Talk:Axiom of Choice

Not really controversial anymore

Any proof which uses the axiom of choice can be transformed into a proof that doesn't. Granted, it will be a somewhat more complicated proof, but it always works, and that's a fact. That is the reason that AC is much less controversial these days than it was, in the early 1900s.

There is a complete explanation of the process and the proof that it's reliable here.

Also, the profoundly intuitive trichotomy is equivalent to AC, so be careful what you call controversial. BenjB 20:29, 27 January 2008 (EST)

Actually, the Axiom of Choice has been proven independent of ZF, so there is no such transformation of a proof. Otherwise, "prove" AC as follows:

---

1. Axiom of Choice | Reason: Axiom of Choice

Then transform it to not need AC. Result: AC proven in ZF,so ZFC=ZF. But AC proved independent of ZF. Therefore, no such transformation exists. QED SamSamson 21:50, 8 June 2008 (EDT)

This axiom actually can prove things which are (provably) unprovable without it; an example is Zorn's lemma. (Funnily enough, although Zorn is a guy's name, it's also a German word meaning "rage" ;-) --AMackenzie 14:29, 2 August 2008 (EDT)

The Axiom of Choice can also be used to "prove" absurdities, as explained by the entry here.--Aschlafly 14:52, 2 August 2008 (EDT)

It is a wonderful thing in mathematics when an absurdity is proven - it expands the consciousness. Either the "absurdity" isn't as absurd as once thought (the Earth is a ball, not a plain; there are just as many whole numbers as there are fractions), or some axiom needs to be rethought, or you've made a mistake. Whatever, you can learn from it.--AMackenzie 17:45, 2 August 2008 (EDT)

Banach-Tarski isn't absurd; it's just counterintuitive. -CSGuy 22:21, 2 August 2008 (EDT)

It is absurd. Indeed, the point of the "proof" was to show how absurd it is.--Aschlafly 23:14, 2 August 2008 (EDT)

It's absurd - no, it's not - yes, it is... As "being absurd" in this context is just a personal opinion, the phrase counter-intuitive seems to fit better. DiEb 10:28, 3 August 2008 (EDT)

Willard quote

I "prettied up" the quote from General Topology, but in context, Willard might not be the best person to quote after saying "many mathematicians reject the Axiom of Choice". Immediately before describing the axiom, he writes:

The following axiom is assumed by most mathematicians when they need it, to the unremitting disgust of a few.

And after the text quoted in the article:

The status of the axiom of choice bears some resemblance to that of the continuum hypothesis, with some differences. It, too, is known to be independant of the other axioms of set theory (that is, it or its negation can be consistently assumed), but it enjoys the status of an accepted part of the theory of sets in the minds of most modern mathematicians; that is, the intuition of almost all mathematicians now is that the axiom of choice should be assumed when needed without hestitation. Moreover, it is usually clearer that, where it is used, it is needed, so that its presence does not usually provoke the same frenzy of attempt to eliminate it.

Wandering 23:57, 2 August 2008 (EDT)

My edits

• I was unable to find references saying that the existence of a basis for every vector space and the existence of subsets of the real line without well-defined Lebesgue measure are equivalent to the Axiom of Choice. I did find references saying that the Axiom of Choice implies these, but that's not the same thing. -CSGuy 21:35, 26 November 2008 (EST)

The Axiom of Choice is not equivalent to the existence of non-Lebesgue measurable subsets of the real line and so I have removed this claim from the article. Intuitively one would not expect these to be equivalent. The existence of a non-Lebesgue measurable subset of the real line only says something about cardinals up to and including the continuum. However the Axiom of Choice makes a statement about all cardinals (in fact it is typically invoked when describing cardinals). AndyJM 14:27, 27 November 2008 (EST)

I got curious and actually looked into this. The Axiom of Choice does imply the existence of non-Lebesgue measurable subsets of the real line. However, the existence of non-Lebesgue measurable subsets of the real line doesn't imply the Axiom of Choice. I found a paper called "The Hahn-Banach theorem implies the existence of a non-Lebesgue measurable set" and will quote the abstract for those who don't want to download a 500+kb document:

“

Abstract. In this paper we show that the Axioms of Zermelo-Fraenkel set theory together with the Hahn-Banach theorem imply the existence of a non-Lebesgue measurable set. Our construction does not make any use of the Axiom of Choice.

”

So the other direction of the claim is apparently wrong, and AndyJM was right to remove it. He was blocked for "removing information from articles" by the same user who also re-inserted the apparently wrong claim. This is exactly why I oppose this "Block first" approach: How can we improve this encyclopedia when we can get blocked and banned by whoever thinks we're wrong? I suggest an unblock. --AlanS 18:38, 27 November 2008 (EST)

Alan, the Axiom of Choice yields many strange behaviors in mathematics. That's why it is still questioned and debated. If the statement is not equivalent to AC (which I will have to look deeper into to believe), then the correct behavior is not to remove the statement altogether but to change it to an implication. Edits that whitewash the Axiom of Choice's questionable role in mathematics are the sort of behavior you see at Wikipedia, not here (see Conservapedia:Critical Thinking in Math for a guide to our positions). I am reinstating the fact, but will tentatively change it to just a consequence of AC instead of an equivalence. -Foxtrot 18:50, 27 November 2008 (EST)

I have no problems with stating that it's a consequence of AC, but that doesn't change (1) that you first re-inserted the apparently wrong information, (2) that you claimed on your talk page that "[i]t is well known that you need AC to create a non-Lebesgue measurable set of real numbers" and (3) that you blocked two users over this content dispute, one of them for five years. --AlanS 18:57, 27 November 2008 (EST)

I reinserted valuable information that was lost by an editor intent to misrepresent the Axiom. If you persist in misrepresnting my actions in this matter, you may find yourself as the third Musketeer. -Foxtrot 19:01, 27 November 2008 (EST)