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# Families of sets not belonging to algebras and combinatorics of finite sets of ultrafilters

Journal of Inequalities and Applications20152015:116

https://doi.org/10.1186/s13660-015-0578-7

• Accepted: 27 January 2015
• Published:

## Abstract

This article is a part of the theory developed by the author in which the following problem is solved under natural assumptions: to find necessary and sufficient conditions under which the union of at most countable family of algebras on a certain set X is equal to $$\mathcal{P}(X)$$. Here the following new result is proved. Let $$\{\mathcal{A}_{\lambda }\}_{\lambda \in \Lambda }$$ be a finite collection of algebras of sets given on a set X with $$\# (\Lambda ) =n>0$$, and for each λ there exist at least $$\frac{10}{3}n+\sqrt{\frac{2n}{3}}$$ pairwise disjoint sets belonging to $$\mathcal{P}(X)\setminus\mathcal{A}_{\lambda }$$. Then there exists a family $$\{U^{1}_{\lambda }, U^{2}_{\lambda }\}_{\lambda \in \Lambda }$$ of pairwise disjoint subsets of X ($$U^{i}_{\lambda }\cap U^{j}_{\lambda '}=\emptyset$$ except the case $$\lambda =\lambda '$$, $$i=j$$); and for each λ the following holds: if $$Q\in \mathcal{P}(X)$$ and Q contains one of the two sets $$U^{1}_{\lambda }$$, $$U^{2}_{\lambda }$$, and its intersection with the other set is empty, then $$Q\notin \mathcal{A}_{\lambda }$$.

## Keywords

• algebra of sets
• σ-algebra
• ultrafilter
• pairwise disjoint sets

• 03E05
• 54D35

## 1 Introduction

The present article is a further development of the theory formulated in . The topic studied in these articles, as well as in the present paper, is sets not belonging to algebras of sets.

### Definition 1.1

An algebra $$\mathcal{A}$$ on a set X is a non-empty family of subsets of X possessing the following properties: (1) if $$M\in \mathcal{A}$$, then $$X \setminus M\in \mathcal{A}$$; (2) if $$M_{1}, M_{2} \in \mathcal{A}$$, then $$M_{1} \cup M_{2} \in \mathcal{A}$$.

It is clear that if $$M_{1}, M_{2}\in\mathcal{A}$$, then $$M_{1}\cap M_{2}\in \mathcal{A}$$ and $$M_{1}\setminus M_{2}\in\mathcal{A}$$; also, it is clear that $$X\in\mathcal{A}$$.

### 1.1 Some notation and names

All algebras and measures are considered on some abstract set $$X \neq\emptyset$$. When it is clear from the context, we will not state explicitly that a set belongs to the family $$\mathcal{P}(X)$$ of all subsets of X. By $$\mathbb{N}^{+}$$ we denote the set of natural numbers. If $$n_{1}, n_{2} \in \mathbb{N}^{+}$$ and $$n_{1} \leq n_{2}$$, then $$[n_{1}, n_{2}] = \{k \in \mathbb{N}^{+}\mid n_{1} \leq k \leq n_{2} \}$$. Let ρ be a real number. By $$\lfloor\rho\rfloor$$ we denote the maximum integer ≤ρ. By $$\lceil\rho\rceil$$ we denote the minimum integer ≥ρ. The symbol $$\#(M)$$ denotes the cardinality of the set M. A set M is countable if $$\#(M) = \aleph_{0}$$.

The following concept was used in .

### Definition 1.2

An algebra $$\mathcal{A}$$ has κ lacunae, where κ is a cardinal number, if there exist κ pairwise disjoint sets not belonging to $$\mathcal{A}$$.

Let $$\{\mathcal{A}_{\lambda }\}_{\lambda \in \Lambda }$$ be a family of algebras and $$\mathcal{A}_{\lambda }\neq \mathcal{P}(X)$$ for each $$\lambda \in \Lambda$$. The following natural question arises: what are possible conditions that distinguish between the cases $$\bigcup_{\lambda \in \Lambda } \mathcal{A}_{\lambda }\neq \mathcal{P}(X)$$ and $$\bigcup_{\lambda \in \Lambda } \mathcal{A}_{\lambda }= \mathcal{P}(X)$$? Let $$\# (\Lambda )\leq\aleph_{0}$$, and let us assume that $$\mathcal{A}_{\lambda }$$ are σ-algebras if $$\# (\Lambda ) = \aleph_{0}$$. In  we obtained necessary and sufficient conditions for the equality $$\bigcup_{\lambda \in \Lambda } \mathcal{A}_{\lambda }= \mathcal{P}(X)$$ to hold. The first publication connected with this topic was that of Erdös  (this paper contains the well-known theorem of Alouglu-Erdös). Some information about the history of the subject after the publication of  and before the publication of  is presented in . In fact, Alouglu and Erdös studied non-measurable sets with respect to families of measures. Let $$\aleph_{1}\le\#(X) \le2^{\aleph_{0}}$$. Let a σ-additive measure μ be defined on X. Here $$\mu(X) = 1$$, the measure of a one-point set equals 0, and the measure of each μ-measurable set equals 0 or 1. Such a measure μ is called a σ-two-valued measure. Clearly, there exist μ-non-measurable sets. The Alouglu-Erdös theorem states that if $$\#(X)=\aleph_{1}$$, then for any countable family of σ-two-valued measures $$\mu_{1}, \ldots, \mu_{k}, \ldots$$ there exists a set which is non-measurable with respect to all these measures. The proof of the Alouglu-Erdös theorem is very simple and is based on the possibility of constructing the well-known Ulam matrix (see ). The non-trivial Gitik-Shelah theorem (see ) asserts the validity of the Alouglu-Erdös theorem if $$\#(X)=2^{\aleph_{0}}$$. Obviously, the Gitik-Shelah theorem is a generalization of the Alouglu-Erdös theorem. The Gitik-Shelah theorem can be reformulated in our language. As before, let us consider the σ-two-valued measures $$\mu_{1}, \ldots, \mu_{k}, \ldots$$ . For each measure $$\mu_{k}$$, we examine the algebra $$\mathcal{A}_{k}$$ of all $$\mu_{k}$$ measurable sets. The Gitik-Shelah theorem asserts that $$\bigcup_{k \in \mathbb{N}^{+}} \mathcal{A}_{k} \neq \mathcal{P} (X)$$. We note that here each algebra $$\mathcal{A}_{k}$$ has $$\aleph_{0}$$ lacunae. If $$\#(X) = \aleph_{1}$$, then the situation is much simpler: each algebra $$\mathcal{A}_{k}$$ has $$\aleph_{1}$$ lacunae. The Gitik-Shelah theorem is used in the proofs of our theorems for countable families of σ-algebras.

### Definition 1.3

Let $$\{\mathcal{A}_{\lambda }\}_{\lambda \in \Lambda }$$ be a family of algebras, and $$\{U^{1}_{\lambda }, U^{2}_{\lambda }\}_{\lambda \in \Lambda }$$ be a family of sets with the following properties:
1. (1)

$$U^{i}_{\lambda }\cap U^{j}_{\lambda '} = \emptyset$$ except when $$\lambda = \lambda '$$, $$i = j$$;

2. (2)

for any $$\lambda \in \Lambda$$, the following holds: if a set Q contains one of the two sets $$U^{1}_{\lambda }$$, $$U^{2}_{\lambda }$$ and its intersection with the other set is empty, then $$Q \notin \mathcal{A}_{\lambda }$$.

Then we say that the family $$\{\mathcal{A}_{\lambda }\}_{\lambda \in \Lambda }$$ has the full set of lacunae $$\{U^{1}_{\lambda }, U^{2}_{\lambda }\}_{\lambda \in \Lambda }$$.

Now we give a simple proposition.

### Proposition 1.4

If a family of algebras $$\{\mathcal{A}_{\lambda }\}_{\lambda \in \Lambda }$$ has the full set of lacunae $$\{U^{1}_{\lambda }, U^{2}_{\lambda }\}_{\lambda \in \Lambda }$$, then there exists a family of pairwise distinct sets $$\{Q_{\vartheta}\}_{\vartheta\in\Theta}$$ such that the following holds:
1. (1)

$$Q_{\vartheta}\notin\bigcup_{\lambda \in \Lambda } \mathcal{A}_{\lambda }$$ for any $$\vartheta\in \Theta$$;

2. (2)

any set $$Q_{\vartheta}$$ is a union of sets $$U^{i}_{\lambda }$$;

3. (3)

$$Q_{\theta_{1}} \setminus Q_{\vartheta_{2}} \notin \bigcap_{\lambda \in \Lambda } \mathcal{A}_{\lambda }$$ for any pair $$\vartheta_{1} \neq\vartheta_{2}$$;

4. (4)

$$\#(\Theta) = 2^{\#(\Lambda )}$$.

### Proof

Put $$\Theta= \mathcal{P}(\Lambda )$$. If $$\vartheta\in \mathcal{P}(\Lambda )$$, put
$$Q_{\vartheta}= \biggl( \bigcup_{\lambda \in\vartheta} U^{1}_{\lambda }\biggr) \cup \biggl(\bigcup _{\lambda\in \Lambda \setminus\vartheta} U^{2}_{\lambda }\biggr).$$
□

In this paper we deal mostly with the following problem: under which conditions a family of algebras $$\{\mathcal{A}_{\lambda }\}_{\lambda \in \Lambda }$$ has a full set of lacunae. We assume that $$\#(\Lambda ) \leq \aleph_{0}$$. This was studied in . The proof of the two following theorems can be found in .

### Theorem 1.5

Let $$\mathcal{A}_{1}, \ldots , \mathcal{A}_{n}$$ be a finite family of algebras, and assume that for each $$k \in[1,n]$$ the algebra $$\mathcal{A}_{k}$$ has $$4k-3$$ lacunae. Then this family has a full set of lacunae.

It is easy to prove (see , Chapter 14) that the estimate $$4k-3$$ is the best possible in some sense.

### Theorem 1.6

Let $$\{\mathcal{A}_{k}\}_{k \in \mathbb{N}^{+}}$$ be a family of σ-algebras, and assume that for each k the algebra $$\mathcal{A}_{k}$$ has $$4k-3$$ lacunae. Then this family has some full set of lacunae.

