All subsets of $\mathbb{R}$ which are countable or the complement of a countable set Let C be the class of all subsets of $\mathbb{R}$ which are either countable or are the complement of a countable set. Show that C is a $\sigma$ field.
This is what I have right now:
(i) $\varnothing$ $\in$ $C$ since $\emptyset$ is countable .
(ii) A $\in$ C since A is countable since given C is countable.
(iii) $\bigcup^{\infty}_{i=1}A_i$ $\in$ C, is true since $A_1,A_2$, are countable then so is $\cup^{}_{n}A_n$ since a countable union of countable sets is countable.
I need help understanding (ii) and (iii).
 A: Let me point out a minor nitpick before answering:

"Countable" here means "at most countable", i.e. "finite or countable". If we were to take "countable" as strictly meaning "in bijection with the natural numbers", this would not produce a $\sigma$-algebra, since $\varnothing$ is finite (cardinality 0) end its complement is uncountable, and hence $\varnothing$ would not be in our class.

That said, I take most of what I say below from comments.
$$\text{(i: The empty set)}$$
$\varnothing$ is finite, hence at most countable, hence $\varnothing\in C$. I am assuming $\varnothing\in\emptyset$ is a typo for $\varnothing\in C$, otherwise that "formula" is Gibberish to me.
$$\text{(ii: Closure under taking complements)}$$
This point is a bit unclear, but it seems to be trying to show closure of $C$ under complement. The other interpretation would be that it's trying to show the "universe set" is in $C$, but that "universe" is $\mathbb{R}$ here, and it has never been called $A$ in the question. So I assume $A\subseteq\mathbb{R}$, and in fact $A\in C$.
We then have two cases:


*

*$A$ is at most countable;

*$A$ is the complement of an at most countable set.


In the first case, $\mathbb{R}\smallsetminus A$ (alias $A^C$) is the complement of an at most countable set, hence $A^C\in C$. In the second case, $A^C$ must be at most countable, since $A$ is not at most countable and hence must be the complement of an at most countable set, and $A=(A^C)^C$;
$$\text{(iii: Closure under countable unions)}$$
If $A_i\in C$ for all $i$, we have three cases:
a) All $A_i$'s are at most countable;
b) All $A_i$'s are complement of at most countable sets;
c) There are some $A_i$'s that are at most countable, and some that are complements of at most countable sets;
in case a), a countable union of countable sets is countable, as you remarked; so your (iii) works for case a), but neglects the other cases. In case b), the union of the $A_i$'s is:
$$\bigcup_iA_i=\bigcup_i(A_i^C)^C=\left(\bigcap_iA_i^C\right)^C,$$
and that intersection is contained in, e.g., $A_1^C$, which is at most countable, and hence is at most countable. In case c), we remark that the union operator is associative and commutative, so we reduce $\bigcup_iA_i$ to:
$$\bigcup_iA_i=\left(\bigcup_{i:|A_i|\leq|\mathbb N|}A_i\right)\cup\left(\bigcup_{i:|A_i^C|\leq|\mathbb N|}A_i\right).$$
So this union is the union of two sets, the first of which is at most countable as shown in case a), and the second of which is the complement of an at most countable set as shown in case b). Hence, all that is left to prove is that, if $|A|\leq|\mathbb N|$ and $|B^C|\leq|\mathbb N|$ (i.e. $A$ is at most countable as is $B^C$), then $A\cup B$ is either at most countable or the complement of an at most countable set. Now, $B\subseteq A\cup B$, so if $B$ is not at most countable, $A\cup B$ cannot be either. Let us look at the complement:
$$(A\cup B)^C=A^C\cap B^C\subseteq B^C,$$
and $B^C$ is at most countable. Hence, $A\cup B$ is the complement of an at most countable set, which completes the proof that $C$ is a $\sigma$-algebra (or $\sigma$-field, as you call it).
I hope this is clear.
