Talk:Set of Finite Subsets of Countable Set is Countable

This result needs Axiom:Axiom of Countable Choice, IIRC. We can construct a set of finite subsets of $A$ not in any $A^{(n)}$ by picking an infinite sequence in $A$ and taking its initial segments. --Lord_Farin 15:41, 4 August 2012 (UTC)

Okay, feel free to take this one on. --prime mover 17:04, 4 August 2012 (UTC)

Done; took the lazy route. --Lord_Farin 17:16, 4 August 2012 (UTC)

@WIP

Cf. Set of Finite Subsets of Countable Set is Countable/Proof 2. I thought that people may want the ACC-independent proof at spot 1, so I haven't finished the subpage exercise yet. --Lord_Farin (talk) 15:30, 1 November 2012 (UTC)