### Remark 1.7

Using the notion of absolute introduced by Gleason in , we can construct a family of algebras $$\{\mathcal{B}_{k}\}_{k \in \mathbb{N}^{+}}$$ with the following properties: each algebra $$\mathcal{B}_{k}$$ has $$\aleph_{0}$$ lacunae, is not a σ-algebra, and $$\bigcup_{k \in \mathbb{N}^{+}} \mathcal{B}_{k} = \mathcal{P}(X)$$ (see , Chapter 5). Hence, Theorem 1.6 and Theorem 2.4 below do not hold if we claim them for algebras which are not assumed to be σ-additive. Therefore, we suppose that all algebras of a countable family of algebras are σ-algebras.

The following definition was given in .

### Definition 1.8

For each $$n \in \mathbb{N}^{+}$$, denote by $$\frak{v}(n)$$ the minimal cardinal number such that if $$\{\mathcal{A}_{\lambda }\}_{\lambda \in \Lambda }$$, $$\#(\Lambda ) = n$$, is a family of algebras, and for each $$\lambda \in \Lambda$$ the algebra $$\mathcal{A}_{\lambda }$$ has $$\frak{v}(n)$$ lacunae, then the family $$\{\mathcal{A}_{\lambda }\}_{\lambda \in \Lambda }$$ has a full set of lacunae.

In  we proved that:
1. (1)

$$\frak{v}(n) = 4n-3$$ for $$n \leq3$$;

2. (2)

$$\frak{v}(n) \leq4n-5$$ for $$n > 3$$;

3. (3)

$$\frak{v}(n) \leq4n - \lfloor\frac{n+3}{2} \rfloor$$ for any n;

4. (4)

$$3n-2 \leq \frak{v}(n)$$ for any n.

In this paper we will improve the upper bound of $$\frak{v}(n)$$.

From here until the end of Section 1 we present propositions and notions which form the method of proofs of our theorems. This method first appeared in  and was later used in . Let βX be the Stone-Čech compactification of X with the discrete topology; βX is the family of all ultrafilters on X.

Consider an algebra $$\mathcal{A}$$. We say that $$a,b\in\beta X$$ are $$\mathcal{A}$$-equivalent iff $$a\cap\mathcal{A}=b\cap\mathcal{A}$$. Let $$[b ]_{\mathcal{A}}$$ denote the $$\mathcal{A}$$-equivalence class of b, and define the kernel of the algebra $$\mathcal{A}$$:
$$\ker\mathcal{A}= \bigl\{ b\in\beta X \mid \# \bigl( [b ]_{\mathcal{A}} \bigr)>1 \bigr\} .$$
If $$\mathcal{A}=\mathcal{P}(X)$$, then $$\ker\mathcal{A}=\emptyset$$. From now on, when we say a and b are $$\mathcal{A}$$-equivalent ultrafilters, we always assume that $$a\neq b$$. If a, b are $$\mathcal{A}$$-equivalent ultrafilters, then we say that a has an $$\mathcal{A}$$ -equivalent ultrafilter b, or a is $$\mathcal{A}$$-equivalent to b.

### Statement 1.9

Consider an algebra $$\mathcal{A}$$ and sets $$U,V \in \mathcal{P}(X)$$ such that $$U \cap V = \emptyset$$. The following two conditions are equivalent. (1) Each set Q containing one of the sets U, V and being disjoint from the other does not belong to $$\mathcal{A}$$. (2) There exist $$\mathcal{A}$$-equivalent ultrafilters a, b such that $$U \in a$$, $$V \in b$$.

### Proof

It is obvious that (1) follows from (2). Let us prove that (2) follows from (1). Let us assume the contrary. We fix an ultrafilter $$q\ni U$$. For any ultrafilter $$r \ni V$$, we choose a set $$W(r) \in r$$ such that $$W(r) \in \mathcal{A}$$ and $$W(r) \notin q$$. Since the set of all ultrafilters which contain V is a compact subset of βX, there exists a finite sequence of sets $$W(r_{1}), \ldots , W(r_{m})$$ with the following properties:
1. (1)

$$W(r_{k}) \in \mathcal{A}$$ for any $$k \in [1,m]$$;

2. (2)

$$W(r_{k}) \notin q$$ for any $$k \in[1,m]$$;

3. (3)

$$V \subseteq \bigcup^{m}_{k=1} W (r_{k})$$.

Let
$$\widetilde{W}(q) = X \Big\backslash \bigcup^{m}_{k=1} W (r_{k}) .$$
It is clear that $$\widetilde{W}(q) \in q$$, $$\widetilde{W}(q) \in \mathcal{A}$$, and $$\widetilde{W}(q) \cap V = \emptyset$$. Since the set of all ultrafilters which contain U is a compact subset of βX, there exists a finite sequence of sets $$\widetilde{W}(q_{1}), \ldots, \widetilde{W}(q_{n})$$ such that $$\widetilde{W}(q_{k}) \in \mathcal{A}$$ for any $$k \in[1,n]$$, $$\bigcup^{n}_{k=1} \widetilde{W}(q_{k}) = \widetilde{W} \supseteq U$$, and $$\widetilde{W} \cap V = \emptyset$$. We have $$\widetilde{W} \in \mathcal{A}$$, a contradiction. □

The following crucial claim is a direct consequence of Statement 1.9.

### Claim 1.10

Consider an algebra $$\mathcal{A}$$ and $$U\in\mathcal{P}(X)$$. Then $$U\notin\mathcal{A}$$ if and only if there exist $$\mathcal{A}$$-equivalent ultrafilters p and q such that $$U\in p$$ and $$U\notin q$$.

### Proof

The sufficiency is obvious. If $$U\notin\mathcal{A}$$, then the sets U and $$V=X\setminus U$$ satisfy the condition (1) of Statement 1.9. Therefore, there exist the corresponding ultrafilters p and q. □

It is clear that if $$\mathcal{A}\neq \mathcal{P}(X)$$, then $$\# (\ker\mathcal{A} )\geqslant2$$. It is rather easy to show that an algebra $$\mathcal{A}$$ has k lacunae, where $$2\leqslant k\leqslant\aleph_{0}$$, if and only if $$\# (\ker\mathcal{A} )\geqslant k$$.1

### Definition 1.11

A set $$M \subseteq\beta X$$ is said to be $$\mathcal{A}$$-equivalent if $$\#(M) > 1$$, any two distinct ultrafilters in M are $$\mathcal{A}$$-equivalent, and there exist no $$\mathcal{A}$$-equivalent ultrafilters a, b such that $$a \in M$$, $$b \notin M$$.

Obviously, an $$\mathcal{A}$$-equivalent set has the form $$[b ]_{\mathcal{A}}$$ (see above). Also it is obvious that an $$\mathcal{A}$$-equivalent set is closed in βX.

### Remark 1.12

Consider algebras $$\mathcal{A}$$, . It is very easy to prove that the following statements are equivalent.
1. (1)

$$\mathcal{A}\supseteq\mathcal{B}$$.

2. (2)

If a and b are $$\mathcal{A}$$-equivalent ultrafilters, then a and b are -equivalent ultrafilters.

3. (3)

If M is an $$\mathcal{A}$$-equivalent set, then M is contained in a certain -equivalent set.

### Remark 1.13

If $$M \subseteq\beta X$$ (in particular, if $$M \subseteq X$$), then by $$\overline{M}$$ we denote the closure M in βX. The following arguments will be used later in this paper. Let $$A \subseteq\beta X$$, $$2 \leq\# (A) < \aleph_{0}$$. The set A is divided into pairwise disjoint sets $$A_{1},\ldots , A_{m}$$ and $$\# (A_{k}) > 1$$ for each $$k \in[1,m]$$. Two different ultrafilters are called a-equivalent if and only if they belong to the same set $$A_{k}$$. We can construct the algebra $$\mathcal{A}$$ such that the a-equivalence relation is in fact the $$\mathcal{A}$$-equivalence relation, $$\ker \mathcal{A}=A$$, and $$A_{1}, \ldots, A_{m}$$ are all $$\mathcal{A}$$-equivalent sets. Indeed, by definition $$M \in \mathcal{A}$$ if and only if for each $$k \in[1,m]$$ either $$A_{k} \cap\overline{M} = \emptyset$$, or $$A_{k} \subseteq\overline{M}$$.

### Remark 1.14

Let us recall that an algebra which does not have $$\aleph_{0}$$ lacunae is called ω-saturated. So, an algebra $$\mathcal{A}$$ is ω-saturated if and only if $$\# (\ker\mathcal{A} )<\aleph_{0}$$. The algebra $$\mathcal{A}$$ from Remark 1.13 is ω-saturated.

### Remark 1.15

Further we use two following very simple statements. (1) By Statement 1.9 a finite family of algebras $$\mathcal{A}_{1},\ldots,\mathcal{A}_{n}$$ has a full set of lacunae if and only if there exist 2n pairwise distinct ultrafilters $$a_{1},\ldots,a_{n},b_{1},\ldots,b_{n}$$ such that $$a_{k}$$, $$b_{k}$$ are $$\mathcal{A}_{k}$$-equivalent ultrafilters for each $$k\in[1,n]$$. (2) Let $$\mathfrak{A}= \{\mathcal{A}_{\lambda}\}_{\lambda \in\Lambda}$$ and $$\mathfrak{A}'= \{\mathcal{A}_{\lambda}' \} _{\lambda\in\Lambda}$$ be two non-empty families of algebras, and $$\mathcal{A}_{\lambda}'\supseteq\mathcal{A}_{\lambda}$$ for every $$\lambda\in \Lambda$$. Assume that the family $$\mathfrak{A}'$$ has a full set of lacunae $$\{U^{1}_{\lambda},U^{2}_{\lambda}\}_{\lambda\in\Lambda}$$. Then the family $$\mathfrak{A}$$ has the same full set of lacunae $$\{ U^{1}_{\lambda},U^{2}_{\lambda}\}_{\lambda\in\Lambda}$$.

## 2 Main results. An open problem

The following result was announced in : $$\frak{v}(n) \leq \lceil\frac{10}{3} n + \frac{2}{\sqrt{3}} \sqrt{n} \rceil$$ for any n. In this paper a stronger theorem is proved.

### Theorem 2.1

$$\frak{v}(n) \leq \lceil\frac{10}{3} n + \sqrt{\frac{2n}{3}} \rceil$$.

### Remark 2.2

The combinatorial nature of Theorem 2.1 is discussed in Section 4. Also in Section 4 the proof of Theorem 4.5 uses the classical Ramsey theorem.

### Problem 2.3

We know that $$\frak{v}(n) \geq3 n-2$$ for any n, and $$\frak{v}(n) > 3n-2$$ if $$n = 2,3$$ since $$\frak{v}(2) = 5$$, $$\frak{v}(3) = 9$$ (see Section 1). Is it true that $$\frak{v}(n) = 3n-2$$ for any $$n \neq2,3$$? This result is obviously true for $$n=1$$.

