How to show the co-countable topology is a topology?

Question

I am trying to show the co-countable topology $\tau$ is a topology on a set $X$.

The first point to show $\emptyset,X\in\tau$ is not difficult.

Now let $S_i\in\tau$. From the definition of co-countable, $S_i=\emptyset$ or $X\backslash S_i$ is countable.

I struggle to prove the second point $\cup S_i\in\tau$:

If $S_i=\emptyset$, then $\cup S_i=\cup\emptyset=\emptyset\in\tau$. The part I got stuck is what if $S_i\neq\emptyset$? So far I have not used $X\backslash S$?

The thing that troubles me a lot is that we are not told beforehand whether $X$ is countable or uncountable. Is it necesary to know?

How are we going to work on the condition being countable?

The third point $\cap S_i\in\tau$ should be similar but I am also not very sure about how can we make use of $X\backslash S_i$?

Thanks for any help on this matter.

Answer 1

Suppose $S_{i} \in \tau$. First prove that

$$ X \setminus (\bigcup S_{i}) = \bigcap (X \setminus S_{i})$$

Now there are two cases. Either all of the $S_{i}$ are the empty set, or at least one is not the empty set. The first case is straightforward. For the second case, ask yourself what the intersection of two countable sets looks like.