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\begin{document}
\title{Theorems on sets not belonging to algebras}
\author{L. \v S. Grinblat}
\address{Department of Mathematics, The College of Judea and Samaria, 
P.O.Box 3, Ariel 44837, Israel}
\email{grinblat@yosh.ac.il}
\date{February 15, 2004}
\commby{David Kazhdan}
\subjclass[2000]{Primary 03E05; Secondary 54D35}
\keywords{Algebra on a set, almost $\sigma$-algebra, ultrafilter, 
pairwise disjoint sets}
\begin{abstract}
Let $\mathcal{A}_1,\dots, \mathcal{A}_n, \mathcal{A}_{n+1}$ be a finite 
sequence of algebras
of sets given on a set $X$, $\bigcup_{k=1}^n \mathcal{A}_k \ne 
\mathfrak{P}(X)$, with more than $\frac{4}{3}n $ pairwise disjoint sets not
belonging to $\mathcal{A}_{n+1}$. 
It has been shown in the author's previous articles that in
this case $\bigcup_{k=1}^{n+1} \mathcal{A}_k \ne \mathfrak{P}(X)$. Let us
consider, instead of $\mathcal{A}_{n+1}$, a finite sequence of algebras
$\mathcal{A}_{n+1}, \dots, \mathcal{A}_{n+l}$. 
It turns out that if for each natural
$i \le l$ there exist no less than $\frac{4}{3}(n+l)- \frac{l}{24}
$ pairwise disjoint sets not belonging to $\mathcal{A}_{n+i}$, then
$\bigcup_{k=1}^{n+l} \mathcal{A}_k \ne \mathfrak{P}(X)$. Besides this
result, the article contains: an essentially important theorem on
a countable sequence of almost $\sigma$-algebras (the concept of
almost $\sigma$-algebra was introduced by the author in 1999),
a theorem on a
family of algebras of arbitrary cardinality (the proof of this
theorem is based on the beautiful idea of Halmos and Vaughan from
their proof of the theorem on systems of distinct
representatives), a new upper estimate of the function $\mathfrak{v}(n)$ 
that was introduced by the author in 2002, and other new results.
\end{abstract}
\maketitle

\section{Introducton}
\begin{defn}
By an {\em algebra} ${\A}$ on a
set $X$ we mean a nonempty system
of subsets 
of $X$ possessing the following property: if $M_1,M_2 \in {\A}$,
then $M_1 \cup M_2$, $M_1 \setminus M_2 \in {\A}$.
\end{defn}

