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Given a family $(X_i)_{i \in I}$ of non-empty sets, why cannot one introduce a function $f : I \to \bigcup_{i \in I} X_i$ as follows :

for all $i \in I$, there exists $x \in X_i$ — let $f(i) := x$,

thus producing a "choice function" ?

Why isn't the above construction valid in $\sf {ZF}$ ?

user8171079
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  • That is a choice. There are infintely many $i$. For each one, "there exists $x$" ... see the choice? From the usual ZF axioms, this is OK for a finite number of choices. But not for an infinite number of them. – GEdgar Jan 31 '22 at 16:36
  • @JoséCarlosSantos Not really — the answer to the question you are referring to assumes that the fact that the Axiom of Choice isn't needed when considering finite sets was proved by induction. My question is rather why the construction I suggested is not correct when dealing with finite sets. – user8171079 Jan 31 '22 at 16:41
  • @GEdgar What are the specific ZF axioms you are referring to? – user8171079 Jan 31 '22 at 16:45
  • ZF axioms: https://en.wikipedia.org/wiki/Zermelo–Fraenkel_set_theory – GEdgar Jan 31 '22 at 16:46
  • I still don't understand what restrictions arise when trying to select a member from an infinite set. My issue here is why we can't introduce a variable $x$ denoting an arbitrary member of a possibly infinite set $X_i$, given that there exists $y$ such that $y \in X_i$. – user8171079 Jan 31 '22 at 16:52
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    Your "definition" of $f$ doesn't define a function. Specifically, since a function is represented (in ZF) by a set of ordered pairs, which ordered pairs $(i,x)$ are in your $f$? – Andreas Blass Jan 31 '22 at 16:53
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    https://math.stackexchange.com/questions/1839913/axiom-of-choice-where-does-my-argument-for-proving-the-axiom-of-choice-fail-he – spaceisdarkgreen Jan 31 '22 at 17:01
  • @AndreasBlass Ok, I think I'm starting to see where the problem is. If we were to formally use the same type of construction as I did but for a finite family of cardinality $p$ of sets then we would unwind a collection of $p$ $\exists x_i \in X_i$ and introduce a function ${(i, x_i) : i \in {1, ..., p}}$, which would be a valid construction. But the proof I wrote would induce an infinite sequence of formulae, which does not represent a proof. Is my intuition correct? – user8171079 Jan 31 '22 at 17:07
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    @user8171079 sorta… there’s a subtlety even in the finite case, in that you can’t just transfer the natural number $p$ into a concrete natural number in the metatheory. You need to use induction inside the theory. – spaceisdarkgreen Jan 31 '22 at 17:14
  • I think my question should be marked as a duplicate of https://math.stackexchange.com/questions/1839913/axiom-of-choice-where-does-my-argument-for-proving-the-axiom-of-choice-fail-he rather than https://math.stackexchange.com/questions/783022/confused-about-axiom-of-choice. Is it possible to change this? – user8171079 Jan 31 '22 at 18:12

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