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}$$\:\:\blacksquare$
How can $\bigcup_\limits{n\in\mathbb{N}}^{}\mathscr{A}_n$ not be a $\sigma-\text{Algebra}$ if the sequence is increasing? It seems a contradiction to me.
Thanks in advance. 
 A: Hint: Find a sequence $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{N}$ such that both $$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.
A: 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.