### Theorem 2.4

It is possible to construct nondecreasing functions $$\varphi: \mathbb{N}^{+} \rightarrow\mathbb{N}^{+}$$ such that the following conditions hold:
1. (1)

$$\underline{\lim}_{n \rightarrow\infty} \frac{\varphi(n) - \frac{10}{3}n }{\sqrt{n}} = \sqrt{\frac{2}{3}}$$;

2. (2)

if $$\{ {\mathcal{A}}_{k} \}_{k\in\mathbb{N}^{+}}$$ is a family of σ-algebras and each algebra $${\mathcal{A}}_{k}$$ has $$\varphi(k)$$ lacunae, then this family has a full set of lacunae.

## 3 Finite families of algebras. Proof of Theorem 2.1

The following lemma is used in the proof of Lemma 3.2.

### Lemma 3.1

Consider an algebra $$\mathcal{A}$$ which is not ω-saturated;2 let a number $$\xi\in\mathbb{N}^{+}$$ be given. Then it is possible to construct an ω-saturated algebra $$\mathcal{A}'$$ such that $$\# (\ker\mathcal{A}' )\geqslant\xi$$ and $$\mathcal{A}'\supset\mathcal{A}$$.

### Proof

Take two distinct $$\mathcal{A}$$-equivalent ultrafilters $$s_{1}$$, $$t_{1}$$. Consider two distinct ultrafilters $$a_{1},a_{2}\in\ker \mathcal{A}\setminus\{s_{1},t_{1}\}$$. If $$a_{1}$$ has an $$\mathcal{A}$$-equivalent ultrafilter in $$\{s_{1},t_{1}\}$$, and $$a_{2}$$ has an $$\mathcal{A}$$-equivalent ultrafilter in $$\{s_{1},t_{1}\}$$, then $$a_{1}$$ and $$a_{2}$$ are $$\mathcal{A}$$-equivalent ultrafilters. Denote $$s_{2}=a_{1}$$, $$t_{2}=a_{2}$$. If, for example, $$a_{1}$$ does not have an $$\mathcal{A}$$-equivalent ultrafilter in $$\{s_{1},t_{1}\}$$, then take an ultrafilter c such that $$a_{1}\neq c$$ and $$a_{1}$$, c are $$\mathcal{A}$$-equivalent ultrafilters. In this case denote $$s_{2}=a_{1}$$, $$t_{2}=c$$. Now take three pairwise disjoint ultrafilters $$b_{1}$$, $$b_{2}$$, $$b_{3}\in\ker\mathcal{A}\setminus\{ s_{1},t_{1},s_{2},t_{2}\}$$. If every ultrafilter $$b_{i}$$ has an $$\mathcal{A}$$-equivalent ultrafilter in $$\{s_{1},t_{1},s_{2},t_{2}\}$$, then in the set $$\{ b_{1},b_{2},b_{3}\}$$ we can choose two distinct $$\mathcal{A}$$-equivalent ultrafilters, for example, $$b_{1}$$ and $$b_{2}$$. Put $$s_{3}=b_{1}$$, $$t_{3}=b_{2}$$. If, for example, $$b_{1}$$ does not have an $$\mathcal{A}$$-equivalent ultrafilter in $$\{s_{1},t_{1},s_{2},t_{2}\}$$, then take an ultrafilter d such that $$b_{1}\neq d$$ and $$b_{1}$$, d are $$\mathcal{A}$$-equivalent ultrafilters. Denote $$s_{3}=b_{1}$$, $$t_{3}=d$$. It is clear that for every $$\ell\in\mathbb{N}^{+}$$ it is possible to construct a sequence of pairwise distinct ultrafilters $$s_{1} , t_{1} , \ldots , s_{\ell}, t_{\ell}$$ such that $$s_{i}$$ and $$t_{i}$$ are $$\mathcal{A}$$-equivalent ultrafilters for all $$i\in[1,\ell]$$. Let $$2\ell\geqslant\xi$$. Define $$M_{1}=\{s_{1},t_{1}\} , \ldots , M_{\ell}=\{s_{\ell},t_{\ell}\}$$. By Remark 1.13 it is possible to construct an algebra $$\mathcal{A}'$$ such that $$\ker\mathcal{A}'=\bigcup_{i=1}^{\ell}M_{i}$$ and $$M_{1},\ldots ,M_{\ell}$$ are $$\mathcal{A}'$$-equivalent sets. □

The following lemma is given in  without proof.

### Lemma 3.2

$$\frak{v}(n) \in \mathbb{N}^{+}$$, and $$\frak{v}(n+1) - \frak{v}(n) \leq 4$$.

### Proof

It is obvious that $$\frak{v}(1) = 1$$. Let $$n \in \mathbb{N}^{+}$$ and assume that $$\frak{v}(n) \in \mathbb{N}^{+}$$. Consider a family of algebras $$\mathcal{A}_{1}, \ldots , \mathcal{A}_{n+1}$$ with $$\# (\ker \mathcal{A}_{k}) \geq \frak{v}(n) + 4$$ for each $$k \in[1,n+1]$$. We must prove that this family has a full set of lacunae. By Lemma 3.1 and the arguments in Remark 1.15 we can assume that the algebras $$\mathcal{A}_{1},\ldots,\mathcal{A}_{n+1}$$ are ω-saturated. We choose $$\mathcal{A}_{n+1}$$-equivalent ultrafilters $$s^{(1)}_{n+1}$$, $$s^{(2)}_{n+1}$$. Put $$B_{k} = \ker \mathcal{A}_{k} \setminus\{s^{(1)}_{n+1} , s^{(2)}_{n+1} \}$$ for each $$k \in[1,n]$$. Put
\begin{aligned} B'_{k} ={}& \bigl\{ q \in B_{k} \mid q \mbox{ does not have an }A_{k}\mbox{-equivalent ultrafilter in } B_{k} \\ &{}\mbox{and has an }A_{k}\mbox{-equivalent ultrafilter in } \bigl\{ s^{(1)}_{n+1} , s^{(2)}_{n+1} \bigr\} \bigr\} . \end{aligned}
It is clear that $$\# (B'_{k}) \leq2$$. Put $$B''_{k} = B_{k} \setminus B'_{k}$$. Clearly, each ultrafilter in $$B''_{k}$$ has an $$\mathcal{A}_{k}$$-equivalent ultrafilter in $$B''_{k}$$. Therefore, by Remark 1.13, we can construct an algebra $$\mathcal{A}'_{k}$$ such that $$\ker \mathcal{A}'_{k} = B''_{k}$$ and the $$\mathcal{A}'_{k}$$-equivalent relation in $$\ker \mathcal{A}'_{k}$$ is in fact the $$\mathcal{A}_{k}$$-equivalent relation. We have $$\# (\ker \mathcal{A}'_{k}) \geq \frak{v}(n)$$ for each $$k \in[1,n]$$. Therefore, there exist 2n pairwise distinct ultrafilters $$s^{(1)}_{1}, s^{(2)}_{1},\ldots,s^{(1)}_{n}, s^{(2)}_{n}$$, and $$s^{(1)}_{k}$$, $$s^{(2)}_{k}$$ are $$\mathcal{A}_{k}$$-equivalent ultrafilters from $$\ker \mathcal{A}'_{k}$$. We have pairwise distinct ultrafilters $$s^{(1)}_{1}, s^{(2)}_{1}, \ldots ,s^{(1)}_{n+1}, s^{(2)}_{n+1}$$, and $$s^{(1)}_{k}$$, $$s^{(2)}_{k}$$ are $$\mathcal{A}_{k}$$-equivalent ultrafilters for each $$k \in[1,n+1]$$. □

### Remark 3.3

It is obvious that $$\frak{v}(1) = 1$$. Therefore, by Lemma 3.2 we have $$\frak{v}(n) \leq4n -3$$ for any n. In Chapter 14, , we proved that $$\frak{v}(4) \leq11$$. Therefore, by Lemma 3.2, we have that $$\frak{v}(n) \leq4n -5$$ for any $$n\geq4$$.

We now turn to the proof of Theorem 2.1. This proof is a strong improvement of the proposition $$\mathfrak{v}(n) \leq 4n - \lfloor \frac{n+3}{2} \rfloor$$ mentioned above (see , Chapter 14).

