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First of all, I am a beginner in the field of Set Theory and my knowledge about Infinitary Logic is only at the informative level, so my questions might be nonsense.

One could think that, for any set $S$ whose elements are non-empty sets $S_i$ (with $i\in I$), there is a function taking any element of $S$ and giving an element of $S_i$. This is clear, because $S_i\not=\varnothing$, so, we can take some element from $S_i$.

That is the naive way of thinking, because it is useful only for finite sets. This is because first order logic is finitary. In particular, you can't ask your butler to pick a sock arbitrarily from each box if you have infinitely many boxes, it would take (at least) all of eternity to have one set of socks, one from each box. This is where the axiom of choice comes in, stating that it is possible to obtain such a set (although there is no way to explain it to the butler in finite time).

I needed that very basic introduction to be able to continue my reasoning. It seems natural to me that an infinitary logic would allow a rule to obtain a choice function for any set. For example, applying "$S_i$ is non-empty, so take some element from it" infinitely many times. But I read on wikipedia something about infinitary logic and they talk about the axiom of choice being required in some cases. So, my reasoning above is surely wrong.

Now I can formulate my questions. Why isn't it possible to do something like the above? Does infinitary logic work otherwise? Is there an infinitary logic in which it is possible? And, if so, is it an extension of usual first-order logic or something bizarre?

Thanks for your answers.

Edit: after the dialogue with Asaf, it seems convenient to clarify some things.

First, the goal is to obtain the axiom of choice as a theorem from a few axioms (such as $\mathsf{ZF}$) and the fact that an infinitary logic allows infinite formulas and this can be exploited to obtain a choice function for infinite sets (something similar to what is done with finite formulas to obtain choice functions for finite sets).

Second, the problem is perhaps too open, since a specific logic would have to be defined and some axioms introduced and, even if it were possible to do the above, the problems involved in using an infinitary logic are enormous.

Yester
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    In what sense is that impossible, introducing a rule to infinitary logic? The question is, what is the point? If the point is to "circumvent the axiom of choice", you still need to assume it. If the point is to "clarify how choice works", your proofs will have to be infinitely long, and then what? And consequently of the latter, the proof and model theories of infinitary logics are kinda wonky compared to first-order logic. So, again, why are we doing this exercise in using the axiom of choice? – Asaf Karagila Jan 02 '23 at 12:48
  • @AsafKaragila, the point is circumvent the axiom of choice, building Set Theory from some infinitary logic. That is, to extend the usual Set Theory to a theory that uses a logic that extends to the usual first-order logic, this is, some infinitary logic. Since first-order logic does not allow infinite proofs or infinite formulas, my intuition tells me that one could formulate the axioms in a different way and perhaps avoid some, like the axiom of choice. – Yester Jan 02 '23 at 13:03
  • But you're not circumventing it. This is like saying that by walking around you're circumventing the need to hop like a bunny. – Asaf Karagila Jan 02 '23 at 13:06
  • Do you mean that in a set theory based on infinitary logic the axiom of choice will, vacuously, not appear? – Yester Jan 02 '23 at 13:11
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    No, this means that you're sweeping AC under the rug, you'll either need it in the meta-theory to make sense of things, or you're going to end up with a restricted version that might as well be developed in an inner model of your meta-theory where AC holds. To make matters worse, you'll want to use something akin to $\cal L_{\infty,\infty}$ if you want these arguments to work at all, which is a terrible logic, since it's a proper class and is hard to work with. – Asaf Karagila Jan 02 '23 at 13:14
  • And all in all, you haven't told me how you are going to develop set theory. What is the logic exactly, what are the inference rules, what proofs can you have, what are the axioms of set theory that you write. After you've done all that, we can start talking about whether or not AC comes in through the back door as a theorem, as a meta-theorem, not at all, or if there is some other problem. But, and I cannot stress this enough, this is hardly a way to "circumvent" AC. – Asaf Karagila Jan 02 '23 at 13:16
  • "The Axiom of Choice is necessary to select a set from an infinite number of pairs of socks, but not an infinite number of pairs of shoes."— Bertrand Russell. So if every set has a distinct element (like left shoe) then AoC not needed; that every set has some distinguished (least) element is the well-ordered theorem=AoC. – David Raveh Jan 02 '23 at 13:25
  • @AsafKaragila, In your last comment I can find what I wanted to express. I don't know if it is possible to build a set theory (extending the usual set theory) from a logic and a few "simple" axioms (like union, extension...) whereby the axiom of choice is a theorem. In particular, my intuition told me that an infinitary logic together with some axioms would allow it to be done. But I don't know if anyone has done it and if a theory like that can present problems, as you mention for the case of $\mathcal{L}_{\infty,\infty}$. – Yester Jan 02 '23 at 13:29
  • @DavidRaveh, your answer is not in line with my question. The problem in choosing socks from each box is not just that the socks are indistinguishable, because for a finite number of boxes that doesn't matter. The second problem is that there are infinitely many boxes, but the first order logic is finitary. – Yester Jan 02 '23 at 13:38
  • @David: I suspect that the goal here is to extend the naive "iterated Existential Instantiation" argument. Of course, the naive argument doesn't even work for proving finite AC from ZF, but that's a different problem. – Asaf Karagila Jan 02 '23 at 13:46
  • Yes, something like a generalization of EI in an infinitary logic, which allows obtaining AC as a theorem. – Yester Jan 02 '23 at 13:54
  • Just add AC to your logic, and you'll get it for free as a theorem. I don't see why we need to jump through infinitary logic. What's the goal? Just to prove AC as a theorem? – Asaf Karagila Jan 02 '23 at 14:24
  • The objective would be to obtain AC with the "minimum cost". That is, allow infinite formulas, use, for example, ZF, and obtain AC as a theorem. In particular, the use of infinite formulas could allow (as far I can imagine) a more general use of EI, to obtain choice functions for infinite sets. – Yester Jan 02 '23 at 14:32
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    Minimum cost? You've paid for it by losing all the tools we have for FOL, like good model theory and good proof theory. You've paid for it by effectively requiring a set theoretic universe to pre-exist, setting the notion of a foundation for set theory on fairly shaky grounds. You've paid for it by making it impossible to discuss independence results, or properly understand what this set theory gives or not give you. This is like throwing the solar system into a black hole because you can't be bothered to pick up a sock from the floor. – Asaf Karagila Jan 02 '23 at 15:12
  • OK, I see. I knew that model theory is basically destroyed in infinitary logic(s). But I didn't know it could give all those problems. Thank you very much for your answers. – Yester Jan 02 '23 at 15:45

