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→Diagonalization
::::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)
::::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)
:::::Got edit-blocked twice here. We're both replying quickly :-) Basically there is no natural way of enumerating [0,1] like with the rational numbers. Even if you assume the reals are countable (as we are doing in the proof), you still have countably many decimal numbers that need to be enumerated by the natural numbers. There is a natural way to do this for the integers--alternate 0, 1, -1, 2, etc.--and there's also a way for the rational numbers using the fact that each one is natural number divided by natural number. However with the reals, not even the [[Archimedean property]] is enough to help you. You can't say: First decimal is 0, next decimal number is what? The least real number we haven't enumerated yet? But there doesn't have to be a least one, even if the reals are countable (there is not even a least rational number bigger than zero, which is why you have to do this trickery with the enumeration of the rationals). The enumeration of the decimal numbers is where you need choice. Admittedly, you don't need full choice since the set is assumedly countable, so I've fixed the entry to reflect that it's countable choice you use. [[User:Foxtrot|Foxtrot]] 22:45, 14 June 2008 (EDT)