Proving the co-countable topology $\tau$:= {$O \subset X: O^c$ is (at most) countable} $\cup $ { $\varnothing$} defines a topology Let X be a nonempty set. We define on X the co-countable topology $\tau$:= {$O \subset X: O^c$ is (at most) countable} $\cup $ { $\varnothing$}.
Show that is a topology.
To show that a topology exists, one must check three properties:
Formally, let X be a set and let τ be a family of subsets of X. Then τ is called a topology on X if:
1. Both the empty set and X are elements of τ.
2. Any union of elements of τ is an element of τ.
3. Any intersection of finitely many elements of τ is an element of τ.

If τ is a topology on X, then the pair (X, τ) is called a topological space.
The members of τ are called open sets in X. A subset of X is said to be closed if its complement is in τ (that is, its complement is open). A subset of X may be open, closed, both (a clopen set), or neither. The empty set and X itself are always both closed and open. An open subset of X which contains a point x is called a neighborhood of x.
The first property is immediately satisfied since the empty set is open, as it is in the definition of the topology above.
The set X is also open, since the complement of X is the empty set, which is countable.
But I fail to write down the other properties nicely in a formal way. Can anyone help me with this?
 A: For the arbitrary union of sets $U_i, i \in I$ we can also assume they're all of the cocountable type (as $\emptyset$ contributes nothing to a union) and use
$$(\bigcup_{i\in I} U_i)^\complement = \bigcap_{i \in I} U_i^\complement \subseteq U_{i_0}^\complement$$ where $i_0 \in I$ is arbitrary. So the complement is also at most countable and so $\bigcup_{i \in I} U_i$ is open by definition.
To show that the intersection of finitely many open sets $W_1, \ldots W_n$ is open, note first that we can assume WLOG that none of the $W_i$ is empty (or the intersection is empty and thus open anyway), so all $W_i^\complement$ are at most countable and then
$$(\bigcap_{i=1}^n W_i)^\complement = \bigcup_{i=1}^n W_i^\complement$$ by de Morgan so the complement of $\bigcap_{i=1}^n W_i $ is a finite union of at most countable sets so at most countable. It follows that by definition, $\bigcap_{i=1}^n W_i $ is open, as required.
Henno Brandsma (https://math.stackexchange.com/users/4280/henno-brandsma), Co-countable topology, URL (version: 2021-07-16): https://math.stackexchange.com/q/4200126
