Talk:Axiom of choice/Archive 2

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Do Wikipedia articles assume the axiom of choice unless otherwise mentioned?

A question on User Talk:CSTAR led to a long discussion of the precise meaning of the independence of the axiom of choice (among other things). By consent of the participants, it was moved here. Gadykozma 20:59, 2 Sep 2004 (UTC). After dying out, it was archived. Gadykozma 13:17, 22 Sep 2004 (UTC)

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Do Wikipedia articles assume the axiom of choice unless otherwise mentioned? Or should results which rely on choice be marked as such? --Matthew Woodcraft

They should be marked as such. If they're not, fix them. --Taw

Pretty much anything I write assumes AC. --Zundark, 2001 Dec 16

I think it is polite, when writing about a central theorem like existence of prime ideals or Hahn-Banach, to mention that it depends on AC, but it is really too much to ask to do the detailed bookkeeping and mention AC for every result which depends on one of those theorems. AC is an accepted axiom in mathematics. --AxelBoldt

This last position seems to be modern practice, at least in the areas of mathematics I'm familiar with. Should we have a note to this effect on the main AC page? --Matthew Woodcraft

Yes, we should. If you're feeling in the mood, this article really needs an overhaul. I moved some paragraphs here from the set theory article months ago, but no-one has yet attempted to merge it all into a coherent whole. --Zundark, 2001 Dec 16

Is there any treatment of "life without AC"? I would find this an interesting topic. --Jeff

I agree with Jeff. It could be nice to have, say, something about "dedekind-finite" sets, etc. --Alex, 2004 Dec.

I would really like to see at least a paragraph in this article about the current status of the controversy regarding the AC. Thanks to the specialist(s) who would add this update! olivier 03:59 Nov 27, 2002 (UTC)

Of course, linguistic abuses can lead to many so-called "paradoxes". For example, "This is not a true statement." seems paradoxical. Yet, the problem is the use of the pronoun "This" which makes the statement self-referential. When things reference themselves, they are no longer first order. The statement is never "within the system" if it references itself. The reference generates a copy of the system, putting the statement outside of the system. Many higher order systems have no known solutions. J. Todd Chapman

it's true that it's not true. lysdexia 22:05, 12 Nov 2004 (UTC)

There isn't much work in this direction (as opposed to continuum hypothesis). Too much of mathematics relies on this axiom. The proofs which do not rely on it are considered to be constructive, and there is some interest in seing if there is a constructive proof in this sense, but there is no controversy about the axiom of choice—there isn't really much to say about it except that it is independent, nonconcstructive but makes life much easier. Continuum hypothesis is almost irrelevant to everyday mathematics and that is perhaps what makes a big difference.

Might help to mention that AC is provably false in Quine's NF though it holds for all the sets one would encounter in "everyday" maths. I'd do it myself but it's much too long since I studied logic. Tom Holbrook 14:39 10 Jul 2003 (UTC)

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The recent edit about stating whether AC is used, and preferring not to use it, is definitely POV. Logicians, and those in constructive mathematics—and of course anyone who wants to compute anything, may agree; but it's not the mainstream approach. Post-Bourbaki, one uses Zorn's Lemma and forgets about it.

Charles Matthews 15:46, 20 Oct 2003 (UTC)

It was my edit, and agreed it's tinged by my set-theorist POV. It is equally POV to state without qualification that AoC is accepted as true (since that's not the point), or that it is of no import whether it is used or not. How about something like "Despite this, a minority of mathematicians..." instead of "Despite this, it is common to..." Onebyone 16:50, 20 Oct 2003 (UTC)

Can be made NPOV by a bit of expansion: eg anyone who wants results true in all toposes etc etc.

Charles Matthews 17:04, 20 Oct 2003 (UTC)

How's that? I haven't mentioned topoi because (I don't understand them and) "system" is fine for the general reader. Onebyone 16:09, 25 Oct 2003 (UTC)

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From the article: Jerry Bona once said: "The Axiom of Choice is obviously true...

Who is Jerry Bona? Kevin Saff 18:29, 23 Mar 2004 (UTC)

I've never heard of him either, but the joke itself is famous, and he invented it, then he deserves the credit. -- Walt Pohl 02:48, 24 Mar 2004 (UTC)

I agree, though the statement seems phrased to imply that Jerry Bona is more famous than the joke, if that makes sense. I guess this is Jerry Bona. Maybe it could read something like Jerry Bona, a UIC math professor, famously cracked, "The Axiom of Choice..."

How about renaming that section to 'jokes', and adding the (perhaps anonymous) question 'what's yellow and equivalent to the Axiom of Choice?'? No reason to remove Jerry's name, though. 84.9.18.135 14:34, 15 January 2006 (UTC)

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Something which might be worth mentioning is that at least some of these consequences of the axiom of choice have been shown to imply the axiom as well (Tychonoff's theorem, for example). I'm not sure which other ones do or don't though. Kevinatilusa

Additionally, is conjectures really the right category to put this under? Calling it a conjecture seems to imply that it could be proven someday, which isn't the case here. Kevinatilusa

I agree; it should be removed from that category (or is it topoi? :)). --66.219.81.52 00:23, 16 Jun 2004 (UTC)

I've removed the conjectures tag, since there are independence results making that misleading. Charles Matthews 07:40, 16 Jun 2004 (UTC)

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This doesn't seem right to me... first the article says:

For example, a proof could use a set S that was previously demonstrated to be non-empty and claim "because S is non-empty, let x be one of the members of S." Here, the use of x requires the axiom of choice.

This doesn't require AC, does it? X consists of one set, S, and as stated in example 1.:

1. Let X be any finite collection of non-empty sets.

Then f can be stated explicitly (out of set A choose a, ...), since the number of sets is finite.

--66.219.81.52 00:23, 16 Jun 2004 (UTC)

The example does require the axiom of choice. I'm not sure what your second question is talking about, though. Little x in the example has nothing to do with the big X, which is a set, that you are talking about. Anyway, from a set theory book this is just about as classic as an example can get.

Given a set X that is nonempty, there exists x in X. This requires only predicate calculus, not the axiom of choice.

To use the element in X requires the axiom of choice. Check out this quote from Hayden and Kennison:

"Choose ab\in Ab [...] This fallacious argument looks reasonable because of a subtle ambiguity in the phrase 'choose ab\in Ab'. It is certainly true that given any non-empty set S, one can 'choose' an element s\in S. This follows from the fact that there exists and x such that x\in S (as S is non-empty), and from our convention about choosing an entity e for which P(e) whenever it is true that there exists an x such that P(x). Thus, for any one of the sets, Ab, we can 'choose' ab\in Ab. But there is no way we can legitimately 'choose' a dependent variable ab for all b\in B. What is needed is a 'choice function' that will enable us to make many choices simultaneously." (Zermelo-Fraenkel Set Theory, 1968, p. 50f)

Note, the above quote is from a book and is not covered by FDL. MShonle 05:33, 25 Jun 2004 (UTC)

I hope to clear this up very precisely. Given a nonempty set X1, producing an element x in X1 requires only existential instantiation. Similarly, given any finite family of disjoint nonempty sets X = {X1, X2, X3, ...XN}, it does not require the axiom of choice to construct a set that contains one element form each XJ in X. This is only finite reiteration all of which can be done in ZF. Now suppose X is an infinite family of disjoint nonempty sets. If we know of a rule by which we can choose a unique element from each set ("choose the least element" if they are so ordered or something like that) then again we do not need the axiom of choice. It is when we can not construct the set that we need the axiom of choice, which is existential rather than constructive. Thus he says we need the axiom to make many (infinitely many) choices simultaneously.

) By the way, assuming there are only finitely many nonempty boxes, I can easily choose a single element from each using only ZF. This article is misleading.

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If no one has a problem with this, I would like to add another equivalent formulation of the AC. Stated in its most compact form: "Given infinite sets A and B, every bijective mapping from A to B has an inverse." It's a different take on the principle. Any objections? Tim 19:54, 2004 Oct 8 (UTC)

Huh?? Maybe you meant that for every surjective mapping f : A --> B there is an "inverse" g : B --> A such that f(g(x)) = x for x ∈ B? As you state it, it doesn't seem equivalent to AC at all. Michael Hardy 20:24, 8 Oct 2004 (UTC)

Arrg. I always get "bijective" "surjective" and "injective" mixed up. I'll have to go back and look at it a little longer. This isn't my formulation, so I'll have to get a little for clarification before I add it... Tim 00:04, 2004 Oct 10 (UTC)

Epimorphisms split. Well, that's more compact.

Actually there are books full of equivalents.

Charles Matthews 20:25, 8 Oct 2004 (UTC)

Baire's theorem

Correct me if I'm wrong, but Baire's theorem requires only inductive choice. That's quite different than the full axiom of choice. I suggest that it be removed from the list of dependencies. (the Baire category theorem page requires checking too). Gady 22:10, 12 Nov 2004 (UTC)

The page says "Several central theorems in various branches of mathematics require the axiom of choice (or one of its weaker versions, such as the Boolean prime ideal theorem, the axiom of countable choice, or the axiom of dependent choice).", though, so while it doesn't explicitely list inductive choice as a weaker version, it does state that full AC may not be necessary to prove the heorems listed. -- Schnee 02:05, 14 Nov 2004 (UTC)

So the page is not wrong, just misleading. These things cannot be wrapped up together — maybe a logician thinks they are all of equal standing (because they are all independent of ZF), but in the rest of mathematics, the axiom of choice has a standing quite distinct from the others. Gady 02:42, 14 Nov 2004 (UTC)

Feel free to update it so it's more clear. :) -- Schnee 16:39, 14 Nov 2004 (UTC)

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Another dfn.

I once heard another definition of the AoC which I quite liked. I haven't researched it or checked it, but it seems to be equiv.:

The cross product of non-empty sets is a non-empty set

Or abusing notation:

Given sets $S_{1},S_{2},\ldots$, the set $S_{1}\times S_{2}\times \ldots$ contains an element.

I guess that by using integer subscripts it looks like there are countably many sets. But you get the idea.

--Sean Kelly 01:09, 21 Dec 2004 (UTC)

Law of the Excluded Middle

The following was recently added to the article:

Interestingly, the axiom of choice implies the law of excluded middle, another principle which is also rejected by constructive mathematicians (and, in particular, by the intuitionists). To see this, for any proposition $P\,$, let $U\,$ be the set $\{x=0\vee P:x\in \{0,1\}\}$ and let $V\,$ be the set $\{x=1\vee P:x\in \{0,1\}\}$. By the axiom of choice, there will exist a choice function $f\,$ for the set $\{U,V\}\,$. Since $f(U)\in U$ and $f(V)\in V$, this implies $[f(U)=0\vee P]\wedge [f(V)=1\vee P]$, which implies $f(U)\neq f(V)\vee P$. Since $P\,$ implies $U=V=\{0,1\}\,$, it must be that $P\,$ implies $f(U)=f(V)\,$, so $f(U)\neq f(V)\vee P$ would imply $\neg P\vee P$. As this could be done for any proposition $P\,$, this completes the proof that the axiom of choice implies the law of the excluded middle.

I've removed this because it doesn't make sense to me. For starters, the axiom of choice is not necessary for the existence of a choice function for a finite set of finite sets. I apologize if I'm being slow here; if someone knows better, it can always be put back. Josh Cherry 03:20, 5 Jan 2005 (UTC)

It doesn't make sense to me either. Dbenbenn 03:50, 5 Jan 2005 (UTC)

Nor to me. What in the world is the set

$\{\,x=0\vee P:x\in \{\,0,1\,\}\,\}\ \mathrm {?}$

By usual notation, it means the same thing as

$\{\,0=0\vee P,\,1=0\vee P\,\}.$

What does that mean? Michael Hardy 20:33, 5 Jan 2005 (UTC)

I think it's supposed to mean the set of all $x\in \{\,0,1\,\}$ such that x=0 or P, i.e., {0, 1} if P is true and {0} if P is false. But I'm not sure, and it took me a while to arrive at that. Josh Cherry 01:09, 6 Jan 2005 (UTC)

Or perhaps the original author meant to say $\{x\in \{0,1\}:(x=0)\vee P\}$. --Sean Kelly 02:20, 6 Jan 2005 (UTC)

The theorem itself is true. The argument provided is hard to follow. -- Walt Pohl 01:23, 6 Jan 2005 (UTC)

Um, the law of excluded middle is just true.

