Difference between revisions of "User talk:DanielB"
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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) | ::::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) | + | :::::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) |
Revision as of 02:45, June 15, 2008
It's not me avoiding the question; it is you who are to rephrase it in a more respectful manner. Karajou 23:51, 27 April 2008 (EDT)
Hadn't you better re-phrase...
Laddie, hadn't you better re-phrase this?
Contempt of the administration is one of the oldest blockable offenses in the book. And you've just shown it.--TerryHTalk 23:05, 6 June 2008 (EDT)
- Can you be arrested for showing contept of the president? No contempt for leadership is on of the hallmarks of freedom. DanielB 19:29, 14 June 2008 (EDT)
Quote
Love that Galilleo quote.CraigC 01:22, 14 June 2008 (EDT)
- Cheers I rather like it. DanielB 19:28, 14 June 2008 (EDT)
Diagonalization
First, thank you for the full proof of the naturals vs. reals case.
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).
As for the Diagonalization and God section, this is a valid philosophical argument and has citations to back it up.Foxtrot 22:14, 14 June 2008 (EDT)
- Okay I might leave the philisophical point, but where is the axiom of choice used in that proof? DanielB 22:17, 14 June 2008 (EDT)
- 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. Foxtrot 22:25, 14 June 2008 (EDT)
- 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. DanielB 22:30, 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. 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. Foxtrot 22:45, 14 June 2008 (EDT)