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I've read people's responses to this question but they are not convincing to me.

For example, someone said that you can't just choose an element from each set of the family because it's not clear how to choose them. But don't we have the same problem when the the family is finite? For non-empty sets $A$ and $B$, to prove that $A\times B$ is non-empty, we also have to choose an element $a\in A$ and $b\in B$. How come we don't need an axiom for that?

In another post, @Arthur wrote,

the axiom of choice takes you from the statement $∀i(∃x_i∈X_i)$ to the statement $∃f(∀i(f(i)∈X_i))$

I try to understand what prevents us to go from the first statement to the second without axiom of choice. (As I'm learning set theory with the book of Halmos, I stick to his definitions. ) By his definition a function $f$ from $I$ to $X$ is a subset of $I\times X$ such that $\forall x,y\in X, \forall i\in I, (i,x)=(i,y)\implies x=y$. Assuming the first statement, we consider the "collection" (for want of a better word) $C$ of elements in $I\times \cup_{i\in I} X_i$, such that for all $(i,x)\in I\times \cup_{i\in I} X_i$, $(i,x)$ is in $C$ if and only if $x=x_i$. As far as I can see, the only reason why $C$ might not be a function that qualifies for the second statement is that the "collection" $C$ may not be a subset of $I\times \cup_{i\in I} X_i$. (I just realized today that a "subcollection" may not be a subset.) Is that where the problem is?

Finally, can we say that the syntax of first order logic can handle the case where the family is finite, but not in the infinite case?

baby bunny
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    SO MANY words have been poured on this topic here. Have you looked through all the similar threads? – Asaf Karagila Aug 08 '17 at 19:29
  • Fortunately, after that introductory sentence, not many will try to convince you. Unfortunately, there'll always be people upvoting such "questions". –  Aug 08 '17 at 19:32
  • Yes, @AsafKaragila is actually an expert on the topic, so I suggest you peruse any links he gives you. Those are going to be about as authoritative as you can hope for here :P – Chris Aug 08 '17 at 19:36
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    "The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" (Jerry Bona) – md2perpe Aug 08 '17 at 19:36
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    This is ultimately just a matter of what the rules of deduction in first-order logic are defined to be. Henning Makholm's answer at https://math.stackexchange.com/questions/1083039/axiom-of-choice-what-exactly-is-a-choice-and-when-and-why-is-it-needed explains it well. – Eric Wofsey Aug 08 '17 at 19:36
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    See also the answers to https://math.stackexchange.com/questions/1839913/axiom-of-choice-where-does-my-argument-for-proving-the-axiom-of-choice-fail-he. – Eric Wofsey Aug 08 '17 at 19:42
  • I've closed this as a duplicate since this question is addressed in many other places. To have this opened, you should explain exactly what it is that the many other questions' answers do not address. – Noah Schweber Aug 08 '17 at 20:59
  • @NoahSchweber I think that actually that question should be marked as a duplicate of this one and not the other way around because that one asks more than one question one of which this one asks but this one has more details related to the question explaining why they think the axiom of choice is obviously true. – Timothy Jul 15 '18 at 03:15

1 Answers1

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The problem is that in order to remedy the problems and paradoxes of naive set theory, the mathematicians around the turn of the century realised that you can't let just any collection be a set. You have to have a list of axioms defining specific sets and set operations, and declare that any collection you can reach in a finite number of steps using the axioms is deemed a set, and nothing more. The axiom of choice lets you access some sets you can't with only the others, specifically functions fulfilling certain properties.

Arthur
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  • So do you agree that the problem is not that such a collection $C$ (described in my question) may not exist, but rather that it may not be a set? So what the axiom of choice actually says is that among such collections, at least one of them is a set? – baby bunny Aug 08 '17 at 20:44
  • Or is it meaningless to talk about the existence of an object other than sets in this framework? – baby bunny Aug 08 '17 at 20:53
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    That is one way of looking at it, yes. I suspect there might be purists who complain (saying that it is meaningless as you say), but I find it tedious and unintuitive to cater to the whims of purists at every turn. – Arthur Aug 08 '17 at 20:55