You miss the point. The author clearly meant within constructivist logic AC entails excluded middle. Excluded middle is not "just true" within constructivist logic. (But whoever wrote what appears there was clear as mud, even if his bottom-line conclusion turns out to be right.) Michael Hardy 23:10, 6 Jan 2005 (UTC)

It doesn't depend on AC. It's not even a statement in mathematics, but rather a law that is used by mathematics. Of course, it's also a statement in various formalizations of logic, some of which, I expect, depend on AC. But in that sense you'd have to say precisely what formalization you're talking about. Dbenbenn 02:41, 6 Jan 2005 (UTC)

Intuitionists reject the law of excluded middle. They're apparently the ones interested in the connection with the axiom of choice. I'm not defending any of this--it seems like madness to me too--but there's a real strain of thought about this out there. The anonymous editor didn't just make this up, despite the fact that it makes no sense in the context of the article. Josh Cherry 03:31, 6 Jan 2005 (UTC)

Hi. I originally added the above apparently controversial material, though from a different computer and without logging in. Josh Cherry and Sean Kelly are both correct about what I meant by the set comprehension, though I'll admit that what I posted may have been quite unclear. I apologize for that, but I do think the material should remain, in a cleaned-up form. The purpose was to demonstrate that even those who follow constructive forms of logic which do not assume the law of the excluded middle are forced to accept this law if they accept the axiom of choice, a well-known and, I think, important theorem. I admit to writing the original proof in a hurry, but hopefully we'll eventually be able to get a much easier-to-follow version onto the page. -Chinju 06:57, 10 Jan 2005 (UTC)

I definitely think it should be included, but the argument you gave was pretty impenetrable. Care to explain the argument, and we'll go from there? -- Walt Pohl 08:15, 10 Jan 2005 (UTC)

Also explain where the all-important choice function is given. I can see you provide the function f, but I don't see how it uses the AoC, as U and V seem to be finite sets. --Sean Kelly 08:35, 10 Jan 2005 (UTC)

The argument I gave was based off of my memory of the one found at the bottom of http://publish.uwo.ca/~jbell/Axiom%20of%20Choice%20and%20Zorn.pdf. I don't know if that makes it clearer or not; if not, perhaps you could mention which part is unclear and I could attempt to explain that more thoroughly. On a different note, Sean Kelly's objection that the choice function is only used on a finite collection of finite sets is a good one, and it took me a while to mull over. I think the resolution is something like this: the Axiom of Choice is classically unnecessary for a choice function in such a case; however, this is not the case intuitionistically.

To see this, consider how one creates a choice function without the axiom of choice for a finite set {X_1, X_2, ..., X_n} of nonempty sets X_1 through X_n. Using finitely many existential instantiations, one can obtain elements e_1 in X_1, e_2 in X_2, ..., through e_n in X_n. Then, naively, one might attempt to create a function f such that f(X_1) = e_1, f(X_2) = e_2, etc. However, this would not necessarily work, for if X_1 = X_2, one needs to make sure e_1 = e_2. So, really, what one might do classically is construct the function f inductively, letting f(X_1) = e_1, letting f(X_2) = e_1 if X_1 = X_2, letting f(X_2) = e_2 if X_1 != X_2, etc. With the law of the excluded middle, this could be shown to be a well-defined function, since [(X_1 = X_2) OR (X_1 != X_2)] would be provable. However, without the law of the excluded middle, such a definition wouldn't work, and so, perhaps, it would not necessarily always be possible to create a choice function for a finite set without the axiom of choice, in intuitionistic logic. That seems to me to be the answer for now, anyway, but I'll think about it some more. -Chinju 18:12, 10 Jan 2005 (UTC)

Suppose, in the case at hand, that I define my choice function by f({0}) = 0, f({1}) = 1, and f({0, 1}) = 0. I presume that the existence of this function is uncontroversial. It does have a larger domain than the elements of {U, V}. However, 1. it is not clear to me that the usual statement of the axiom of choice requires that the domain be minimal and 2. even if it does, I imagine that a modified version that allowed for a larger domain would be equally unacceptable to intuitionists and equally useful to everybody else. Josh Cherry 02:06, 11 Jan 2005 (UTC)

I only dabble in intuitionistic thinking, so I'm not sure, but I think the intuitionistic response would be that your proposed function hasn't actually been demonstrated to be a choice function for {U, V}, since you haven't demonstrated U to be {0}, {0, 1}, or {1}, even though this disjunction is, of course, classically easy to show from the fact that U is a non-empty subset of {0, 1} (and similarly for V). So your proposed choice function wouldn't have been shown to be applicable in this case. But I have to think about this some more. -Chinju 03:43, 11 Jan 2005 (UTC)

Rephrasing Chinju's remark:

The set U is defined as { x in {0,1}: x=0 or P }. The domain of your function f is the set whose elements are {0}, {1} and {0,1}. You say

f has a larger domain than the elements of {U, V};

I do not think that an intuitionist would agree. How would you convince an intuitionist that U is in the domain of f? --Aleph4 13:07, 20 January 2006 (UTC)

Correct. You must embrace madness to see why f is not a choice function. If U = {0} or U = {0,1}, you've already got P or not-P. If in your head you thought of the power set of {0,1} as 2^{0,1}, you've already assumed the law of the excluded middle, and thereby lost your grip on insanity. The section is still a bit opaque though, so I'm changing it. -Mad Dan 17:25, 29 April 2006 (UTC)

Banach-Tarski paradox nonconstructive?

Is the statement that the proof of the Banach Tarski paradox is nonconstructive relly true? I saw a proof (albeit a long time ago), but I had a definite idea that the proof actually constructed the a finite partition of the sphere... Schnolle 12:36, 16 Feb 2005 (UTC)

The nonconstructivity is in step 3 of Banach-Tarski paradox#A sketch of the proof, where you use the axiom of choice to pick one element from every orbit of a particular action. dbenbenn | talk 13:19, 16 Feb 2005 (UTC)

non-empty

Why does X have to consist of non-empty sets? Or is this only in the first formulation? It doesnt seem necessary in the second def. -MarSch 14:58, 20 Apr 2005 (UTC)

Save that one for April 1? Charles Matthews 15:31, 20 Apr 2005 (UTC)

No this is serious. There exists a (unique) function from the empty set to itself, right? -MarSch 15:54, 20 Apr 2005 (UTC)

Let X be a set of non-empty sets. There exists a choice function f defined on X such that for each set S in X, f(S) is an element of S. So, if S is the empty set, how can we have f(S) an element of S? A choice function f here presumably means a function from X to Y, where Y is the union of all the elements Z of X. If the empty set were an element of X, no such function could exist. Charles Matthews 17:53, 20 Apr 2005 (UTC)

It's one of those things that's too obvious to come up with on your own... we all make mistakes. =) —Sean κ. ⇔ 21:46, 20 Apr 2005 (UTC)

right, my bad :) thanks for clearing up my confusion. MarSch 12:50, 21 Apr 2005 (UTC)

Is this appropriate

from the article:

On the other hand, the negation of the axiom of choice is also bizarre. For example, the statement that for any two sets S and T, either the cardinality of S is less than or equal to the cardinality of T or the cardinality of T is less than or equal to the cardinality of S is equivalent to the axiom of choice. Put differently, if the axiom of choice fails, then there are sets S and T of incomparable size: Neither can be mapped in a one-to-one fashion onto a subset of the other.

Perhaps, I'm not following the point here, but it would seem to me that if the axiom of choice is equivalent to the statement "for any two sets S and T, either the cardinality of S is less than or equal to the cardinality of T or the cardinality of T is less than or equal to the cardinality of S" then rejecting the axiom of choice as an axiom does not rendered the statement about cardinalities false but merely renders it unprovable one way or the other (since obviously it must be independent from the other axiom if it is equivalent to the axiom of choice). It would seem to me that the quoted section should rather conclude, "if the axiom of choice is not accepted then there exists sets S and T such that the existance of a one-to-one mapping from one into a subset of the other cannot be shown to exists." Which doesn't really seem all that bizarre.--Heathcliff 03:04, 9 May 2005 (UTC)

If on the other hand the quoted section means to refer to bizarre consequences of postulating the negation of the axiom of choice then it seems to me that no one has ever been in favor of doing that. As far as I know those who oppose the axiom of choice either want it rejected as an axiom or replaced with a weaker axiom, but no one want its negation taken as an axiom in its place.--Heathcliff 06:14, 7 May 2005 (UTC)

Certainly the quote is talking about considering the negation of AC. It says as much in the first phrase. I don't personally know if there are people who want to use the negation of AC, but the first sentence of the article claims that there are people who at least explore the possibility of negation of AC. Furthermore, in the context of the quotation, where it is discussing how counterintuitive eg the Banach-Tarski paradox may seem, it may be natural to not only want the Banach-Tarski result to be unprovable, but to actually want it to be false. Certainly there is something appealing about a world where Banach-Tarski is not possible. Negation of AC will give you this. -Lethe | Talk 06:24, May 7, 2005 (UTC)

If we let C(S,T) be the sentence "the cardinality of S and T can be compared", the quoted section states, "AoC iff. ∀S and T, C(S,T) ". So if not AoC, then the negation of the right side follows, i.e. ∃S and T such that not C(S,T).

As for your last statement, rejecting the AoC is the exact same thing as assuming the AoC is false, wouldn't you agree? Since the AoC is an independent axiom, you either have to assume it is true or assume it is false; it can't be proven or disproven from the other standard axioms. If by rejecting the AoC you mean ignoring it, you still probably would want to replace it with new axiom(s), so the question of whether it is true or false is still an interesting one. Perhaps you can add in interesting axioms where the AoC becomes false. Either way, I agree that the quoted section may need some rewording. —Sean κ. ⇔ 06:54, 7 May 2005 (UTC)

Furthermore, in the context of the quotation, where it is discussing how counterintuitive eg the Banach-Tarski paradox may seem, it may be natural to not only want the Banach-Tarski result to be unprovable, but to actually want it to be false. Possibly, but I've never heard of anyone arguing that the axioms of set theory should be constructed with the specific intent of making the Banach-Tarski paradox false. I think the problem in the minds of some people is that it should not be provably true. I'm certainly no expert on this, and would welcome any quotes from mathematicians or logicians who feel that the negation of the axiom of choice should be postulated, but I don't think it's fair to simply assume that there are any. I personally will be surprised if there's even one.--Heathcliff 16:17, 7 May 2005 (UTC)

Well you said you couldn't see the point of the paragraph, and the point is this: AC is counterintuitive, but ~AC is also counterintuitive. As to whether there really are practicing mathematicians who assume ~AC, I've never heard of any. According to the intro paragraph, the schools that explore axiom systems with ~AC exist primarily in the field of axiomatic set theory. That explains why I've never met them, all the mathematicians I know are geometers and topologists. But the paragraph you quoted never claimed that anyone uses ~AC, it only mentions that ~AC may be considered counterintuitive, which is a useful statement, in my opinion. I'm not really sure what you're complaining about. We don't need to find a collections of mathematicians who take ~AC to be true just to justify mentioning that ~AC is counterintuitive (even though I believe it possible that such a collection may exist). -Lethe | Talk 16:37, May 7, 2005 (UTC)

As for your last statement, rejecting the AoC is the exact same thing as assuming the AoC is false, wouldn't you agree? I completely disagree. Rejecting it is refusing to accept it as an axiom; assuming that it is false is acceptiing its negation as an axiom. These are fundemently different things. If the axiom of choice is simply rejected then it cannot be said to be either true nor false since it cannot be deduced one way or the other from the other axioms. If it is postulated to be false then obviosuly it is false, but so far as I know no one has ever wanted to do that. I think the section should simply need to be removed because it seems to me to be written from the point of view that rejecting the axiom of choice is the same as postulating that it is false which (unless I'm missing something obvious) is completely incorrect.--Heathcliff 16:17, 7 May 2005 (UTC)

I also disagree. Constructivists do not assume the negation of AC. For them, statements which rely on AC don't become false, they are simply nonconstructable. -Lethe | Talk 16:37, May 7, 2005 (UTC)

I think we're really just bickering over what we mean by "rejecting". I agree that it's not useful to take the negation of the AoC as an axiom, but it is useful to assume that the AoC is false, in order to explore the consequences. Can't we just rewrite the quote to make that clear?

