# Algebra on infinite sets is not closed under countable union

this proof came up during self study, rather than looking at the other proofs out there, I would like to correct if necessary the following one...

Let $X$ be an infinite set and $A$ the collection of sets that are finite or with finite complement in $X$. $A$ is obviously an algebra. When trying to prove that $A$ is not closed under countable union, the following argument popped up.

Let $x_i$ be points in $X$ with $i\neq j \Rightarrow x_i\neq x_j$, the singletons $\{x_i\}\in A, \forall i$ since they are trivially finite. Also $\cup_{i\in \mathbb{N}} \{x_i\}\subset X$ since each of them belong to $X$.

Define $$y_n:=x_{2n}$$ and $$z_n:=x_{2n+1}$$, then $\cup_{i\in \mathbb{N}}\{y_i\}$ and $\cup_{i\in \mathbb{N}}\{z_i\}$are both strictly included in $X$ and are countable , also $$\cup_{i\in \mathbb{N}}\{z_i\}\subset X \backslash \cup_{i\in \mathbb{N}}\{y_i\}$$

So $\cup_{i\in \mathbb{N}}\{y_i\}$ is a countable union of sets in $A$ whose complement is infinite, hence $\cup_{i\in \mathbb{N}}\{y_i\}\notin A$ and so $A$ is not stable under countable union.

Does that work ?

"Also $\cup_{i\in \mathbb{N}} \subset X$ since $A$ is an algebra."
seems like a typo? You don't need it, just that $\{x_i\} \in A$.
The rest of the arument is fine, as you show a concrete countable union of members of $A$ is not in $A$ as it is infinite and its complement as well.