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If we have an infinite collection of sets, where each set contains two elements, 0 and 1, without using axioms of choice, we can define a choice function that will always pick the element 0 from each set. However, it seems that if we only assume the set contains two elements, without specifying them in detail, we can not show the existence of a choice function without using the axiom of choice.

Since all the sets has a cardinality of two, we can define bijections between the two elements in each set and {0,1}. So regardless of how the bijection is defined, we can always choose the element that maps to 0.

I don't understand the difference between this operation compares to always choosing 0 in each set, but one requires the axiom of choice, while the other doesn't.

I have read many answers on stack exchange about axioms of choice. Like this one. And I saw answers that used the word "uniform" selection. I'm really confused about what "uniform" means in this case, and no questions seem to answer my confusion.

EDIT: The comment section is getting too long. So I posted a new question.

wsz_fantasy
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    " define a bijection between each set in the collection and {0,1}" requires the axiom of choice. – Anne Bauval Oct 10 '22 at 23:16
  • @AnneBauval Thank you. That’s a typo. I’ve fixed it. Clearly the bijection between all the set in the collection and {0,1} does not exist. – wsz_fantasy Oct 10 '22 at 23:22
  • You can find the bijection for any one set, but you can't necessarily find a set of bijections that work for all sets. There are two bijections for any one set, so you are are now just trying to pick one element from each of those $2$-element sets of bijections. – Thomas Andrews Oct 10 '22 at 23:24
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    That was much worse than a typo, and cannot be fixed. Simultaneously "define a bijection between the two elements in each set and {0,1}" requires the axiom of choice. – Anne Bauval Oct 10 '22 at 23:25
  • The key is that the language of set theory doesn't let you do things in general "simultaneously," without the axiom of choice. – Thomas Andrews Oct 10 '22 at 23:27
  • @ThomasAndrews But how can you “simultaneously” pick 0 instead of 1 in all sets? – wsz_fantasy Oct 10 '22 at 23:28
  • @wsz_fantasy by "simultaneously" we mean (given ${X_i\mid i\in\omega}$ a set of $2$-elements sets) "you can't find a set $F$ of functions such that each $f\in F$ is a bijection $f:X_i→{0,1}$ and for each $i\in \omega$ there exists such function and there are no $2$ such functions for specific $i$ ", indeed for a specific $i\in ω$ we can find $f: X_i\to ω$, but we can't talk about infinitely many such functions – ℋolo Oct 10 '22 at 23:42
  • Think of it this way: The axiom of choice lets to do something non-constructive. It asserts the existence of a function that we cannot define or evaluate explicitly. Simultaneously picking $0$ is not reallysimtaneous, it is just defining a constant function, one of the most easy constructive functions to define. Any function, I guess, defines all values "simultaneously" in some sense. But some functions do so without making a choice. We can define $f(n)=n^2,$ but that doesn't have the simultaneous problem because we can write down the function - how to compute it for any input. – Thomas Andrews Oct 11 '22 at 00:09
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    In any event, there is nothing in ZF without choice which lets you prove such a choice exists. "Simultaneously" is not a word in formal set theory. There are a lot of things you might want to say in formal systems that you can't say in those systems. The axioms of set theory is designed to allow you to say those things. While we can sometimes do stuff that remotely resembles "simultaneously" doing something, the axiom of choice is the rule that let's us do that in a lot of contexts. The axiom of choice is the codification of your intuitive notion of "simultaneously" making a decision. – Thomas Andrews Oct 11 '22 at 00:24
  • @ThomasAndrews I understand what you mean intuitively, but can you explain in a more rigorous language the difference between these functions? For instance, taking the smallest value for each set given it's total ordered would not require the axiom of choice – wsz_fantasy Oct 11 '22 at 00:50
  • Well-ordering, not total. There is no intuition for why you can't prove Choice from ZF. People tried to for years, until it was proven to be an independent axiom. When you say that you have a set of well-ordered non-empty sets, it might seem like those well-orderings are independent, but it actually means we've already made a choice - there are lots of well-orderings of each set. But we've already got, for each set $S,$ a well ordering $<_S.$ The only way to write that is as a function, $S\mapsto <_S.$ So the existence of the well ordering of all the sets is already a function. – Thomas Andrews Oct 11 '22 at 01:07
  • And yet again, you would have to first (ahem) simultaneously choose a total ordering on each of those infinitely many sets. – Lee Mosher Oct 11 '22 at 01:07
  • Essentially, we use $S\mapsto <_S$ to define our function, not just assert the function exists. – Thomas Andrews Oct 11 '22 at 01:08
  • Indeed, our tendency to write orderings as pairs $(P,<)$ means that a set of ordered sets is almost already a function. It is a function if the underlying sets are distinct. – Thomas Andrews Oct 11 '22 at 01:12
  • Let me just point out one really cool thing about the axiom of choice: It raises one's perception and awareness of what's going on in various proofs, namely, the awareness of when one is making choices. Some choices don't need the axiom of choice, e.g. choices made by applying existential instantiation. Other choices can be made constructively, e.g. pick the smallest element out of each of a collection of totally ordered sets. – Lee Mosher Oct 11 '22 at 01:14
  • It's interesting to learn how to recognize when a choice you are making cannot be done constructively or by existential instantiation, but only by applying the axiom of choice. – Lee Mosher Oct 11 '22 at 01:15

1 Answers1

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Since all the sets has a cardinality of two, we can define bijections between the two elements in each set and {0,1}.

