A N Niel
Posts:
2,248
Registered:
12/7/04


Re: another boring critisism of Cantor's Theorem
Posted:
Mar 29, 2004 1:04 PM


In article <b671fc3e.0403290950.3b31631a@posting.google.com>, Craig Feinstein <cafeinst@msn.com> wrote:
> Everyone I am sure is acquainted with Cantor's theorem that the number > of reals is greater than the number of natural numbers and the > diagonal argument which proves this. Basically, it says aleph_0 < > 2^aleph_0, where aleph_0 is the cardinality of the set of natural > numbers (and 2^aleph_0 is the cardinality of the reals). > > I have no problem with his argument, except that I can give another > equally convincing argument that aleph_0=2^aleph_0: Suppose that > 2^aleph_0>aleph_0: This implies that aleph_0 = log (2^aleph_0) >log > (aleph_0), where "log" (base 2) is the inverse of the "2 to the power" > function generalized for sets of infinite cardinality.
and why should such an inverse exist?
> Then if log > (aleph_0) is an infinite cardinal number, then aleph_0 is not the > least infinite cardinal number, which contradicts the definition of > aleph_0. And if log (aleph_0) is a finite number, then 2^(log > (aleph_0))=aleph_0 is a finite number, which contradicts its > definition too. Therefore, aleph_0=2^aleph_0. But Cantor's theorem > says the opposite of this!
actually, we conclude log(aleph_0) does not exist. That is, there is no cardinal x such that 2^x = aleph_0.
> > Therefore, by playing around with infinite numbers, one is playing > with fire: One is opening up pandora's box to lots of contradictions > and paradoxes. > > Dr. Ben Zona, Ph.D. M.D. Hon.Doc.
Actually, the problem is "playing" with log before defining it and proving its properties.

