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| | ==Diagonalization== | | ==Diagonalization== |
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| − | First, thank you for the full proof of the naturals vs. reals case.
| + | Moved to [[Talk:Diagonalization]] |
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| − | Diagonalization is classically applied to show that the reals are bigger than the natural numbers, but the diagonalization argument also works to take any infinite set and construct a set of larger cardinality (namely the power set).
| + | == spelling == |
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| − | As for the Diagonalization and God section, this is a valid philosophical argument and has citations to back it up.[[User:Foxtrot|Foxtrot]] 22:14, 14 June 2008 (EDT)
| + | Daniel: |
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| − | :Okay I might leave the philisophical point, but where is the axiom of choice used in that proof? [[User:DanielB|DanielB]] 22:17, 14 June 2008 (EDT) | + | There's really no polite way to put this, but could you please use a spell checker for your contributions? By the way, I really do respect many of your contributions. [[User:SamHB|SamHB]] 22:41, 21 June 2008 (EDT) |
| | + | :SamHB there probably is a polite way of saying this, see the button at the top that says edit, click it. [[User:DanielB|DanielB]] 22:54, 21 June 2008 (EDT) |
| | + | ::He was politely trying to say, "improve your spelling or else you will be mocked eternally" o.O [[User:NathanG|Nate]] <sup>[[User_talk:NathanG|my opinion matters?]]</sup> 15:00, 5 October 2008 (EDT) |
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| − | ::You had to enumerate all the decimal representations of the reals (i.e. well-order them by the natural numbers). There is no natural way to do this without the axiom of choice. [[User:Foxtrot|Foxtrot]] 22:25, 14 June 2008 (EDT)
| + | ==Probation== |
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| − | :::And that is why the contradiction occurs, because there is no way of ordering the without the axiom of choice, you assume you can and watch it fail. [[User:DanielB|DanielB]] 22:30, 14 June 2008 (EDT)
| + | Daniel, be very careful what advice you give other contributors, especially newbies like [[User:Lemonpeel]]. See his talk for my corrections. |
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| − | ::::Okay I am starting to see the problem here, the well-order theorem is dependent on the axiom of choice. However countable sets are well ordered. [[User:DanielB|DanielB]] 22:37, 14 June 2008 (EDT)
| + | You really would be better off '''checking''' your notions with me or the project director before spouting off like that. Consider yourself on probation. --[[User:Ed Poor|Ed Poor]] <sup>[[User talk:Ed Poor|Talk]]</sup> 20:46, 19 August 2008 (EDT) |