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Families of sets not belonging to algebras and combinatorics of finite sets of ultrafilters
Journal of Inequalities and Applications volume 2015, Article number: 116 (2015)
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 \(Q\notin \mathcal{A}_{\lambda }\).
1 Introduction
The present article is a further development of the theory formulated in [1–7]. 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 nonempty 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 [5].
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 σalgebras if \(\# (\Lambda ) = \aleph_{0}\). In [6] 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 [8] (this paper contains the wellknown theorem of AlougluErdös). Some information about the history of the subject after the publication of [8] and before the publication of [1] is presented in [2]. In fact, Alouglu and Erdös studied nonmeasurable sets with respect to families of measures. Let \(\aleph_{1}\le\#(X) \le2^{\aleph_{0}}\). Let a σadditive measure μ be defined on X. Here \(\mu(X) = 1\), the measure of a onepoint set equals 0, and the measure of each μmeasurable set equals 0 or 1. Such a measure μ is called a σtwovalued measure. Clearly, there exist μnonmeasurable sets. The AlougluErdös theorem states that if \(\#(X)=\aleph_{1}\), then for any countable family of σtwovalued measures \(\mu_{1}, \ldots, \mu_{k}, \ldots\) there exists a set which is nonmeasurable with respect to all these measures. The proof of the AlougluErdös theorem is very simple and is based on the possibility of constructing the wellknown Ulam matrix (see [9]). The nontrivial GitikShelah theorem (see [10]) asserts the validity of the AlougluErdös theorem if \(\#(X)=2^{\aleph_{0}}\). Obviously, the GitikShelah theorem is a generalization of the AlougluErdös theorem. The GitikShelah theorem can be reformulated in our language. As before, let us consider the σtwovalued 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 GitikShelah 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 GitikShelah theorem is used in the proofs of our theorems for countable families of σalgebras.
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)
\(U^{i}_{\lambda }\cap U^{j}_{\lambda '} = \emptyset\) except when \(\lambda = \lambda '\), \(i = j\);

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

(2)
any set \(Q_{\vartheta}\) is a union of sets \(U^{i}_{\lambda }\);

(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)
\(\#(\Theta) = 2^{\#(\Lambda )}\).
Proof
Put \(\Theta= \mathcal{P}(\Lambda )\). If \(\vartheta\in \mathcal{P}(\Lambda )\), put
□
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 [1–3]. The proof of the two following theorems can be found in [2].
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 \(4k3\) lacunae. Then this family has a full set of lacunae.
It is easy to prove (see [2], Chapter 14) that the estimate \(4k3\) is the best possible in some sense.
Theorem 1.6
Let \(\{\mathcal{A}_{k}\}_{k \in \mathbb{N}^{+}}\) be a family of σalgebras, and assume that for each k the algebra \(\mathcal{A}_{k}\) has \(4k3\) lacunae. Then this family has some full set of lacunae.
Remark 1.7
Using the notion of absolute introduced by Gleason in [11], 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 σalgebra, and \(\bigcup_{k \in \mathbb{N}^{+}} \mathcal{B}_{k} = \mathcal{P}(X)\) (see [2], 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 σadditive. Therefore, we suppose that all algebras of a countable family of algebras are σalgebras.
The following definition was given in [2].
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 [2] we proved that:

(1)
\(\frak{v}(n) = 4n3\) for \(n \leq3\);

(2)
\(\frak{v}(n) \leq4n5\) for \(n > 3\);

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

(4)
\(3n2 \leq \frak{v}(n)\) 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 [1] and was later used in [2–7]. Let βX be the StoneČech 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 \(\mathcal{A}\):
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)
\(W(r_{k}) \in \mathcal{A}\) for any \(k \in [1,m]\);

(2)
\(W(r_{k}) \notin q\) for any \(k \in[1,m]\);

(3)
\(V \subseteq \bigcup^{m}_{k=1} W (r_{k})\).
Let
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 \(\# (\ker\mathcal{A} )\geqslant k\).^{Footnote 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 \(\mathcal{A}\), ℬ. It is very easy to prove that the following statements are equivalent.

(1)
\(\mathcal{A}\supseteq\mathcal{B}\).

(2)
If a and b are \(\mathcal{A}\)equivalent ultrafilters, then a and b are ℬequivalent ultrafilters.

