±2⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ). The inclusion 𝑆±2⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ)βŠ†β†‘π›Όfollows by the definition of 𝑆±2(𝑛,π‘Ÿ)and of 𝑆+2(𝑛,π‘Ÿ). On the other side, if π‘€βˆˆβ†‘π›Ό, it follows that π›ΌβŠ‘π‘€; that is, 𝑀=𝑖1β‹―π‘–π‘Ÿβˆ£0𝑗2β‹―π‘—π‘›βˆ’π‘Ÿ, with 𝑖1≻0Β§and 𝑗2β‹―π‘—π‘›βˆ’π‘Ÿβˆˆ{0Β§,1,…,π‘›βˆ’π‘Ÿβˆ’1}. Therefore π‘€βˆˆπ‘†Β±2⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ), and this proves the other inclusion.(iv)Let us consider the maximum of the sublattice 𝑆±1(𝑛,π‘Ÿ); that is 𝑑1=π‘Ÿ(π‘Ÿβˆ’1)β‹―20∣0β‹―0(π‘›βˆ’π‘Ÿ). Since ↓𝑆±1(𝑛,π‘Ÿ)βŠ†β†“π›½, it is sufficient to show that ↓𝑑1=𝑆±1⋃𝑆(𝑛,π‘Ÿ)βˆ’1(𝑛,π‘Ÿ); this proof is similar to (iii).(v)If π‘€βˆˆπ‘†Β±1(𝑛,π‘Ÿ); it has the form 𝑀=𝑖1β‹―π‘–π‘Ÿβˆ’10βˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1(π‘›βˆ’π‘Ÿ), with 𝑖1≻0Β§ and 𝑗1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1∈{0Β§,1,…,π‘›βˆ’π‘Ÿβˆ’1}; therefore its complement has the form 𝑀𝑐=π‘–ξ…ž1β‹―π‘–ξ…žπ‘Ÿβˆ’10∣0π‘—ξ…ž2β‹―π‘—ξ…žπ‘›βˆ’π‘Ÿ, with 𝑖1′≻0Β§ and π‘—ξ…ž2β‹―π‘—ξ…žπ‘›βˆ’π‘Ÿβˆˆ{0Β§,1,…,π‘›βˆ’π‘Ÿβˆ’1}; hence π‘€π‘βˆˆπ‘†Β±2(𝑛,π‘Ÿ). This shows that (𝑆±1(𝑛,π‘Ÿ))π‘βŠ†π‘†Β±2(𝑛,π‘Ÿ).
Now, if π‘€βˆˆπ‘†Β±2(𝑛,π‘Ÿ), we write 𝑀 in the form (𝑀𝑐)𝑐; then π‘€π‘βˆˆ(𝑆±2(𝑛,π‘Ÿ))π‘βŠ†π‘†Β±1(𝑛,π‘Ÿ); therefore π‘€βˆˆ(𝑆±1(𝑛,π‘Ÿ))𝑐. This shows that 𝑆±2(𝑛,π‘Ÿ)βŠ†(𝑆±1(𝑛,π‘Ÿ))𝑐.

At this point let us recall the definition of basis for 𝑆(𝑛,π‘Ÿ), as given in [3] in a more general context. In the same way as an antichain uniquely determines a Boolean order-preserving map, a basis uniquely determines a Boolean map that has the properties (BM1) and (BM2) (see [3] for details). Hence the concept of basis will be fundamental in the sequel of this proof.

Definition 4.3. A basis for 𝑆(𝑛,π‘Ÿ) is an ordered couple βŸ¨π‘Œ+βˆ£π‘Œβˆ’βŸ©, where π‘Œ+ and π‘Œβˆ’ are two disjoint antichains of 𝑆(𝑛,π‘Ÿ) such that(B1)(β†“π‘Œ+)β‹‚(π‘Œπ‘βˆ’)=βˆ…, (B2)((β†‘π‘Œ+)⋃↑(π‘Œπ‘βˆ’β‹‚))β†“π‘Œβˆ’=βˆ…, (B3)𝑆(𝑛,π‘Ÿ)=((β†‘π‘Œ+)⋃↑(π‘Œπ‘βˆ’β‹ƒ))β†“π‘Œβˆ’. In the proof of Theorem 4.5 we will construct explicitly such a basis.
We will also use the following result that was proved in [3].