\subsection{}
Here is some information necessary for understanding
the article. All algebras
are considered on some abstract set $X \neq \emptyset$. Unless the 
contrary follows from the context, by a set we always mean a
subset of $X$. As usual, $\mathfrak{P}(M)$ denotes the set of all
subsets of $M$. The symbol $\#(M)$ denotes the cardinality of the
set $M$. By $\mathbb{N}^+$ we denote the set of natural numbers. By
${\mathbb{N}}$ we denote the set of nonnegative integers. If $n_1, n_2
\in \mathbb{N}^+$ and $n_1 \le n_2$, then
\[ [n_1, n_2] = \lbrace k \in \mathbb{N}^+ \mid
n_1 \le k \le n_2 \rbrace.\]
By $\lceil \rho \rceil$ we denote the
maximum integer $\le \rho$. By $\lfloor \rho \rfloor$ we denote
the minimum integer $\ge \rho$.
\subsection{}
A few words about the results of the present article.
The theorems deduced further develop the theory introduced in
\cite{Gr1} and \cite{Gr2}. Note that \cite{Gr1} is fully comprised in 
\cite{Gr2}.
Suppose $n \in \mathbb{N}^+$ and $\psi:[1,n]\rightarrow
\mathbb{N}^+$ is a function such that if a sequence of algebras
$\A_1, \dots, \A_n$ is given and for each $k \in [1,n]$ there exist
$\psi(k)$ pairwise disjoint sets not
belonging to $\A_k$, then $\bigcup_{k=1}^n \A_k \ne \mathfrak{P}(X)$. Let us construct the following function:
\[\psi'(k)=
  \begin{cases}
    \psi(k) & \text{if $k \in [1,n]$}, \\
    \lceil \frac{4}{3}n \rceil +1 & \text{if $k=n+1$}.
  \end{cases}\]
In both \cite{Gr1} and \cite{Gr2} it is shown that if a sequence of algebras
$\A_1, \dots, \A_{n+1}$ is given and for each $k \in [1,n+1]$ there
exist $\psi'(k)$ pairwise disjoint sets not belonging to $\A_k$,
then $\bigcup_{k=1}^{n+1} \A_k \ne \mathfrak{P}(X)$. Note that the
estimate $\lceil \frac{4}{3}n \rceil+1$ is the best possible in a 
certain sense (see Theorem 2.2(2) in \cite{Gr2}). Now, take $l \in
\mathbb{N}^+$ and construct the function as follows:
\[\psi''(k)=
  \begin{cases}
    \psi(k) & \text{if $k \in [1,n]$}, \\
    \lfloor \frac{4}{3}(n+l) \rfloor - \lceil \frac{l+24}{24} \rceil
    & \text{if $k \in [n+1,n+l]$}.
  \end{cases}\]
  It turns out that if a sequence of algebras
$\A_1, \dots, \A_{n+l}$ is given and for each $k \in [1,n+l]$ there
exist $\psi''(k)$ pairwise disjoint sets not belonging to $\A_k$,
then $\bigcup_{k=1}^{n+l} \A_k \ne \mathfrak{P}(X)$. If $l=1$, we
arrive at the above-mentioned result from \cite{Gr1} and \cite{Gr2}.

Let $\psi_*:\mathbb{N}^+ \rightarrow \mathbb{N}^+$ be a
function such that if $n \in \mathbb{N}^+$, a sequence of
algebras $\A_1, \dots, \A_n$ is given and for each $k \in [1,n]$
there exist $\psi_*(k)$ pairwise disjoint sets not belonging to
$\A_k$, then $\bigcup_{k=1}^n \A_k \ne \mathfrak{P}(X)$. Now, let
${\mathcal B}_1, \dots, {\mathcal B}_k, \dots $ be a countable
sequence of algebras and for each $k \in \mathbb{N}^+$ there
exist $\psi_*(k)$ pairwise disjoint sets not belonging to
${\mathcal B}_k$. Can we assert that $\bigcup {\mathcal B}_k \ne
\mathfrak{P}(X)$? Generally speaking, no (see \cite{Gr2}). However, for
${\mathcal B}_k$ being $\sigma$-algebras, it is true. We are
interested in the situation when ${\mathcal B}_k$ are almost
$\sigma$-algebras. (The concept of an almost $\sigma$-algebra was
introduces in \cite{Gr1}. Here, it is given in Section 2.5.) We have
deduced an essentially important Theorem 2.6 on a countable
sequence of almost $\sigma$-algebras.

We consider the families of algebras $\lbrace \A_\lambda
\rbrace_{\lambda \in \Lambda}$ such that for each $\lambda$ there
exist two sets $U_1^\lambda, U_2^\lambda \not\in \A_\lambda$ and
$U_{i_1}^{\lambda_1} \cap U_{i_2}^{\lambda_2}= \emptyset$ unless
$\lambda_1 = \lambda_2, i_1 = i_2$. Interesting results have been
achieved here: Theorem 2.10, when the cardinality of $\Lambda$ is not
restricted in any way, and Theorem 2.14, when $\#(\Lambda)=
\aleph_0$ and $\A_\lambda$ are almost $\sigma$-algebras. It should
be pointed out that, as was shown in \cite{Gr1} and \cite{Gr2},
$\bigcup_{\lambda \in \Lambda} \A_\lambda \ne \mathfrak{P}(X)$ if
$1 \le \#(\Lambda) < \aleph_0$ and the algebras $\A_\lambda$ are not
restricted in any way.

