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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?

Egli
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    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 – Maxime Ramzi Apr 02 '17 at 08:58
  • @Max Did the set exist before the enunciation of the axiom or not? – Egli Apr 02 '17 at 08:59
  • That question doesn't make sense. The enunciation of the axiom changes nothing : it just allows you to prove that this set exists. – Maxime Ramzi Apr 02 '17 at 09:29
  • 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). – Egli Apr 02 '17 at 10:33
  • 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? – Egli Apr 02 '17 at 10:43
  • Your last comment is very unclear. And while I think that I might understand the question itself, it is nevertheless vague and inaccurate because of the unfortunate reason, that in order to correctly formulate this question, you need to have some basic understanding in logic and set theory, in which case you are likely to answer it yourself. – Asaf Karagila Apr 02 '17 at 11:38
  • Then I will make a very simple question: may I create the power set of the real numbers with a set theory that includes the negation of AC? And a second question: if anybody could answer their questions with some 'basic knowledge' of a theory, why does this website exist? – Egli Apr 02 '17 at 11:58
  • Of course, I meant 'does the power set of the real number exist?' – Egli Apr 02 '17 at 12:05
  • (1) Yes, because the power set axiom tells you that you can do that. Choice has absolutely nothing to do with that. (2) There are a lot of simple-to-state questions in mathematics, many of which do not have an answer with our current mathematical technology, and others have very complicated and hard answers. "All the king's horses and all the king's men, couldn't put Humpty Dumpty back together again". – Asaf Karagila Apr 02 '17 at 19:57
  • Also, I wasn't trying to belittle you or anything. I'm merely pointing out the difficult reality in which mathematics takes place: it is easy to come up with good questions that have very complicated answers. And often times in order to understand a very complicated answer, you'd need the tools which make it easy to come up by yourself. – Asaf Karagila Apr 03 '17 at 06:55

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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.

Asaf Karagila
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  • I modified my question so that the word 'creation' is not included anymore. Thank you for your answer. – Egli Apr 02 '17 at 09:04
  • One more question: does the choice of the infimum from the sets that have an infimum require the axiom of choice? – Egli Apr 02 '17 at 09:07
  • @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. – Arthur Apr 02 '17 at 09:12
  • Thank you. That is all. Have a good day. – Egli Apr 02 '17 at 09:16