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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?

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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 $A_1\cap A_2 \in \tau$.

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  • $\begingroup$ Is there a way we can do this without resorting to cardinal Arithematic? $\endgroup$ – Atif Farooq Oct 10 '18 at 20:55
  • $\begingroup$ @AtifFarooq For a general-topology course, you don't need that. I put them for the sake of rigor. Simply stating that "a finite union of countable sets is countable" is enough. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Oct 10 '18 at 21:03
  • $\begingroup$ Thanks for pointing that out. $\endgroup$ – Atif Farooq Oct 10 '18 at 21:07

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