Difference between revisions of "Talk:Axiom of Choice"

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(Not really controversial anymore)
(My edits: new section)
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<blockquote>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.</blockquote>
 
<blockquote>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.</blockquote>
 
[[User:Wandering|Wandering]] 23:57, 2 August 2008 (EDT)
 
[[User:Wandering|Wandering]] 23:57, 2 August 2008 (EDT)
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== My edits ==
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*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. -[[User:CSGuy|CSGuy]] 21:35, 26 November 2008 (EST)

Revision as of 02:35, November 27, 2008

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)