This actually does not help at all, although it's a little subtle to see why: it's true that for each such set, such a bijection exists, but in order to write down a choice function you need to choose a collection of such bijections for each set simultaneously. This means you need a choice function for the collection of bijections! And there are two for each set! So you're in exactly as bad a position as you were before.

To say this more explicitly, suppose $I$ is an index set and $X_i, i \in I$ is a collection of $2$-element sets indexed by $I$. You are saying: well, since by hypothesis each $X_i$ has cardinality $2$, for each $i$ there exists a bijection $X_i \cong \{ 0, 1 \}$. And yes, that's true. But to write down the choice function you want using these bijections, you need to choose one such bijection for each $i \in I$, and there are two. In other words, you need a choice function for the collection of sets $Y_i = \text{Iso}(X_i, \{ 0, 1 \})$, each of which also has two elements!

Coming at this from another direction, here is a more-or-less explicit example of a collection of sets of cardinality two where I challenge you to write down an actual choice function: consider the collection of all fields of characteristic $\neq 2$ in which $-1$ has a square root (if you want to cut this down to a set then take isomorphism classes and place some restriction on the cardinality). There are always exactly two such square roots, one of which is the negative of the other, which in the complex numbers are $i$ and $-i$. Can you tell me how to choose, for each such field, one of the two square roots of $-1$? Note that there are "surprising" examples of such fields, such as the field of $p$-adic numbers $\mathbb{Q}_p$ for $p \equiv 1 \bmod 4$ a prime.

Another example in a similar vein: consider the collection of all connected orientable manifolds. Any such manifold has exactly two orientations, one of which is the negative of the other. Can you tell me how to choose, for each such manifold, one of its two orientations?

Qiaochu Yuan
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  • I understand this case now. Previously, I was also thinking about doing existential instantiation for each individual set, can you elaborate on why that doesn’t work as well on an infinite set but does work in finite cases? So I think a choice function has to be constructively in some sense, but is there a general way for us to determine what is a constructive choice? – wsz_fantasy Oct 11 '22 at 01:49
  • Btw, I forgot to add, in the case of the finite axiom of choice, which can be deduced from ZF, the choice function doesn’t have to be chosen constructively. – wsz_fantasy Oct 11 '22 at 02:56
  • @wsz_fantasy: existential instantiation only lets you make one choice at a time, so you can only use it to make finitely many choices (since a proof has finite length). The rest of the ZF axioms, without choice, simply do not let you make infinitely many choices arbitrarily. It takes a little familiarity with what the ZF axioms do and don't let you do to see what kinds of choice functions you can write down; basically they have to be "explicit" or "constructive" or "computable" in some sense. – Qiaochu Yuan Oct 11 '22 at 03:54
  • I see. I found this (link)[https://math.stackexchange.com/questions/2687765/confusion-about-axiom-of-choice] which mentioned the term “finitary rule”, see the paragraph before the last paragraph. Does that mean with the ZF axioms, we can not prove the existence of such a choice function, like the function mentioned in this question and linked post? Nonetheless, are they still by definition choice functions, where we just don’t know whether they exist or not by ZF? – wsz_fantasy Oct 11 '22 at 04:11
  • @wsz_fantasy This is a bit misleading, because the proof of "finite choice" does not actually proceed by doing some requisite finite number of existential instantiations. Rather, it is proven by induction on the size of the finite set, so just invokes existential instantiation once when arguing the induction step (and perhaps once for the base case, depending on how exactly you formulate things). The former approach doesn't actually make any sense when you try to do it in detail... one way to think about it is there's no guarantee a finite set in a model is actually (externally) finite. – spaceisdarkgreen Oct 11 '22 at 04:16
  • @QiaochuYuan Just to add on my previous comment, if these are both indeed choice functions, (if not let me know), can we prove their existence with ZFC? I think for the one mentioned in this function, we can always choose a bijective function and thus the choice function exist, but what about the one about “infinite existential instantiation”? I’m not too familiar with trans finite induction, but can it be proved to exist using ZFC as well? – wsz_fantasy Oct 11 '22 at 04:22
  • (I wrote that comment before I saw you added the previous comment with the link... note that the linked answer's point 1 is the same point I made.) – spaceisdarkgreen Oct 11 '22 at 04:29
  • @spaceisdarkgreen No worries, I have understand what you mean. Btw, can you clarify the difference between internally and externally finite. And I should probably post the above question in a new post, unless the answer can be explained in the comment. – wsz_fantasy Oct 11 '22 at 04:37
  • @wsz_fantasy I think it's probably more instructive to try and fail to make sense of a proof by "repeated existential instantiation" (and yes, if still curious, that's probably best for a different post, though I'm sure variations have been asked before). The issue that should jump out is that if that were really the way it went you would need a different proof for each finite number... but you don't want infinitely many proofs, you want one proof. (The quantifier "for all natural numbers" is inside the statement you're trying to prove, not in the metatheory.) – spaceisdarkgreen Oct 11 '22 at 04:43