Are both $\tau_{1}$ and $\tau_{2}$ are topologies on $\mathbb{R}$?

Question

Let $\tau_{1}=\{G \subseteq \mathbb{R} : G \ \text {is finite or}\ \mathbb{R} \setminus G \text{ is finite} \}$ and $\tau_{2}=\{G \subseteq \mathbb{R} : G \ \text {is countable or}\ \mathbb{R} \setminus G \text{ is countable } \}$.

Are $\tau_1$ and $\tau_2$ both topologies on $\Bbb R$?

I think yes they both are, since $\tau_1$ is cofinite topology and $\tau_2$ is co-countable topology.

Any hints/solution will be appreciated.

Answer 1

The axioms for a set $\tau$ of subsets of $X$ to form a topology are:

1. $\emptyset, X\in\tau$
2. $A, B\in\tau\implies A\cap B\in\tau$
3. For any index set $I$, if $A_i\in\tau$, then $\bigcup_{i\in I}A_i\in\tau$.

Now, 1. and 2. clearly hold for both $\tau_1$ and $\tau_2$, but 3. fails for both.