(3)
If M is an \(\mathcal{A}\)equivalent set, then M is contained in a certain ℬequivalent 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 aequivalent if and only if they belong to the same set \(A_{k}\). We can construct the algebra \(\mathcal{A}\) such that the aequivalence 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 ωsaturated. So, an algebra \(\mathcal{A}\) is ωsaturated if and only if \(\# (\ker\mathcal{A} )<\aleph_{0}\). The algebra \(\mathcal{A}\) from Remark 1.13 is ωsaturated.
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 nonempty 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 [3]: \(\frak{v}(n) \leq \lceil\frac{10}{3} n + \frac{2}{\sqrt{3}} \sqrt{n} \rceil\) 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 \(\frak{v}(n) \geq3 n2\) for any n, and \(\frak{v}(n) > 3n2\) if \(n = 2,3\) since \(\frak{v}(2) = 5\), \(\frak{v}(3) = 9\) (see Section 1). Is it true that \(\frak{v}(n) = 3n2\) for any \(n \neq2,3\)? This result is obviously true for \(n=1\).
The final section of this article is devoted to the proof of the following theorem, which is a generalization of theorems of AlaougluErdös and GitikShelah.
Theorem 2.4
It is possible to construct nondecreasing functions \(\varphi: \mathbb{N}^{+} \rightarrow\mathbb{N}^{+}\) such that the following conditions hold:

(1)
\(\underline{\lim}_{n \rightarrow\infty} \frac{\varphi(n)  \frac{10}{3}n }{\sqrt{n}} = \sqrt{\frac{2}{3}}\);

(2)
if \(\{ {\mathcal{A}}_{k} \}_{k\in\mathbb{N}^{+}}\) is a family of σalgebras 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 ωsaturated;^{Footnote 2} let a number \(\xi\in\mathbb{N}^{+}\) be given. Then it is possible to construct an ωsaturated 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 [2] 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 ωsaturated. 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
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, [2], 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 [2], 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
Let \(\mathcal{A}_{1} , \ldots , \mathcal{A}_{n}\) be algebras such that
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 ωsaturated. We will prove that there exist pairwise distinct ultrafilters
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
Then there exists a set of pairwise distinct ultrafilters
such that \(a_{k}\), \(b_{k}\) are \(\mathcal{A}_{k}\)equivalent ultrafilters for each \(k \in[1,n1]\). Consider \(\ker\mathcal{A}_{n}\). It is clear that
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 \(k_{1}\in[1,n1]\) 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
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 nonexistence 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 \(k_{2}\in[2,n1]\) 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
Suppose also that there exists a set of pairwise distinct ultrafilters
and the following holds:

(A)
\((a_{i+1},c_{i} )\) and \((b_{i+1},d_{i} )\) are two pairs of \(\mathcal{A}_{i}\)equivalent ultrafilters for each \(i\in [2,\eta1]\);

(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
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 nonexistence 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 \(\{ a_{\eta+1},\ldots,a_{n1},b_{\eta+1},\ldots,b_{n1} \}\). We have
Therefore, there exist distinct ultrafilters \(c_{\eta},d_{\eta}\in L_{\eta}\) and \(k_{\eta+1}\in[\eta+1,n1]\) 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 \(c_{1},\ldots ,c_{\rho1}, c_{n}, d_{1},\ldots, d_{\rho1}, d_{n}\) such that the following holds:

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

(b)
\((a_{i+1}, c_{i})\) and \((b_{i+1}, d_{i})\) are two pairs of \(\mathcal{A}_{i}\)equivalent ultrafilters for each \(i \in[1, \rho1]\);

