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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?


marked as duplicate by Noah Schweber set-theory Aug 8 '17 at 20:58

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    $\begingroup$ SO MANY words have been poured on this topic here. Have you looked through all the similar threads? $\endgroup$ – Asaf Karagila Aug 8 '17 at 19:29
  • $\begingroup$ Fortunately, after that introductory sentence, not many will try to convince you. Unfortunately, there'll always be people upvoting such "questions". $\endgroup$ – Professor Vector Aug 8 '17 at 19:32
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    $\begingroup$ "The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" (Jerry Bona) $\endgroup$ – md2perpe Aug 8 '17 at 19:36
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    $\begingroup$ 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 math.stackexchange.com/questions/1083039/… explains it well. $\endgroup$ – Eric Wofsey Aug 8 '17 at 19:36
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    $\begingroup$ See also the answers to math.stackexchange.com/questions/1839913/…. $\endgroup$ – Eric Wofsey Aug 8 '17 at 19:42

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.

  • $\begingroup$ 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? $\endgroup$ – baby bunny Aug 8 '17 at 20:44
  • $\begingroup$ Or is it meaningless to talk about the existence of an object other than sets in this framework? $\endgroup$ – baby bunny Aug 8 '17 at 20:53
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    $\begingroup$ 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. $\endgroup$ – Arthur Aug 8 '17 at 20:55

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