Last but not least, we consider an interesting function $\mathfrak{v}(n)$, which was introduced in \cite{Gr2}. The problem on the upper
estimate is both complicated and interesting. Given here is a new
upper estimate of $\mathfrak{v}(n)$.
\subsection{}
This section deals with the Main Idea which first appeared 
in \cite{Gr1} and on which the investigation in 
\cite{Gr1} and \cite{Gr2} was based. We will consider 
ultrafilters on $X$. Each ultrafilter is a point $\beta X$, and vice versa, each 
point $\beta X$ is an ultrafilter on $X$. (Here, as usual, $\beta X$ is the 
Stone-\u Cech compactification of $X$ in discrete topology.)

Consider an algebra ${\A}$ and a set $U \notin {\A}$. There are two possible cases:
\begin{itemize}
\item[1)] If $U \subset D$, then $D\notin {\A}$; and then
there exists an ultrafilter $a
\ni U$ such that if $M \in a$, then $M \notin {\A}$.
\item[2)] There exists a set $D \supset U$ such that $D\in {\A}$; then there exist 
ultrafilters $a, b$ such that $a \ni U, b \ni D \setminus U$, and if $M$ 
belongs to one of those ultrafilters but does not belong to the other one, then 
$M \notin {\A}$.
\end{itemize}

These arguments lead one to accept the following concepts. An ultrafilter $a$ is said to be $\A$-
{\em special} if $a \cap \A = \emptyset$. Two ultrafilters $a, b$ are said 
to be $\A$-{\em similar} 
if $a \neq b$, there exists $D \in \A, a, b$, and, whenever $M$ belongs to 
one of these ultrafilters but not to the other one, we have $M
\notin \A$.
\begin{nota}
If $M\subset \beta X$ , 
then we denote by $\overline{M}$ the closure of $M$ in $\beta X$.
\end{nota}
\begin{main}
{\em Let $\lbrace {\A}_\lambda \rbrace$ be a family of algebras. Then 
$\bigcup \A_\lambda \ne \mathfrak{P}(X)$ if and only if there exist sets $S,T \subset 
\beta X$ such that $\overline{S} \cap \overline{T} = \emptyset$ and for
each $\lambda$ at least one of the
following two conditions holds:
\begin{itemize}
\item[(1)] there exists an ${\A}_\lambda$-special ultrafilter $z_\lambda \in S$;
\item[(2)] there exist ${\A}_\lambda$-similar ultrafilters
$s_\lambda, t_\lambda$ such that $s_\lambda \in S$ and $t_\lambda
\in T$.
\end{itemize}}
The considerations given in this section are used in the proofs of
all the theorems below.
\end{main}

\section{Formulating the results}
\begin{defn}
For each pair of numbers $n, l
\in \mathbb{N}^+$ define the minimum number $\mathfrak{q}_n^l
\in \mathbb{N}$ such that if $\A_1, \dots, \A_n, \A_{n+1},\dots,
\A_{n+l}$ is a sequence of algebras and (1) $\bigcup_{k=1}^n \A_k
\ne \mathfrak{P}(X)$, (2) for each $i \in [1, l]$ there exist
$\lceil \frac{4}{3}n \rceil + \mathfrak{q}_n^l$ pairwise disjoint
sets not belonging to $\A_{n+i}$, then $\bigcup_{k=1}^{n+l} \A_k
\ne \mathfrak{P}(X)$.
\end{defn}

\subsection{}
Estimation of numbers $\mathfrak{q}_n^l$ is, no doubt,
a very complicated and interesting problem. From what was said in
Section 1.3, it follows that \[\mathfrak{q}_n^l \le \Big\lfloor
\frac{4}{3}(n+l)\Big\rfloor - \Big\lceil \frac{l+24}{24}\Big\rceil
- \Big\lceil \frac{4}{3}n \Big\rceil.\]

We have managed to prove the following theorem.

