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I previously asked a question, asking whether we can define a choice function from ZF on a collection of sets with each sets having a cardinality of two. I stated the below in the original question.

Since all the sets have a cardinality of two, we can define bijections between the two elements in each set and {0,1}. So regardless of how the bijection is defined, we can always choose the element that maps to 0.

However, it turns out that choosing from one of the two bijections itself requires the axiom of choice, and thus ZF cannot prove the existence of such a function.

Similarly, I mentioned in the comment section that if use existential instantiation and induction, we can prove the finite axiom of choice, but I was wondering why I cannot define a choice function by applying existential instantiation on all the sets. I think the reason is that such a function can also not be proven to exist purely with ZF.

Nonetheless, are both these two examples still choice functions? And if they are, can they be proven to exist in ZFC?

I think with the axiom of choice, we can choose one bijection out of the two possible, and thus the first one can be proven to exist. Is that correct? However, for the second part with existential instantiation, can we also prove such a choice function to exist in ZFC? I am not familiar with transfinite induction, and I cannot really observe a way to do so.

wsz_fantasy
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    To do existential instantiation with infinitely many sets you would need to work in an infinitary logic, not vanilla first-order ZF. – blargoner Oct 11 '22 at 15:47
  • The choice function can be proven to exist in ZFC. Exactly how depends on your precise wording for the Axiom of Choice. If your version of the Axiom of Choice is that product of a nonempty family of nonempty sets is itself nonempty, then you can do a single application of existential instantiation in the product of your sets to get your Choice Function for the family of sets (because the product is precisely the collection of all choice functions). – Arturo Magidin Oct 11 '22 at 16:53
  • Likewise, if you have a family ${A_i}{i\in I}$ in which $I\neq\varnothing$ and $A_i$ has two elements for each $i\in I$, then you can create the family ${B_i}{i\in I}$, where $B_i$ is the set of all bijections from $A_i$ to ${0,1}$; the Axiom of Choice guarantees that $\prod_{i\in I} B_i$ is nonempty, and existential instantiation on that set gives you a choice of bijection for each set in your original family. – Arturo Magidin Oct 11 '22 at 16:55
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    Definitions/descriptions in set theory are inherently finite. The reason you cannot define a function by existential instantiation on infinitely many sets one set at a time is that this would require an infinite formula, which specifies infinitely many existential instantiations. – Arturo Magidin Oct 11 '22 at 16:58

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