Lemma 4.4. Let βŸ¨π‘Š+βˆ£π‘Šβˆ’βŸ© be a basis for 𝑆(𝑛,π‘Ÿ). Then the map ξ‚»ξ€·π‘Šπ΄(π‘₯)𝑃ifπ‘₯βˆˆβ†‘+ξ€Έξšβ†‘ξ€·π‘Šπ‘βˆ’ξ€Έ,𝑁ifπ‘₯βˆˆβ†“π‘Šβˆ’(4.7) is such that π΄βˆˆπ’²+(𝑆(𝑛,π‘Ÿ),𝟐).

Theorem 4.5. If π‘ž is a fixed integer with 2π‘›βˆ’1β‰€π‘žβ‰€2π‘›βˆ’2π‘›βˆ’π‘Ÿ, then there exists a boolean map π΄π‘žβˆˆπ’²+(𝑛,π‘Ÿ) such that |𝑆+π΄π‘ž(𝑛,π‘Ÿ)|=π‘ž.

Proof. Let π‘ž be a fixed integer such that 2π‘›βˆ’1β‰€π‘žβ‰€2π‘›βˆ’2π‘›βˆ’π‘Ÿ. We determine a specific Boolean total map π΄βˆˆπ’²+(𝑛,π‘Ÿ) such that |𝑆+𝐴(𝑛,π‘Ÿ)|=π‘ž. We proceed as follows.
The case π‘Ÿ=1 is proved in the previous Theorem 4.1. Let us assume then π‘Ÿ>1. Since 𝑆±1(𝑛,π‘Ÿ) is a finite distributive sublattice of graded lattice 𝑆(𝑛,π‘Ÿ), also 𝑆±1(𝑛,π‘Ÿ) is a graded lattice. We denote by 𝑅 the rank of 𝑆±1(𝑛,π‘Ÿ) and with 𝜌1 its rank function. Note that the bottom of 𝑆±1(𝑛,π‘Ÿ) is 𝑏1=10β‹―0∣1β‹―(π‘›βˆ’π‘Ÿβˆ’1)(π‘›βˆ’π‘Ÿ) and the top is 𝑑1=π‘Ÿ(π‘Ÿβˆ’1)β‹―20∣0β‹―0(π‘›βˆ’π‘Ÿ).
We write π‘ž in the form π‘ž=2π‘›βˆ’1+𝑝 with 0≀𝑝≀2π‘›βˆ’2π‘›βˆ’π‘Ÿβˆ’2π‘›βˆ’1=2nβˆ’1βˆ’2π‘›βˆ’π‘Ÿ=|𝑆±1(𝑛,π‘Ÿ)|. We will build a map π΄βˆˆπ‘Š+(𝑛,π‘Ÿ) such that |𝑆+𝐴(𝑛,π‘Ÿ)|=2π‘›βˆ’1+𝑝.
If 𝑝=0, we take 𝐴=𝐴𝑓, with 𝑓 as in Theorem 4.1, and hence we have |𝑆+𝐴(𝑛,π‘Ÿ)|=2π‘›βˆ’1.
If 𝑝=2π‘›βˆ’1βˆ’2π‘›βˆ’π‘Ÿ, we take 𝐴=𝐴𝑔, with 𝑔 as in Theorem 4.1, and we have |𝑆+𝐴(𝑛,π‘Ÿ)|=2π‘›βˆ’2π‘›βˆ’π‘Ÿ.
Therefore we can assume that 0<𝑝<2π‘›βˆ’1βˆ’2π‘›βˆ’π‘Ÿ.
If 0≀𝑖≀𝑅, we denote by β„œπ‘– the set of elements π‘€βˆˆπ‘†Β±1(𝑛,π‘Ÿ) such that 𝜌1(𝑀)=π‘…βˆ’π‘–, and we also set π›½π‘–βˆΆ=|β„œπ‘–| to simplify the notation. We write each β„œπ‘– in the following form: β„œπ‘–={𝑣𝑖1,…,𝑣𝑖𝛽𝑖}. If 0≀𝑙≀𝑅, we set π”˜lΜ‡β‹ƒβˆΆ=𝑖=0,…,π‘™β„œπ‘–.
