A few weekends ago, I got a chance to chat with a "Cantor crackpot". This is not pejorative; it is the term he used to describe himself. S, as I'll call him, is a pleasant and educated person, but he is convinced that Cantor's proof of the uncountability of the real numbers is wrong.
Here is what he recently wrote to me (paraphrased): Cantor's proof is wrong because the diagonal method that he used fails to produce a number not on the list. He illustrated this with the following example, in which S purports to give a 1-1 correspondence between the integers and the real numbers:
integer <-> real 23456 <-> 0.65432 23457 <-> 0.75432 23458 <-> 0.85432 23459 <-> 0.95432 23460 <-> 0.06432
This is a common misunderstanding among people when they first see Cantor's proof. I think this misunderstanding is essentially rooted in the following misconception: either that the only real numbers are those with terminating expansions, or that the set of integers contains objects with infinitely long base-10 representations. In this case, having talked with S, I know his misunderstanding is of the latter type.
In his example above of the purported bijection, we can ask, what integer corresponds to the real number 1/3? Its decimal expansion is 0.33333... where the 3's go on forever to the right. This must correspond to the integer ....3333333 where the 3's go on forever to the left. But this is not an integer!
So in this case the misunderstanding is really of a trivial nature. I would be interested in speaking to people who deny the correctness of Cantor's proof based on more elaborate misunderstandings.
I think I've seen the idea that basically because the proof works, it's flawed. Taken to its logical extreme, they don't believe in proof by contradiction.
ReplyDeleteThe proof they're thinking of goes "Suppose we have an enumerated list of all the real numbers. I make a number not on the list. Therefore the list is incomplete." But they think that because you made a number not on the list, you shouldn't have assumed it existed in the first place. I think I've seen this argument most recently in the comments to one of Vi Hart's recent videos. https://www.youtube.com/watch?v=lA6hE7NFIK0&list=UUOGeU-1Fig3rrDjhm9Zs_wg
(Of course, the proof isn't actually by contradiction in the first place. You just show that any list is complete, you don't need to assume it's a complete list in the first place.)
"But this is not an integer! "
ReplyDeleteWhile I agree, I'm not sure why I agree. Do you have a proof?
The integers are defined recursively, you start at zero and add one (or subtract one) successively and each step generates another integer.
ReplyDeleteThe number of steps, and therefore the result, is chosen to be finite. Each integer is finite, but there's "always one more" so the count of the integers is infinite.
It's possible to define a number system that includes infinite representations to the left, but they aren't integers. (The GoodMath BadMath blog described one such system previously. The author of that blog covers Cantor cranks with some regularity!)
MarkCC over at Good Math/Bad Math has a recurring series on Cantor cranks:
ReplyDeletehttp://www.goodmath.org/blog/category/bad-math/cantor-crankery/