TITLE: Existence in variants of the Axiom of Choice
QUESTION [7 upvotes]: Let $\{A_i: i \in I\}$ be a nonempty family of nonempty sets. Why is it allowed to prove the Axiom of Choice using the Well Ordering Principle as follows:

There is a well-ordering of $\cup_{i \in I} A_i$. Using this order, there is a first element in each $A_i$. Let $f(i)$ be this first element. Then $f$ is a choice function for this family.

But not to `prove' the Axiom of Choice directly as follows:

Each $A_i$ is nonempty: there is an element of $A_i$. Let $f(i)$ be this element. Then $f$ is a choice function for this family.

Both arguments appeal to the existence of some unspecified object (what well-order? what element of $A_i$?); why is that fine in the first case, but not the second? (The former approach, for instance, is from Aliprantis and Border, Infinite Dimensional Analysis, 3rd ed, p. 20.)
I must admit that I have only read about naive set theory, not axiomatic set theory: perhaps I just haven't learned the rules of the game.

REPLY [2 votes]: Below is an excerpt from an old answer of mine that I think addresses the difference between making one choice and making an infinity of choices well.
Note that the kind of reasoning you use in your suggested proof of AC is actually allowed in mathematics in general. But in order to formalize that reasoning in axiomatic set theory we need the axiom of choice. Or in other words, the purpose of the axiom of choice is to state that this kind of reasoning is allowed.
I don't think there's any real way around being at least somewhat familiar with the workings of formal logic and axiomatic set theory for this; otherwise you will have to make do with analogies and semi-intuitive appeals to what you can and cannot imagine doing or what can be told in finite time. Do not let that fool you into thinking that the need for AC is a matter of dogma and opinion; once you have the right prerequisites there is a precise technical way in which it is needed.

The ability to choose single elements during a proof is built into the rules of first-order logic; we don't need a set-theoretic axiom for doing so. More precisely, if we know $\exists x.\varphi(x)$, then it is admissible to pick an $x$ that $\varphi$ holds for. Either the $\exists x$ was proved earlier in the proof -- in which case we actually know a specific thing with this property, because that is how you prove an $\exists$ -- or the $\exists x$ comes from an axiom, in which case the ability to pick such an $x$ is "morally" part of what the axiom promises us.
So when the Axiom of Choice promises us that "$\exists$" a choice function, it doesn't only say that there are choice functions out there, but also that we're allowed to pick one.

Now, why would we want to have a choice function exist inside set theory? The main reason is such that we can wrap up the many choices it represents and handle them as a single object -- particularly in the axiom schemas that allow us to supply a formula to specify what we're doing. The formula may contain set parameters, but syntactically it can only contain a finite number of parameters.  The Axiom of Choice allows us to parameterize an instance of the axiom schema with an infinity of choices when we need it.
The two axiom schemas this is relevant for is the Axiom of Replacement and the Axiom of Selection.
The Axiom of Replacement allows us to form $\{F(x)\mid x\in A\}$ for any set $A$. The function $F$ does not a priori have to exist as a set within the set-theoretic universe; it can be given by a logical formula. But we need to be able to prove that there is only (or at most) one $F(x)$ for each $x$ -- otherwise the axiom might produce a set so large that it needs to be a proper class instead. However, often we actually want to use an underspecified $F$ that just says "choose something with such-and-such property", where we can prove that at least one something always exists. For this we need to wrap up all of the choices as a set parameter to the formula that represents $F$; the Axiom of Choice is what we need to do that. First-order logic allows us to make single choices one by one during a proof, but not to write down choose something as part of a formula.
The Axiom of Selection allows us to form $\{x\in A\mid \varphi(x)\}$. Here, $\varphi$ can be an arbitrary formula (with parameters), but sometimes we want $\varphi$ to depend on some arbitrary choices that need to be the same for all the $x$s in $A$ we consider. Again the problem -- of part of it -- is that the formula $\varphi$ cannot itself speak about making choices in a controlled way, but we can parameterize $\varphi$ by a choice function that we select once and for all outside the application of the Selection axiom.

(CW since the answer is mostly self-plagiarized).