If 0β‰€π‘™β‰€π‘…βˆ’2, we set π”…π‘™Μ‡β‹ƒβˆΆ=𝑖=𝑙+2,…,π‘…β„œπ‘– and π”…π‘…βˆ’1∢=π”…π‘…βˆΆ=βˆ…. We can then write 𝑝 in the form βˆ‘π‘=π‘˜π‘–=0|β„œπ‘–|+𝑠=|π”˜π‘˜|+𝑠, for some 0≀𝑠<|β„œπ‘˜+1| and some 0β‰€π‘˜β‰€π‘…βˆ’1. Depending on the previous number 𝑠 we partition β„œπ‘˜+1 into the following two disjoint subsets: β„œπ‘˜+1={𝑣(π‘˜+1)1,…,𝑣(π‘˜+1)𝑠}̇⋃{𝑣(π‘˜+1)(𝑠+1),…,𝑣(π‘˜+1)π›½π‘˜+1}, where the first subset is considered empty if 𝑠=0.
In the sequel, to avoid an overload of notations, we write simply 𝑣𝑖 instead of 𝑣(π‘˜+1)𝑖, for 𝑖=1,…,π›½π‘˜+1.
Let us note that 𝑆±1(𝑛,π‘Ÿ)=π”˜π‘˜Μ‡β‹ƒβ„œπ‘˜+1Μ‡β‹ƒπ”…π‘˜.
We define now the map π΄βˆΆπ‘†(𝑛,π‘Ÿ)β†’πŸ: ⎧βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺ⎩𝐴(𝑀)=𝑃ifπ‘€βˆˆπ‘†Β±2Μ‡ξšπ‘†(𝑛,π‘Ÿ)+1Μ‡ξšπ‘†(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ),𝑃ifπ‘€βˆˆπ”˜π‘˜Μ‡ξšξ€½π‘£1,…,𝑣𝑠,𝑁ifπ‘€βˆˆπ”…π‘˜Μ‡ξšξ‚†π‘£π‘ +1,…,π‘£π›½π‘˜+1,𝑁ifπ‘€βˆˆπ‘†βˆ’1(Μ‡ξšπ‘†π‘›,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ).(4.8) Let us observe that |𝑆+𝐴(𝑛,π‘Ÿ)| = |𝑆±2̇⋃𝑆(𝑛,π‘Ÿ)+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ)|+|π”˜π‘˜Μ‡β‹ƒ{𝑣1,…,𝑣𝑠}| = (2π‘›βˆ’1βˆ’2π‘›βˆ’π‘Ÿ)+2π‘›βˆ’π‘Ÿβˆ’1+2π‘›βˆ’π‘Ÿβˆ’1+|π”˜π‘˜|+𝑠 = 2π‘›βˆ’1+𝑝 = π‘ž.
Therefore, if we show that π΄βˆˆπ’²+(𝑛,π‘Ÿ), the theorem is proved.
We write β„œπ‘˜ in the following way: β„œπ‘˜={𝑀1,…,𝑀𝑑}̇⋃{𝑀𝑑+1,…,π‘€π›½π‘˜},where {𝑀1,…,𝑀𝑑} = β„œπ‘˜β‹‚β†‘{𝑣1,…,𝑣𝑠} and {𝑀𝑑+1,…,π‘€π›½π‘˜} = β„œπ‘˜β§΅{𝑀1,…,𝑀𝑑}. Analogously β„œπ‘˜+2 = {𝑧1,…,π‘§π‘ž}̇⋃{π‘§π‘ž+1,…,π‘§π›½π‘˜+2}, where {π‘§π‘ž+1,…,π‘§π›½π‘˜+2} = β„œπ‘˜+2⋂↓{𝑣𝑠+1,…,π‘£π›½π‘˜+1} and {𝑧1,…,π‘§π‘ž} = β„œπ‘˜+2⧡{π‘§π‘ž+1,…,π‘§π›½π‘˜+2}.
We can see a picture of this partition of the sublattice 𝑆±1(𝑛,π‘Ÿ) in Figure 3.
Depending on 𝑠 and π‘˜, we build now a particular basis for 𝑆(𝑛,π‘Ÿ).
To such aim, let us consider the minimum 𝛼=10β‹―0∣01β‹―(π‘›βˆ’π‘Ÿβˆ’1) of 𝑆±2(𝑛,π‘Ÿ) and the subsets 𝑇+∢={𝑣1,…,𝑣𝑠,𝑀𝑑+1,…,π‘€π›½π‘˜} and π‘‡βˆ’βˆΆ={𝑣𝑠+1,…,π‘£π›½π‘˜+1,𝑧1,…,π‘§π‘ž}. Let us distinguish two cases:(a1)π›Όβˆˆβ†‘π‘‡+,(a2)π›Όβˆ‰β†‘π‘‡+. We define two different couples of subsets as follows:
In the case (a1) we set π‘Œ+∢=𝑇+andπ‘Œβˆ’βˆΆ=π‘‡βˆ’Μ‡β‹ƒ{πœƒ};in the case (a2) we set π‘Œ+∢=𝑇+⋃{𝛼} and π‘Œβˆ’βˆΆ=π‘‡βˆ’Μ‡β‹ƒ{πœƒ}.
