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I don't mean why is it important. I mean why can't we just define the "selector function" like $S\colon \mathbb{F} \to A, $ such that $S(X) = x \in X$, without an axiom?

Why can't we do that but we can, for example, take some set that satisfies a condition from an uncountable family?

Jessica Singer
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    I would say it is because, to prevent we are talking about an empty theory, it is required in the first place that the existence of a "selector function" is established! – Yes Jan 23 '18 at 03:09
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    You only need the axiom of choice when you want to formalize some mathematical proof inside ZFC set theory. If you just want to work in normal, informal mathematics, you don't need to worry about the axioms, and as you say you can use form a selector function by referring to it. But in ZFC set theory, the axioms available for constructing sets are more limited - in order to avoid set theoretic paradoxes - so some of the constructions in informal math end up corresponding to the axiom of choice once they are formalized in ZFC. – Carl Mummert Jan 23 '18 at 03:21
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    Also related https://math.stackexchange.com/questions/6489/can-you-explain-the-axiom-of-choice-in-simple-terms/1243483#1243483 – Carl Mummert Jan 23 '18 at 03:27

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As I understand your proposal, the problem is that it requires you to have a condition to construct that set.

In many situation it's certainly possible to do so and in those situation you need not to resort to the axiom of choice to guarantee the existence of a selector function. A vast amount of mathematics can get by without the need of the axiom of choice.

However if you have a situation when you can't guarantee that existence by a concrete example you will have a bit more trouble. In some special cases you could perhaps prove the existence without the use of AoC, but in general you would need to use the AoC to prove that.

There is proof that the AoC can't be proven from the other axioms.

skyking
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