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This answer to this question contains a proof that every set can be linearly ordered, but doesn't prove choice along the way as this comment points out.

I'm trying to figure out why the axiom of choice is not a consequence of this argument.

Here's my best guess as to what's happening, but it feels like I'm missing something obvious.

Every finite linear order is a well-order, so we have the existence of a well-order for every finite subset.

However, given a relation symbol $R$, $R$ being a well-order is not a first-order property ... therefore when we invoked compactness we didn't get the well-order property because we were not promised it.

I'm having trouble understanding this comment, however.

Yes. I gave my students this question (with a generous hint) recently. After that I sat to understand exactly where it fails to well-order $A$, and it's a very subtle point indeed!

This suggests that, in some sense, we actually need to do something to get our combined well-order to "break", or that we can get pretty far almost-constructing a well-order by applying compactness.

Greg Nisbet
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    You are asking "why doesn't this argument prove the axiom of choice?", but you haven't actually said what "this argument" is. How, specifically, do you want to modify the argument for linear orders in order to prove every set can be well-ordered? – Eric Wofsey May 19 '21 at 05:03
  • @EricWofsey I think there's an argument that almost works but fails for a specific, subtle reason that is heavily implied by the linked answer and the comments associated with it. Unfortunately, I can't tell you what that argument is, specifically, because I can't see it. – Greg Nisbet May 19 '21 at 05:20
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    There is not a unique such argument and in particular there is no specific argument that is heavily implied by that answer and its comments. There are various different arguments you might try to make which fail for different reasons. – Eric Wofsey May 19 '21 at 05:21

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If you had an infinitary logic, you could add, for every $S \subseteq A$ nonempty, the sentence $\bigvee\limits_{a \in S} \bigwedge\limits_{b \in S \setminus \{a\}} c_a < c_b$. But this sentence cannot be expressed in finitary logic because of the infinitary $\bigvee$.

Alternately, you could add for each nonempty $S \subseteq A$ a symbol $b_S$ and, for every $a \in S$, the sentence $b_S < c_a \lor b_S = c_a$.

One runs into two problems: first, it is not guaranteed that $b_S = c_a$ for some $a \in A$, and second, even if this were true, it would not imply that $b_S = c_a$ for some $a \in S$. At best, it would show that all sets can be linearly ordered where each set has a lower bound. But if one considers $\mathbb{Z} \cup \{-\infty\}$ under the usual order, we see that this is a linear order in which all sets have a lower bound $-\infty$ but not all sets have a minimal element, particularly $\mathbb{Z}$ itself.

Mark Saving
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