\begin{theorem}
$\mathfrak{q}_n^l
\le \lfloor \frac{4}{3}(n+l) \rfloor - \lceil \frac{l+21}{21}
\rceil - \lceil
\frac{4}{3}n \rceil$ if $l \ge 42$;\\
\hspace*{1.8cm}\vspace{0.1cm}$\mathfrak{q}_n^l \le \lfloor
\frac{4}{3}(n+l) \rfloor - \lceil \frac{l+18}{18} \rceil - \lceil
\frac{4}{3}n \rceil$ if $l \ge 72$;\\
\hspace*{1.8cm}\vspace{0.1cm}$\mathfrak{q}_n^l \le \lfloor
\frac{4}{3}(n+l) \rfloor - \lceil \frac{l+17}{17} \rceil - \lceil
\frac{4}{3}n \rceil$ if $l \ge 85$;\\
\hspace*{1.8cm}\vspace{0.1cm}$\mathfrak{q}_n^l \le \lfloor
\frac{4}{3}(n+l) \rfloor - \lceil \frac{l+16}{16} \rceil - \lceil
\frac{4}{3}n \rceil$ if $l \ge 112$;\\
\hspace*{1.8cm}\vspace{0.1cm}$\mathfrak{q}_n^l \le \lfloor
\frac{4}{3}(n+l) \rfloor - \lceil \frac{l+15}{15} \rceil - \lceil
\frac{4}{3}n \rceil$ if $l \ge 195$.
\end{theorem}
\subsection{}
This section contains concepts and assertions from
\cite{Gr2}. Denote by $\Psi^n$ ($n \in \mathbb{N}^+$) {\em the totality
of all functions} $\psi:  [1, n]\rightarrow \mathbb{N}^+$ {\em
such that if $\psi \in \Psi^n$, a sequence of algebras $\A_1, \dots,
\A_n$ is given, and for each }$k \in [1, n]$ {\em there exist
$\psi(k)$ pairwise disjoint sets not belonging to $\A_k$, then}
$\bigcup_{k=1}^n \A_k \ne \mathfrak{P}(X)$. For each $n \in
\mathbb{N}^+$, denote by $\mathfrak{g}(n)$ {\em the minimum
natural number such that the function $\psi(k)=\mathfrak{g}(n)$
for all $k \in [1,n]$ belongs to $\Psi^n$.} Let us fix $n \in
\mathbb{N}^+$.  By definition, $\xi(n; k)$ {\em is any function
(of the variable $k$) defined on }[1,$n$], {\em taking values in
$\mathbb{N}^+,$ and satisfying}
\[\sum_{k=1}^n 2^{-\big\lceil \frac{\xi(n;k) + 1}{2}\big\rceil}
\le 1. \] It is easy to show that $\xi (n; k) \in \Psi^n$. Assume
that $n \in \mathbb{N}^+$ and for each $k   \in [1, n]$ a
nonempty class $\tilde{\Psi}^k \subset \Psi^k$ is chosen. Let $m
\in \mathbb{N}, p \in \mathbb{N}^+$, and $n \ge p-m > 0$.
Define the function (of the variable $k$) on $[m+1, p]$ taking
values in $\mathbb{N}^+$ as follows:
\[ \psi (m,p; k) = 2m + \psi (k - m), \mbox{ where } \psi \in
\tilde{\Psi}^{p-m}.\] Let $m \le n$ be natural numbers. One can
show that the function
\[ \psi(k)=
  \begin{cases}
    \psi (0,m;k) & \text{if $k \in [1,m]$}, \\
    \psi (m,n+1; k) & \text{if $k \in [m+1, n+1]$}
  \end{cases}
\] belongs to $\Psi^{n+1}$.
\subsection{}
Fix $n \in \mathbb{N}^+$. Let $m \in [1,n]$. Using
the numbers ${\mathfrak{q}}_m^l$, the considerations from \cite{Gr2}, and knowing
that for each $k \in [1,n]$ a nonempty class
$\tilde{\Psi}^k \subset \Psi^k$ is chosen, one can construct
the class $\tilde{\Psi}^{n+1} \subset \Psi^{n+1}$. The class $\tilde{\Psi}^{n+1}$ is formed by:\\
\hspace*{0.5cm}1) the functions of the type \[ \psi(k)=
  \begin{cases}
    \psi (0,m ;k) & \text{if $k \in
[1,m]$}, \\
    \Big\lceil \frac{4}{3}m \Big\rceil + \mathfrak{q}_m^{n+1-m} & \text{if $ k \in [m+1,n+1]$};
  \end{cases}
\]\\
\hspace*{0.5cm}2) the function $ \psi(k) \equiv \mathfrak{g}(n+1)$;\\
\hspace*{0.5cm}3) the functions of the type $\xi (n+1;k)$;\\
\hspace*{0.5cm}4) the functions of the type
\[ \psi(k)=
 \begin{cases}
    \psi (0,m;k)  & \text{if $k \in [1,m]$}, \\
    \psi (m,n+1; k) & \text{if $k \in [m+1, n+1]$}.