Step 1. βŸ¨π‘Œ+βˆ£π‘Œβˆ’βŸ© is a couple of two disjoint antichains of 𝑆(𝑛,π‘Ÿ).
In both cases (a1) and (a2) it is obvious that π‘Œ+β‹‚π‘Œβˆ’=βˆ….
Case (a1)
The elements {𝑣1,…,𝑣𝑠} are not comparable between them because they have all rank π‘…βˆ’(π‘˜+1) and analogously for the elements {𝑀𝑑+1,…,π‘€π›½π‘˜} that have all rank π‘…βˆ’π‘˜. Let now π‘£βˆˆ{𝑣1,…,𝑣𝑠} and let π‘€βˆˆ{𝑀𝑑+1,…,π‘€π›½π‘˜}; then π‘€βˆ‰β†“π‘£ because 𝜌1(𝑣)<𝜌1(𝑀), and π‘€βˆ‰β†‘π‘£ because {𝑀𝑑+1,…,π‘€π›½π‘˜}⋂↑{𝑣1,…,𝑣𝑠}=βˆ… by construction. For the elements in π‘Œβˆ’ different from πœƒ, we can proceed as for π‘Œ+. On the other side, we can observe that πœƒ is not comparable with none of the elements 𝑣𝑠+1,…,π‘£π›½π‘˜+1,𝑧1,…,π‘§π‘ž since these are all in 𝑆±1(𝑛,π‘Ÿ) while πœƒβˆˆπ‘†βˆ’2(𝑛,π‘Ÿ). Thus π‘Œ+ is an antichain.
Case (a2)
In this case we only must show that 𝛼 is not comparable with none of the elements 𝑣1,…,𝑣𝑠,𝑀𝑑+1,…,π‘€π›½π‘˜. At first from the fact that π›Όβˆ‰β†‘π‘‡+ it follows π‘£π‘–βˆ‰β†“π›Ό for each 𝑖=1,…,𝑠, and π‘€π‘—βˆ‰β†“π›Ό for each 𝑗=𝑑+1,…,π›½π‘˜. Moreover, the elements 𝑣1,…,𝑣𝑠 and 𝑀𝑑+1,…,π‘€π›½π‘˜ are all in 𝑆±1(𝑛,π‘Ÿ); hence they have the form 𝑖1β‹―π‘–π‘Ÿβˆ’10βˆ£π‘—1β‹―π‘—π‘›βˆ’π‘Ÿβˆ’1(π‘›βˆ’π‘Ÿ), with 𝑖1≻0Β§, while 𝛼=10β‹―0∣01β‹―(π‘›βˆ’π‘Ÿβˆ’1), so π›Όβˆ‰β†“π‘£π‘– and π›Όβˆ‰β†“π‘€π‘— for each 𝑖=1,…,𝑠 and each 𝑗=𝑑+1,…,π›½π‘˜.
Step 2. βŸ¨π‘Œ+βˆ£π‘Œβˆ’βŸ© is a basis for 𝑆(𝑛,π‘Ÿ).
We must see that (B1), (B2), and (B3) hold in both cases (a1) and (a2).
