Re: Zenkin's paper on Cantor (reply of Dr. Zenkin)

From: Eray Ozkural exa (examachine_at_gmail.com)
Date: 11/14/04 

Date: 14 Nov 2004 01:36:24 -0800

cbrown@cbrownsystems.com (Chas Brown) wrote in message news:<ba7dcfd7.0411131541.5975da42@posting.google.com>...
> examachine@gmail.com (Eray Ozkural exa) wrote in message news:<320e992a.0411091141.38c7158b@posting.google.com>...
> > "Shmuel (Seymour J.) Metz" <spamtrap@library.lspace.org.invalid> wrote in message news:<4181a33d$1$fuzhry+tra$mr2ice@news.patriot.net>...
> > > In <320e992a.0410271136.4479a210@posting.google.com>, on 10/27/2004
> > > at 12:36 PM, examachine@gmail.com (Eray Ozkural exa) said:
> > >
> > > >However, if he can settle the following metamathematical theorem:
> > >
> > > That's not a Metamathematical theorem absent definitions of
> > > "infinitary reasoning" and "abstraction of actual infinity". Without
> > > the definitions it's just Philosophy.
> >
>
> Not even Philosophy - without definitions, it's just Wanking.

If you understand the terms, it can be mathematics.

> > The first is a piece of cake as it means that there is no halting
> > program which can output a 0 or 1 for falsehood or truth of the
> > proposition under consideration. (I cannot see why Brown thought it
> > was hard)
>
> But Zenkin's claim is that _Cantor's proof_ is an _example_ of
> "infinitary reasoning". Since Cantor's proof can easily be verified as
> true (under the usual axiomatic system) in a finite number of steps,
> this would contradict your (natural) definition, and his argument must
> fail immediately.

Hmmm. Given of course the truth of "the usual axiomatic system", by
which you probably mean ZFC. Zenkin's argument is more like, well,
this is a finite nonhalting program, that never says anything.

While the truth of Cantor's theorem follows from ZFC in a finite
number of steps (this is easily seen!), I must admit that I cannot
readily observe how Cantor's diagonal *proof* terminates in a finite
number of steps. I think if you explained a bit, it would be
interesting. (The argument others made was something like, "I can
finish reading the proof in finite time", which is not quite relevant.
I can finish reading "Let A be 0. Add infinity to A, infinite times.
Return A." in a finite number of reading steps, but unfortunately the
algorithm it describes never ends, therefore it is not an algorithm.)
Just to be painstakingly precise.

Regards, 

--
Eray Ozkural

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