# Does the Axiom of choice allow the existence of a choice set before the end of its enunciation?

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?

• 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 – Max Apr 2 '17 at 8:58
• @Max Did the set exist before the enunciation of the axiom or not? – Egli Apr 2 '17 at 8:59
• That question doesn't make sense. The enunciation of the axiom changes nothing : it just allows you to prove that this set exists. – Max Apr 2 '17 at 9:29
• 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 2 '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 2 '17 at 10:43