Case (a1)
(B1)Let us begin to observe that π‘‡π‘βˆ’={𝑣𝑐𝑠+1,…,π‘£π‘π›½π‘˜+1,𝑧𝑐1,…,π‘§π‘π‘ž}βŠ†(𝑆±1(𝑛,π‘Ÿ))𝑐=𝑆±2(𝑛,π‘Ÿ). Since πœƒπ‘=Θ, we have then π‘Œπ‘βˆ’=π‘‡π‘βˆ’Μ‡β‹ƒ{πœƒπ‘}βŠ†π‘†Β±2̇⋃(𝑛,π‘Ÿ){Θ}βŠ†π‘†Β±2̇⋃𝑆(𝑛,π‘Ÿ)+1(𝑛,π‘Ÿ).On the other hand, since β†“π‘Œ+βŠ†β†“π‘†Β±1(𝑛,π‘Ÿ), by Lemma 4.2(iv) we have also that β†“π‘Œ+βŠ†π‘†Β±1(𝑛,π‘Ÿ).β‹ƒπ‘†βˆ’1(𝑛,π‘Ÿ). Hence (β†“π‘Œ+)β‹‚(π‘Œπ‘βˆ’)=βˆ…. This proves (B1).(B2)We show at first that β†‘π‘Œπ‘βˆ’β‹‚β†“π‘Œβˆ’=βˆ…. Since π‘Œπ‘βˆ’βŠ†π‘†Β±2̇⋃(𝑛,π‘Ÿ){Θ}, we have that β†‘π‘Œπ‘βˆ’βŠ†β†‘(𝑆±2⋃(𝑛,π‘Ÿ))β†‘Ξ˜. By Lemma 4.2(i) and (iii) we have then β†‘π‘Œπ‘βˆ’βŠ†π‘†Β±2̇⋃𝑆(𝑛,π‘Ÿ)+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ). On the other side, since β†“π‘‡βˆ’βŠ†β†“π‘†Β±1(𝑛,π‘Ÿ), by Lemma 4.2(iv) it follows that β†“π‘‡βˆ’βŠ†π‘†Β±1(𝑛,π‘Ÿ).β‹ƒπ‘†βˆ’1(𝑛,π‘Ÿ). By Lemma 4.2(ii), we have β†“πœƒ=π‘†βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ). Hence it holds β†“π‘Œβˆ’βŠ†π‘†Β±1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ). This proves that β†‘π‘Œπ‘βˆ’β‹‚β†“π‘Œβˆ’=βˆ…. We show now that also β†‘π‘Œ+β‹‚β†“π‘Œβˆ’=βˆ…; we proceed by contradiction. Let us suppose that there exists an element π‘§βˆˆβ†‘π‘Œ+β‹‚β†“π‘Œβˆ’, then there are two elements 𝑀+βˆˆπ‘Œ+ and π‘€βˆ’βˆˆπ‘Œβˆ’ such that 𝑀+βŠ‘π‘§βŠ‘π‘€βˆ’, hence 𝑀+βŠ‘π‘€βˆ’. We will distinguish the following five cases, and in each of them we will find a contradiction.(a)𝑀+∈{𝑣1,…,𝑣𝑠} and π‘€βˆ’βˆˆ{𝑣𝑠+1,…,π‘£π›½π‘˜+1}. In this case 𝑀+ and π‘€βˆ’ are two distinct elements having both rank π‘…βˆ’(π‘˜+1) and such that 𝑀+βŠ‘π‘€βˆ’; it is not possible.(b)𝑀+∈{𝑣1,…,𝑣𝑠} and π‘€βˆ’βˆˆ{𝑧1,…,π‘§π‘ž}. In this case 𝑀+ has rank π‘…βˆ’(π‘˜+1) while π‘€βˆ’ has rank π‘…βˆ’(π‘˜+2)<π‘…βˆ’(π‘˜+1), and this contradicts the condition 𝑀+βŠ‘π‘€βˆ’.(c)𝑀+∈{𝑀𝑑+1,…,π‘€π›½π‘˜} and π‘€βˆ’βˆˆ{𝑣𝑠+1,…,π‘£π›½π‘˜+1}.This case is similar to the previous because 𝑀+ has rank π‘…βˆ’π‘˜ while π‘€βˆ’ has rankπ‘…βˆ’(π‘˜+1).(d)𝑀+∈{𝑀𝑑+1,…,π‘€π›½π‘˜} and π‘€βˆ’βˆˆ{𝑧1,…,π‘§π‘ž}.This is similar to the previous because 𝑀+ has rank π‘…βˆ’π‘˜ while π‘€βˆ’ has rank π‘…βˆ’(π‘˜+2).(e)𝑀+∈{𝑣1,…,𝑣𝑠,𝑀𝑑+1,…,π‘€π›½π‘˜} and π‘€βˆ’=πœƒ. In this case the condition 𝑀+βŠ‘π‘€βˆ’ implies that 𝑀+βˆˆβ†“πœƒ; since β†“πœƒ=π‘†βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ) this is not possible since 𝑀+βˆˆπ‘†Β±1(𝑛,π‘Ÿ).(B3)Since 𝛼 is the minimum of 𝑆±2(𝑛,π‘Ÿ) we have that 𝑆±2(𝑛,π‘Ÿ)βŠ†β†‘π›Ό, moreover, π›Όβˆˆβ†‘π‘Œ+; therefore 𝑆±2(𝑛,π‘Ÿ)βŠ†β†‘π›ΌβŠ†β†‘π‘Œ+. Since Θ=πœƒπ‘βˆˆπ‘Œπ‘βˆ’, it follows that β†‘Ξ˜βŠ†β†‘π‘Œπ‘βˆ’. By Lemma 4.2(i) we have then that (𝑆+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ))=β†‘Ξ˜βŠ†β†‘π‘Œπ‘βˆ’. By Lemma 4.2(ii) we also have that (π‘†βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ))=β†“πœƒβŠ†β†“π‘Œβˆ’.