(c)
\(\mathfrak{F} \cap\{c_{1}, \ldots , c_{\rho1} , c_{n},d_{1}, \ldots , d_{\rho 1} , d_{n} \} = \emptyset\), see Figure 1.
(2) Put
Clearly,
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 onetoone. (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 nonexistence of ultrafilters \(a^{*}_{1}, \ldots, a^{*}_{n}, b^{*}_{1}, \ldots, b^{*}_{n}\) is yielded immediately.) Put
Obviously, \(\frak{I}_{1}^{\rho}\cap \frak{I}_{2}^{\rho}=\emptyset\). Since \(\#(L_{\rho})\#([\rho+1,n1])>0\), we have \(\#(\frak{I}_{1}^{\rho})=\tau>0\). Clearly,
Put
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:
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
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
The above constructions clearly show that \(\alpha\leq\beta+ \gamma\). Put
Clearly,
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 onetoone. 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 nonexistence of ultrafilters \(a^{*}_{1}, \ldots, a^{*}_{n}, b^{*}_{1}, \ldots, b^{*}_{n}\) is yielded immediately. Put
Obviously, \(\hat{\frak{I}}_{1}\cap\hat{\frak{I}}_{2}=\emptyset\). Since \(\#(\hat{L})\#([\rho+1,n1])>0\), we have \(\#(\hat{\frak{I}}_{1})=\hat{\tau}>0\). Clearly,
If \(\tau\ge\hat{\tau}\), put
If \(\tau<\hat{\tau}\), put
Clearly,
(3) We fix \(\nu\in \frak{I}\). A number \(k\in[1,\rho]\) is called νmarked 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 \(k>1\): \((a_{k},d_{k1})\) and \((b_{k},c_{k1})\) are pairs of \(\mathcal{A}_{\nu}\)equivalent ultrafilters.
Put
Our aim is to prove that
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
and \((a_{k}, a'_{k})\), \((b_{k}, b'_{k})\) are pairs of \(\mathcal{A}_{\rho+1}\)equivalent ultrafilters for each \(k \in[\rho+2, n1 ]\). 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 11. \(q_{\rho+ 1} = q'_{\rho+ 1}\).
We consider only two subcases of Case 11.
Case 111. There exists an ultrafilter \(q^{*}\notin Z_{\rho}\) such that \(q^{*}\), \(q_{\rho+ 1}\) are \(\mathcal{A}_{\rho+ 1}\)equivalent ultrafilters.
Case 112. There exists an ultrafilter \(q^{*} \in Z_{\rho}\) such that \(q^{*}\), \(q_{\rho+ 1}\) are \(\mathcal{A}_{\rho+ 1}\)equivalent ultrafilters.
Case 12. \(q_{\rho+ 1} \neq q'_{\rho+ 1}\).
We consider only two subcases of Case 12.
Case 121. \(q_{\rho+ 1}\), \(q'_{\rho+ 1}\) are \(\mathcal{A}_{\rho + 1}\)equivalent ultrafilters.
Case 122. 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 \(\mathcal{R}_{k}=\{a_{k},b_{k},c_{k1}, d_{k1}\}\) if \(k\in[2,\rho]\).
First we consider Cases 111 and 121. 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^{∗}):

\(a_{k}\), \(d_{k1}\) are \(\mathcal{A}_{\rho+ 1}\)equivalent ultrafilters and \(\# (\ker \mathcal{A}_{\rho+ 1} \cap \mathcal{R}_{k}) = 2\);
 (3^{∗}):

\(b_{k}\), \(c_{k1}\) 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
Recall that \(\rho= \lfloor\frac{n}{3} \rfloor\). Therefore we have
Now consider Case 112. 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 122. 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 112 and 122 we have
It is clear that in Case 1 there may be subcases which we have not considered. But always
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:

(i)
\((a_{\rho1}, b_{\rho 1})\), \((c_{\rho 2}, d_{\rho 2})\), \((a_{2}, q_{\rho+ 1})\), \((b_{2}, c_{1})\) are pairs of \(\mathcal{A}_{\rho+ 1}\)equivalent ultrafilters;

(ii)
\(a_{\rho+1}\), \(q_{\rho+ 1}\) are \(\mathcal{A}_{\rho}\)equivalent ultrafilters;

