# Infinite countable Union of $\sigma-\text{Algebras}$ exercise

Show that if $\{\mathscr{A}_n\}_{n\in\mathbb{N}}$ is an increasing sequence of $\text{Algebras}$ of subsets of a set $X$. Show by example that even if $\{\mathscr{A}_n\}$ is a $\sigma-\text{Algebras}$ for every ${n\in\mathbb{N}}$ the union still may not be a $\sigma-\text{Algebra}$. Solution:Let $X=\mathbb{N}$, and $\mathscr{A}_n=$ the family of subsets of $\{1,2,3...n\}$ and their complements. Clearly, $\mathscr{A}_n$ is a $\sigma-\text{Algebras}$ and $\mathscr{A}_1\subset\mathscr{A}_2\subset\mathscr{A}_3...$ However $\bigcup_\limits{n\in\mathbb{N}}^{}\mathscr{A}_n$ is the family of all finite and co-finite subsets of $\mathbb{N}$ which is not a $\sigma-\text{Algebra}$$\:\:\blacksquare How can \bigcup_\limits{n\in\mathbb{N}}^{}\mathscr{A}_n not be a \sigma-\text{Algebra} if the sequence is increasing? It seems a contradiction to me. Thanks in advance. • What about the example is eluding you? It seems perfectly clear to me. Commented Apr 16, 2017 at 13:35 • All the \mathscr{A}_n are finite algebras, so trivially \sigma-algebras. The first time infinite unions can appear at all is after taking the final union. Commented Apr 16, 2017 at 13:37 • It is the fact the sequences are increasing which means the union of the sigma-algebras must contain \mathbb{N} which countable infinite, that is why I think the union is a sigma-algebra too. What I do not get is the reason behind the family of finite and co-finite subsets of \mathbb{N} . Commented Apr 16, 2017 at 13:38 • Already \mathscr{A}_0 contains \mathbb{N} as the complement of \emptyset. Nothing to do with increasingness. Commented Apr 16, 2017 at 13:41 • Why does the union is not a sigma-algebra. It must contain\mathbb{N},right? Commented Apr 16, 2017 at 13:42 ## 2 Answers A set A is in \mathscr{A}_n iff A \subseteq \{1,\ldots n\} or X\setminus A \subseteq \{1,\ldots,n\}. This is just the definition of \mathscr{A}_n. A \in \mathscr{A}:= \bigcup_n \mathscr{A}_n iff \exists n: A \in \mathscr{A}_n iff \exists n: A \subseteq \{1,\ldots, n\} \text{ or } X\setminus A \subseteq \{1,\ldots,n\} iff A is finite or A is cofinite (i.e. has finite complement ) Now, A_k = \{2k\} for k \in \mathbb{N} is in \mathscr{A}_{2k} \subseteq \mathscr{A}, but the union of the A_k is the set of even numbers E which is not in \mathscr{A} by the above criterion: it's neither finite nor cofinite. So \mathscr{A} is indeed not a \sigma-algebra, exactly for the reason claimed. • Comments are not for extended discussion; this conversation has been moved to chat. Commented Nov 10, 2018 at 0:08 Hint: Find a sequence (a_n)_{n \in \mathbb{N}} \subseteq \mathbb{N} such that both$$A := \bigcup_{n \in \mathbb{N}} \{a_n\} \qquad \text{and} \qquad A^c$$are infinite sets. Show that$A \notin \bigcup_k \mathcal{A}_k$and conclude that$\bigcup_k \mathcal{A}_k$is not closed under countable unions, hence not a$\sigma$-algebra. • How can$A\notin\bigcup_k \mathcal{A}_k$if$\mathcal{A}_k$, contains$\mathbb{N}$? Commented Apr 16, 2017 at 15:16 • @PedroGomes why do you think that$\mathbb{N} \in \mathcal{A}_k$implies$A \in \bigcup_k \mathcal{A}_k$.....? You have to choose the sequence$(a_n)_n$in such a way that$A^c$is infinite (in particular$A \neq \mathbb{N}$because otherwise$A^c = \emptyset\$ wouldn't be an infinite set).
– saz
Commented Apr 16, 2017 at 15:19