Clearly, you can imagine any logician writing down the axioms he wants to work with. If he writes down ZF + ~AC, this is a different set of axioms than ZF. Some statements that are undecidable in ZF will be false in ZF + ~AC. So rejecting AC as an axiom is different than assuming the negation. The former is constructive mathematics, and the latter is some funny exploration in axiomatic set theory that is nonstandard, but still interesting. -Lethe | Talk 17:27, May 7, 2005 (UTC)

To address Heathcliff's comments, we might also considering adding a note about the incompleteness of set theory, and how it is entirely possible that some things are better left unprovable. —Sean κ. ⇔ 17:05, 7 May 2005 (UTC)

Better? How about mathematically required! Hilbert would have wanted all true statements to be provable. For him, it's worse that some things are unprovable, but still a fact of life. For us, it gives us the freedom to take AC or leave it. -Lethe | Talk 17:23, May 7, 2005 (UTC)

Anyways, I think it is interesting and appropriate that the negation of the AoC implies something as unusual and unappealing as Banach-Tarski. —Sean κ. ⇔ 19:47, 7 May 2005 (UTC)

Actually, Banach-Tarski is a consequence of (the positive of) AC. The negation of AC will actually disallow such nonmeasurable constructions. -Lethe | Talk 20:27, May 7, 2005 (UTC)

Sorry for going out of chronological order, but the above is a sufficiently widespread misconception that it should be addressed in situ. The negation of AC does not "disallow such constructions"; it's consistent with ¬AC that the reals can be wellordered, which is all you need to get Banach-Tarski. --Trovatore 20:33, 23 July 2005 (UTC)

I think it is interesting and appropriate that the negation of the AoC implies something as unusual and unappealing as Banach-Tarski. I personally see nothing interesting about it at all. It simply means that there are theoretical axioms which have never been suggested by anyone (in this case ~AC) that if adopted would lead to unappealing results. Every single statement which is currently unprovable from ZF could be taken as a postulate and all sorts of strangess might result, but unless there's an actual proposal for such an axiom there isn't much point in offering arguments against it.

I have a problem with the section as it is because it will lead many people to believe that simply rejecting AC leads to the counter-intuitive result given. Imagine in the last US election campaign if someone had said, "Vote for Bush because Nader will make a terrible president." Such an argument misses the real debate because it ignores the real alternative (in this example Kerry) and replaced it with an alternative that is more easily rejected but ultimately irrelevent.

The article doesn't say anything about "rejecting" AC, it says assuming the negation. I don't know why you keep trying to think they're the same thing; theyr'e not. Therefore the quote should not lead anyone to believe something happens when "rejecting" AC. -20:23, May 8, 2005 (UTC)

If people believe the result is interesting then it is not neccessary to delete it (and someone must have found intersting if they bother to prove it). But I think the article needs to make absolutely clear that it cannot be used as an argument against rejecting AC. (against rejecting? perhaps I should just say for accepting)--Heathcliff 21:55, 7 May 2005 (UTC)

I think you should stop using the word "rejecting" altogether. One uses axioms which then become true. And that's all. You can assume an axiom or its negation. If you do neither, then the "axiom" you didn't assume isn't an axiom, and has no relevance to your model. -Lethe | Talk 20:23, May 8, 2005 (UTC)

Okay, I guess I'm going to make one more argument for it and then not contribute to this discussion, since I believe we all hold the same opinion but just keep contradicting each other for some reason. The quote is poorly worded as is, since Heathcliff believes it could be misinterpreted. I agreed. So instead, I propose the following,

Those uncomfortable with the axiom of choice might be curious about the consequence of its negation, "There exists some collection of sets such that it is impossible to choose one member from each set." It has been shown that the axiom of choice is equivalent to the statement that for any two sets, the cardinality of one is less than or equal to the cardinality of the other. Therefore, if the axiom of choice is false, then there are sets S and T of incomparable size: Neither can be mapped in a one-to-one fashion onto a subset of the other. Any set theory in which either the axiom of choice or its negation is provable must accept either this counterintuitive result or the Banach-Tarski theorem. —Sean κ. ⇔ 23:49, 7 May 2005 (UTC)

I suggest we add something like to the end of what you suggest: It should be noted that there is no movement among mathematicians nor logicians to postulate the negation of the axiom of choice to set theory, but rather only small movements to either replace it with a weaker form or elimate it without replacing it.--Heathcliff 01:22, 8 May 2005 (UTC)

That should only be noted if it is true, whereas this article seems to suggest that there are indeed people who investigate systems containing ~AC. -Lethe | Talk 20:23, May 8, 2005 (UTC)

I think that the consequences of ~AC should definitely be covered. Yes, merely not taking AC is different from taking ~AC, and the distinction must be made clear. However, the only reason for not taking AC is to allow the possibility that ~AC is true. In any particular model of ZF, either AC is true or ~AC is true. Which type of model should we be interested in? If we were convinced that the kind in which ~AC is true are not what we're interested in, we would take AC as an axiom. Note also that from ZF we can prove that either the Banach-Tarski paradox holds or there exist incomparable sets. There is no possibility of avoiding all such strangeness by ducking the issue by taking neither AC nor ~AC. Josh Cherry 13:51, 8 May 2005 (UTC)

However, the only reason for not taking AC is to allow the possibility that ~AC is true. The reason a handful of logicians and mathemticians have for not taking AC is that they do not think that it is obviously true. Axioms must be so clearly an undeniably true that they can simply be assumed without proof. Axioms are not created by pairing a statement with its negation and deciding which one we'd rather be true.--Heathcliff 03:04, 9 May 2005 (UTC)

Not so. As the article on axiom insists, an axiom need not be obviously true. Is the parallel postulate obviously true? How about alternative axioms that contradict it that are used in non-Euclidean geometry? But even in the "obviously true" way of thinking about things, implications of the negation of AC are relevant to whether we should take AC as an axiom. If a proposition (such as ~AC) leads to results that are considered absurd, this can be turned into an argument that it is obviously false. If it is obviously false, then its negation (such as AC) is obviously true. Josh Cherry 04:06, 9 May 2005 (UTC)

In that case we could then apply the same logic to AC. Show that it leads to a result at least as seemingly absurb as ~AC and conclude by the same stadards you have adopted above that AC should obviously be false. Thereby demonstrating that AC should be both true and false!--Heathcliff 13:30, 9 May 2005 (UTC)

And people do apply similar logic to AC. "Anti-choice" people say "look, AC leads to bizarre things like BT". "Pro-choice" people say "oh yeah, well ~AC leads to bizarre things like incomparable sets". In fact, if you think it's obviously true that any two sets are comparable, then you think that something equivalent to AC is obviously true. These are real arguments that real mathematicians have made, and should be covered in an article about AC. Josh Cherry 03:33, 10 May 2005 (UTC)

That should only be noted if it is true, whereas this article seems to suggest that there are indeed people who investigate systems containing ~AC. I believe that it is true which is why I suggested adding it. I certainly could be wrong and would welcome any examples of mathematicians or logicians who believe ~AC should be accept as an axiom. And if they exist the article should note this. In any event the article needs to make it clear the counter-intuitive result described only comes about if ~AC is taken as a axiom and that it does not result from simply rejecting AC. This may seem obvious to you, but it will not be obvious to some people reading the article because many people will not realize that there are actually 3 posibilities being discussed: accept AC, accept ~AC, reject both. In this very discussion Sean has already made the mistake of equating the lost two, and I think that will be very common if the artcle is not made clearer. It's not enough for the entry to be accurate it must also be clear.--Heathcliff 03:04, 9 May 2005 (UTC)

Yes, I agree that it must be clear in that paragraph that what we're discussing is the ramifications of the negation of AC, not the "rejection" of AC. The paragraph currently states "on the other hand, the negation is bizarre". So the quote states explicitly that it's dealing with the negation. How much clearer to you want it to be? Do you want it to say what the concept of negation actually means? Will that help? Or maybe we can just wikify negation -Lethe | Talk 06:34, May 10, 2005 (UTC)

As I believe I've already said it is accurate; it is simply not clear. As I believe I have also already said, the reason it is not clear is that many people will not understand that rejection and negation are two different things (as we've seen in this discussion). Another thing that makes it unclear (and what made it unclear to me) is that I still don't undertand the purpose of it. So while I know that rejection and negation are two different things since rejection is as far as I know the only one of the two anyone has ever proposed I wondered when I read it if the author of that section might himself have confused the two concepts since to me considering the effect of an axiom no one wants seems pointless. Yes I think explaining what negation means will help. I also think epxlicitly stating that it is not the same is rejection and that simply rejecting AC does not lead to the result given would help as well. I'll make the changes myself soon unless someone beats me to it. I wound still prefer that the article also point out that no one actually wants to postulate ~AC (unless of course I'm incorrect and someone does.) But I'll hold off on adding that since there seems to be disagreement over it.--Heathcliff 23:03, 11 May 2005 (UTC)

I guess this is getting kind of repetitive. I'll just state again: yes, you want clarity, not just accuracy. But the article explicitly states that it's talking about assuming the negation of AC. That should be crystal clear. There is no reason to think it means anything other than the negation of AC.

Also, about wanting ~AC to be true, that's not how axioms work. Axioms are things from which theorems follow, not things that are true or false. -Lethe | Talk 00:45, May 12, 2005 (UTC)

I agree, as my comments above would suggest. However, I think that the section on consequences of ~AC is broken. It lists, for example, There is a model of ZF in which there is a vector space with no basis. That's not a consequence of ~AC, it's a fact, right (assuming consistency of ZF)? The connection is that in such models ~AC is true. Josh Cherry 04:04, 12 May 2005 (UTC)

The existence of a vector space with no basis is: 1. undecidable in ZF. 2. False in ZFC. 3. True in ZF~C. The statement is definitely a consequence of ~AC, and not just a true statement following from the consistency of ZF, right? -Lethe | Talk 06:08, May 12, 2005 (UTC)

My point is that then there is a vector space with no basis should be given as a consequence of ~AC. There is a model of ZF in which there is a vector space with no basis is a fact (assuming consistency). Whatever we do concerning AC and ~AC, the fact remains that ZF, which contains neither, has models with basis-less vector spaces and models without them. Josh Cherry 13:17, 12 May 2005 (UTC)

OK, I see the distinction now. thanks Josh and Sean. -Lethe | Talk 18:21, May 12, 2005 (UTC)

It may be helpful to point out that (I assume), you're talking about model theory, where one constructs some definition of a "set" and "set inclusion" such that there is a vector space with no basis. I don't know if that's clear. —Sean κ. ⇔ 16:23, 12 May 2005 (UTC)

There is no reason to think it means anything other than the negation of AC. It is not clear or it would have not already have been misunderstood. I intend to change it at some point unless there is an objection.

Axioms are things from which theorems follow, not things that are true or false I think we still use the words true and false even if they do not reflect any form of ultimate or absolute truth. Otherwise instead of saying "Statement P is true given ZF" We would say "State P follows from ZF but isn't really true since ZF isn't really true). But I see no point in arguing semantics. The real question I'm getting at is, are there any logicians and mathematicians who want to use ~AC as an axiom? Obviously someone must have taken it as an axiom and investigate the result or we wouldn't have the result quoted on the page, but in my mind that doesn't tell us if there really is any interest in it as an axiom.--Heathcliff 12:41, 12 May 2005 (UTC)

Well it sounds to me like you have in mind to remove or at least weaken the discussion about the consequences of assuming ~AC as an axiom, based on a mistaken notion about how axioms have to be somehow "obviously true". I do have objections to that. -Lethe | Talk 18:21, May 12, 2005 (UTC)

As I've already said, I'm not going to take anything out of the aticle. I'm just going to explicitely state that postulating the the negation of AC is not the same as rejecting AC and that rejecting AC does not lead to the result sited. You agree that these facts are true so adding them will not weaken the discussion. As for axioms being obviously true: I'll retract that statement since you are clearly correct that anything can be accepted as an axiom if it suits the individual though I'm not sure why you keep coming back to this issue since it has no baring on what I'm proposing to add. I'd hoped that my pointing out that the real question I have is not, is AC or ~AC obviously true, but are there any logicians and mathematicians who want to use ~AC as an axiom? But as I've already said I'm holding off on adding anything about that since there are unresolved objections.--Heathcliff 22:13, 12 May 2005 (UTC)

Here's what I added to the artcile:

(It should be noted by those unfamiliar with formal logic that accepting the negation of the axiom of choice is not the same as rejecting the axiom of choice and that simply adopting the Zermelo-Fraenkel axioms without the axiom of choice will not lead to any of the following results.)