### Proof of Theorem 2.1

(1) By Remark 3.3 our theorem is true for all $$n \leq13$$. (This can be verified by a simple computation.) Fix a natural number $$n \geq14$$ and a real number
$$\omega(n) \geq\sqrt{\frac{2n}{3}}.$$
Let $$\mathcal{A}_{1} , \ldots , \mathcal{A}_{n}$$ be algebras such that
$$\# (\ker \mathcal{A}_{k} ) \geq\frac{10}{3} n + \omega(n)$$
for each $$k \in[1,n]$$. By Lemma 3.1 and arguments in Remark 1.15, we can assume that the algebras $$\mathcal{A}_{1},\ldots, \mathcal{A}_{n}$$ are ω-saturated. We will prove that there exist pairwise distinct ultrafilters
$$a^{*}_{1}, \ldots , a^{*}_{n}, b^{*}_{1}, \ldots , b^{*}_{n}$$
such that $$a^{*}_{k}$$, $$b^{*}_{k}$$ are $$\mathcal{A}_{k}$$-equivalent ultrafilters for each $$k \in[1,n]$$. Our goal is to contradict the assumption that ultrafilters $$a^{*}_{1}, \ldots , a^{*}_{n}, b^{*}_{1}, \ldots , b^{*}_{n}$$ do not exist. Inductively assume that
$$\frak{v}(n-1) \leq \biggl\lceil \frac{10}{3} (n - 1)+ \sqrt{\frac{2n-2}{3}} \biggr\rceil .$$
Then there exists a set of pairwise distinct ultrafilters
$$\mathfrak{F}= \{a_{1}, \ldots , a_{n-1} , b_{1}, \ldots , b_{n-1} \}$$
such that $$a_{k}$$, $$b_{k}$$ are $$\mathcal{A}_{k}$$-equivalent ultrafilters for each $$k \in[1,n-1]$$. Consider $$\ker\mathcal{A}_{n}$$. It is clear that
$$\# (\ker\mathcal{A}_{n}\setminus\mathfrak{F} )\geqslant \frac {10}{3}n+\omega(n)-2n+2=\frac{4}{3}n+\omega(n)+2 .$$
If there exist two $$\mathcal{A}_{n}$$-equivalent ultrafilters in $$\ker \mathcal{A}_{n}\setminus\mathfrak{F}$$, we immediately obtain the required construction yielding the existence of ultrafilters $$a_{1}^{*},\ldots ,a_{n}^{*}, b_{1}^{*},\ldots,b_{n}^{*}$$. Therefore, each ultrafilter from $$\ker \mathcal{A}_{n}\setminus\mathfrak{F}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\mathfrak{F}$$. Therefore, there exist distinct ultrafilters $$c_{n}, d_{n}\in\ker\mathcal{A}_{n}\setminus\mathfrak{F}$$ and $$k_{1}\in[1,n-1]$$ such that $$(a_{k_{1}},c_{n} )$$ and $$(b_{k_{1}},d_{n} )$$ are two pairs of $$\mathcal{A}_{n}$$-equivalent ultrafilters. For simplicity, say $$k_{1}=1$$. Now consider $$\ker\mathcal{A}_{1}$$. It is clear that
$$\# \bigl(\ker\mathcal{A}_{1}\setminus \bigl(\mathfrak{F}\cup \{ c_{n},d_{n} \} \bigr) \bigr)\geqslant\frac{10}{3}n+ \omega(n)-2n=\frac {4}{3}n+\omega(n) .$$
If there exist two $$\mathcal{A}_{1}$$-equivalent ultrafilters in $$\ker \mathcal{A}_{1}\setminus (\mathfrak{F}\cup \{c_{n},d_{n} \} )$$, we immediately obtain the required construction yielding the existence of ultrafilters $$a_{1}^{*},\ldots,a_{n}^{*}, b_{1}^{*},\ldots ,b_{n}^{*}$$. Similarly, if an ultrafilter in $$\ker\mathcal{A}_{1}\setminus (\mathfrak{F}\cup \{c_{n},d_{n} \} )$$ has an $$\mathcal{A}_{1}$$-equivalent ultrafilter in $$\{a_{1},b_{1},c_{n},d_{n} \}$$, then the construction which contradicts the non-existence of ultrafilters $$a_{1}^{*},\ldots,a_{n}^{*}, b_{1}^{*},\ldots,b_{n}^{*}$$ is yielded immediately. So, each ultrafilter in $$\ker\mathcal{A}_{1}\setminus (\mathfrak{F}\cup \{c_{n},d_{n} \} )$$ has an $$\mathcal{A}_{1}$$-equivalent ultrafilter in the set $$\mathfrak{F}\setminus \{a_{1},b_{1} \}$$. Therefore, there exist distinct ultrafilters $$c_{1},d_{1}\in\ker\mathcal{A}_{1}\setminus (\mathfrak{F}\cup \{c_{n},d_{n} \} )$$ and $$k_{2}\in[2,n-1]$$ such that $$(a_{k_{2}},c_{1} )$$ and $$(b_{k_{2}},d_{1} )$$ are two pairs of $$\mathcal{A}_{1}$$-equivalent ultrafilters. For simplicity, say $$k_{2}=2$$. This process can be continued. Suppose that there exists a natural number η such that
$$3\leqslant\eta\leqslant\frac{1}{3}n+\omega(n)+2.$$
Suppose also that there exists a set of pairwise distinct ultrafilters
$$\mathfrak{E}= \{c_{1},\ldots,c_{\eta-1},c_{n},d_{1}, \ldots,d_{\eta -1},d_{n} \}$$
and the following holds:
1. (A)

$$(a_{i+1},c_{i} )$$ and $$(b_{i+1},d_{i} )$$ are two pairs of $$\mathcal{A}_{i}$$-equivalent ultrafilters for each $$i\in [2,\eta-1]$$;

2. (B)

$$\mathfrak{F}\cap\mathfrak{E}=\emptyset$$.

Let us recall what we have said above: $$(a_{1},c_{n})$$ and $$(b_{1},d_{n})$$ are two pairs of $$\mathcal{A}_{n}$$-equivalent ultrafilters; $$(a_{2},c_{1})$$ and $$(b_{2},d_{1})$$ are two pairs of $$\mathcal{A}_{1}$$-equivalent ultrafilters.

Define $$L_{\eta}=\ker\mathcal{A}_{\eta}\setminus (\mathfrak{F}\cup \mathfrak{E} )$$. It is clear that
$$\# (L_{\eta})\geqslant\frac{10}{3}n+\omega(n)-(2n-2)-2\eta = \frac{4}{3}n+\omega(n)-2\eta+2 .$$
If there exist two $$\mathcal{A}_{\eta}$$-equivalent ultrafilters in $$L_{\eta}$$, we immediately obtain the required construction yielding the existence of ultrafilters $$a_{1}^{*},\ldots,a_{n}^{*},b_{1}^{*},\ldots,b_{n}^{*}$$. Similarly, if an ultrafilter in $$L_{\eta}$$ has an $$\mathcal{A}_{\eta}$$-equivalent ultrafilter in $$\{a_{1},\ldots,a_{\eta},b_{1},\ldots,b_{\eta}\}\cup\mathfrak{E}$$, then the construction which contradicts the non-existence of ultrafilters $$a_{1}^{*},\ldots,a_{n}^{*}$$, $$b_{1}^{*},\ldots ,b_{n}^{*}$$ is yielded immediately. Therefore, every ultrafilter from $$L_{\eta}$$ has an $$\mathcal{A}_{\eta}$$-equivalent ultrafilter in $$\{ a_{\eta+1},\ldots,a_{n-1},b_{\eta+1},\ldots,b_{n-1} \}$$. We have
$$\# (L_{\eta})-\# \bigl( [\eta+1,n-1 ] \bigr)\geqslant \frac{4}{3}n+\omega(n)-2\eta+2-n+\eta+1=\frac{1}{3}n+\omega(n)+3-\eta >0 .$$
Therefore, there exist distinct ultrafilters $$c_{\eta},d_{\eta}\in L_{\eta}$$ and $$k_{\eta+1}\in[\eta+1,n-1]$$ such that $$(a_{k_{\eta+1}},c_{\eta})$$ and $$(b_{k_{\eta+1}},d_{\eta})$$ are two pairs of $$\mathcal{A}_{\eta}$$-equivalent ultrafilters. For simplicity, say $$k_{\eta +1}=\eta+1$$. We have that $$(a_{i+1},c_{i})$$ and $$(b_{i+1},d_{i})$$ are two pairs of $$\mathcal{A}_{i}$$-equivalent ultrafilters for each $$i\in[1,\eta]$$.
Put $$\rho=\lfloor\frac{n}{3}\rfloor$$. In view of the above, we can assume that there exist pairwise distinct ultrafilters $$c_{1},\ldots ,c_{\rho-1}, c_{n}, d_{1},\ldots, d_{\rho-1}, d_{n}$$ such that the following holds:
1. (a)

$$(a_{1}, c_{n})$$ and $$(b_{1}, d_{n})$$ are two pairs of $$\mathcal{A}_{n}$$-equivalent ultrafilters;

2. (b)

$$(a_{i+1}, c_{i})$$ and $$(b_{i+1}, d_{i})$$ are two pairs of $$\mathcal{A}_{i}$$-equivalent ultrafilters for each $$i \in[1, \rho-1]$$;

3. (c)
$$\mathfrak{F} \cap\{c_{1}, \ldots , c_{\rho-1} , c_{n},d_{1}, \ldots , d_{\rho- 1} , d_{n} \} = \emptyset$$, see Figure 1. Figure 1 Ultrafilters $$\pmb{a_{i}}$$ , $$\pmb{b_{i}}$$ , $$\pmb{c_{i}}$$ , and $$\pmb{d_{i}}$$ .

(2) Put
\begin{aligned}& Z_{\rho}= \{a_{1}, \ldots , a_{\rho}, b_{1}, \ldots , b_{\rho}, c_{1}, \ldots , c_{\rho-1} , c_{n}, d_{1} , \ldots , d_{\rho-1}, d_{n} \} , \\& Z'_{\rho}= \{a_{1}, \ldots , a_{n-1}, b_{1}, \ldots , b_{n-1}, c_{1}, \ldots , c_{\rho-1} , c_{n}, d_{1} , \ldots , d_{\rho-1}, d_{n} \} , \\& Z''_{\rho}=\{a_{\rho+1}, \ldots,a_{n-1},b_{\rho+1},\ldots,b_{n-1}\} , \\& L_{\rho}= \ker \mathcal{A}_{\rho}\setminus Z'_{\rho}. \end{aligned}
Clearly,
\begin{aligned}& \begin{aligned}[b] \# (L_{\rho}) & \geq \frac{10}{3} n + \omega(n) - 4 \cdot \biggl\lfloor \frac{n}{3} \biggr\rfloor - 2 \biggl(n-1 - \biggl\lfloor \frac{n}{3} \biggr\rfloor \biggr) \\ & = \frac{4}{3} n - 2 \cdot \biggl\lfloor \frac{n}{3} \biggr\rfloor + \omega (n) + 2 \geq\frac{2}{3} n + \omega(n) + 2, \end{aligned} \\& \begin{aligned}[b] \# (L_{\rho}) - \# \bigl([ \rho+1, n-1]\bigr) &\geq \frac{2}{3} n + \omega(n) + 2 - \biggl(n-1 - \biggl\lfloor \frac{n}{3} \biggr\rfloor \biggr) \\ &= \biggl\lfloor \frac{n}{3} \biggr\rfloor - \frac{n}{3} + \omega(n) + 3 > 0. \end{aligned} \end{aligned}
The above arguments show that the following assumption should be made: for each ultrafilter $$q\in L_{\rho}$$, there exists an ultrafilter $$\tilde{q}\in Z_{\rho}''$$ such that q and $$\tilde{q}$$ are $$\mathcal{A}_{\rho}$$-equivalent ultrafilters. In general, there can be such q for which the number of corresponding $$\tilde{q}$$ is greater than 1. We choose in an arbitrary way only one $$\tilde{q}$$ for each $$q\in L_{\rho}$$. We obtain the mapping $$f:L_{\rho}\rightarrow Z_{\rho}''$$, $$f(q)=\tilde{q}$$. The map f is one-to-one. (If $$f(q_{1})=f(q_{2})$$ and $$q_{1}\ne q_{2}$$, then $$q_{1}$$, $$q_{2}$$ are $$\mathcal{A}_{\rho}$$-similar ultrafilters, and the construction which contradicts the non-existence of ultrafilters $$a^{*}_{1}, \ldots, a^{*}_{n}, b^{*}_{1}, \ldots, b^{*}_{n}$$ is yielded immediately.) Put
\begin{aligned}& \begin{aligned}[b] \frak{I}_{1}^{\rho}={}&\bigl\{ k\in[\rho+1,n-1] \mid \mbox{there exist ultrafilters } q_{k}^{a},q_{k}^{b} \in L_{\rho} \\ &{}\mbox{such that } f\bigl(q_{k}^{a}\bigr)=a_{k}, f\bigl(q_{k}^{b}\bigr)=b_{k}\bigr\} , \end{aligned} \\& \begin{aligned}[b] \frak{I}_{2}^{\rho}={}&\bigl\{ k\in[\rho+1,n-1] \setminus \frak{I}_{1}^{\rho} \mid \mbox{there exists an ultrafilter } q_{k}^{*}\in L_{\rho} \\ &{}\mbox{ such that } f\bigl(q_{k}^{*}\bigr)\in\{a_{k},b_{k} \}\bigr\} . \end{aligned} \end{aligned}
Obviously, $$\frak{I}_{1}^{\rho}\cap \frak{I}_{2}^{\rho}=\emptyset$$. Since $$\#(L_{\rho})-\#([\rho+1,n-1])>0$$, we have $$\#(\frak{I}_{1}^{\rho})=\tau>0$$. Clearly,
$$\#\bigl(\frak{I}_{2}^{\rho}\bigr)=\#(L_{\rho})-2\tau\ge \frac{2}{3}n+\omega(n)+2-2\tau.$$
Put
$$L_{n} = \ker \mathcal{A}_{n} \setminus Z'_{\rho}.$$
We have obtained above the estimate $$\# (L_{\rho})\geqslant\frac {2}{3}n+\omega(n)+2$$. In exactly the same way, the following estimate can be obtained:
$$\# (L_{n}) \geq\frac{2}{3} n + \omega(n) + 2 .$$
If there exist two $$\mathcal{A}_{n}$$-equivalent ultrafilters from $$L_{n}$$, we immediately obtain the required construction regarding the existence of ultrafilters $$a^{*}_{1}, \ldots, a^{*}_{n}, b^{*}_{1}, \ldots, b^{*}_{n}$$. Similarly, if an ultrafilter in $$L_{n}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\{a_{1}, b_{1}, c_{1}, \ldots, c_{\rho- 1}, c_{n}, d_{1}, \ldots, d_{\rho- 1}, d_{n}\}$$, then it is easy to find the corresponding ultrafilters $$a^{*}_{1}, \ldots, a^{*}_{n}, b^{*}_{1}, \ldots, b^{*}_{n}$$.
We are interested in the following situation: let $$q \in L_{n}$$, and q has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\{a_{2},\ldots,a_{\rho},b_{1},\ldots,b_{\rho}\}$$. Let, for instance, q and $$a_{2}$$ be $$\mathcal{A}_{n}$$-equivalent ultrafilters. Then let us consider $$d_{1}$$. For $$d_{1}$$ there are four possible cases:
$$\langle1\rangle$$