    \end{cases}\](In \cite{Gr2}, we present the inductive metod for constructing
    functions from $\Psi^n.$ Yet, it is less perfect since it makes
    no use of the numbers ${\mathfrak{q}}_m^l$.)

\subsection{}
{\em An algebra ${\A}$ is called a $\sigma$-algebra if for any countable 
sequence $M_1, \dots$, $M_k, \ldots \in {\A}$ we have }${\A} \ni
\bigcup_{k=1}^{\infty}M_k$. In \cite{Gr1}, there was introduced an
essentially important concept: {\em an algebra $\A$ is said to be
an almost $\sigma$-algebra if for any countable sequence of sets
$M_1, \dots, M_k, \dots$ such that $\mathfrak{P}(M_k)\subset \A_k$ for
each $k$, we have $\bigcup M_k \in \A$.} Denote by $\Psi$ the
totality of all functions $\psi$: $\mathbb{N}^+ \rightarrow
\mathbb{N}^+$ possessing the following property: for each $n \in
\mathbb{N}^+$ there exists a function $\psi_n \in \Psi^n$ such
that $\psi_n (k)= \psi (k)$ for all $k \in [1, n]$.

\subsection{}
Let $\psi \in \Psi$, and let $\A_1, \dots, \A_k, \dots$ be
a countable sequences of $\sigma$-algebras such that for each $k$
there exist $\psi(k)$ pairwise disjoint sets not belonging to
$\A_k$. Then, as was mentioned in Section 1.3, $\bigcup
\A_k \ne \mathfrak{P}(X)$. It is not known whether this statement
holds true when $\sigma$-algebras are replaced with almost
$\sigma$-algebras.

 Let $\eta \in \mathbb{N}^+$.  If $n \in [1, \eta]$,
put $ \Psi^n (\eta)=\Psi^n$. If $n = \eta +1$, $ \Psi^n (\eta)$
will denote the class of all functions from $\Psi^n$ derived with
the help of the functions of classes $ \Psi^k (\eta)$, where $k
\le n-1$, and by means of the inductive method from Section 2.4.
On the analogy, by induction, a class of functions $\Psi^n (\eta)$
is built for each natural $n \ge \eta + 2$. Consider the class of
functions $ \Psi (\eta) \subset \Psi$: {\em a function $\psi \in
\Psi (\eta)$ if and only if for each $n \in \mathbb{N}^+$ there
exists a function $\psi_n \in \Psi^n (\eta)$ such that $\psi_n (k)
\le \psi (k)$ for all $k \in [1,n]$}. Put
\[\tilde \Psi=\bigcup_{\eta=1}^\infty \Psi (\eta).\] The next theorem
generalizes in depth some of the main results
from \cite{Gr1} and \cite{Gr2}.
\begin{theorem}
{Let $\psi \in \tilde \Psi$, and
let $\A_1, \dots, \A_k, \dots$ be a countable sequence of almost
$\sigma$-algebras such that for each $k$ there exist $\psi(k)$
pairwise disjoint sets not belonging to $\A_k$. Then $\bigcup \A_k
\ne \mathfrak{P}(X).$}
\end{theorem}

