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Mar 30, 2020 at 21:31 comment added Pace Nielsen It appears that someone is going through my old posts and downvoting them. If you don't like the question for some reason, please add a comment explaining why.
Oct 25, 2017 at 7:50 comment added David Roberts If you're worried about picking an element from a family consisting of one set, then you should check out "The world's simplest axiom of choice", which deals with (in non-classical logic) a family consisting of at most one set... The fact you noticed it comes down to logic is perhaps not coincidental.
Jul 20, 2017 at 14:58 comment added Christopher King What if you allow one to introduce function symbols in such cases?
Jul 11, 2017 at 14:28 comment added post.as.a.guest "it has often confused me how many authors find full choice so much different from finite choice." "We are picking elements in either case--why treat the infinite case as fundamentally different from the finite one?" Perhaps you should back up. Do you have a problem with the axiom of infinity (or large cardinal axioms in general)? I think the "opposing position" (if you want to call it that) is that there is (for both socio-historical reasons and philosophical ones) a difference between finite/infinite. Exactly how this is manifested (as in the rest of your post) is a different question.
Jul 11, 2017 at 5:28 comment added Pace Nielsen @Andreas How about the more modest goal of interpreting our use of the axiom of choice (rather than the statement of the axiom itself) as us exercising the ability to choose? This seems to be how (semantically speaking) many authors (even prominent set theorists) write about the axiom.
Jul 11, 2017 at 5:19 comment added Pace Nielsen ...continued... To put it another way, instead of talking about "chosing" I'll start using Joel's more expressive terminology of "uniform choosing", which informally speaking means having access to enough information to work with all the choices at once. [Formally speaking, it must mean asserting the existence of some set, which gets back to Andreas' point.]
Jul 11, 2017 at 5:15 comment added Pace Nielsen @bof Ultimately, I think your question boils down to what one means by "choosing" infinitely often. For example, how do you interpret your own phrase "given any elements $x_1,x_2,\ldots$"? I took that to mean we have access to a function $f:\omega\to V$ such that $f(i)=x_i$. [Otherwise, what does it mean to be given an infinite list?] Now the fact that $\{x_1,x_2,\ldots\}$ is a set, rather than a class, follows from an axiom of replacement.
Jul 11, 2017 at 4:26 comment added Benjamin Dickman Having read the question and comments (and answer) I think that the following paper may be worth a look, of interest, etc: Velleman, D. J. (1993). Constructivism liberalized. *The Philosophical Review, 102*(1), 59-84. jstor.org/stable/2185653.
Jul 11, 2017 at 4:20 comment added bof The axiom of union guarantees the existence of the union of any set of sets. I wonder what set you're applying the axiom of union to, to get your choice function.
Jul 11, 2017 at 4:11 comment added Andreas Blass @PaceNielsen It might be possible to develop a semantics of the sort you describe, but I'm not aware of any such semantics. In fact, I'm not aare of any attempt to develop a semantics that explicitly deals with us and our abilities; the closest thing I've seen is Brouwer's theory of the creative subject. Developing a semantics that explicitly involves our ability to choose things looks like a huge task.
Jul 11, 2017 at 4:10 vote accept Pace Nielsen
Jul 11, 2017 at 4:06 comment added Pace Nielsen @AndreasBlass I did say "I'm being a bit informal here." That said, I'd like to delve into your statement a bit more. Are you asserting that there is no sound semantics for which the axiom of choice is interpreted as the ability for us to choose things?
Jul 11, 2017 at 4:03 comment added Pace Nielsen @bof The axioms of pairing and union should do the trick. First, for each integer $i$ form the ordered pair $<a_i,x_i>$. Then take the union over $i\in \omega$ of these sets. This is a choice function.
Jul 10, 2017 at 22:41 comment added bof If we want to prove the existence of a choice set for a set $\{a,b\}$ of two nonempty sets $a$ and $b,$ it's not enough that we can "choose" elements $x_a\in a$ and $x_b\in b;$ we still need an axiom, the axiom of pairing, to get from the chosen elements $x_a$ and $x_b$ to the choice set $\{x_a,x_b\}.$ Now suppose you have a denumerable set $\{a_1,a_2,\dots\}$ of nonempty sets $a_i.$ Given that you can "choose" elements $x_i\in a_i,$ then what? Do you have an axiom saying "given any elements $x_1,x_2,\dots$ the set $\{x_1,x_2,\dots\}$ exists"?