$$d_{1} \notin\ker \mathcal{A}_{n}$$;

$$\langle2\rangle$$

$$b_{2}$$, $$d_{1}$$ are $$\mathcal{A}_{n}$$-equivalent ultrafilters;

$$\langle3\rangle$$

$$d_{1}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\{a_{3}, \ldots, a_{\rho}, b_{3}, \ldots, b_{\rho}\}$$;

$$\langle4\rangle$$

$$d_{1}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$Z''_{\rho}$$.

In case $$\langle2\rangle$$ let us consider $$c_{1}$$. For $$c_{1}$$ the possible corresponding cases are:
$$\langle \mathrm{i}\rangle$$

$$c_{1} \notin\ker \mathcal{A}_{n}$$;

$$\langle \mathrm{ii}\rangle$$

$$c_{1}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\{a_{3}, \ldots , a_{\rho}, b_{3}, \ldots , b_{\rho}\}$$;

$$\langle \mathrm{iii}\rangle$$

$$c_{1}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$Z''_{\rho}$$.

Consider case $$\langle3\rangle$$ for $$d_{1}$$. Let $$d_{1}$$, $$b_{3}$$ be $$\mathcal{A}_{n}$$-equivalent ultrafilters. Let us consider $$c_{2}$$. For $$c_{2}$$ there are four possible cases:
$$\langle1\rangle$$

$$c_{2} \notin\ker \mathcal{A}_{n}$$;

$$\langle2\rangle$$

$$a_{3}$$, $$c_{2}$$ are $$\mathcal{A}_{n}$$-equivalent ultrafilters;

$$\langle3\rangle$$

$$c_{2}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\{a_{4}, \ldots , a_{\rho}, b_{2}, b_{4}, \ldots , b_{\rho}\}$$;

$$\langle4\rangle$$

$$c_{2}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$Z''_{\rho}$$.

Consider case $$\langle3\rangle$$ for $$c_{2}$$. Let $$b_{2}$$, $$c_{2}$$ be $$\mathcal{A}_{n}$$-equivalent ultrafilters. Let us consider $$c_{1}$$. For $$c_{1}$$ the possible corresponding cases are:
$$\langle \mathrm{i}\rangle$$

$$c_{1} \notin\ker \mathcal{A}_{n}$$;

$$\langle \mathrm{ii}\rangle$$

$$c_{1}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\{a_{3}, \ldots , a_{\rho}, b_{4}, \ldots , b_{\rho}\}$$;

$$\langle \mathrm{iii}\rangle$$

$$c_{1}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$Z''_{\rho}$$.

Continuing these constructions in an obvious way, we find an ultrafilter
$$q_{*} \in\{ c_{1}, \ldots , c_{\rho- 1} , d_{1}, \ldots , d_{\rho- 1} \}$$
such that one of the following two statements is true: (1) $$q_{*} \notin \ker \mathcal{A}_{n}$$; (2) $$q_{*}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$Z''_{\rho}$$. Let us put
\begin{aligned}& \alpha= \# \bigl( \bigl\{ q \in L_{n} \mid q \mbox{ has an } \mathcal{A}_{n}\mbox{-equivalent ultrafilter in } \{a_{2}, \ldots , a_{\rho}, b_{2}, \ldots , b_{\rho}\} \bigr\} \bigr) , \\& \beta= \# \bigl( \{ c_{1}, \ldots , c_{\rho- 1}, d_{1}, \ldots , d_{\rho- 1} \} \setminus\ker \mathcal{A}_{n} \bigr) , \\& \begin{aligned}[b] \gamma={}& \# \bigl(\bigl\{ q_{*} \in\{ c_{1}, \ldots, c_{\rho- 1}, d_{1}, \ldots, d_{\rho- 1} \} \mid q_{*} \mbox{ has an } \mathcal{A}_{n}\mbox{-equivalent} \\ &{}\mbox{ultrafilter in } Z''_{\rho}\bigr\} \bigr). \end{aligned} \end{aligned}
The above constructions clearly show that $$\alpha\leq\beta+ \gamma$$. Put
$$\hat{L}=\bigl\{ q\in L_{n}\cup\{c_{1},\ldots c_{\rho-1},d_{1},\ldots,d_{\rho-1}\} \mid q \mbox{ has } \mathcal{A}_{n}\mbox{-similar ultrafilter in }Z''_{\rho} \bigr\} .$$
Clearly,
$$\#(\hat{L})\ge\#(L_{n})+\gamma-\alpha\ge\frac{2}{3}n+\omega(n)+2 + \beta+ \gamma-\alpha\ge\frac{2}{3}n+\omega(n)+2 .$$
So, for every ultrafilter $$q\in\hat{L}$$, there exists an ultrafilter $$\bar{q}\in Z_{\rho}''$$ such that q and $$\bar{q}$$ are $$\mathcal{A}_{n}$$-similar ultrafilters. In general it can happen that for some q there exist more than one corresponding $$\bar{q}$$. Choose arbitrarily only one ultrafilter $$\bar{q}$$ for each $$q\in\hat{L}$$. We obtain a mapping $$\hat{f}:\hat{L}\rightarrow Z_{\rho}''$$, $$\hat{f}(q)=\bar{q}$$. Consider the corresponding map $$\hat{f}:\hat{L}_{\rho}\to Z''_{\rho}$$. It is one-to-one. Indeed, if $$\hat{f}(q_{1})=\hat{f}(q_{2})$$ and $$q_{1}\ne q_{2}$$, then $$q_{1}$$, $$q_{2}$$ are $$\mathcal{A}_{n}$$-similar ultrafilters, and the construction which contradicts the non-existence of ultrafilters $$a^{*}_{1}, \ldots, a^{*}_{n}, b^{*}_{1}, \ldots, b^{*}_{n}$$ is yielded immediately. Put
\begin{aligned}& \begin{aligned}[b] \hat{\frak{I}}_{1}={}&\bigl\{ k\in[\rho+1,n-1] \mid\mbox{there exist ultrafilters } \frak{q}_{k}^{a},\frak{q}_{k}^{b} \in\hat{L} \\ &{}\mbox{such that } \hat{f}\bigl(\frak{q}_{k}^{a} \bigr)=a_{k}, \hat{f}\bigl(\frak{q}_{k}^{b} \bigr)=b_{k}\bigr\} , \end{aligned} \\& \begin{aligned}[b] \hat{\frak{I}}_{2}={}&\bigl\{ k\in[\rho+1,n-1]\setminus\hat{\frak{I}}_{1}\mid \mbox{there exists an ultrafilter } \frak{q}_{k}^{*}\in\hat{L} \\ &{}\mbox{such that } \hat{f}\bigl(\frak{q}_{k}^{*}\bigr)\in \{a_{k},b_{k}\}\bigr\} . \end{aligned} \end{aligned}
Obviously, $$\hat{\frak{I}}_{1}\cap\hat{\frak{I}}_{2}=\emptyset$$. Since $$\#(\hat{L})-\#([\rho+1,n-1])>0$$, we have $$\#(\hat{\frak{I}}_{1})=\hat{\tau}>0$$. Clearly,
$$\#(\hat{\frak{I}}_{2})=\#(\hat{L})-2\hat{\tau}\ge\frac{2}{3}n+ \omega(n)+2-2\hat{\tau}.$$
If $$\tau\ge\hat{\tau}$$, put
$$\frak{I}= (\hat{\frak{I}}_{1}\cup\hat{\frak{I}}_{2} )\cap \frak{I}_{1}^{\rho}.$$
If $$\tau<\hat{\tau}$$, put
$$\frak{I}= \bigl(\frak{I}_{1}^{\rho}\cup \frak{I}_{2}^{\rho} \bigr)\cap\hat{\frak{I}}_{1}.$$
Clearly,
$$\#(\frak{I})\ge\frac{2}{3}n+\omega(n)+2-n+1+ \biggl\lfloor \frac{n}{3} \biggr\rfloor >\omega(n)+2.$$
(3) We fix $$\nu\in \frak{I}$$. A number $$k\in[1,\rho]$$ is called ν-marked if the following is true:
• for $$k=1$$: $$(a_{1},d_{n})$$ and $$(b_{1},c_{n})$$ are pairs of $$\mathcal{A}_{\nu}$$-equivalent ultrafilters;

• for $$k>1$$: $$(a_{k},d_{k-1})$$ and $$(b_{k},c_{k-1})$$ are pairs of $$\mathcal{A}_{\nu}$$-equivalent ultrafilters.

