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Talk:Axiom of Choice (view source)
Revision as of 19:27, November 27, 2008
, 19:27, November 27, 2008no edit summary
*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)
*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)
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). [[User:AndyJM|AndyJM]] 14:27, 27 November 2008 (EST)