Jul 10, 2017 at 22:00 comment added Andreas Blass I'm pretty sure I've written this on MO before, but it's easier to repeat it than to find it: The axiom of choice is, despite its name, not about our ability to choose things. It's not about us at all. It's exclusively about the existence of certain sets.
Jul 10, 2017 at 20:54 answer added Joel David Hamkins timeline score: 17
Jul 10, 2017 at 20:40 comment added Burak @PaceNielsen: I think our intuition does treat that sentence the way you wish. $\forall y \exists x P(x,y)$ means that there are $x_y$ around in the universe that you can choose for each $y$ such that $P(x_y,y)$ holds. The only problem is that, if you really want to make that choice, then you have to show that the cartesian product (indexed by the class of $y$'s) of those sets containing $x_y$'s is non-empty. I think if you work with set theories including classes and assume global choice, then you get exactly what you want.
Jul 10, 2017 at 20:38 comment added Goldstern @PaceNielsen: You are right, global choice fits better. Or - more natural, or less natural, depending on your taste - ZF with an additional predicate which gives a global set-like well-order (which may be used in separation and replacement axioms). Or class theories.
Jul 10, 2017 at 20:30 comment added Pace Nielsen @Burak The question really isn't about the axiom of choice, per se. My question is whether or not the proposed semantic strengthening of existential elimination appears in the literature, in a natural and nice way; say in ZF+GC for instance.
Jul 10, 2017 at 20:27 review Close votes
Jul 10, 2017 at 21:25
Jul 10, 2017 at 20:25 comment added Burak I don't really understand what the question is after. (This is not a criticism, I literally didn't understand.) Instead of considering the axiom of choice as a principle that allows you to "choose" elements, consider it as a principle that says certain cartesian products of non-empty sets are non-empty. (The choice really happens when you pick an element from this non-empty cartesian product.) Under this interpretation, what exactly is the question after?
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Jul 10, 2017 at 20:02 comment added Pace Nielsen @Goldstern Are you sure about that? First, since this choice is being made for all sets $y$ simultaneously, I think you need at least the stronger axiom of global choice. Second, this semantics would be applied to more properties than merely $P(x,y)\equiv x\in y$.
Jul 10, 2017 at 19:59 comment added Pace Nielsen @AsafKaragila I don't disagree with you. If you read my post, I was very careful to say that the ability to "pick a single element from a single non-empty set" was a consequence of the metalogical assumption. You are right that to go further you need induction in ZF which does require some of the axioms of ZF (or a stronger metatheory).
Jul 10, 2017 at 19:59 comment added Gerhard Paseman You might find enlightenment in considering the multiplicative axiom which is equivalent to the axiom of choice. I'm sure the correct form is on Wikipedia and is something like 'The infinite Cartesian product of nonempty sets is nonempty'. I think early scholars of AC came up with this version just for people with your question. (For the previous comment of mine, I need the words 'nonempty y' in another paragraph as well.) Gerhard "Who Were Very Likely Themselves" Paseman, 2017.07.10.
Jul 10, 2017 at 19:47 comment added Gerhard Paseman You need to be a bit less informal. Logic allows such picks from a nonempty set y. If y is empty then you run into a problem. For n from 1 to infinity, consider : throw balls labeled 10n+I for I from 1 through 10 into an infinite urn and a) for sequence 1, draw out the ball with largest label 2) for sequence 2, draw out the ball with smallest label. Now try one more pick after this. How do you know if the last pick is allowed? How can you guarantee that a certain infinite system of choices exist? For some models of ZF, it doesn't. Gerhard "Hopefully You Get This Idea" Paseman, 2017.07.10.
Jul 10, 2017 at 19:40 comment added Goldstern Yes, there is a natural semantics/language where the sentence $\forall y\exists x\ P(x,y) $ is interpreted as the ability to CHOOSE for each $y$ some $x=x(y) $ such that $P(x(y),y)$ holds. Namely, set theory with choice, e.g. ZFC. (I write this as a comment and not as an answer, because I assume that you do not consider ZFC natural enough.)
Jul 10, 2017 at 19:30 comment added Asaf Karagila It is not a rule of logic which allows us to prove finite choose. It is the combination of the inference rule and the fact that ZF proves induction. See, if you work over a model with non-standard integers, then inference rules can only get you through the standard integers, but not further. But the fact ZF proves induction for the natural numbers (or whatever it perceives as those) allows us to actually prove finite choice.
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