Say I index a countably-infinite set $A$ bijectively with the positive integers so that $$A=\{a_1, a_2, a_3,\dots\} $$ The indexing gave an order to the set. Was the choice axiom used?
Do we need the axiom of choice to well-order a set with countably many elements?
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3No. The function that verifies that the set is countably infinite suffices. – 2012-08-27
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0@MichaelGreinecker: Would you mind elaborating, please? I don't quite follow what you mean. – 2012-08-27
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1What Asaf wrote. – 2012-08-27
1 Answers
No. There is no need for the axiom of choice. This is essentially by definition.
The definition of countability is to have an injection into $\omega$. Generally speaking if $A$ is a set, $\alpha$ is well-ordered and $f\colon A\to\alpha$ is an injection then $A$ can be well-ordered.
Proof. Fix a well-ordering of $\alpha$, $\prec$ and define $a. Since $f$ is injective we can easily see this is an order-embedding and therefore $<$ is a well-ordering of $A$.
In the particular case of a countable set, we can write $A=\{a_n\mid n\in\mathbb N\}$ so we can define an order on $A$ as follows: $$a_m\prec a_n\iff m
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0What if we define countable to mean "surjection from $\omega$"? – 2012-08-27
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2@ZhenLin: These two are of course equivalent. If $A$ is a set, $\alpha$ is well-ordered and $f\colon\alpha\to A$ is a surjective function, define $g\colon A\to\alpha$ by $g(a)=\min\{\beta\in\alpha\mid f(\beta)=a\}$. This is an injective function and we return to the previous case. – 2012-08-27
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0@ZhenLin: The empty set will feel so proud for being uncountable. – 2012-08-27
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0@MichaelGreinecker: Sometimes even nonempty finite sets don't get to be countable. I don't think authors who define "countable" as "equinumerous with $\mathbb N$" go as far as to consider finite sets to be uncountable, though. – 2012-08-27
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1@HenningMakholm: I'm aware that many peple use *countable* and *countably infinite* synonymously. But defining countable to be countably infinite or both finite and nonempty seems odd to me. – 2012-08-27