Date: Feb 24, 2013 2:56 PM
Subject: Re: Matheology § 222 Back to the roots

On 24 Feb., 15:14, Jim Burns <> wrote:
> On 2/23/2013 4:48 PM, Nam Nguyen wrote:

> > On 23/02/2013 2:38 PM, Virgil wrote:
> >> In article
> >> In mathematics [...] proofs of existence do
> >> not always require that one find an example of the thing claimed to
> >> exist.

> > So, how would one prove the existence of the infinite set of
> > counter examples of Goldbach Conjecture, given that it does not
> > "not [...] require that one find an example" of such existences?

> Your use of the subjunctive mood ("how would one prove") requires
> anyone wishing to answer your question (challenge?) to assume
> the existence of an infinite set of counter-examples to the
> Goldbach conjecture, whether it truly exists or not.
> I strongly suspect that you do not intend your challenge to be
> read this way. However, on the slight chance that I am wrong on
> this point, I will answer your challenge *as you asked it* :
> Let A stand for "There exists an infinite set of counter-examples
> to the Goldbach conjecture". The proof your are looking for is
> "From A, we conclude A".
> If the subjunctive reading gives a less-than-trivial question,
> then what did you mean instead?
> Did you mean to assert, as part of your question, that there exists
> an infinite set of counter-examples to the Goldbach conjecture?
> How that is proven would be helpful information to anyone
> attempting to answer you, if you had that information. Do you?
> If your intended question was intended to challenge someone's
> assertion that a proof of existence NEVER needs an example,
> then that would have made some sense. However, let me remind you
> that Virgil did not assert that. (See above.)
> All that is needed to support what Virgil ACTUALLY wrote is
> a single example of a proof of existence that does not
> require one to find an example of the thing claimed to
> exist. One example, and so "not always".
> Examples come to mind such as the Banach-Tarski paradox,
> in which something can be shown to exist by using the
> Axiom of Choice, which asserts the existence of a choice function
> axiomatically but does not provide the choice function.
> Do you see a problem with this?

It is not more and not less a problem than the axiom of sum (AS):
There are 10 different natural numbers the sum of which is 5.

Remember: AC has been objected because the opponents thought a well-
ordering of |R could not be defined.

Remember: Zermeolo and Fraenkel had supposed that a well-ordering of |
R was possible, not yet defined though.

Remember: The result 1 = 2 (for measurable volumes of spheres)
contradicts AC.

So only the gang of blinded matheologians can really belive in the
idea that healty humans with sober brains are willing to accept AC or
AS and the "mathematics" derived from it.

Regards, WM