Put
$$\chi_{\nu}=\#\bigl(\bigl\{ k\in[1,\rho]\mid k\mbox{ is a }\nu\mbox{-marked number}\bigr\} \bigr) .$$
Our aim is to prove that
$$\chi_{\nu}>\frac{\omega(n)}{2}.$$
We have the following options:
$$\langle1\rangle$$

There exist ultrafilters $$q_{\nu}^{a},q_{\nu}^{b}\in \frak{I}_{1}^{\rho}$$ and an ultrafilter $$\frak{q}_{\nu}^{*}\in\hat{\frak{I}}_{2}$$.

$$\langle2\rangle$$

There exists an ultrafilter $$q_{\nu}^{*}\in \frak{I}_{2}^{\rho}$$ and ultrafilters $$\frak{q}_{\nu}^{a},\frak{q}_{\nu}^{b}\in\hat{\frak{I}}_{1}$$.

$$\langle3\rangle$$

There exist ultrafilters $$q_{\nu}^{a},q_{\nu}^{b}\in \frak{I}_{1}^{\rho}$$ and ultrafilters $$\frak{q}_{\nu}^{a},\frak{q}_{\nu}^{b}\in\hat{\frak{I}}_{1}$$.

Denote $$q_{\nu}^{*}$$ by $$q_{\nu}$$. Denote $$\frak{q}_{\nu}^{*}$$ by $$q'_{\nu}$$. Choose one of two ultrafilters $$q_{\nu}^{a}$$, $$q_{\nu}^{b}$$ and denote it by $$q_{\nu}$$; at this step we do not consider the second ultrafilter. Choose one of two ultrafilters $$\frak{q}_{\nu}^{a}$$, $$\frak{q}_{\nu}^{b}$$ and denote it by $$q'_{\nu}$$; at this step we do not consider the second ultrafilter. If possible, the ultrafilter $$q'_{\nu}$$ is taken from $$\hat{L}\setminus L_{n}$$ . Let $$\nu=\rho+1$$. We know that there exists a corresponding ultrafilter $$q_{\rho+1}\in L_{\rho}$$ which has an $$\mathcal{A}_{\rho}$$-equivalent ultrafilter in $$\{a_{\rho+1},b_{\rho+1}\}$$. We also know that there exists a corresponding ultrafilter $$q'_{\rho+1}\in\hat{L}$$ which has $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\{a_{\rho+1},b_{\rho+1}\}$$.
When the number $$\chi_{\rho+ 1}$$ attains its minimal value, we must assume the following: there exist pairwise distinct ultrafilters
$$a'_{\rho+2}, \ldots, a'_{n-1} , b'_{\rho+2}, \ldots , b'_{n-1} \in\ker \mathcal{A}_{\rho+1} \Big\backslash \Bigl(Z'_{\rho}\cup \bigl\{ q_{\rho+1} , q'_{\rho + 1} \bigr\} \Bigr) ,$$
and $$(a_{k}, a'_{k})$$, $$(b_{k}, b'_{k})$$ are pairs of $$\mathcal{A}_{\rho+1}$$-equivalent ultrafilters for each $$k \in[\rho+2, n-1 ]$$. We will only consider the cases where finding ultrafilters $$a_{1}^{*}, \ldots, a_{n}^{*}, b_{1}^{*}, \ldots, b_{n}^{*}$$ is not immediate.

Case 1. $$q'_{\rho+ 1} \in L_{n}$$.

Case 1-1. $$q_{\rho+ 1} = q'_{\rho+ 1}$$.

We consider only two subcases of Case 1-1.

Case 1-1-1. There exists an ultrafilter $$q^{*}\notin Z_{\rho}$$ such that $$q^{*}$$, $$q_{\rho+ 1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters.

Case 1-1-2. There exists an ultrafilter $$q^{*} \in Z_{\rho}$$ such that $$q^{*}$$, $$q_{\rho+ 1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters.

Case 1-2. $$q_{\rho+ 1} \neq q'_{\rho+ 1}$$.

We consider only two subcases of Case 1-2.

Case 1-2-1. $$q_{\rho+ 1}$$, $$q'_{\rho+ 1}$$ are $$\mathcal{A}_{\rho + 1}$$-equivalent ultrafilters.

Case 1-2-2. There exists an ultrafilter $$q^{*} \in\{a_{1}, \ldots , a_{\rho}, b_{1}, \ldots , b_{\rho}\}$$ such that $$q^{*}$$, $$q_{\rho+ 1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters.

Before we consider these cases, let us denote $$\mathcal{R}_{1}=\{a_{1},b_{1},c_{n},b_{n}\}$$, and $$\mathcal{R}_{k}=\{a_{k},b_{k},c_{k-1}, d_{k-1}\}$$ if $$k\in[2,\rho]$$.

First we consider Cases 1-1-1 and 1-2-1. For the situation where the number $$\chi_{\rho+ 1}$$ attains its the minimum value, we have the following options for the set $$\mathcal{R}_{1}$$:
(1):

$$a_{1}$$, $$b_{1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters and $$\# (\ker \mathcal{A}_{\rho+ 1} \cap \mathcal{R}_{1}) = 2$$;

(2):

$$a_{1}$$, $$d_{n}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters and $$\# (\ker \mathcal{A}_{\rho+ 1} \cap \mathcal{R}_{1}) = 2$$;

(3):

$$b_{1}$$, $$c_{n}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters and $$\# (\ker \mathcal{A}_{\rho+ 1} \cap \mathcal{R}_{1}) = 2$$;

(4):

the number 1 is $$(\rho+ 1)$$-marked.

If $$k \in[2,\rho]$$, by analogy, we have the following options for the set $$\mathcal{R}_{k}$$:
(1):

$$a_{k}$$, $$b_{k}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters and $$\# (\ker \mathcal{A}_{\rho+ 1} \cap \mathcal{R}_{k}) = 2$$;

(2):

$$a_{k}$$, $$d_{k-1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters and $$\# (\ker \mathcal{A}_{\rho+ 1} \cap \mathcal{R}_{k}) = 2$$;

(3):

$$b_{k}$$, $$c_{k-1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters and $$\# (\ker \mathcal{A}_{\rho+ 1} \cap \mathcal{R}_{k}) = 2$$;

(4):

the number k is $$(\rho+ 1)$$-marked.

So we have
$$4(n-1-\rho) + 4 \cdot\chi_{\rho+ 1} + 2 (\rho- \chi_{\rho+1}) = \# ( \ker \mathcal{A}_{\rho+1}) \geq\frac{10}{3} n + \omega(n) .$$
Recall that $$\rho= \lfloor\frac{n}{3} \rfloor$$. Therefore we have
$$\chi_{\rho+1} > \frac{\omega(n)}{2} + 1 .$$
Now consider Case 1-1-2. The situation is as follows:
$$\langle \mathrm{a}\rangle$$

$$c_{\rho- 1}$$, $$q_{\rho+ 1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters;

$$\langle \mathrm{b}\rangle$$

$$a_{\rho}$$, $$d_{\rho- 1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters;

$$\langle \mathrm{c}\rangle$$

for $$\mathcal{R}_{1}$$ one of the options (1)-(4) is fulfilled;

$$\langle \mathrm{d}\rangle$$

for $$\mathcal{R}_{k}$$, where $$k \in[2,\rho- 1]$$, one of the options (1)-(4) is fulfilled.

Now consider Case 1-2-2. The situation is as follows: $$b_{\rho}$$, $$q_{\rho+ 1}$$ are $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters, and the conditions $$\langle \mathrm{b}\rangle$$, $$\langle \mathrm{c}\rangle$$, $$\langle \mathrm{d}\rangle$$ are fulfilled. It is clear that in Cases 1-1-2 and 1-2-2 we have
$$\chi_{\rho+1} > \frac{\omega(n)}{2} + 1 .$$
It is clear that in Case 1 there may be subcases which we have not considered. But always
$$\chi_{\rho+1} > \frac{\omega(n)}{2} + 1 .$$
Case 2. $$q'_{\rho+ 1} \in\hat{L} \setminus L_{n}$$. Suppose that $$q'_{\rho+ 1} = c_{\rho- 2}$$ and $$c_{\rho- 2}, a_{\rho+ 1}$$ are $$\mathcal{A}_{n}$$-equivalent ultrafilters. For the number $$\chi_{\rho+1}$$ to be minimal and the situation to be nontrivial, we assume the following:
1. (i)

$$(a_{\rho-1}, b_{\rho- 1})$$, $$(c_{\rho- 2}, d_{\rho- 2})$$, $$(a_{2}, q_{\rho+ 1})$$, $$(b_{2}, c_{1})$$ are pairs of $$\mathcal{A}_{\rho+ 1}$$-equivalent ultrafilters;

2. (ii)

$$a_{\rho+1}$$, $$q_{\rho+ 1}$$ are $$\mathcal{A}_{\rho}$$-equivalent ultrafilters;

3. (iii)
$$\ker \mathcal{A}_{\rho+ 1} \subset Z'_{\rho}\cup \{a'_{\rho+2} , \ldots , a'_{n-1}, b'_{\rho+ 2} , \ldots , b'_{n-1} \} \cup\{ q_{\rho+ 1} \}$$, see Figure 2. Figure 2 Illustration for Case 2.