I had wanted to add something at the end like, "However, understanding some of the results from ~AC may provide insight into the debate over AC" or something like that, but I couldn't get the wording right (nothing I wrote really sounded entirely accurate nor reflected completely what I was trying to get at). It also seemed a little unencyclopedic (is that a word?) so I gave up on it.--Heathcliff 22:24, 12 May 2005 (UTC)

P or ~P

All this talk about AC and ~AC has got me to thinking about the statement (P or ~P) which is taken to be true regardless of what P is. But suppose we are using ZF (w/o AC) and P = AC. Then is it really fair to say that AC or ~AC is true. Obviously, I'm not the first person to wonder about this and that's why there are people who reject th law of excluded middle, but it's the first time I've ever thought, of it and it is interesting. Any thoughts?--Heathcliff 03:40, 9 May 2005 (UTC)

Yep, in standard logic "AC or ~AC" is a tautology, and always provable. I don't find this troublesome. Either there exists a choice function for every set of nonempty sets, or there doesn't. Josh Cherry 04:12, 9 May 2005 (UTC)

Which statement is more interesting, "For any P, P or ~P" or "There is P st. P or ~P, but neither is provable." I find the first intuitive, since the law of the excluded middle is logically equivalent to proof by contradiction, which I couldn't imagine living without. Does your comment concern the second or first statement with P=AC? —Sean κ. ⇔ 05:18, 9 May 2005 (UTC)

Yep, in standard logic "AC or ~AC" is a tautology, and always provable. I don't find this troublesome. Either there exists a choice function for every set of nonempty sets, or there doesn't. I'm not sure I agree with this last statement in ZF. Since we cannot prove either statement to be true or false. I also believe that the statement you have made (P is True) or (~P is True) is stonger that (P or ~P) is True. As for P or ~P being a tautology in standard logic, well obviously. I was merely noting that it is not as obviously correct to me to accept it as a tautology as it once was. In other words perhaps standard logic is "wrong". As for Sean's question, I'm afraid I'm not sure what you're asking could you ask it a different way.--Heathcliff 13:28, 9 May 2005 (UTC)

Feel free to become an intuitionist and reject the law of excluded middle and much of the type of reasoning that most of us take for granted. Josh Cherry 13:21, 9 May 2005 (UTC)

I think you answered my question by showing us you're an intuitionist ;). Which is fine, just remember you're giving up proof by contradiction. As far as how this discussion applies to the AoC, I would claim (without evidence) that ZF always assumes the axioms of standard propositional logic, which includes the law of the excluded middle. (So if you want to talk about ZF without the LoEM, you should probably make note of it, ZF\LoEM er... maybe not). —Sean κ. ⇔ 17:44, 9 May 2005 (UTC)

I can't tell if this is directed at me or not. If it is. I am not an intuitionist and I'm not sure why you assume that I am. I hope that intuitionist are not the only people who ponder the validity of their own basic assumptions. If you were addressing Josh, I'm sorry to have butted in.--Heathcliff 23:08, 11 May 2005 (UTC)

It's funny this discussion came up. At the AMS convention in Atlanta, Paul Cohen and some other were giving a panel on the CH, and I think someone in the audience asked, "What do you think of the following proof: case 1: CH holds, blah, blah, blah; case 2: ~CH holds, blah, blah, blah; in each case, you prove X. Is X true?" I'm not sure what the response was, but the law of excluded middle seems to generate some emotional reactions. The way I understand it, those who work in logic and set theory often take an attitude akin to how we view the group axioms or the parallel postulate. Is the parallel postulate "true"? The question has no meaning. Sometimes, the mathematics that comes about from assuming it is more useful, other times from assuming its negation. Sometimes, you just want to see what happens in neutral geometry. Same thing here. MOST of the time, mathematics will need AC, or some weaker version of it to get by. You're discarding a lot by not availing yourself of it. On the other hand, it's possible to see how much is provable from ZF alone. And it's also possible to see what happens by assuming ~AC or ~(weaker versions of AC). We can do all this without getting caught up in trying to figure out "whether it's really true or not". This is one thing I didn't really "get" at that panel discussion in January...they really seemed to figure that CH or ~CH was some empirical question that we could "figure out" if we just thought about it long enough or something. Or maybe it was all just way over my head. In any case, set theorists and logicians aren't the ONLY people who think about AC and ~AC and various weakened forms of it...people interested in the foundations of analysis (real number system, TVS, functional analysis, descriptive set theory, construction of certain types of real functions) have to deal with these issues directly. Revolver 18:37, 11 May 2005 (UTC)

And of course when we say AC implies such and such we really means ZFA implies such and such. Could someone tell me how much work has been done in significantly altering the other axioms of set theory?--Heathcliff 23:12, 11 May 2005 (UTC)

Take a gander at non-well-founded set theory for one example. Revolver 02:33, 12 May 2005 (UTC)

Thanks. That is interesting. Years ago I ask a couple of math professors why it was that a set could not contain itself. They seemed to think it was obvious and didn't really seem to understand why anyone would ask such a question. Neither ever mentioned non-well-founded set theory.--Heathcliff 12:48, 12 May 2005 (UTC)

Perhaps they were trying to prove "CH or ~CH" without the law of the excluded middle? —Sean κ. ⇔ 05:43, 12 May 2005 (UTC)

This is one thing I didn't really "get" at that panel discussion in January...they really seemed to figure that CH or ~CH was some empirical question that we could "figure out" if we just thought about it long enough or something. I suppose it dependeds a lot on how you think of a set. Or how do you define an undefined term? :) It seems to me that some people (perhaps because set is not defined) see it simply as something (anything) to which the axioms are applied. But others tend to think of sets in a more concrete sense. For them I think the question of is there a set whose cardinality is greater that aleph-0 and less than c is not quite as abstract a question. Perhaps I'm talking about constructionist, but I think it's subtler than that (though related). I just think that whether a person is a pure constructionist or not the way he thinks about sets has a big impact on the way he views most questions. I've always thought of sets as physical things. Things that you can actually build with your hands. Like putting numbers in a box. This does not make me a constructionist, but it means contemplating sets that cannot be constructed goes against the natural way I conceive of sets in my mind. And it slants my judgement. But then I think everyone has to deal with this in one way or another.--Heathcliff 13:08, 12 May 2005 (UTC)

new additions by 207.75.180.91

I like quite a lot what 207.75.180.91 has done with the article, and explaining the three possibilities. What do you think, User:Heathcliff? -Lethe | Talk 02:38, May 13, 2005 (UTC)

Progress, though I think it needs some work. It would be easier to do than explain. However, I'm lacking in some knowledge. In which places can There is a model of ZF¬C in which... be replaced with In all models of ZF¬C...? Josh Cherry 03:17, 13 May 2005 (UTC)

Ummm.. so in the past few days, the following statement has been changed three times:

There is a model of ZF in which the real numbers are a countable union of countable sets.

There is a model of ZF¬C in which the real numbers are a countable union of countable sets.

[In any model of ZF¬C,] the real numbers are a countable union of countable sets.

I'm willing to believe that the last statement is the most accurate, but it'd be nice to have a reference for all of these things, since the last two statements are saying very different things. —Sean κ. ⇔ 11:32, 13 May 2005 (UTC)

I have added the reference I originally found those claims in. 141.211.63.35 22:11, 13 May 2005 (UTC)

The new revisions by 207.75.180.91 make the artcile less clear rather than more. Particularly, "Note that any model of ZF¬C is also a model of ZF, so for each of the following statements, there exists a model of ZF in which that statement is true:" Why does the artcile not make the statement "Note that any model of ZFC is also a model of ZF." The intention seems to be to imply to those not familiar with how logic works that ZF~C is somehow more closely connected to ZF than ZFC is to ZF. My revision was perfectly clear and I only made it after many days of discussion and after it was determined that no one disagreed with the accuracy of what I wanted to add. I did not add additional things which were disputed and then someone shows up out of the blue and makes unecessary revisions that not only eliminate what I clarified but make it more confusing. And I find it interesting that the first editting that 207.75.180.91 does on wikipedia is this! It looks like 208.75.180.91 is someone's sock puppet and I hope he'll have enough respect for the rest of us to let us know who he really is.--Heathcliff 12:05, 13 May 2005 (UTC)

I am a graduate student at a well-known university, and I value my anonymity very, very highly, because anything I do here will be recorded forever; consequently it may later affect my professional reputation. Because of this I refuse to use a username. My IP changes because I use different computers in different labs and dialup at home. Despite all of this, I am not a sock puppet.

If you desire anonymity, you are actually better off with a user account: without an account, your IP address is recorded, which can be used to figure out where you are. For example, I can tell that you are at the university of michigan. If you edit from your office, it can be traced back to your person. Whereas if you have a user account, then only your account name is recorded. And if your account name gives no indication of your real name, then this is actually more anonymous. For example, Lethe is not my real name. Anyway, I don't care if you want to edit anonymously, I think you've addressed Heathcliff's concerns very well. -Lethe | Talk 23:19, May 13, 2005 (UTC)

Ah, but I have no computer in my office. This is why I use computer labs.  :-)

And, in response to Michael Hardy below, I think that perhaps I am overly protective of my professional reputation, because I don't have one yet. This is related to the fact that I do not have a high opinion of Wikipedia's advanced mathematics articles; there are so few people out there who really understand such things that it is unsurprising that many of Wikipedia's articles are of doubtful quality. This is less true of the more elementary articles, such as this one; yet the reason I made my edit of 22:34, 28 Apr 2005 was because I thought the article was POV. This is probably a bad example, since the axiom of choice tends to attract cranks; but also, Riemann integral was factually incorrect before my edits of 1 Aug and 2 Aug 2004. (Thanks, by the way, for cleaning up the typography.) Perhaps I am a pessimist, but I can imagine someone reading an article I had edited and imagining that I was responsible for the entire thing, when in fact someone else may have come along later and messed it up, as happened in Riemann integral when someone added the I_Q example. That's what I have in mind when I want to avoid having my edits attributed to me. I concede that I could do as Lethe suggests, but as of right now, I don't want to. I suppose that I want to be anonymous even among Wikipedians. 207.75.180.159 01:23, 14 May 2005 (UTC)

This is related to the fact that I do not have a high opinion of Wikipedia's advanced mathematics articles; there are so few people out there who really understand such things that it is unsurprising that many of Wikipedia's articles are of doubtful quality.

I'm not so sure it's a result of there are so few people out there who really understand such things as it is that many of the mathematical people here at wikipedia tend to "branch out" into subjects beyond their primary research or what they studied in-depth in graduate school, and so they are often venturing into areas in which they are say, knowledgeable, but not experts. If everyone contributed only to things they were 100% expert in, very little math articles would get written. The result is that mistakes do occur. At this point in time, I would not recommend that someone use wikipedia's math articles as a reference comparable to textbooks or standard references. Maybe this will change some time in the future, but that's not the way I use it now. I use it to get a good read on things, realising that if I want definites on particulars that I want to stake my life to, I should go somewhere else. I might say that a similar problem infests PlanetMath, I hardly trust anything I read there. Revolver 05:56, 14 May 2005 (UTC)

"More elementary" could be construed as subjects that zillions of undergraduates are required to take in assembly-line courses, such as calculus, differential equations, vector analysis, linear algebra, etc. Many of those are clumsily written. By that standard, this article is "more advanced". Perhaps most Wikipedia math articles are on a level comparable to this one, and are pretty good for things not intended as textbooks. Michael Hardy 01:41, 14 May 2005 (UTC)

If you dispute my changes, be bold! and change them yourself. I removed what you had written because I thought that your statement was better made earlier, in the "Independence of AC" section. I also thought that it was a worthwhile point and deserved expansion, which is why I tried to make the same point over a full paragraph. If you think I did a bad job, please fix it. Remember the motto of Bob the Builder: "Can we fix it?" "YES WE CAN!" 141.211.63.35 22:11, 13 May 2005 (UTC)

If Wikipedia is reaching the point where it affects people's professional reputations (or maybe it will reach that point some day?), then writing here with a username would seem to be a desirable thing since it would be another opportunity to enhance your reputation — you could add your edits to your CV. Saying you'd rather be anonymous under those circumstances almost makes it sound as if you expect to add bad stuff here. Do you refuse to let students in classes you teach fill out evaluations of the instructor because it can affect your professional reputation? Sure, you'll make a mistake from time to time, but fallibility is not incompetence and nobody thinks it is. Michael Hardy 22:24, 13 May 2005 (UTC)

Actually I see that the artcile does point out that "any model of ZFC is also a model of ZF", but it's two sections above the section it refers to in a paragraph about proving theroems in ZF without AC nor ~AC. I think part of the problem is just poor layout of the article. Also it seems that someone or some poeple find these two facts--any model of ZFC is also a model of ZF and any model of ZF~C is also a model of ZF--important. Perhaps both should be explained in a new section to make it clear what people are trying to get at. For instance, what is the relevance of these facts and how do they impact the discussion of AC? (I'm not asking for an answer here, I'm suggesting a topic for the article itself.)--12.74.51.1 13:37, 13 May 2005 (UTC)

models

I think I see why I was wrong in saying "in all models of ZF¬C"...e.g. there may be a model where countable choice holds, but not arbitrary choice, so in this model, statements which only require countable choice WILL be true, although choice doesn't hold in general. Is this the right reasoning? Can this be explained by someone who maybe understands it better? Revolver 04:43, 14 May 2005 (UTC)

To elaborate, it may be useful to point out which, if any, (I don't know for certain) statements DO hold in all models. E.g. the statement about sequences involving continuous functions seems only to involve countable choice, so it probably doesn't hold in all models, but the statement about vector spaces would seem to hold in all models (isn't it equivalent to choice??) Revolver 04:44, 14 May 2005 (UTC)