To complete the proof of (B3) let us observe that 𝑆±1⋃(𝑛,π‘Ÿ)=(π‘˜π‘–=0β„œπ‘–)Μ‡β‹ƒβ„œπ‘˜+1̇⋃(⋃𝑅𝑗=π‘˜+2β„œπ‘—), where (β‹ƒπ‘˜π‘–=0β„œπ‘–)βŠ†β†‘β„œπ‘˜βŠ†β†‘π‘Œ+, (⋃𝑅𝑗=π‘˜+2β„œπ‘—)βŠ†β†“β„œπ‘˜+2βŠ†β†“π‘Œβˆ’.
Moreover, since β„œπ‘˜+1={𝑣1,…,𝑣𝑠}̇⋃{𝑣𝑠+1,…,π‘£π›½π‘˜+1},with {𝑣1,…,𝑣𝑠}βŠ†π‘Œ+βŠ†β†‘π‘Œ+ and {𝑣𝑠+1,…,π‘£π›½π‘˜+1}βŠ†π‘Œβˆ’βŠ†β†“π‘Œβˆ’, then β„œπ‘˜+1βŠ†(β†‘π‘Œ+β‹ƒβ†“π‘Œβˆ’). This shows that 𝑆±1(𝑛,π‘Ÿ)βŠ†(β†‘π‘Œ+β‹ƒβ†“π‘Œβˆ’). Since the six sublattices 𝑆+𝑖(𝑛,π‘Ÿ), 𝑆±𝑖(𝑛,π‘Ÿ) and π‘†βˆ’π‘–(𝑛,π‘Ÿ) for 𝑖=1,2 are a partition of 𝑆(𝑛,π‘Ÿ), the property (B3) is proved.
Case (a2)
(B1)We begin to show that π›Όβˆ‰π‘Œπ‘βˆ’. In fact, it holds that 𝛼=(π‘Ÿ(π‘Ÿβˆ’1)β‹―20∣0β‹―0(π‘›βˆ’π‘Ÿ))𝑐; suppose that π›Όβˆˆπ‘Œπ‘βˆ’, and this shows that we obtain a contradiction. By π›Όβˆˆπ‘Œπ‘βˆ’ it follows 𝛼𝑐=𝑑1βˆˆπ‘Œβˆ’. But 𝑑1 is the top of 𝑆±1(𝑛,π‘Ÿ), so π‘Œβˆ’=𝑑1, since π‘Œβˆ’ is antichain. This means that β„œπ‘˜+1=β„œ0 and this is not possible, of course. Since π‘Œ+=𝑇+̇⋃{𝛼}, we have (β†“π‘Œ+)β‹‚π‘Œπ‘βˆ’=((↓𝑇+)β‹‚π‘Œπ‘βˆ’)β‹ƒβ‹‚π‘Œ((↓𝛼)π‘βˆ’), and, moreover, as in the proof of (B1) in the case (a1), we also have that (↓𝑇+)β‹‚π‘Œπ‘βˆ’=βˆ…; therefore, to prove (B1) it is sufficient to show that β‹‚π‘Œ(↓𝛼)π‘βˆ’=βˆ…; As in the proof of (B1) in the case (a1), we have π‘Œπ‘βˆ’βŠ†π‘†Β±2̇⋃(𝑛,π‘Ÿ){Θ}. Since 𝛼 is the minimum of 𝑆±2(𝑛,π‘Ÿ) and Ξ˜βˆ‰β†“π›Ό, it follows that the unique element of ↓𝛼 that can belong to π‘Œπ‘βˆ’ is 𝛼, but we have shown before that this is not true.(B2)As in the previous case we have (β†‘π‘Œπ‘βˆ’)β‹‚β†“π‘Œβˆ’=βˆ…; moreover (β†‘π‘Œ+)β‹‚(β†“π‘Œβˆ’)=((↑𝑇+)β‹‚β‹‚(↑𝛼))(β†“π‘Œβˆ’)((↑𝑇+)β‹‚(β†“π‘Œβˆ’β‹ƒβ‹‚))((↑𝛼)(β†“π‘Œβˆ’)). As in the case (a1) we have ((↑𝑇+)β‹‚(β†“π‘Œβˆ’))=βˆ…; therefore, to prove (B2) also in the case (a2), it is sufficient to show that β‹‚((↑𝛼)(β†“π‘Œβˆ’))=βˆ…. As in the proof of (B2) in the case (a1) it results that β†“π‘Œβˆ’βŠ†π‘†Β±2̇⋃𝑆(𝑛,π‘Ÿ)βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ), and, by definition of 𝛼, it is easy to observe that β‹‚(↑𝛼)(𝑆±1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ))=βˆ…. Hence β‹‚((↑𝛼)(β†“π‘Œβˆ’))=βˆ….(B3)It is identical to that of case (a1).