We assume that one of the cases (1)-(4) holds for $$\mathcal{R}_{1}$$ and that one of the cases (1)-(4) holds for $$\mathcal{R}_{k}$$, where $$k \in[3, \rho] \setminus\{\rho- 1 \}$$. We have
$$4(n-1-\rho) + 4 \cdot\chi_{\rho+ 1} + 2 (\rho- 2 - \chi_{\rho+ 1}) + 6 = \# ( \ker \mathcal{A}_{\rho+1} ) \geq\frac{10}{3} n + \omega(n) .$$
Recall that $$\rho= \lfloor\frac{n}{3} \rfloor$$. Therefore we have
$$\chi_{\rho+1} > \frac{\omega(n)}{2} .$$
Analyzing the other situations in Case 2, we come to the same conclusion: $$\chi_{\rho+1} > \frac{\omega(n)}{2}$$; and we can assume that if k is a $$(\rho+ 1)$$-marked number, then $$q'_{\rho+ 1} \notin \mathcal{R}_{k}$$. It is obvious that the same conclusion is true in Case 1.
(4) It is obvious that for each $$\nu\in\mathfrak{I}$$ we have $$\chi _{\nu}>\frac{\omega(n)}{2}$$, and $$q'_{k}\notin \mathcal{R}_{k}$$ if k is a ν-marked number. We know that
$$\omega(n)\geqslant\sqrt{\frac{2n}{3}} ,\qquad \#(\mathfrak{I})>\omega (n)+2 ,\qquad \rho= \biggl\lfloor \frac{n}{3} \biggr\rfloor .$$
Therefore we have
$$\frac{\omega(n)}{2}\cdot\#(\mathfrak{I})>\frac{\omega(n)}{2}\cdot\bigl(\omega (n)+2 \bigr)>\rho .$$
Therefore there exist distinct numbers $$\nu_{1},\nu_{2}\in\mathfrak{I}$$ and $$k_{0}\in[1,\rho]$$ such that $$k_{0}$$ is a $$\nu_{1}$$-marked number and $$\nu_{2}$$-marked number. Let $$\nu_{1}=\rho+1$$, $$\nu_{2}=\rho+2$$. Consider the ultrafilters $$q_{\rho+1}$$, $$q'_{\rho+2}$$. If $$q_{\rho+1}\neq q'_{\rho +2}$$, put $$z_{\rho+1}=q_{\rho+1}$$, $$z'_{\rho+2}=q'_{\rho+2}$$. Let $$q_{\rho+1}= q'_{\rho+2}$$. There are two possible cases.

I. There exist the ultrafilters $$q^{a}_{\rho+1}$$ , $$q^{b}_{\rho+1}$$ , and assume that $$q_{\rho+1}= q_{\rho+1}^{b}$$. Put $$z_{\rho+1}=q^{a}_{\rho+1}$$, $$z'_{\rho+2}=q'_{\rho+2}$$.

II. The ultrafilters $$q^{a}_{\rho+1}$$ , $$q^{b}_{\rho+1}$$ do not exist. Then there exist the ultrafilters $$\mathfrak{q}^{a}_{\rho+1}$$, $$\mathfrak {q}^{b}_{\rho+1}$$, and assume that $$q'_{\rho+1}=\mathfrak{q}_{\rho+1}^{b}$$. Put $$z_{\rho+2}=q_{\rho+2}$$. If $$q'_{\rho+1}\neq z_{\rho+2}$$, put $$z'_{\rho+1}=q'_{\rho+1}$$. Otherwise we have $$\mathfrak{q}^{a}_{\rho +1}\in L_{n}$$ since $$q'_{\rho+1}= q_{\rho+2}\in L_{n}$$ (see in the part (3) of our proof how we have chosen the ultrafilter $$q'_{\nu}$$); and put $$z'_{\rho+1}=\mathfrak{q}^{a}_{\rho+1}$$.

Thus, we consider either the pair of ultrafilters $$z_{\rho+1}$$, $$z'_{\rho +2}$$, or the pair of ultrafilters $$z'_{\rho+1}$$, $$z_{\rho+2}$$. These two pairs have the same properties. We will consider the pair $$z_{\rho +1}$$, $$z'_{\rho+2}$$. We have the following:
1

$$z_{\rho+1}$$ has an $$\mathcal{A}_{\rho}$$-equivalent ultrafilter in $$\{ a_{\rho+1},b_{\rho+1}\}$$;

2

$$z'_{\rho+2}$$ has an $$\mathcal{A}_{n}$$-equivalent ultrafilter in $$\{ a_{\rho+2},b_{\rho+2}\}$$;

3

$$z_{\rho+1}\neq z'_{\rho+2}$$;

4

$$z_{\rho+1}\notin Z'_{\rho}$$;

5

$$z'_{\rho+2}\notin\mathfrak{F}\cup \mathcal{R}_{k_{0}}$$.

Suppose that $$a_{\rho+1}$$ and $$z_{\rho+1}$$ are $$\mathcal{A}_{\rho}$$-equivalent ultrafilters, $$a_{\rho+2}$$ and $$z'_{\rho+ 2}$$ are $$\mathcal{A}_{n}$$-equivalent ultrafilters, and $$k_{0} = 3$$. It is possible that
$$z'_{\rho+ 2} \in \{c_{3}, \ldots, c_{\rho- 1}, d_{3}, \ldots, d_{\rho- 1} \} .$$
Suppose that
$$z'_{\rho+ 2} \notin\{d_{3}, \ldots , d_{\rho- 1} \} .$$
Now it is easy to construct the corresponding ultrafilters $$a^{*}_{1}, \ldots, a^{*}_{n}, b^{*}_{1}, \ldots, b^{*}_{n}$$. Let us list them in pairs:$$(a^{*}_{1}, b^{*}_{1}) = (a_{1}, b_{1})$$, $$(a^{*}_{2}, b^{*}_{2}) = (a_{2}, b_{2})$$, $$(a^{*}_{3},b^{*}_{3}) = (b_{4}, d_{3})$$, …, $$(a^{*}_{\rho-1 },b^{*}_{\rho- 1}) = (b_{\rho}, d_{\rho- 1})$$, $$(a^{*}_{\rho},b^{*}_{\rho}) = (a_{\rho+ 1},z_{\rho+ 1})$$, $$(a^{*}_{\rho+ 1}, b^{*}_{\rho+ 1}) = (a_{3}, d_{2})$$, $$(a^{*}_{\rho+ 2}, b^{*}_{\rho+ 2}) = (b_{3}, c_{2})$$, $$(a^{*}_{\rho+ 3},b^{*}_{\rho+ 3}) = (a_{\rho+3}, b_{\rho+ 3})$$, …, $$(a^{*}_{n-1},b^{*}_{n-1} ) = (a_{n-1}, b_{n-1})$$, $$(a^{*}_{n}, b^{*}_{n}) = (a_{\rho+ 2}, z'_{\rho+ 2})$$, see Figure 3. Figure 3 Construction of ultrafilters $$\pmb{a^{*}_{1}, \ldots, a^{*}_{n}}$$ , $$\pmb{b^{*}_{1}, \ldots, b^{*}_{n}}$$ .  □

## 4 Combinatorial theorems

In this section we consider for each $$n \in \mathbb{N}^{+}$$ a matrix $$\frak{M}(n)$$ which has n rows and $$\aleph_{0}$$ columns. We denote by $$\alpha^{k}_{i}$$ the element of $$\frak{M}(n)$$ in the ith row and the kth column. The following holds:
1. (1)

$$\alpha^{k}_{i} \in\mathbb{N}$$;

2. (2)

for any $$\alpha^{k}_{i} > 0$$, there exists $$\alpha^{k'}_{i}$$ such that $$\alpha^{k}_{i} = \alpha^{k'}_{i}$$ and $$k \neq k'$$.

We denote by $$w(\frak{M}(n),i)$$ the number of nonzero elements in the ith row of $$\frak{M}(n)$$. It is clear that
$$0 \leq w\bigl( \frak{M}(n),i\bigr) \leq\aleph_{0} .$$

### Definition 4.1

A matrix $$\frak{M}(n)$$ is said to be saturated if there exist pairwise distinct natural numbers $$k_{1}, k'_{1}, \ldots , k_{n}, k'_{n}$$ such that $$\alpha^{k_{i}}_{i} = \alpha^{k'_{i}}_{i} > 0$$ for each $$i \in[1,n]$$.

### Definition 4.2

For each $$n \in \mathbb{N}^{+}$$, denote by $$\frak{v}'(n)$$ the minimal natural number such that if for some matrix $$\frak{M}(n)$$ we have $$w(\frak{M}(n),i) \geq \frak{v}'(n)$$ for each $$i \in[1,n]$$, then $$\frak{M}(n)$$ is saturated.

We suppose that $$\frak{v}'(n) \in \mathbb{N}^{+}$$ since, obviously, $$\frak{v}'(n) < \aleph_{0}$$.

It is easy to prove that $$\frak{v}(n) = \frak{v}'(n)$$. Therefore, by Theorem 2.1, the following theorem is true.

### Theorem 4.3

If for some matrix $$\frak{M}(n)$$ we have
$$w \bigl(\frak{M}(n), i\bigr) \geq\frac{10}{3} n + \sqrt{\frac{2n}{3}}$$
for each $$i \in[1,n]$$, then $$\frak{M}(n)$$ is saturated.

The following theorem is a particular case of the well-known theorem of Ramsey .

### Theorem 4.4

Consider a set S, $$\# (S) = n \in \mathbb{N}^{+}$$, and let T be the family of all two-element subsets of S. We divide T into two disjoint sub-families $$T_{1}$$, $$T_{2}$$. Fix a natural number $$\mu\geq2$$. We claim that there exists the minimal number $$R(\mu) \in \mathbb{N}^{+}$$ such that if $$n \geq R(\mu)$$, then there exists a set $$S' \subset S$$, $$\# (S') = \mu$$, and either all two-element subsets of $$S'$$ belong to $$T_{1}$$ or they all belong to $$T_{2}$$.

In the formulation of the following theorem, we use the number $$R(\mu)$$ from Theorem 4.4.

