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→Diagonalization
:::::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|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)
:::::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|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)
::::::Foxtrot you seem to be missing the essential part of the proof in that it is a proof by contaradiction. If a set is countable then it can be well ordered without using the axiom of choice. I have assumed it is countable and then created an arbritrary ordering assuming I could. I then found an number that was not in my well ordered list. Why? Because they are uncountable and you can't do this, hence a contradiction occurs and so we have proved that the numbers between [0,1] are uncountable. If you use the axiom of choice then my number would be on the list somewhere and the proof fails. [[User:DanielB|DanielB]] 22:52, 14 June 2008 (EDT)