I'm not sure if this question has any sense. Either way, hopefully someone can help me see either the right question or the right way of thinking about this. I don't have any special background in set theory, myself.(adsbygoogle = window.adsbygoogle || []).push({});

A set is countably infinite if there is a bijection between it and the natural numbers. Right?

Suppose I tell you set [itex]A[/itex] is countably infinite, but we don't know anything else about set [itex]A[/itex]. So we know there are bijections [itex]A \to \mathbb{N}[/itex]. I ask you to give me a concrete, example bijection [itex]\phi : A \to \mathbb{N}[/itex].

To construct one such [itex]\phi[/itex]... do you have to use AC? I ask because it seems like you'll have to make an infinite number of choices, arbitrarily choosing and then mapping each [itex]x \in A[/itex] to an [itex]i \in \mathbb{N}[/itex].

Put another way...

Ordinarily I feel completely comfortable listing the elements of an arbitrary countable set [itex]A[/itex] as [itex]A = \{x_1, x_2, ...\}[/itex] when I need to use the elements. But given a concrete countable set with no possible, "natural" decision rule to assign each element to a natural number... if I then want to list the elements of the set as [itex]\{x_1, x_2, ...\}[/itex], am I not implicitly using AC?

Moreover--and this might be closer to the heart of the question, but I don't know--todream upa countably infinite set that has no "natural" decision rule to assign each element to a natural number, do I need to invoke AC?

Sorry again if this is inane or just silly.

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# Countably Infinite Set, Axiom of Choice

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