It's just that "in all models of ZF¬C" is just a more complicated way of saying "¬AC implies". It's a pretty strong statement. —Sean κ. ⇔ 04:53, 14 May 2005 (UTC)

I understand my mistake. Thinking aloud again, I really don't like the wording "there exists a model", even if it is now corrected from what I put which was incorrect. The reason as I think was mentioned above, is that it makes it sound like this statement requires ZF~C. We don't say, "there exists a model of ZFC in which every vector space has a basis" is a consequence of ZFC, we simply say "every vector space has a basis" is a consequence of ZFC. If something is really equivalent to choice, then it seems it would hold in any model, and we should say this. On the other hand, the statement about sequences involving continuous functions only holds in certain models (namely, if I think right, those in which countable choice does not hold); it is possible to imagine a model where countable choice holds but not arbitrary choice, in this case, the statement is certainly NOT a "consequence of ZF~C"!, to my mind, a "consequence of ZF~C" SHOULD be a statement that holds in ALL models. Whereas AC has weaker versions, ~AC has _stronger_ versions (the negations of the weaker versions of AC), and it's kind of misleading the way it's worded, it makes it sounds like the statements follow from ZF~C, when really there just a model where it doesn't hold. Revolver 05:01, 14 May 2005 (UTC)

Okay, it doesn't say "consequences" it say "results requiring", this is technically true, I suppose (the results imply ~AC), but not in the spirit of the first part "results requiring AC". I say this because the first part on results requiring AC allows weaker versions, (not all results there require AC). Moreover, all the results there ARE consequences of AC, since the weaker versions are implied by arbitrary choice. This doesn't happen with ~AC...not all the results are consequences of ~AC...the result about sequences is one example. But the way the two sections seem to mirror each other, the reader is left with this impression. Revolver 05:23, 14 May 2005 (UTC)

I think I finally hit what bothers me most. Not all the statements under "results requiring AC" actually require AC...some require weaker versions, so the heading here is misleading. It is true that all statements in the ~AC part, without the qualifier "there exists a model", DO require ~AC, in the sense that in every model where the statement holds, ~AC is true. Again, there seems to be a lack of consistency between the two sections. Revolver 05:23, 14 May 2005 (UTC)

I see I was the one who originally titled the section "results requiring AC", so my bad! Revolver 05:23, 14 May 2005 (UTC)

Yeah, the logical status of various propositions is unclear from the article. There are three cases for propositions P of interest: 1. $ZFC\vdash P$ and $ZF\lnot C$ leaves $P$ undecided. 2. $ZFC\vdash P$ and $ZF\lnot C\vdash \lnot P$ 3. $ZFC$ leaves $P$ undecided and $ZF\lnot C\vdash \lnot P$ (three additional cases result from substituting $\lnot P$ for $P$, but that doesn't really give us anything new). How about a two-column chart showing the status (provably true, provably false, or undecided) of each proposition under $ZFC$ and $ZF\lnot C$? Josh Cherry 01:56, 18 May 2005 (UTC)

Dewdney link

There is a quote in this article from A. K. Dewdney's "famous" April Fool's Day article in the Scientific American of April 1989, but after following up all the links and trying google, I could find no reference to the substance of the article. I may have missed something, but could someone who knows what this article said give a brief explanation of it either in this or the Dewdney article? Thanks --RMoloney 20:27, 17 May 2005 (UTC)

Results requiring ¬AC

This section is still problematic. It asserts that various propositions are true in some, but not necesarily all, models of ZF¬C. I presume it also means to assert that they are false in all models of ZFC. One problem is that this section does not bear the same relationship to ¬AC that "Results requiring AC" bears to AC. But there is a bigger problem than the title. It doesn't make sense to have a section like this in addition to "Results requiring AC". Listing a statement here is equivalent to lising its negation in "Results requiring AC". In fact there is outright redundancy here. For example, saying "There exists a model of ZF¬C in which there is a vector space with no basis" is redundant with listing "Every vector space has a basis" in "Results requiring AC". I'm tempted to remove this section altogether.

What would be useful is a list of important statements that are true in all models of ZF¬C but false in some models of ZF.Josh Cherry

That's just another way of saying "negations of equivalents of AC". I think the way to handle it is to add an "equivalents of AC" section (I was surprised to see that there's not one already), and let people figure out the negations on their own. --Trovatore 16:15, 23 July 2005 (UTC)

Not so. Only a proper subset of such statements, namely those that are false in all models of ZFC, are negations of equivalents of AC. Josh Cherry 16:24, 23 July 2005 (UTC)

You're right, my bad. Should be "propositions inconsistent with AC". Still, they're just the negations of consequences of AC; should be listed only if they're more naturally stated than their negations are. Which is more natural statement--"every set of reals is measurable" or "there is a nonmeasurable set"? --Trovatore 16:28, 23 July 2005 (UTC)

No, they're not (necessarily) inconsistent with AC, and not everything inconsitent with AC is one of them. An example of a statement in the set I have in mind, but not equivalent to ¬AC or inconsistent with AC, is the negation of the generalized continuum hypothesis. Anyway, I take it that you agree that 1. the "Results requiring ¬AC" should not exist in its current form and 2. stronger statements should be made somewhere about important propositions that are true in all models of ZF¬C but not in all models of ZF. Josh Cherry 16:44, 23 July 2005 (UTC)

Although we seem to have moved on, just for the record I'd better concede that you're right here. I got a little confused counting negations and existence-of-model quantifiers and such.--Trovatore 19:52, 23 July 2005 (UTC)

As I understand it, "there is a vector space with no basis" is is such a statement. If so, this is worth noting, but the article fails to do so. I'll reiterate my suggestion above for a table showing the status of various propositions (true/false/undecided/unknown) in ZF augmented with various choice-related axioms. Josh Cherry 16:07, 23 July 2005 (UTC)

---

One other thing to note here: I wonder if the person who started this section didn't have in mind something more like "Consequences of ¬AC". The problem is, of course, that there almost aren't any. Roughly speaking, ¬AC is consistent with any particular set being wellorderable, so any "local" consequence of AC that you like, is still consistent with ¬AC. --Trovatore 16:34, 23 July 2005 (UTC)

Someone may well have had this in mind. I can only interpret "Consequences of ¬AC" to mean "things that follow from ZF¬C but are undecided by ZF". That's exactly what I'm saying would be useful, unlike what's there. I wouldn't say that there "almost aren't any" such things. "There exists a set that cannot be well-ordered" is such a thing, and an important one.Josh Cherry

But (in some sense that I'm not making extremely precise here), it can't be any particular set.--Trovatore 17:51, 23 July 2005 (UTC)

It's some (well, many) particular set in any particular model. It may not be the same set in different models of ZF¬C, but the proposition "there exists..." remains a consequence of ZF¬C. And it's still important. After all, we consider it important that AC implies that specific infinite-dimensional vector spaces a basis, even though we can't say what that basis is (perhaps it's even different in different models of ZFC; I don't know). Josh Cherry 18:08, 23 July 2005 (UTC)

Yeah, but that's not the same thing. Given a particular vector space, we may not be able to say what a basis for it is, but we can say that it has a basis of von Neumann rank not higher than, I don't know, two more than the rank of the vector space itself. But given ZF+¬AC, there's no way to put a bound on the least rank of a non-wellorderable set. It could be way up above anything we're interested in. --Trovatore 18:20, 23 July 2005 (UTC)

I find it ironic that we're talking about whether consequences of the negation of AC are interesting because they don't clearly pick out the objects whose existence they assert. But that's neither here nor there. The article currently states that, e.g., the existence of a basis for every vector space is undecided by ZF but decided in the affirmative by ZFC. It leaves the reader in the dark about what happens in ZF¬C. Does ZF¬C decide that statement in the negative, or does it leave it undecided? In fact it decides it in the negative. A reader might want to know that. Do you object to informing the reader of that fact? Josh Cherry 18:34, 23 July 2005 (UTC)

That fact (assuming it's true) should be mentioned. If you're right about it, it makes the existence of a basis for every vector space an equivalent of AC, and could be stated there. OTOH I think your table idea is not bad; it might give some organization that's lacking at the moment. --Trovatore 19:07, 23 July 2005 (UTC)

"There exists an uncountable set whose cardinality is less than that of the power set of the natural numbers" is another,Josh Cherry

No, that's incorrect. It's consistent with ¬AC that there's a wellordering of the reals in order type ℵ₁ --Trovatore 17:51, 23 July 2005 (UTC)

Oops again--what I said is true but doesn't refute your point. But this does: It's consistent with ¬AC that there's a wellordering of the reals in order type ℵ₂ --Trovatore 17:54, 23 July 2005 (UTC)

Hm, I'm not sure about this. I had in mind the negation of the generalized continuum hypothesis, but what I wrote had to do with the negation of the plain old continuum hypothesis. So strike that and replace it with "the generalized continuum hypothesis is false". A "consequence of ¬AC" in the sense given above, yes? And important, right? Josh Cherry 18:12, 23 July 2005 (UTC)

Yeah, I suppose. Note again though that any bounded part of GCH is consistent with ¬AC. That is, for any ordinal α, there's a model of ZF+¬AC in which GCH holds up to α. --Trovatore 18:27, 23 July 2005 (UTC)

as is "there exists a vector space with no basis". Josh Cherry 16:52, 23 July 2005 (UTC)

Where's the flaw?

"Any union of countably many countable sets is itself countable" requires AoC. So I suppose there must be a flaw in this totally constructive proof. Where is it?

Let A₀, A₁, A₂ etc. be countably infinitely many countably infinite sets, and let B be their union. Let x_0;0, x_0;1, x_0;2 etc. be the elements of A₀; let x_1;0, x_1;1, x_1;2 etc. be the elements of A₁, and so far.

This is the flaw right here, no need to look at the rest of the proof. The fact that A₀ is countable does not by itself give you a particular enumeration x_0i of A₀, it just tells you there is one. So how do you pick such an enumeration for each A_n? If AC fails, each A_n has an enumeration, but there may not be a sequence of such enumerations, one for each A_n. --Trovatore 17:52, 16 August 2005 (UTC)

Oh, BTW -- you don't need full AC for this. Countable AC, written AC_ω, is enough; it says that, given countably many nonempty sets, you can choose one element from each. Therefore AD actually implies that the union of any countable collection of countable sets of reals is countable. --Trovatore 17:58, 16 August 2005 (UTC)

N×N is countable. (A simple bijection is between an ordered pair of natural numbers to the natural number formed interleaving digits from the members of the pair, e.g. (1234; 567) maps to 10253647. Let's call it n(a; b)=n.) Now simply map x_a;b to n(a; b) and that's a function from N to B, let's call it f(n)=x_a;b. Obviously this is not a bijection if the sets A_n aren't all disjoint. For example, if x_12;34=x_56;78=5, then f(1324)=f(5768)=5. Now let's skip duplicates. That is, let's take a set S⊆N such as that y is in S iff it is the smallest possible number for that peculiar value of f(n). For example, f(0), f(1), f(2) etc. are, in order, 8, 1, 4, 2, 2, 5, 8, 1, 7, 6, 1, 9…, S will be {0, 1, 2, 3, 5, 6, 8, 9, 11…}. S, being a subset of N, can be ascendingly ordered. In this case, the first element of S is 0, the second element of S is 1 and so on. Let's call the smallest element of S y₀, the next smallest y₁, and so on. In the example, y₀=0, y₁=1, y₄=5. If B is finite, it is countable by definition, Q.E.D.; else, reading the series f(0), f(1)… you will encounter new elements infinitely often, and the series y₀, y₁…, indicating when you encounter them, is countably infinite, and so is S. Therefore, given any n∈N there is an unique y_n. Now let's define the function SUCK_MINE(n)=f(y_n). It is obviously a bijection between N and B, Q.E.D. If the sets A_n were finitely many, or some of them were finite, B would be countable a fortiori.--Army1987 23:18, 15 August 2005 (UTC)

Finite sets

The theorem

AC_fin: For every finite set X (all of whose members are nonempty) there is a choice function.

is provable in ZF without AC. An informal proof is given in the article on AC: pick an element of the first element of X, then one from the next, etc. I claim that a formalisation of this proof (in particular: a formalisation of etc) would use the principle of mathematical induction, because this proof really says: whenever I have defined a partial (non-total) choice function, I can extend it to one more element.

Somewhat more formally, let AC_n be the statement:

AC_n: For every set X (all of whose members are nonempty) of size n there is a choice function.

Then one can prove that AC_n implies AC_n+1, and use mathematical induction to get AC_fin.