Step 3. The map A defined in (4.8) is such that π΄βˆˆπ’²+(𝑛,π‘Ÿ).
Since we have proved thatβŸ¨π‘Œ+βˆ£π‘Œβˆ’βŸ© is a basis for 𝑆(𝑛,π‘Ÿ), by Lemma 4.4 it follows that the map π΄βˆˆπ’²+(𝑆(𝑛,π‘Ÿ),𝟐) if the two following identities hold: ξ€·β†‘π‘Œ+ξ€Έξšξ€·β†‘π‘Œπ‘βˆ’ξ€Έ=𝑆±2Μ‡ξšS(𝑛,π‘Ÿ)+1Μ‡ξšπ‘†(𝑛,π‘Ÿ)+2Μ‡ξšπ”˜(𝑛,π‘Ÿ)π‘˜Μ‡ξšξ€½π‘£1,…,𝑣𝑠,(4.9)β†“π‘Œβˆ’=π”…π‘˜Μ‡ξšξ‚†π‘£π‘ +1,…,π‘£π›½π‘˜+1ξ‚‡Μ‡ξšπ‘†βˆ’1Μ‡ξšπ‘†(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ).(4.10)
We prove at first (4.10). By definition of π”…π‘˜ and π‘‡βˆ’ it easy to observe that π”…π‘˜Μ‡β‹ƒ{𝑣𝑠+1,…,π‘£π›½π‘˜+1}βŠ†β†“π‘‡βˆ’, and, moreover, by Lemma 4.2(ii) we also have that β†“πœƒ=π‘†βˆ’1⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ); hence, since β†“π‘Œβˆ’β‹ƒ=β†“πœƒβ†“π‘‡βˆ’, it results that π”…π‘˜Μ‡β‹ƒ{𝑣𝑠+1,…,π‘£π›½π‘˜+1}β‹ƒπ‘†βˆ’1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’2(𝑛,π‘Ÿ)βŠ†β†“π‘Œβˆ’. On the other hand, by Lemma 4.2(iv) we have that β†“π‘‡βˆ’βŠ†π‘†Β±1̇⋃𝑆(𝑛,π‘Ÿ)βˆ’1(𝑛,π‘Ÿ) because π‘‡βˆ’ is a subset of 𝑆±1(𝑛,π‘Ÿ). At this point let us note that the elements of β†“π‘‡βˆ’ that are also in 𝑆±1(𝑛,π‘Ÿ) must belong necessarily to the subset π”…π‘˜Μ‡β‹ƒ{𝑣𝑠+1,…,π‘£π›½π‘˜+1}. This proves the other inclusion and hence (4.10).
