Showing a set involving complement is a topological space. 
Question:
Let $X$ be a set. Let $\tau$ consists of $\varnothing$ , and any set $U \subseteq X$ such that $X\setminus U$ is either finite or countable.
Show that $\tau$ is a topology on $X$.

Looking only for $\mathbf{hints}$.
Let $\tau=\left \{ \varnothing,U \right \}$.
Obviously, $X \subseteq X$ so $X \in \tau$
Now, $\tau=\left \{ \varnothing, \left \{ U_{i} \right \}_{i \in I} \right \}$ with $X \in \left \{ U_{i} \right \}_{i \in I}: \exists i \in I$
By hypothesis, $X\setminus U$ is countable so it is expected that $\exists$ a bijection
$f:\mathbb{Z}^{+}\rightarrow X\setminus U$
I would like to take this further. My intuition says that if I can show that the elements in $U$ are open sets, then the elements in $U$ are elements of $\tau$ which then satisfies one of the axiom for topological space.
However, I am unable to do so.
Any Hint is appreciated.
Thanks in advance.
 A: This $\tau$ is the topology of co-countable sets.
Recall the axioms:


*

*Empty set and full space are in $\tau$. (You've shown this.)

*Arbitrary union of open sets are open.

*Finite intersection of open sets are open.


The second is simple. If you have two set that contain all but countably many elements, then their union must as well since the complement cant get any larger. This extends to arbitrary unions.
The third is where you need to get a bit more precise, but it essentially boils down to that a finite (or even countably infinite) union of countable sets is countable. (These countable sets are the complements of the sets in your topology.)
A: Use De Morgan's Laws and that the countable union of countable sets is countable.
$\tau$ define a topology in $X$ if:
$i)$ $\emptyset,X\in\tau$;
$ii)$ Arbitrary union of sets in $\tau$ is in $\tau$;
$iii)$ Finite intersection of sets in $\tau$ is in $\tau$.
$i)$ is trivial. Now, if $\{U_{\lambda}\}_{\lambda\in \Lambda}$ is a family of element of $\tau$, from De Morgan's Laws, $$X\setminus\bigcup_{\lambda\in\Lambda}U_{\lambda\in\Lambda}=\bigcap_{\lambda\in\Lambda}(X\setminus U_{\lambda})\subseteq X\setminus U_{\mu},$$ for a $\mu\in\Lambda$ fixed. Since $X\setminus U_{\mu}$ is either finite or countable, $X\setminus\bigcup_{\lambda\in\Lambda}U_{\lambda\in\Lambda}$ is too, i.e, $\bigcup_{\lambda\in\Lambda}U_{\lambda\in\Lambda}\in\tau.$ Then $ii)$ holds.
For $iii)$, if $U_{1},...U_{n}\in\tau$, we have $$X\setminus\bigcap_{i=1}^{n}U_{i}=\bigcup_{i=1}^{n}(X\setminus U_{i}),$$which is either finite or countable because countable union of countable sets is countable. The proof is done.