\begin{rem}
It should be pointed out that,
besides the Main Statement, the proofs of the result on
$\sigma$-algebras given in Section 2.6 and Theorem 2.6 use the
deep and nontrivial theorem of Gitik-Shelah from \cite{G-S}. The
authors' proof of this theorem is metamathematical and uses the
forcing method. Purely mathematical proofs have been proposed by
Fremlin in \cite{F1} and \cite{F2}, and by Kamburelis in \cite{K}. In 
\cite{Gr2} we
give a purely mathematical proof of the Gitik-Shelah theorem
following \cite{K}.
\end{rem}
\subsection{}
In \cite{Gr1} and \cite{Gr2}, the following theorem was proved.

\begin{theorem}
{Let $\A_1, \dots, \A_k, \dots$ be a countable
sequence of $\sigma$-algebras, and let for each $k$ there exist
two sets $U_1^k, U_2^k \notin \A_k;\ U_i^k \cap U_j^l = \emptyset$ unless
$k=l, i=j$. Then $\bigcup \A_k \ne \mathfrak{P}(X)$.}
\end{theorem}
\begin{rem}
As was mentioned in Section
1.3, if considered in Theorem 2.8 is a finite sequence of
algebras, the condition of $\sigma$-additivity of algebras can be
ignored.
\end{rem}
\subsection{}
The following theorem deals with a family of algebras
having arbitrary cardinality. Its proof uses the idea of the proof
of Theorem 2.8 (which, in turn, uses the Main Statement) as
well as the idea of applying the Tychonoff theorem to proving a
theorem on a system of distinct representatives (see \cite{H-V}).
\begin{theorem}
{Let $\lbrace {\A}_\lambda \rbrace_{\lambda \in \Lambda}$ be a family of 
$\sigma$-algebras, and $\lbrace U_1^\lambda, U_2^\lambda \rbrace _{\lambda \in \Lambda}$ 
a family of sets such that
\begin{itemize}
  \item[1)] $U_i^\lambda \notin {\A}_\lambda$,
  \item[2)] $U_{i_1}^{\lambda_1} \cap U_{i_2}^{\lambda_2} = \emptyset$ unless $\lambda_1=\lambda_2,
  i_1=i_2$.
\end{itemize}
Suppose, for each $\lambda \in \Lambda$ there exists $\Lambda'(\lambda) \subset \Lambda$ such that
\begin{itemize}
\item[a)] $\lambda \notin \Lambda'(\lambda)$,
\item[b)] $\# (\Lambda \setminus \Lambda'(\lambda)) \le \aleph_0$,
\item[c)] ${\A}_\lambda \ni \bigcup_{\lambda' \in
\Lambda'(\lambda)}(U_1^{\lambda'} \cup U_2^{\lambda'}) $.
\end{itemize}
Then $\bigcup_{\lambda \in \Lambda} \A_\lambda \ne \mathfrak{
P}(X)$.}
\end{theorem}
\begin{rem}
Taking into account Remark 2.9, one
can prove the following statement. Theorem 2.10 remains true if the
algebras $\A_\lambda$ are not required to be $\sigma$-additive and
condition b) is replaced by a stronger condition
\begin{itemize}
\item[b$^*)$] \  $\# (\Lambda \setminus \Lambda'(\lambda)) < \aleph_0$.
\end{itemize}
\end{rem}
\subsection{}
We do not know whether Theorem 2.8 is true if
$\sigma$-algebras are replaced by almost $\sigma$-algebras. Yet the
following theorem, the proof of which is based on the idea of the
proof of Theorem 2.8, is true.
\begin{theorem}
{Let $\A_1, \dots, \A_k, \dots$ be a countable
 sequence of almost $\sigma$-algebras and let for each $k$ there
 exist three sets $U_1^k, U_2^k, U_3^k \notin \A_k$; $U_i^k \cap
 U_j^l= \emptyset$ unless $k=l,\ i=j$, and
 \[ \#(\{l \in \mathbb{N}^+|U_i^k \notin \A_l\})<\aleph_0\]
 for each $U_i^k$. Then $\bigcup \A_k \ne \mathfrak{
P}(X)$.}
\end{theorem}
\subsection{}
Before proceeding to the next theorem, it is
necessary to mention a definition used in \cite{Gr2}.