Now I would be curious to see a formal proof that does not (explicitly or implicitly) use the principle of induction.... (I am not claiming it does not exist, only that the "usual" proof uses induction).

-- Aleph4 14:38, 6 September 2005 (UTC)

Ah, my apologies. I thought that by your statement, you were saying something like,

Clearly we can do this: We start at the first box, choose an item; go to the second box, choose an item; and so on. There are only finitely many boxes, so eventually we stop. This gives us an explicit choice function. It takes the first box to the first element we chose, the second box to the second element, and so on. This process would formally use induction.

When instead you're saying something like:

Clearly we can do this: We start at the first box, choose an item; go to the second box, choose an item; and so on. There are only finitely many boxes, so eventually we stop. This gives us an explicit choice function. It takes the first box to the first element we chose, the second box to the second element, and so on. Formally, we can prove: for any n, a choice function exists for sets of size n, using induction.

I'll put it back —Sean κ. + 15:07, 6 September 2005 (UTC)

interpenetration

I haven't read the whole section yet, and perhaps I won't understand it all when I do, but it looks the problem with the axiom of choice may be summarized in the idea of 'interpenetration'.

The problem as I see it is, you can pick 1 object from infinitely many bins, but if some bins overlap (as some definitions or sets of defined numbers may sometimes overlap) you end up with many bins, not empty, but sharing '1' object - therefore non-empty, but, between them, they should only have a 'fractional' portion to give, if they were to each give equally to choosing function. So, if you have an 'infinite' set of bins, but a finite set of objects, for instance, you run into a problem, unless the 'portion' of the 'give' of each object is run down to a limit or something.

Other ways to solve the problem might involve 'recursive' or fractal type phenomenon, things that take into account holism and self-similarity, so that reproduction of the same thing again and again doesn't occur - reuse of functions with 'function libraries' and function calls in programs might be an example of this - I mean, something existing only once in a 'direct' form and having other instances occuring as 'links' or 'reflections' of the original. (Note Indra's Net for further info on self-similarity, and Fritjof Capra's "Tao of Physics" for other metaphors created to show similarity between intuitive and rational views of the universe.) Another example might be the idea of 'templates', where other instances are in fact 'transformations' of this template, the transformations not being inherent in the object, but for instance, stored separately, using the 'symmetries' of/between a given template and an object in order to 'compress information'. (something most useful with a complex 'image' type scenario, where only minor changes occur, so mapping the change as a function symmetrical to the original saves on 'raw data' at the expense of additional 'processing').

An example showing this effect might be the idea of decreasing probability of primes occuring as one progresses towards ever larger numbers. ie (2*3*5*7*11*13*17*19...*n)^-1 = probability of there being primes greater than n ... if you instead added 1/2 (probability of number being even) + 1/3 +1/5 +1/7 etc, you would end up with the overlapping probabilities added in, ie even numbers divisible by 3 included in 1/2, again when you consider 1/3 of numbers being divisible by 3 . I hope I'm making sense, I'm not a math wiz, just 'math interested' - I sort of decided to randomly drop a post, since I got caught up in trying to learn some math stuff in wikipedia, and thought there might be something I might actually have a relevant idea about. I'll cross link this post with my blog. Might remove later, depending.

m_G, http://www.livejournal.com/users/onfeynyuan_shen/78070.html

I don't think this has too much bearing on the axiom of choice (AoC). It is simply an axiom of set theory that you could assume, just like all the other ones. If you have two one-element-sets and you want to make a set containing both elements you will have to do that using some axiom. If there is no axiom (or sequence of axioms) which allows it, then it is impossible. In the same way the AoC allows the construction of some set, only more complicated. --MarSch 09:53, 23 October 2005 (UTC)

determinacy requiring AC

The following is listed as a result requiring AC:

Every infinite game $G(T,X)$ where $X$ is either open or closed is determined.

The claim is true, but there are a few problems with it:

1. I believe T here is supposed to be a tree, and X a set of branches of T. That's not clearly stated.
2. No one has yet bothered to give a definition of such games G(T,X); there's a to-be-written section Determinacy#Games played on trees that someone might fill in.
3. The claim is maybe a little misleading; if T is a tree on the naturals (or other wellordered set), no choice is required. On the other hand, in the completely general situation, you don't need infinite games at all— AC is equivalent to the determinacy of all length-2 games (which of course are trivially both open and closed). --Trovatore 20:30, 1 December 2005 (UTC)

Abbreviation "CC"

I don't believe I've ever encountered this abbreviation in the literature; in my experience it's always called AC_ω. Lethe, do you have a source? --Trovatore 01:22, 9 March 2006 (UTC)

Sure. The first book I picked up, Schechter's HAF uses CC. -lethe ^{talk +} 01:24, 9 March 2006 (UTC)

I've also got LM, HB and UF now. Do you feel that they shouldn't go in? It doesn't really matter much whether they go in or not, but it's hard to write things like ZF + DC + BP without some abbreviations, right? -lethe ^{talk +} 01:51, 9 March 2006 (UTC)

Now I'm reading a paper where the property "every set of reals has the Baire property" is represented with exactly that sentence, and not with an abbreviation. Yuck! But I guess BP is not standard. -lethe ^{talk +} 01:58, 9 March 2006 (UTC)

I guess I don't have any problem with using the abbreviation to simplify the exposition; I rewrote the sentence to try to avoid the implication that everyone in the field would recognize it. --Trovatore 02:01, 9 March 2006 (UTC)

BP = "Boolean Prime" ideal theorem? Very few of these are standard. In fact, CC may be AC_ω or AC^ω. — Arthur Rubin | (talk) 19:26, 8 May 2006 (UTC)

Solovay model

OK, first, again I'm not familiar with the abbreviation "BP" for the claim "every set of reals has the property of Baire" and I don't think this is standard; the text should avoid implying that it's a standard abbreviation.

More importantly, the text I removed claimed that from ZF+DC+"every set has the property of Baire" you can prove that every set is measurable. That's false. It's true that in the Solovay model, ZF+DC holds, and every set of reals is measurable and has the p.o.B., but it's not true that you can recover measurability from the p.o.B. In fact, ZF+DC+"every set has the p.o.B" is equiconsistent with ZFC, but ZF+DC+"every set is Lebesgue measurable" is stronger (I think the same consistency strength as ZFC+"there exists an inaccessible"). --Trovatore 01:29, 9 March 2006 (UTC)

As far as the initialism goes, BP is used in HAF, and now I'm reading the paper of Shelah 1984 where he proves that ZF + DC + BP is consistent to figure out where my mistake was (I'm a little confused at the moment). Obviously he uses the initialism BP as well. -lethe ^{talk +} 01:46, 9 March 2006 (UTC)

What is HAF? --Trovatore 01:49, 9 March 2006 (UTC)

Handbook of Analysis and Foundations. http://math.vanderbilt.edu/~schectex/ccc/index.html. I guess it's not very standard? -lethe ^{talk +} 01:52, 9 March 2006 (UTC)

Weakened AC and strengthened ¬AC

We should put those in order (in ZF), right? Something like: AC → DC → CC and AD → LM → ¬AC, would be nice. But maybe they don't make a total order... -lethe ^{talk +} 02:16, 9 March 2006 (UTC)

I don't think they are linearly ordered, no. For example, I've never heard that either of what you're calling LM and BP implies the other, and since it's a natural question, presumably someone has models of ZF+DC+LM+¬BP and ZF+DC+BP+¬LM. (LM is stronger than BP consistency-wise, but I don't think it outright implies it.) --Trovatore 02:20, 9 March 2006 (UTC)

Intuititively

Intuitively speaking, AC says that given a collection of bins, each containing at least one object, then exactly one object from each bin can be picked and gathered in another bin - even if there are infinitely many bins, and there is no "rule" for which object to pick from each.

This does not make sense to me, as far as picking and gathering. Is there supposed to be a postcondition met as well?

Full Decent 00:26, 29 March 2006 (UTC)

No postcondition. That's the whole bit. Maybe you could mention what postcondition you think is required, and we could figure out how to word it to make it clear. Right now, I'm not sure what you're looking for. -lethe ^{talk +} 06:55, 29 March 2006 (UTC)

Of shoes and socks ...

Russell's quote

The Axiom of Choice is necessary to select a set from an infinite number of socks, but not an infinite number of shoes

seems to imply that the following is equivalent to (AC):

Let S be a set, and ~ an equivalence relation on S such that every equivalence class has two elements. Then there is a a choice function as in (AC).

Now, as far as I can see, it's also entirely possible that the "sock selection theorem" is strictly weaker than (AC). I certainly don't see a proof that it implies (AC).

RandomP 02:35, 21 April 2006 (UTC)

The axiom of finite choice states that every collection of finite sets has a choice function. I expect the finite axiom of choice to be equivalent to a version where every collection of sets of order 2 has a choice function, though I'm not sure. I am sure, however, that the axiom of finite choice is strictly weaker than the proper axiom of choice, so you're right, being able choose one out of every pair of socks does not imply AC. By the way, don't confuse the axiom of finite choice with the finite axiom of choice, which states that for every finite collection of sets there is a choice function. The finite axiom of choice follows from ZF, so is not usually needed as an axiom at all. -lethe ^{talk +} 03:15, 21 April 2006 (UTC)

“a home page for the AXIOM OF CHOICE --- an introduction and links collection by Eric Schechter, Vanderbilt University” might be a useful external link.

Bertrand Russell’s quote is cited there as follows:

Bertrand Russell (more famous for his work in philosophy and political activism, but also an accomplished mathematician) once said:

To choose one sock from each of infinitely many pairs of socks requires the Axiom of Choice, but for [infinitely many pairs of] shoes, the Axiom is not needed. The idea is that the two socks in a pair are identical in appearance, and so we must make an arbitrary choice if we wish to choose one of them. For shoes, we can use an explicit algorithm --- e.g., "always choose the left shoe."

I excerpted the following text from page 572 of E. T. Bell’s classic book “Men of Mathematics”:

A set is characterized by three qualities: it contains all things to which a certain definite property (say redness, or volume, or taste) belongs; no thing not having this property belongs to the set; each thing in the set is recognizable as the same thing and as different from all other things in the set --- briefly, each thing in the set has a permanent recognizable individuality. The set itself is to be grasped as a whole.

Formally, by the generally accepted Zermelo-Fraenkel set theory’s axiom of extensionality --- that is, two sets are equal if and only if they have precisely the same members — a set is a collection of distinct objects without regard for their order as elements.

Thus, it appears to me that the very “definition” of a set as having distinct elements (that is, say, {0,0} = {0}) refutes Russell’s “pair-of-socks” argument as well as the “proof” of Banach-Tarski[-Hausdorff] paradox on axiom of choice.

* In Russel’s “pair-of-socks” example, if the “two” socks in each pair are indistinguishable from each other then there ought to be only “one” sock in each set (“pair”).

* In the “proof” of Banach-Tarski[-Hausdorff] paradox in Wikipedia, the decompositions S(a), S(a⁻¹), S(b) and S(b⁻¹) that “results in four copies of each original element” should be viewed as only one element because, say, {a,a,a,a} = {a}.

On the other hand, the physical “law of conservation of parity” does not hold in weak interactions so “left” could always be distinguished from “right” — which one could successfully communicate even to an alien in another planet anywhere in the universe (the Project Ozma problem) --- please read Martin Gardner’s “The Ambidextrous Universe”. [BenCawaling@Yahoo.com] BenCawaling 09:52, 12 June 2006 (UTC)

Yet another indecipherable comment from Ben. In regard the last, the confusion between "indistinguishable" and "identical" has no place anywhere in Mathematical logic. — Arthur Rubin | (talk) 17:12, 12 June 2006 (UTC)

But this confusion has a place in the Axiom of choice... Well, had, until just now... -Dan 18:52, 12 June 2006 (UTC)

product of cardinal numbers

I've removed

This statement can be reformulated as

An arbitrary product of non-zero cardinal numbers is non-zero. In this form it can be seen as telling us that cardinal arithmetic has no zero divisors even when we consider infinite products; infinite products of non-zero numbers are never zero.

I'm not sure which definition of "cardinal number" is intended here. Since the usual definition requires the axiom of choice, and we do not do more than hint at definitions for cardinal numbers without (AC), it is unlikely to be anything but confusing.

If this goes back, it should go back

• with a definition of cardinal numbers without AC
• explicitly pointing out that cardinal numbers without AC do not contain small ordinals
• in the "results requiring AC" section.