To prove now (4.9) we must distinguish the cases (a1) and (a2). We set Ξ”βˆΆ=𝑆±2̇⋃𝑆(𝑛,π‘Ÿ)+1̇⋃𝑆(𝑛,π‘Ÿ)+2β‹ƒπ”˜(𝑛,π‘Ÿ)π‘˜Μ‡β‹ƒ{𝑣1,…,𝑣𝑠}. Let us first examine the case (a1). Since 𝛼 is the minimum of 𝑆±2(𝑛,π‘Ÿ), we have 𝑆±2(𝑛,π‘Ÿ)βŠ†β†‘π›ΌβŠ†β†‘π‘Œ+. Moreover, since π‘Œβˆ’=π‘‡βˆ’β‹ƒ{πœƒ}, it follows that β†‘π‘Œπ‘βˆ’βŠ‡β†‘πœƒπ‘=↑(Θ)=𝑆+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ) by lemma 3.2(i). Finally, since π”˜π‘˜Μ‡β‹ƒ{𝑣1,…,𝑣𝑠}βŠ†β†‘π‘‡+=β†‘π‘Œ+, the inclusion βŠ‡ in (4.9) is proved. To prove the other inclusion βŠ† we begin to observe that β†‘π‘Œ+β‹‚(π‘†βˆ’1⋃𝑆(𝑛,π‘Ÿ)βˆ’1(𝑛,π‘Ÿ))=βˆ…; therefore the elements of β†‘π‘Œ+ that are not in 𝑆±2̇⋃𝑆(𝑛,π‘Ÿ)+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ) must be necessarily in 𝑆±1(𝑛,π‘Ÿ), and such elements, by definition of 𝑇+, must be necessarily in π”˜π‘˜Μ‡β‹ƒ{𝑣1,…,𝑣𝑠}.This proves that β†‘π‘Œ+βŠ†Ξ”, For ↑(π‘Œπ‘βˆ’), we have ↑(π‘Œπ‘βˆ’)=↑(π‘‡π‘βˆ’β‹ƒ{πœƒ}𝑐)=(β†‘π‘‡π‘βˆ’)⋃(β†‘Ξ˜), where ↑(Θ)=𝑆+1̇⋃(𝑛,π‘Ÿ)𝑆+2(𝑛,π‘Ÿ) by Lemma 4.2(i), and, since π‘‡βˆ’βŠ†π‘†Β±1(𝑛,π‘Ÿ), also π‘‡π‘βˆ’βŠ†(𝑆±1(𝑛,π‘Ÿ))𝑐=𝑆±2(𝑛,π‘Ÿ), by Lemma 4.2(v). Therefore β†‘π‘‡π‘βˆ’βŠ†β†‘π‘†Β±2(𝑛,π‘Ÿ)βŠ†π‘†Β±2⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ) by Lemma 4.2(iii). This shows that ↑(π‘Œπ‘βˆ’)βŠ†π‘†Β±2⋃𝑆(𝑛,π‘Ÿ)+2⋃𝑆(𝑛,π‘Ÿ)+1(𝑛,π‘Ÿ)βŠ†Ξ”, hence, the inclusion βŠ†. The proof of (4.9) in the case (a1) is therefore complete.
Finally, to prove (4.9) in the case (a2), it easy to observe that the only difference with respect to case (a1) is when we must show that β†‘π‘Œ+βŠ†Ξ”. In fact, in the case (a2) it results that π›Όβˆ‰β†‘π‘‡+ and π‘Œ+=𝑇+⋃{𝛼}, while π‘Œβˆ’ is the same in both cases (a1) and (a2). Therefore, in the case (a2), the elements of β†‘π‘Œ+ that are not in 𝑆±2̇⋃𝑆(𝑛,π‘Ÿ)+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ) must be in (↑𝑇+)⋂𝑆±1(𝑛,π‘Ÿ) or in ⋂𝑆(↑𝛼)Β±1(𝑛,π‘Ÿ). As in the case (a1) we have (↑𝑇+)⋂𝑆±1(𝑛,π‘Ÿ)=π”˜π‘˜Μ‡β‹ƒ{𝑣1,…,𝑣𝑠}, and, since 𝛼 is the minimum of 𝑆±2(𝑛,π‘Ÿ), it results that ↑𝛼=↑𝑆±2(𝑛,π‘Ÿ)βŠ†π‘†Β±2⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ) by Lemma 4.2(iii); hence ⋂𝑆(↑𝛼)Β±1(𝑛,π‘Ÿ)=βˆ…. Therefore, also in the case (a2), the elements of β†‘π‘Œ+ that are not in 𝑆±2̇⋃𝑆(𝑛,π‘Ÿ)+1̇⋃𝑆(𝑛,π‘Ÿ)+2(𝑛,π‘Ÿ) must be necessarily in π”˜π‘˜Μ‡β‹ƒ{𝑣1,…,𝑣𝑠}. The other parts of the proof are same as in case (a1). Hence we have proved the identities (4.9) and (4.10). By Lemma 4.4 it follows then that the map π΄βˆˆπ’²+(𝑆(𝑛,π‘Ÿ),𝟐). Finally, by definition of 𝐴, we have obviously 𝐴(πœƒ)=𝑁, 𝐴(πœ‰1)=𝑃, and 𝐴(Θ)=𝑃. This shows that π΄βˆˆπ’²+(𝑛,π‘Ÿ). The proof is complete.

Figure 3

To conclude we emphasize the elegant symmetry of the induced partitions on 𝑆(𝑛,π‘Ÿ) from the boolean total maps π΄π‘žβ€™s constructed in the proof of the Theorem 4.5.


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