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How can one show that Dedekind's cuts are equivalent to Cantor's definition of an irrational number based on Cauchy sequences? The problem hints that it might be useful to use the lemma that, for any positive integers (p,q), there exists a positive integer M, such that Mq > p.

In the way I tried to solve it, I started with the def. of Dedekind's cut: a < y < b, then said a = p/q, and b = M. Then I got y - p/q > 0, and y - M < 0. Then, one can choose p/q and y to be such that the module of the inequalities above will be < epsilon, and I guess this would satisfy a Cauchy's sequence requirement. However, I am definitely not very happy with this solution, especially with the last steps.

Any hints would be useful.

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I am not so sure about that. The problem I am trying to solve gives a hint that was not employed in any of the cited discussions, and it might turn to be a different proof, which could be very interesting. – Raphael R. Mar 22 '11 at 9:27

It suffices to show, that Cantor's real numbers are Dedekind-complete, that is every nonempty subset, which is bounded from above has a supremum. Then by axioms of the real numbers we have that Cantor's real numbers are equivalent to Dedekind's real numbers.

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Ok. That's an elegant proof. Thank you. However, it does not use the lemma given in the problem, and I am very curious as how that lemma would be used to show this relationship between Dedekind cuts and Cauchy sequences, since I am not happy with the way I tried to use the lemma to solve this. – Raphael R. Mar 22 '11 at 9:31
@Raphael: It does make use of the lemma, as it is needed to prove the existence of the supremum. – George Lowther Apr 20 '11 at 13:48

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