RandomP 02:06, 28 April 2006 (UTC)

You don't need AC to have cardinalities; you need it only to identify them with ordinals. Still, I didn't really love the "infinite products" version, as it was previously stated, because it was too easy to misread "infinite product" as "the multiplicative equivalent of an infinite series". There ought to be some wording that gets around this, but I don't care too much one way or the other, as it's a trivial variant of the "Cartesian product" version anyway. --Trovatore 02:14, 28 April 2006 (UTC)

I'm quite willing to accept "the set of sets in bijection with A that are of the lowest possible rank" as the cardinal number of A, but I wouldn't know how to choose a representative in each cardinal number, which is how the product of cardinality numbers is defined. That the product of sets representing nonzero cardinal numbers is nonempty is not true even with (AC) if we define "weird" cardinal numbers.

So if it really is a trivial variant, mind telling me how to do it?

RandomP 02:24, 28 April 2006 (UTC)

Well, first of all I think the coding you suggest is not really of the essence. We know what it means for two sets to have the same cardinality; cardinality itself, informally, is just the abstraction of that. The fact that we can code a complete invariant for the equivalence relation as a set, is nice, but it's not really what we mean by cardinality.

Still, going with the invariant you suggest, I think it's not hard to define arbitrary products of cardinalities. Just take the set of all cartesian products of all choices of representatives, and now close under equinumerosity at minimum rank. If AC holds, this will always be nonzero (provided all input cardinals are nonzero), and conversely, unless I've missed something. I don't see what you mean at all by the sentence about "weird cardinal numbers"; maybe you could give an example. --Trovatore 02:36, 28 April 2006 (UTC)

"close under equinumerosity at minimum rank" what, exactly? As far as I can tell, the set of cartesian products of representatives of the cardinal numbers we want to take the product of is empty, in general. Are you trying to look for a set equinumerous with an element of the empty set? :-)

if you want to say "oh, and if there is no representative, then say it's 0", that's a new rule. even for the product of 0 by 0, there is one choice of representatives (the empty set times the empty set), so the set of all cartesian products is {{}}, and we have a (unique) representative.

RandomP 02:50, 28 April 2006 (UTC)

OK, point taken. It still strikes me as an artifact of coding, though. What we really mean by cardinality is "what's left of a set when you forget everything about it that distinguishes it from a set equinumerous with it". Products are well-defined on those abstractions (which we're not bothering to code as sets), and with that notion of product, the claimed assertion is equivalent to AC. --Trovatore 03:05, 28 April 2006 (UTC)

The reason I added it is that it seems to provide the formulation in which it is clearest that AC is intuitively true. We don't want the product of nonzero quantities to be zero. If someone doesn't come up with a better formulation, and add that, I'm likely to revert. Some of the comments seem to be ignoring how cardinal arithmetic works, it seems to me. Gene Ward Smith 08:02, 28 April 2006 (UTC)

I'm probably missing something, but doesn't it defeat the whole point if you can't write the axiom of choice in the language of ZF? -Dan 15:40, 28 April 2006 (UTC)

It is written in the language of ZF. A cardinal number, you see, is a set, except two different sets might be the same cardinal number. But that doesn't matter, because all we care about is whether it's empty or not. So the current statement

An arbitrary product of non-zero cardinal numbers is non-zero.

really means

An arbitrary cartesian product of nonempty sets is nonempty.

Which is already in the article. Beyond the sheer unneeded redundancy of it, we need an ad-hoc definition of cardinal numbers and their product (which is unnecessarily complicated in requiring disjoint representatives). The sentence about zero divisors is confusing and unnecessary as well. "Cardinality" and "cardinal number" are used interchangeably without comment.

I strongly want this part just to go away. If you absolutely need to discuss cardinal arithmetic in ZF-C, or NBG with the appropriate changes, go write the appropriate article. This is not the place for it.

Anyone who really wants it to stay?

RandomP 16:17, 28 April 2006 (UTC)

I think that in order to stay, there will need to be a notable publication which chooses to use this as an axiom, as opposed to a result. This seems to me to be a result equivalent to the axiom of choice rather than an axiom. Why demand that ordinals and cardinals be defined before being able to state the axiom of choice? Elroch 16:32, 28 April 2006 (UTC)

Okay, I've moved it back here for now.

Removed section:

This statement can be reformulated as

An arbitrary product of non-zero cardinal numbers is non-zero.

Here a product of cardinal numbers is defined in terms of cardinal representatives which are disjoint sets, so that the cardinality of the product of cardinals can be taken as the cardinality of the Cartestian product. In this form it can be seen as telling us that cardinal arithmetic has no zero divisors even when we consider products of infinite quantities of cardinals; infinitely large products of non-zero numbers are never zero. Another version of the axiom of choice may be formulated in terms of cardinal arithmetic as the claim that cardinalities are totally ordered:

Given any two sets, either the sets are of the same size, or one is larger than the other.

Removed section with obvious fixes in case it goes back in

This statement can be reformulated as

An arbitrary product of non-zero cardinal numbers is non-zero.

Here a product of cardinal numbers is defined in terms of cardinal representatives which are disjoint sets, so that the cardinality of the product of cardinals can be taken as the cardinality of the Cartesian product. In this form it can be seen as telling us that cardinal arithmetic has no zero divisors even when we consider products of infinite families of cardinals; products of infinitely many non-zero numbers are never zero. Another version of the axiom of choice may be formulated in terms of cardinal arithmetic as the claim that cardinalities are totally ordered:

Given any two sets, either the sets are of the same size, or one is larger than the other.

Note that "products of infinitely many non-zero numbers are never zero" might confuse some poor real analysis student quite terribly. Fear not, $\prod _{n\in \mathbb {N} }1/2$ is not affected.

Apart from that, it still seems to me to be a weasely way of saying:

Define the cardinality of a set X to be the set X. We say that two cardinalities X and Y are "equal" if X and Y can be put into one-to-one correspondence. The only set of cardinality {} is {}, and we define 0:={}. Define a product of cardinalities to be the cartesian product of the cardinalities considered as sets. Now the statements

Products of arbitrary families of nonempty sets are nonempty

and

Products of arbitrary families of cardinalities none of which are "equal" to zero are not "equal" to zero

are indeed (trivially) correct. However, the seemingly obvious statement

Let {A_i} and {B_i} be two families of cardinalities indexed by I, such that for all i in I, A_i is "equal" to B_i. Then the products of the two families are "equal"

is incorrect.

RandomP 16:47, 28 April 2006 (UTC)

Now that it's removed, what is also removed is the strongest intuitive argument that AC is true. Was that really necessary? Gene Ward Smith 18:20, 28 April 2006 (UTC)

I'm sorry, what exactly is that strongest intuitive argument? That a cartesian product of an infinite family of nonempty sets would be empty is fairly unintuitive; contrived definitions of "cardinal number" don't change this statement, they just obfuscate it. Since the removed section contained mathematically incorrect statements and language that just didn't make sense, yes, it was necessary.

Of course, feel perfectly free to disagree with me; if you think you can write the section so it doesn't have those problems, just put it back. But before you do that, please verify it's actually mathematically consistent, defines any nonstandard terms it uses, and doesn't end up being overly long. I'd suggest just putting it up on the talk page first, since we already know there's disagreement about what should or should not go in, and there's no need to have an edit war when we can just have a discussion.

But I am going to be bold and remove nonsensical and confusing statements from the article.

At present, I'm not convinced that there is a way to construct cardinal numbers in ZF-C (i.e. as sets) and define a product of an infinite family of those without potentially inviting contradictions. Products of finite families are perfectly fine, of course.

If you try to avoid those problems by defining cardinality as cardinality of a specific set, then saying two cardinalities are "equal" if their underlying sets are in one-to-one correspondence, products of equals will no longer provably be equal. So products of infinite families are no longer well-defined.

Maybe I'm just being stupid and there's a simple way around all those problems - but I haven't seen it here, and I'm pretty sure I wouldn't remove it then.

RandomP 07:48, 29 April 2006 (UTC)

It's not a contrived definition of cardinals, it's just the plain ordinary one. And there cannot be a contradiction in ZF and yet not a contradiction in ZFC; that's nonsensical. Gene Ward Smith 19:55, 29 April 2006 (UTC)

I'm sorry, I meant to write "a contrived definition of cardinal numbers and their infinitary product". I still consider both the lowest-rank definition and the one as the proper class of sets in bijection to the given set as a bit weird; as I showed below, being able to define a consistent infinitary product of cardinalities is equivalent to AC. Assuming ~(AC) (i.e. ZF-C, which is what I wrote) and the existence of such a product thus leads to a contradiction, as I conjectured above.

The ordinary definition of cardinals, for me, is the unique smallest von Neumann ordinal in bijection with the given set; in other words, cardinal numbers are those ordinal numbers not in bijection with any smaller ones. That definition is totally unworkable without ZFC, and the other common definition involves proper classes, which I'm reluctant to consider "numbers".

Overall, the only part of the above that I agree with is that contradictions in ZF necessitate contradictions in ZFC. However, it is not appropriate to my post, as I was talking about ZF-C, not ZF.

RandomP 20:27, 29 April 2006 (UTC)

What was wrong with the bit about any two sets are the same size or one is larger than the other? That makes perfect sense without cardinal numbers. If you want we can say: for any two sets, one can be injected into the other. Or am I out to lunch, and that is actually weaker than AC? -Dan 16:56, 28 April 2006 (UTC)

Nothing wrong with that statement. It's listed under "results requiring AC" in the article as equivalent. I wouldn't mind adding "for any two sets, one can be injected into the other" in the statement section, though I'm not sure the non-ZF crowd agrees that that is still the axiom of choice - the proof of the equivalence of WOP, AC, HMP, ZL requires other axioms of ZF, of course.

RandomP 07:48, 29 April 2006 (UTC)

Nah, under results requiring AC is where it belongs. Now that I think about it, even if it is formally equivalent, it doesn't represent the same intuition. The other (remaining) statements have the force of picking one element out of each set. -Dan 15:54, 29 April 2006 (UTC)

Here is the strongest common sense argument I can see for leaving out the cardinality statement. There are some tribes that are known to only use a very small number of cardinalities, such as zero, one, two and "many". In the statements of the axiom of choice, we are essentially simplifying this to talk only about sets that have cardinalities zero or "one or many". This is enough to define the axiom of choice without the complex machinary that distinguishes between the different variants of "one or many". Elroch 19:36, 28 April 2006 (UTC)

I'm not quite sure why that is a "common sense" argument. It can be made precise quite easily. Let the Elroch cardinality |A| of a set A be "empty" if it is the empty set and "nonempty" otherwise. We'd like to define a functorial class "product" from the class of ordered families of Elroch cardinalities to the set of Elroch cardinalities such that

$\prod _{i\in I}|A_{i}|=|\prod _{i\in I}A_{i}|$

but being able to do so is equivalent to AC. Put another way, a "product of an infinite family of cardinal numbers" makes sense only with AC. That's certainly a good argument to the intuition of those who think products of infinite families are somehow natural things to consider. Note, however, that in the special case of countably many finite cardinalities (essentially the case where this definition might overlap with one for countable products of real numbers) which "should" have a finite cardinality as their product, all is still well. That's just the axiom of finite choice.

RandomP 16:59, 29 April 2006 (UTC)

"Put another way, a "product of an infinite family of cardinal numbers" makes sense only with AC." This makes no sense; if you cannot define products you cannot even formulate the axiom of choice in the usual way. If you can define products, then you can define products of cardinals, by using representive cardinals. This is how we would define, for instance, 2 x 3: we take the set of apples and pears, and the set of George, Mike and Alice, and we get (George, apple) ... (Alice, pear)}, six elements. There's nothing mysterious going on here, and we don't need ordinal numbers or the von Neumann assignment. Two sets have the same cardinality iff they are in one-ome relationship, and that's it. Gene Ward Smith 20:05, 29 April 2006 (UTC)

You're confusing products of sets (which can always be defined in ZF) with products of cardinalities.

RandomP 20:13, 29 April 2006 (UTC)

I'm doing no such thing. If you have two lists of sets, each in 1-1 relation with the other, then you can define a 1-1 relation between the products of the sets without AC. Gene Ward Smith 20:27, 29 April 2006 (UTC)

Proof? RandomP 20:29, 29 April 2006 (UTC)

The statement is incorrect. If it were true, the cartesian product of finite nonempty sets would always be nonempty, since every finite set is in 1-1 correspondence with a finite ordinal. However, without AC, that is untrue, in general.

For the general case, note that every nonempty set A can be put into 1-1 correspondence with a set containing {}. The cartesian product of a family of sets all containing {} is nonempty, containing the constant sequence ({}). Application of Gene Ward Smith's statement yields a choice function, in contradiction to "without AC".

RandomP 20:44, 29 April 2006 (UTC)

And if the representative sets for each cardinal are disjoint, as per my proposal, then they don't all contain {}. Gene Ward Smith 21:25, 29 April 2006 (UTC)

Okay, let me be more detailed here (and wrap around to the first column).

