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Talk:Axiom of Choice (view source)
Revision as of 02:35, November 27, 2008
, 02:35, November 27, 2008→My edits: new section
<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)
== 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. -[[User:CSGuy|CSGuy]] 21:35, 26 November 2008 (EST)