# Concerning the Countable-Closed Topology.

I am required to prove the following proposition. The following is my attempt thus far. Is it correct?

Proposition. Let $$X$$ be any infinite set. The countable-closed topology is defined to be the topology having as its closed sets $$X$$ and all countable subsets of $$X$$. Prove that this is indeed a topology on $$X$$.

Proof. Let $$\tau$$ denote the supposed topology in question. Since $$\varnothing$$ is countable and therefore closed, it follows that $$X$$ is open. In addition from hypothesis $$X$$ is closed, implying that $$\varnothing$$ is open.

Now let $$\{\mathcal{A}_j:j\in J\}$$ collection of sets in $$\tau$$, indexed by some set $$J$$. We prove that $$\bigcup_{j\in J}\mathcal{A}_j\in \tau$$. Consider that $$\bigcap_{j\in J}X\backslash A_j = X\backslash\left(\bigcup_{j\in J}\mathcal{A}_j\right)$$. It is therefore sufficient to show that the set $$\bigcap_{j\in J}X\backslash A_j$$ is countable and therefore closed in $$\tau$$.

Now the set $$\bigcap_{j\in J}X\backslash A_j$$ maybe either finite, in which case we are done or infinite in which we proceed as follows. Let $$j_0\in J$$, the set $$\mathcal{A}_{j_0}\in\tau$$, thus $$X\backslash\mathcal{A}_{j_0}$$ is closed and therefore countable. Thus we have a bijection $$H:\mathcal{A}_{j_0}\to\mathbf{N}$$, restricting the domain of $$H$$ to the set $$\bigcap_{j\in J}X\backslash A_j$$ yields the required bijection between $$\bigcap_{j\in J}X\backslash A_j$$ and $$\mathbf{N}$$, proving that $$\bigcap_{j\in J}X\backslash A_j$$ is countable.

I am struggling to prove that $$A_1\cap A_2\in\tau$$ whenever $$A_1,A_2\in \tau$$ any suggestions here?

It suffices to show that $$X \setminus (A_1 \cap A_2) = (X \setminus A_1) \cup(X \setminus A_2)$$ is countable. By hypothesis that $$A_1,A_2 \in \tau$$, $$X \setminus A_1$$ and $$X \setminus A_2$$ are countable. From , we have $$\aleph_0 + \aleph_0 = \aleph_0$$, where $$\aleph_0 = \mathop{\rm{card}}(\Bbb{N})$$. Therefore, $$\mathop{\rm{card}}(X \setminus (A_1 \cap A_2)) = \mathop{\rm{card}}((X \setminus A_1) \cup(X \setminus A_2)) \\ \le \mathop{\rm{card}}(X \setminus A_1)+\mathop{\rm{card}}(X \setminus A_2) = \aleph_0 + \aleph_0 = \aleph_0.$$ Hence $$X \setminus (A_1 \cap A_2)$$ is countable and $$A_1\cap A_2 \in \tau$$.