On sets not belonging to algebras

Abstract

Let $\mathcal{A}_1,..., \mathcal{A}_n, \mathcal{A}_{n+1}$ be a finite sequence of algebras of sets given on a set $X, \cup^n_{k=1}\mathcal{A}_k \neq \mathfrak{B}(X)$, with more than $\frac{4}{3}n$ pairwise disjoint sets not belonging to $\mathcal{A}_n$. It was shown in [4] and [5] that in this case $\cup^{n+1}_{k=1}\mathcal{A}_k \neq \mathfrak{B}(X)$. Let us consider, instead $\mathcal{A}_{n+1}$, a finite sequence of algebras $\mathcal{A}_{n+1},..., \mathcal{A}_{n+1}$. It turns out that if for each natural $i \leq l$ there exist no less than $\frac{4}{3}(n+l)-\frac{l}{24}$ pairwise disjoint sets not belonging to $\mathcal{A}_{n+i}$, then $\cup^{n+1}_{k=1}\mathcal{A}_k \neq \mathfrak{B}(X)$. But if $l \geq$ 195 and if for each natural $i \leq l$ there exist no less than $\frac{4}{3}(n+l)-\frac{l}{15}$ pairwise disjoint sets not belonging to $\mathcal{A}_{n+i}$, then $\cup^{n+1}_{k=1}\mathcal{A}_k \neq \mathfrak{B}(X)$. After consideration of finite sequences of algebras, it is natural to consider countable sequences of algebras. We obtained two essentially important theorems on a countable sequence of almost $\sigma$-algebras (the concept of almost $\sigma$-algebra was introduced in [4]).