We can state the Axiom of Choice as follows: 'If A is a family of nonempty sets, then there is a function f with domain A such that f(a) ∈ a for every a ∈ A. Such a function f is called a choice function for A'.

However, if A may be any family of nonempty sets, it may also include the set that we create through the choice of exactly one element from a known collection of nonempty sets. That is, if we allow any set to exist in the first place, we either do not need a choice function to create a new set, or we tacitly assume that at the beginning the choice function only applies to certain collections of sets. But since the Choice Set is a set, then after its statement of existence we may apply the choice function to the collection including this set and other sets to create a new set that did not exist before.

Am I wrong? My problem is: which sets are allowed to exist before the statement of existence of the Choice set?

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    $\begingroup$ There is no creation of anything. The axiom of choice only gives the existence of a certain set, but this set is not "created" while you use the axiom $\endgroup$ – Max Apr 2 '17 at 8:58
  • $\begingroup$ @Max Did the set exist before the enunciation of the axiom or not? $\endgroup$ – Egli Apr 2 '17 at 8:59
  • $\begingroup$ That question doesn't make sense. The enunciation of the axiom changes nothing : it just allows you to prove that this set exists. $\endgroup$ – Max Apr 2 '17 at 9:29
  • $\begingroup$ Let me try to word the question differently. We are speaking about a family of nonempty sets. Without further specification, the Choice Set C that is stated to exist may be part of a family of nonempty sets as well. (I am modifying this post, wait a minute). $\endgroup$ – Egli Apr 2 '17 at 10:33
  • $\begingroup$ Then we could use the Axiom of Choice again to state the existence of a second Choice Set that has an element from the first Choice Set, and so on. My question is: since it looks like that only through the Axiom of Choice we are allowed to state the existence of what looks like a very large amount of sets, then which sets are allowed to exist if in our set theory we include the negation of AC? $\endgroup$ – Egli Apr 2 '17 at 10:43

Axioms do not create sets. Axioms let us prove that some objects, in this case sets, exist. But existence is not predicated on provability.

Some families of sets admits a choice function, and that much is provable without the axiom of choice. If a family does not admit a choice function, then sets from which you can construct a choice function also do not exist.

  • $\begingroup$ I modified my question so that the word 'creation' is not included anymore. Thank you for your answer. $\endgroup$ – Egli Apr 2 '17 at 9:04
  • $\begingroup$ One more question: does the choice of the infimum from the sets that have an infimum require the axiom of choice? $\endgroup$ – Egli Apr 2 '17 at 9:07
  • $\begingroup$ @IacopoSbrolli No. If each member in the family comes pre-equipped with a single "preferred element" of any kind (be it the smallest, largest, reddest, the left one, or anything else), then using those preferred elements to construct a choice function may be done without the AoC. That being said, if each member of the family has different ways of deciding a preferred element, then you have to choose such a way for each member, which again requires AoC. $\endgroup$ – Arthur Apr 2 '17 at 9:12
  • $\begingroup$ Thank you. That is all. Have a good day. $\endgroup$ – Egli Apr 2 '17 at 9:16

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