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I think the following helps de-mystify things (it did for me, anyways):

To start with, let's forget about logic entirely for a second and look for a "purely mathematical" situation where the axiom of choice makes sense but fails. This is actually much easier to come by than you might expect. For instance, given a family of functions $\mathbb{F}$ on the real numbers, let $\underline{\mathbb{F}}$ be the two-sorted structure consisting of

  • a "numbers sort" ranging over reals and a "functions sort" ranging over elements of $\mathbb{F}$, and

  • the usual field structure on the numbers sort and the application operation relating the numbers and functions sorts.

Specifically, I'm interested in the case where $\mathbb{F}$ is the set of continuous functions; call this $\mathbb{F}_c$. We then have the following "failure of choice inside $\underline{\mathbb{F}_c}$:"

  • the sentence $$(*)\quad\forall x\exists y[(x=0\leftrightarrow y=0)\wedge (x\not=0\leftrightarrow y=1)]$$ is true in $\underline{\mathbb{F}_c}$, but

  • for no $f$ in the "function sort" of $\underline{\mathbb{F}_c}$ do we have $$\underline{\mathbb{F}_c}\models \forall x[(x=0\leftrightarrow f(x)=0)\wedge (x\not=0\leftrightarrow f(x)=1)].$$

What I want to emphasize is that there is no logical subtlety here at all. In particular, there is "in reality" a function $h$ with the property that for each $x\in\mathbb{R}$ we have $\mathbb{F}_c\models (x=0\leftrightarrow h(x)=0)\wedge(x\not=0\leftrightarrow h(x)=1),$ but this $h$ exists "outside" the structure-of-interest $\underline{\mathbb{F}_c}$. And it doesn't matter what logic we decide to use to talk about this structure.

Taking logic back into account, one way to describe the situation that I like is the following:

When we choose to analyze a structure $\mathfrak{A}$ using a logic $\mathcal{L}$, we might choose to think of the logic $\mathcal{L}$ as a giant machine whose behavior is determined by our "metatheory." In particular, depending on the details of $\mathcal{L}$ and our metatheory it is entirely possible that $\mathcal{L}$ can "build choice sequences" for every family of sets coming from $\mathfrak{A}$. However, this has no bearing on whether $\mathfrak{A}$ satisfies (the appropriately-phrased "internal" version of) the axiom of choice: the giant machine $\mathcal{L}$ can build whatever shiny objects it wants to, but those objects might not have counterparts in the possibly-very-limited structure $\mathfrak{A}$. Tension only appears in this context emerges when we try to smuggle in an implicit assumption that the structure we're looking at "does everything our machines can do" - that is, when we conflate theory (= the hypotheses on our structure-of-interest $\mathfrak{A}$) and metatheory (= that part of our metatheory governing $\mathcal{L}$).

(In fact when we're careful to avoid this conflation there is a lot of interesting stuff to do involving versions of $\mathsf{ZF}$ for logics other than first-order logic - see e.g. this old question of mine - but in order to put this on firm ground we have to be very careful about "language level" issues.)

Noah Schweber
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  • As you said in What can I do when I don't understand an answer?, when one does not understand an answer, it is convenient to ask for clarification. Indeed, I do not understand your answer very well, but I know that this is due to a lack of knowledge on my part. Thank you very much for your help and I hope to have the necessary knowledge soon. – Yester Jan 04 '23 at 15:43
  • @Yester Feel free to ask for clarification about any points! There's a lot of material here, and I'm happy to expand on it. – Noah Schweber Jan 04 '23 at 17:08