(iii)
\(\ker \mathcal{A}_{\rho+ 1} \subset Z'_{\rho}\cup \{a'_{\rho+2} , \ldots , a'_{n1}, b'_{\rho+ 2} , \ldots , b'_{n1} \} \cup\{ q_{\rho+ 1} \}\), see Figure 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
Recall that \(\rho= \lfloor\frac{n}{3} \rfloor\). Therefore we have
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 νmarked number. We know that
Therefore we have
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
Suppose that
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:\((a^{*}_{1}, b^{*}_{1}) = (a_{1}, b_{1})\), \((a^{*}_{2}, b^{*}_{2}) = (a_{2}, b_{2})\), \((a^{*}_{3},b^{*}_{3}) = (b_{4}, d_{3})\), …, \((a^{*}_{\rho1 },b^{*}_{\rho 1}) = (b_{\rho}, d_{\rho 1})\), \((a^{*}_{\rho},b^{*}_{\rho}) = (a_{\rho+ 1},z_{\rho+ 1})\), \((a^{*}_{\rho+ 1}, b^{*}_{\rho+ 1}) = (a_{3}, d_{2})\), \((a^{*}_{\rho+ 2}, b^{*}_{\rho+ 2}) = (b_{3}, c_{2})\), \((a^{*}_{\rho+ 3},b^{*}_{\rho+ 3}) = (a_{\rho+3}, b_{\rho+ 3})\), …, \((a^{*}_{n1},b^{*}_{n1} ) = (a_{n1}, b_{n1})\), \((a^{*}_{n}, b^{*}_{n}) = (a_{\rho+ 2}, z'_{\rho+ 2})\), see Figure 3.
□
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)
\(\alpha^{k}_{i} \in\mathbb{N} \);

(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
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
for each \(i \in[1,n]\), then \(\frak{M}(n)\) is saturated.
The following theorem is a particular case of the wellknown theorem of Ramsey [12].
Theorem 4.4
Consider a set S, \(\# (S) = n \in \mathbb{N}^{+}\), and let T be the family of all twoelement subsets of S. We divide T into two disjoint subfamilies \(T_{1}\), \(T_{2}\). 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 twoelement 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
for any \(i \in[1,n]\), and \(n \geq R(\mu)\). Then

(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)
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;

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

(b)
\(\cap D \neq\emptyset\).^{Footnote 3}

(a)
Proof
Let us use the notation of Theorem 4.4. By Theorem 4.3 there exists a corresponding family of segments
Let T be the family of all subsets of S with the exact two elements. Divide T into two disjoint subfamilies \(T_{1}\), \(T_{2}\). 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 wellknown result is given, for example, in [13]:
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 σalgebras
In the first nine subsections we present facts from [1] and [2].
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)
\(\# (Z) \leq\aleph_{0}\);

(2)
if \(Z \neq\emptyset\), all points of Z are irregular;

(3)
\(\ker \mathcal{A}\subseteq\overline{Z}\).
The proof of the following theorem is in [2], 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 σalgebras. 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)
\(\ker \mathcal{A}_{k} \subseteq\overline{W}\) for each k;

(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 GitikShelah 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 GitikShelah theorem but the rather simple AlaogluErdös 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 [2], Chapter 7.
Lemma 5.6
If \(\mathcal{A}\) is a simple σalgebra, then \(\ker \mathcal{A}\subseteq \overline{sp \mathcal{A}}\).
The proof of the lemma below is in [2], Chapter 7.
Lemma 5.7
If \(\mathcal{A}\) is a simple σalgebra and \(a \in sp \mathcal{A}\), then
Remark 5.8
If an ωsaturated algebra \(\mathcal{A}\) is a σalgebra, 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 ωsaturated σalgebra \(\mathcal{A}\) and let \(\nu\in \mathbb{N}^{+}\). We can construct an ωsaturated σalgebra \(\mathcal{A}'\) 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.^{Footnote 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 \(k \in[n_{m1}+1, n_{m} ]\), where \(m \in \mathbb{N}^{+}\), then
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 ωsaturated σalgebras. 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
We can construct (see Remark 1.13) ωsaturated σalgebra \(\mathcal{A}'_{k}\) and

(1)
\(\ker \mathcal{A}'_{k} = E_{k}\);

(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
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
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
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
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). □
Notes
If \(\# (\ker\mathcal{A} )\geqslant\aleph_{0}\), then, as it is shown in [2], \(\# (\ker \mathcal{A} )\geqslant2^{2^{\aleph_{0}}}\).
In footnote a we already noticed that in this case \(\# (\ker\mathcal{A} )\geqslant2^{2^{\aleph_{0}}}\).
It is clear that if \(\cap D \neq\emptyset\), then \(\# (\cap D) \geq2\).
It is clear that \(\mathcal{A}'\supset\mathcal{A}\) (see Remark 1.12).
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Grinblat, L.Š. Families of sets not belonging to algebras and combinatorics of finite sets of ultrafilters. J Inequal Appl 2015, 116 (2015). https://doi.org/10.1186/s1366001505787
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DOI: https://doi.org/10.1186/s1366001505787