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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.

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2 Answers 2

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The easy way is to realize that a countable union of countable sets is countable. You can find a proof of this assertion in any topology textbook.
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.

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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.)

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