Assume that your statement is correct, and

If for all i, A_i can be put into bijection with B_i, then the products of the families are also in 1-1 correspondence

and furthermore, as I claim

every set can be put into 1-1 correspondence with a set containing {}

then, I claim, the axiom of choice is already true. To see this, let {A_i} be an arbitrary family of nonempty sets. Let {B_i} be a family of sets in bijection with the A_i through the maps f_i:A_i &rightarrow B_i Then the product of the B_i is nonempty, containing ({}). Thus the product of the A_i is also nonempty, containing (f^-1_i({})). This proves the axiom of choice.

So, one of the indented statements must fail without the axiom of choice. I think we can agree that it must be the first.

RandomP 21:36, 29 April 2006 (UTC)

Well done. Although, I'm not sure about the last line. Doesn't "every set can be put into 1-1 correspondence with a set containing {}" feel like a choice? Out of any set we can always pick the element which maps to {}. -Dan 23:42, 29 April 2006 (UTC)

It is a choice, but it's not an infinitude of choices happening simultaneously. Choosing n elements from a set you know to contain at least n elements is something you can do without (AC): proving that it has at least n elements means that, precisely.

RandomP 23:57, 29 April 2006 (UTC)

Sure. Then you went from the set A of sets A_i to the set B of sets B_i. Doesn't that seem like what you called an infinitude of choices happening simultaneously? Specifically, I think you are invoking the axiom scheme of replacement on the predicate "can be put into bijection with a set containing {}" to form B from A, right? Can I invoke replacement on a slight permutation of the same predicate to create, from a set of non-empty disjoint sets, a set containing one element from each? -Dan 02:01, 30 April 2006 (UTC)

Hmm, no, obviously, since that would prove the axiom of choice ;-)

The axiom schema of replacement article does claim the scheme holds even without the uniqueness condition, however I do not quite understand how the argument presented there works without (AC). Actually, I don't understand the argument there, full stop.

RandomP 19:37, 30 April 2006 (UTC)

Thanks for pointing me at that article. I hope I fixed it a bit. -Dan 15:40, 1 May 2006 (UTC)

Okay, just to be clear here, I am no longer sure I can prove "an infinitary multiplication of cardinal numbers can be defined if and only if (AC) holds". There's still something vaguely similar that works for models of ZF and defining a product of cardinalities outside of the model, but I'm too ignorant to even state that coherently :-)

RandomP 03:08, 2 May 2006 (UTC)

How was otheruses template offensive?

I don't get it.

RandomP 01:26, 1 May 2006 (UTC)

Yeah, I'm curious about that too. I remember I was pretty annoyed at a dab line at the top of the mathematics article pointing to mathematics (producer), but really, people looking for that article should have some way to find it. I would have liked to use {{otheruses}} at mathematics to help them out, but unfortunately no one seems to have a third article called "mathematics", which is what you're supposed to have before writing mathematics (disambiguation). --Trovatore 01:37, 1 May 2006 (UTC)

Oh, I see now that Michael hasn't removed the hat note, just reworded and reformatted. Can't say I care much either way. --Trovatore 01:42, 1 May 2006 (UTC)

It's offensive because it frequently gets used thoughtlessly with results that are stupid or worse. "For other uses of women, see...". "For other uses of personal lubricant, see...". "For other uses of the axiom of choice" is clearly NOT what was intended, but it's easy for a reader to misconstrue it that way, especially if the reader is not very familiar with Wikipedia customs. The "otheruses" templates are straightjackets that cannot be judiciously adapted to the occasion. They do not even allow the editor to choose intelligently between a capital and a lower-case initial letter in the name of the disambiguation page. Michael Hardy 01:52, 1 May 2006 (UTC)

But it didn't say "for other uses of the axiom of choice"; it just said "For other uses, see Axiom of choice (disambiguation)". Takes a little interpolation on the reader's part, but I don't think that's so terrible. Also the capital letter helps make the use-mention distinction. But as I say, it's not that I care much either way. --Trovatore 02:00, 1 May 2006 (UTC)

...besides, a Wikipedia article titled "zebra" is supposed to be about zebras, not about the WORD "zebra", since this is not supposed to be a dictionary. The term "other uses" is used because people have in mind other uses of the word. Michael Hardy 01:54, 1 May 2006 (UTC)

PS: Now I've solved the problem user:Trovatore wrote about, at mathematics. Michael Hardy 01:58, 1 May 2006 (UTC)

I disagree with the change. In the examples you gave, there would be a legitimate issue, but they don't actually exist like that! Usually the missing article together with the singular form is sufficient to prevent a misreading. This is true for "for other uses of axiom of choice". While I would agree with a change of the template, I don't think this article is the place to do it.

I also disagree with the way the change was done. Template:dablink is quite minimal, and will create a lot of work for people who want it to do something else.

I think Template:otheruses1 would be more appropriate, unless there is existing policy demanding dablink.

RandomP 02:10, 1 May 2006 (UTC)

Consequences of the Axiom of Choice

(Yes, the Jean Rubin mentioned there is my late mother.) My copies of the book are at work, but many of the questions about specific statements related to the axiom of choice are there. Perhaps I'll look them up and comment later. — Arthur Rubin | (talk) 09:01, 7 May 2006 (UTC)

Oh? I saw a Rubin referenced in one of my textbooks a couple months ago, and briefly entertained the idea that it might be you. But it wasn't an A. Rubin. Maybe it was your mother? -lethe ^{talk +} 09:35, 7 May 2006 (UTC)

Equivalents of the Axiom of Choice

I'm not in a position to state which forms are interesting, but, now that JRSpriggs has separated the statements that are stronger, equivalent, or weaker than the Axiom of Choice, I'll make an effort to ensure that that categorization is correct. A few minutes ago, I moved 3 forms from the "weaker" to the "equivalent" section. — Arthur Rubin | (talk) 23:22, 25 May 2006 (UTC)

Thanks. I was merely trying to reorganize the text in a more logical order. I do not know enough to re-evaluate the status of most of those theorems. JRSpriggs 02:50, 26 May 2006 (UTC)

category theory and the axiom of choice

The category theory equivalents of choice have been left in the section for weaker versions of choice. I believe that this isn't entirely accurate. Basically, a category theoretic version of choice (eg every category has a skeleton) may be equivalent to choice or weaker than choice depending on how large you allow your categories to get. I'm not sure though. It depends on whether you define your categories in terms of sets or not, it's a complicated issue. 03:15, 26 May 2006 (UTC)

There are "class" forms of the Axiom of Choice statable in ZF; for example, that there is a formula Φ such that

$(\forall x)((\exists y)(y\in x)\rightarrow (\exists !y)(\Phi (x,y)\land y\in x))$

"If x is non-empty then there is exactly one y ∈ x such that Φ(x,y)."

But I'm not sure we want to go there. — Arthur Rubin | (talk) 16:15, 26 May 2006 (UTC)

— Arthur Rubin | (talk) 16:15, 26 May 2006 (UTC)

I’m not sure we want to go there either, mostly because I don’t know the details very well, so I can’t go there (but wouldn't the class form of choice be stronger than standard set AC?). The point is, I have the impression that describing the categorical choices as weaker may be incorrect. With the previous header, we didn't have to worry about, equivalents and weaker were lumped together. Now that we've separated, we may have a mistake. -lethe ^{talk +} 21:28, 26 May 2006 (UTC)

I'm also not at all sure the way the article looks now is all that great. "Every small category has a skeleton" does imply the axiom of choice (because a skeleton for the category of an equivalence relation is exactly a choice of representatives); I believe the argument is similar for "if two small categories are weakly equivalent, they are equivalent".

Hate to suggest it, but maybe weasel words would be the best way out? Essentially replace "weaker" by "weaker or equivalent", or have a third section for those statements we haven't (yet) sorted?

RandomP 22:59, 26 May 2006 (UTC)

I see that JRSpriggs has recently addressed the issue of category theoretic "equivalents" of choice. I'm not crazy about the solution, mostly because it involves doubling all the category results. And I think these statements can also be arranged to be strictly weaker than choice as well, depending on how you define your category. The real problem is that I don't know this thorny issue well enough to simply state the answer.

But here's my suggestion. Create a new subheading (perhaps subordinate to either the equivalent, weaker, or stronger section), labeled category-theoretic equivalents of choice or some such, and put some text at the top of this section which parametrizes our ignorance. Say something like "Depending on the foundational constructions used to define the categories, the following results may be strictly stronger than, equivalent to, or strictly weaker than the standard axiom of choice for ZFC". Some day someone will come along who knows exactly how these results relate to choice, she can elaborate. -lethe ^{talk +} 04:27, 27 May 2006 (UTC)

Naming issues

Note: I'm going to refer to my mother's survey books as Equivalents, Equivalents II, and Consequences, to avoid retyping "of the Axiom of Choice" each time. If there are any other survey books on forms of the Axiom of Choice or related forms, I'd be happy to use those, as well.

I was wondering if there would be any objection to noting (in an HTML comment) how these forms are denoted in those books, so we (or, at least those of us who have those books) can "easily" see whether we've stated the forms correctly. — Arthur Rubin | (talk) 16:15, 26 May 2006 (UTC)

countable sets

Pardon for interupting, and since no one else has expressed doubt about this yet, I'll assume I'm somehow mistaken and ask for clarification: how does "every countable union of countable sets is countable" depend on AC? It seems to me that, if you have a priori an indexing of natural numbers on the sets and their respective elements, you have a ready-made injection to N x N. And N x N is clearly countable, and explicitly so, by defining, e.g., (i,j) |--> (i)(i-1)/2 + j (by arranging N x N in a 2-dimensional array, and moving down a diagonal until you hit the bottom, start at the top of the next diagonal, etc.) How was choice invoked in any of this?

While the fact that each set is countable means that each set has an indexing by N, it also has infinitely many different indexings by N. You can choose an indexing for finitely many of these sets without choice, but not for all of them at once without choice. The key here is that knowing a set is countable does not give you a privileged indexing. You have to choose one. For infinitely many sets, you have to choose infinitely many times. -lethe ^{talk +} 05:48, 30 May 2006 (UTC)

Anonymous forever: Lethe is correct. Classical first order logic only allows you to choose one thing at a time. By repetition, you can get any finite number of choices. But an infinite number of choices would require an infinite proof which would be invalid. The axiom of choice allows you to do the infinite number of choices in one step. By the way, your pairing function on N should be P(i,j) = ((i+j)(i+j+1)/2)+j, if you use zero-based numbering. JRSpriggs 06:02, 30 May 2006 (UTC)

Also, someone has asked this before. See above at #Where.27s_the_flaw.3F. So if my or JR's explanations don't please you, you can try Trovatore's above. I guess it's not an uncommon reaction. -lethe ^{talk +} 06:18, 30 May 2006 (UTC)

Is this worth commenting on in the article? I'm thinking of:

The reason that we are able to choose least elements from subsets of the natural numbers is the fact that the natural numbers are well-ordered ... it turns out that every set can be well-ordered if and only if the axiom of choice is true.

While this is correct, I'm wondering if we haven't given the impression here that collections of well-orderable (a fortiori countable) sets have choice functions (in ZF). -Dan 13:41, 30 May 2006 (UTC)

I'm not exactly sure what you have in mind for the article, but I'd like to object to your a fortiori above. Countability is not in any way related to well-orderability. For example, the integers are countable but not well-ordered, while omega_1 is well-ordered but not countable. N and R provide examples that are countable and well-ordered and neither countable nor well-ordered, so all combinations are possible, and there is no relation between the two. -lethe ^{talk +} 14:12, 30 May 2006 (UTC)

To add to the confusion — there are models in which the union of a well-ordered family of well-orderable sets is well-orderable, but the union of a countable number of countable sets is not necessarily countable. — Arthur Rubin | (talk) 18:11, 30 May 2006 (UTC)

Digression continued at /Ben — Arthur Rubin | (talk) 18:49, 16 June 2006 (UTC)

Is that dependent choice versus countable choice? -lethe ^{talk +} 20:53, 30 May 2006 (UTC)

The integers are countable and not well-ordered. But aren't they well-orderable, just like any other countable set? Do you pity the fool who reads the paragraph I quoted? (I just went back and added highlights) -Dan 19:19, 30 May 2006 (UTC)

I have revised the paragraph in question in some minor ways that may help. Please tell me what you think. -lethe ^{talk +} 20:37, 30 May 2006 (UTC)

It's better, certainly. I'm still not sure it quite explains the well-order issue. For example: the integers do not come "pre-equipped" with a well-order. But an arbitrary collection of sets of integers will have choice functions. On the other hand an arbitrary collection of arbitrary countable sets won't. This doesn't mean I have a better wording (or else I would just do it). -Dan 18:17, 12 June 2006 (UTC)