\begin{definition}
Assume that $\mathbb{N}^* \subset
\mathbb{N}^+$ and let $n$ be an arbitrary natural number. The
number \[ {\overline{lim}_{n \rightarrow \infty}} \quad \frac{\#\
(\lbrace k \le n \mid k \in \mathbb{N}^* \rbrace)}{n}\] denoted
$p(\mathbb{N}^*)$ is called the {\em density} of $\mathbb{N}^*$. The number
 \[ {\underline{lim}_{n \rightarrow \infty}} \quad \frac{\#\ (\lbrace k 
 \le n \mid k \in \mathbb{N}^* \rbrace)}{n}\] denoted $p_a 
 (\mathbb{N}^*)$ is called the {\em absolute density} of $\mathbb{N}^*$.
\end{definition}
\subsection{}
The idea of the proof of Theorem 2.8 is used in the
 proof of the following interesting theorem.

\begin{theorem}
{Let $\A_1, \dots, \A_k, \dots$ be a countable
 sequence of almost $\sigma$-algebras and let for each $k$ there
 exist two sets $U_1^k, U_2^k \notin \A_k$; $U_i^k \cap
 U_j^l= \emptyset$ unless $k=l,\ i=j$, and
 \[ p(\{l \in \mathbb{N}^+|U_i^k \notin \A_l\})=0\]
 for each $U_i^k$. Then there will be $\mathbb{N}^* \subset \mathbb{N}^+$ such that
 $p (\mathbb{N}^*)=1,\ p_a (\mathbb{N}^*)\ge \frac{2}{3}$, and $\bigcup_{k
\in \mathbb{N}^*} \A_k \ne \mathfrak{P}(X)$.}
\end{theorem}
\begin{rem}
Here, it is appropriate to give a
quotation from \cite{Gr2}. If in the assumptions of Theorem 2.8 one
does not impose any restrictions on the algebras $\A_k$, such as
$\sigma$-additivity or almost $\sigma$-additivity, then, clearly,
one can only assert the following: for any real number $\rho <1$
there exists $\mathbb{N}^* \subset \mathbb{N}^+$ such that
$p_a (\mathbb{N}^*)>\rho$ and $\bigcup_{k \in\mathbb{N}^*} 
\A_k \ne \mathfrak{P}(X)$.
\end{rem}
\subsection{}
Let us proceed to another problem we tackled in \cite{Gr2}.
\begin{definition}
For each $n \in \mathbb{N}^+$, denote by
$\mathfrak{v}(n)$ the minimum natural number such that if ${\A}_1,\dots,
{\A}_n $ is a sequence of algebras, and for each $k \in [1, n]$
there exist $\mathfrak{v}(n)$ pairwise disjoint sets not belonging to
${\A}_k$, then there exist pairwise disjoint sets $U_1, \dots, U_n,
V_1, \dots, V_n$ such that if $Q \supset U_k$ and $Q \cap V_k = \emptyset$,
then $Q \notin {\A}_k$.
\end{definition}
\subsection{}
It is shown in \cite{Gr2} that:
\begin{enumerate}
\item[(1)] $\mathfrak{v}(n) = 4n - 3$ for $n \le 3$.
\item[(2)] $\mathfrak{v}(n) \le 4n - 5$ for $n > 3$.
\item[(3)] $\mathfrak{v}(n) \le 4n - \lceil \frac{n + 3}{2}
\rceil$. 
\item[(4)] $3n - 2 \le \mathfrak{v}(n)$.
\end{enumerate}

\subsection{}
The proof of the following theorem is based on the
considerations from \cite{Gr2}.
\begin{theorem}
${\mathfrak{v}}(n) \le \lfloor \frac{10}{3} n +
\frac{2}{\sqrt 3} \sqrt n  \rfloor.$
\end{theorem}

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\end{document}