# Showing that the set of open sets of the finite-closed topology on $\mathbb{Z}$ is countably infinite

Here is what I have tried so far.

Define $$F = \{X \subseteq \mathbb{Z} \ : \ |X| < \infty\}$$ be the set of all finite subsets of $$\mathbb{Z}$$. Further define $$F_k$$ be the set of subsets of finite subsets of $$\mathbb{Z}$$ containing $$k$$ elements. Consequently, $$F = \bigcup_{k = 1}^\infty F_k$$. Define the mapping $$g_k: F_k \rightarrow \mathbb{Z}^k$$ as follows: $$g_k(X) = (x_1, x_2, \ldots, x_k)$$ where $$X = \{x_1, x_2, \ldots, x_k\} \subseteq \mathbb{Z}$$.

At this point, I would like to show that $$g_k$$ is a bijection putting $$F$$ into 1-1 correspondence with $$\mathbb{N}$$ making it countably infinite and then the number of open sets in the finite-closed topology follows immediately. However, if I do not order the set $$X$$ above, then it seems the mapping is not injective but if I do order it, then the mapping fails to be surjective.

I sort of get the feeling that the bijection construction is overkill and there is an easier way to shwow this, but I am not sure.

For each $$k \in \mathbb{N}$$, $$F_k$$ is clearly an injection into $$\mathbb{Z}^k$$ by the map you defined as $$g_k$$ thus $$|F_k| \leq |\mathbb{Z}|$$ which makes $$F_k$$ a countable set. Then $$F$$ is a countable union of countable sets and thus is countable.
You have a good idea. But your map $$g_k$$ is not well defined (it depends on choosing an order of the $$x_i$$), and as you point out, it doesn't seem to give a bijection. However, you don't really need a bijection. $$F$$ is clearly infinite, so you only need to show that it's countable.
One approach is to show that each $$F_k$$ is countable (think about why $$\Bbb{Z} \times \Bbb{Z}$$ is countable), and then note that $$F$$ is a countable union of countable sets, which must also be countable.
Probably easier is this: how would you put the elements of $$F$$ in order? A natural way to do this should give you the bijection you want.
(By the way, $$F$$ gives the closed sets of the "cofinite topology" on $$\Bbb{Z}$$. You do have to add one more element, the entire set $$\Bbb{Z}$$, to make a topology, but clearly this does not affect countability.)