### Theorem 4.5

Consider a matrix $$\frak{M}(n)$$, and fix a natural number $$\mu\geq2$$. Let
$$w \bigl(\frak{M}(n), i\bigr) \geq\frac{10}{3} n + \sqrt{\frac{2n}{3}}$$
for any $$i \in[1,n]$$, and $$n \geq R(\mu)$$. Then
1. (1)
there exist pairwise distinct natural numbers
$$k_{1}, k'_{1}, \ldots , k_{n}, k'_{n}$$
such that $$\alpha^{k_{i}}_{i} = \alpha^{k'_{i}}_{i} > 0$$ and $$k_{i} < k'$$ for each $$i \in[1,n]$$;

2. (2)
there exists a family of segments
$$D \subset\bigl\{ \bigl[k_{i}, k'_{i} \bigr] \bigr\} _{i \leq n},$$
$$\#(D) = \mu$$, and one of the following two cases holds;
1. (a)

if $$I_{1}, I_{2} \in D$$ are distinct, then $$I_{1} \cap I_{2} = \emptyset$$;

2. (b)

$$\cap D \neq\emptyset$$.3

### Proof

Let us use the notation of Theorem 4.4. By Theorem 4.3 there exists a corresponding family of segments
$$S = \bigl\{ \bigl[k_{i}, k'_{i} \bigr] \bigr\} _{i \leq n} .$$
Let T be the family of all subsets of S with the exact two elements. Divide T into two disjoint sub-families $$T_{1}$$, $$T_{2}$$. Let $$T_{1}$$ be the family of pairs of disjoint segments. Let $$T_{2}$$ be the family of pairs of distinct joint segments. By Theorem 4.4 there exists a family $$D \subset S$$ such that $$\#(D) = \mu$$ and all pairs of distinct segments from D belong either to $$T_{1}$$ or to $$T_{2}$$. If all pairs of distinct segments belong to $$T_{2}$$, then it is easy to see that $$\cap D \neq\emptyset$$. □

### Remark 4.6

The following well-known result is given, for example, in :
$$R( \mu) \leq \left ( \begin{array}{@{}c@{}} 2 \mu- 2 \\ \mu- 1 \end{array} \right ) .$$
Therefore Theorem 4.5 is true if the condition $$n \geq R(\mu)$$ will be exchanged by $$n \geq \bigl ( {\scriptsize\begin{matrix} 2 \mu- 2 \cr \mu- 1 \end{matrix}} \bigr )$$.

## 5 Countable families of σ-algebras

In the first nine subsections we present facts from  and .

### Definition 5.1

A point $$a \in\beta X$$ is said to be irregular if for any countable sequence of sets $$M_{1}, \ldots , M_{k}, \ldots \subset\beta X$$ such that $$a \notin\overline{M}_{k}$$ for all k, we have $$a \notin\overline{\cup M_{k}}$$.

Since a point of βX is an ultrafilter on X and, vice versa, an ultrafilter on X is a point of βX, we will also call an irregular point an irregular ultrafilter. All points of X are irregular.

### Definition 5.2

An algebra $$\mathcal{A}$$ is said to be simple if there exists $$Z \subseteq \beta X$$ such that:
1. (1)

$$\# (Z) \leq\aleph_{0}$$;

2. (2)

if $$Z \neq\emptyset$$, all points of Z are irregular;

3. (3)

$$\ker \mathcal{A}\subseteq\overline{Z}$$.

The proof of the following theorem is in , Chapter 17.

### Theorem 5.3

Let $$\mathcal{A}_{1}, \ldots , \mathcal{A}_{k}, \ldots$$ and $$\mathcal{B}_{1}, \ldots , \mathcal{B}_{k}, \ldots$$ be two countable families of σ-algebras. Let all algebras $$\mathcal{A}_{k}$$ be simple, and among the algebras $$\mathcal{B}_{k}$$ let there be no simple algebras. Then there exist pairwise disjoint sets $$W, U_{1}, \ldots , U_{k}, \ldots , V_{1}, \ldots , V_{k}, \ldots$$ such that:
1. (1)

$$\ker \mathcal{A}_{k} \subseteq\overline{W}$$ for each k;

2. (2)

for each $$k \in \mathbb{N}^{+}$$, the following holds: if a set Q contains one of the two sets $$U_{k}$$, $$V_{k}$$ and intersection with the other set is empty, then $$Q \notin \mathcal{B}_{k}$$.

### Remark 5.4

The Gitik-Shelah theorem is essentially used in the proof of Theorem 5.3. Under the assumption that the continuum hypothesis ($$\aleph_{1}=2^{\aleph _{0}}$$) is true, the proof of Theorem 5.3 essentially uses not the nontrivial Gitik-Shelah theorem but the rather simple Alaoglu-Erdös theorem.

### Definition 5.5

The set $$\{ a \in\ker \mathcal{A}\mid a \mbox{ is an irregular point}\}$$ is called the spectrum of an algebra $$\mathcal{A}$$ and is denoted $$sp \mathcal{A}$$.

It is clear that if $$\mathcal{A}$$ is a simple algebra, then $$\#(sp \mathcal{A}) \leq \aleph_{0}$$.

The proof of the lemma below is in , Chapter 7.

### Lemma 5.6

If $$\mathcal{A}$$ is a simple σ-algebra, then $$\ker \mathcal{A}\subseteq \overline{sp \mathcal{A}}$$.

The proof of the lemma below is in , Chapter 7.

### Lemma 5.7

If $$\mathcal{A}$$ is a simple σ-algebra and $$a \in sp \mathcal{A}$$, then
$$\{ b \in sp \mathcal{A}\mid a \textit{ is } \mathcal{A}\textit{-equivalent to } b \} \neq\emptyset .$$

### Remark 5.8

If an ω-saturated algebra $$\mathcal{A}$$ is a σ-algebra, then $$\mathcal{A}$$ is simple and $$\ker \mathcal{A}= sp \mathcal{A}$$.

The proof of the following lemma is easily derived from Lemma 5.7 and arguments in Remark 1.13.

### Lemma 5.9

Let $$\mathcal{A}$$ be a simple but not ω-saturated σ-algebra $$\mathcal{A}$$ and let $$\nu\in \mathbb{N}^{+}$$. We can construct an ω-saturated σ-algebra $$\mathcal{A}'$$ such that $$\ker \mathcal{A}' \subset sp \mathcal{A}$$, $$\# (\ker \mathcal{A}') \geqslant\nu$$, and two ultrafilters are $$\mathcal{A}'$$-equivalent if and only if they are $$\mathcal{A}$$-equivalent.4

### Proof of Theorem 2.4

Consider a sequence of integers $$n_{0} = 0 < n_{1} < n_{2} < \cdots < n_{m} < \cdots$$. Construct the function $$\varphi: \mathbb{N}^{+}\rightarrow \mathbb{N}^{+}$$ as follows: if $$k \in[n_{m-1}+1, n_{m} ]$$, where $$m \in \mathbb{N}^{+}$$, then
$$\varphi(k) = 4 \cdot n_{m-1} + \biggl\lceil \frac{10}{3} (n_{m} - n_{m-1} ) + \sqrt{\frac{2 (n_{m} - n_{m-1} )}{3} } \biggr\rceil .$$
We can choose numbers $$n_{1}, n_{2}, \ldots , n_{m} , \ldots$$ such that condition (1) of our theorem is true. By Theorem 5.3 and Lemma 5.9 we can suppose that all algebras $$\mathcal{A}_{k}$$ are ω-saturated σ-algebras. Put $$\mathcal{A}'_{k} = \mathcal{A}_{k}$$ if $$k \in[1,n_{1}]$$. By Theorem 2.1 there exists a set of pairwise distinct irregular ultrafilters $$G_{1} = \{ s_{1}, t_{1}, \ldots , s_{n_{1}}, t_{n_{1}} \}$$, and $$s_{k}$$, $$t_{k}$$ are $$\mathcal{A}'_{k}$$-equivalent ultrafilters for each $$k \in [1,n_{1}]$$. Let $$k \in[n_{1} + 1, n_{2} ]$$ and
$$E_{k} = \{ a \in\ker \mathcal{A}_{k} \setminus G_{1} \mid a \mbox{ has }A_{k}\mbox{-equivalent ultrafilter in }\ker \mathcal{A}_{k} \setminus G_{1}\}.$$
We can construct (see Remark 1.13) ω-saturated σ-algebra $$\mathcal{A}'_{k}$$ and
1. (1)

$$\ker \mathcal{A}'_{k} = E_{k}$$;

2. (2)

two ultrafilters are $$\mathcal{A}'_{k}$$-equivalent if and only if they are $$\mathcal{A}_{k}$$-equivalent.

In view of Remark 1.12, $$\mathcal{A}_{k}'\supseteq\mathcal{A}_{k}$$. It is clear that
$$\# \bigl(\ker \mathcal{A}'_{k}\bigr) \geq \biggl\lceil \frac{10}{3} (n_{2} - n_{1} ) + \sqrt{\frac{2 (n_{2} - n_{1} )}{3} } \biggr\rceil .$$
By Theorem 2.1 there exist pairwise distinct irregular ultrafilters $$s_{n_{1}+1}, t_{n_{1}+1},\ldots, s_{n_{2}}$$, $$t_{n_{2}}$$, and $$s_{k}$$, $$t_{k}$$ are $$\mathcal{A}'_{k}$$-equivalent ultrafilters for each $$k \in [n_{1} + 1, n_{2}]$$. Put
$$G_{2} = \{ s_{1}, t_{1}, \ldots , s_{n_{2}} , t_{n_{2}} \} .$$
It is clear that $$\#(G_{2}) = 2 n_{2}$$. Consider algebras $$\mathcal{A}_{n_{2}+1} , \ldots , \mathcal{A}_{n_{3}}$$. We can construct corresponding algebras $$\mathcal{A}'_{n_{2}+1} , \ldots , \mathcal{A}'_{n_{3}}$$, and
\begin{aligned}& \ker \mathcal{A}'_{k} \cap G_{2} = \emptyset, \\& \# \bigl(\ker \mathcal{A}'_{k}\bigr) \geq \biggl\lceil \frac{10}{3} (n_{3} - n_{2} ) + \sqrt{\frac {2 (n_{3} - n_{2} )}{3} } \biggr\rceil \end{aligned}
for each $$k \in[n_{2} + 1, n_{3} ]$$ and so on. Further, we consider algebras $$\mathcal{A}_{n_{3}+1} , \ldots , \mathcal{A}_{n_{4}}$$ and so on. So we can construct pairwise distinct irregular ultrafilters
$$s_{1}, t_{1}, \ldots , s_{k}, t_{k} , \ldots ,$$
such that $$s_{k}$$, $$t_{k}$$ are $$\mathcal{A}_{k}$$-equivalent ultrafilters for each $$k \in \mathbb{N}^{+}$$. We can construct a corresponding family of sets $$\{U^{1}_{k}, U^{2}_{k} \}_{k \in \mathbb{N}^{+}}$$ (see Definition 1.3). □
Footnotes
1

If $$\# (\ker\mathcal{A} )\geqslant\aleph_{0}$$, then, as it is shown in , $$\# (\ker \mathcal{A} )\geqslant2^{2^{\aleph_{0}}}$$.

2

In footnote a we already noticed that in this case $$\# (\ker\mathcal{A} )\geqslant2^{2^{\aleph_{0}}}$$.

3

It is clear that if $$\cap D \neq\emptyset$$, then $$\# (\cap D) \geq2$$.

4

It is clear that $$\mathcal{A}'\supset\mathcal{A}$$ (see Remark 1.12).

## Declarations

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

## Authors’ Affiliations

(1)
Department of Computer Science and Mathematics, Ariel University, Ariel, Israel

## References 