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| Subject: | Re: Cantor's diagonal method |
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| Date: | Mon 02.11.98 11:49 |
| From: | Alex Bogomolny |
Dear Mike:
> In Cantor's Diagonal Proof he assumes a complete list of
> reals between 0 and 1 and then constructs a number that
> isn't listed. He then concludes something from this. :-(
> I think the number isn't listed because THE LIST IS NOT
> COMPLETE. (and never can be)
As you rightly observe, Cantor assumes that the set of real numbers exists AND CAN be listed in a sequence. His diagonal method derives a contradiction from this assumption. Your remark above that "THE LIST IS NOT
COMPLETE. (and never can be)" reminds of the following anecdote from Littlewood's Miscellany:
Teacher: "Assume x is the number of sheep in a flock." After the proof we know that the list is not complete and can't be - but not before. Why do you deny Nator's right to assume that the set of reals can be listed? He does not claim that it can. He only assumes this as a possibility. The proof consists in showing that listing the reals is actually impossible. Why do you think it is easy to find the smallest number listed? Not every list has a smallest number. This is one thing. The second is that you, as Cantor before, list the reals. Why can you do that while denying Cantor his rights?
Your proof only shows that, if and when real numbers are listed, the sequence has no smallest number. But this is obvious to start with: numbers 1/2, 1/3, 1/4, ... are included somewhere on your list, right? The smallest number > 0 on the list would be less than any of these. Which is clearly impossible.
> Is Cantor's Diagonal Proof sound?
Absolutely - in so far as you accept the notion of the set of all real numbers. In the beginning of the century, there was a group of mathematicians (Intuitionists) who did not accept the totality of all reals as a given. They never had any serious influence on the development of mathematics.
All the best,
Student: "But Teacher, assume that x is not the number of sheep in the flock."
> isn't complete by finding the smallest number > 0 listed.
> If this number is divided by 10 it will still be between 0
> and 1 but clearly not on the list since it is now the
> smallest number.
Alexander Bogomolny
Copyright © 1996-2008 Alexander Bogomolny
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