Re: (sketch of a) Proof that the set of Real Numbers doesn't exist

From: Shmuel (Seymour J.) Metz (spamtrap_at_library.lspace.org.invalid)
Date: 02/25/05 

Date: Fri, 25 Feb 2005 15:31:17 -0500

In <cvjg8o$92$1@gander.coarse.univie.ac.at>, on 02/25/2005
   at 03:10 AM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:

>well, I do have a problem with the word "equivalent", especially
>since I do not properly know the meaning of the word "canonically".

I was going to put it in the language of category theory, but instead
I'll make a stronger statement. Given two distinct models of R[1],
there exists a unique order-preserving isomorphism between them.

>fact is that in maths there is the concept of "equivalence-relation"

It's not connected to that usage.

>several models of R might be isomorphic and homeomorphic, but this
>does merely guarantee a certain set of properties to be in common
>(most probably them all being complete archimedean fields and Q
>being dense), this does merely guarantee some kind of
>equivalence-relation for each pair of models.

The fact that a model of R is a complete Archimedean filed guaranties
that it is equivalent to any other model of R. It has nothing to do
with equivalence relations.

>That's why one needs to ask if your set is an open set in all of
>them, no matter how you choose a and b!

What do you mean by "open set"? The standard topology for an ordered
field is the order topology, and the canonical isomorphism between
complete archimedean fields is a homeomorphism. An open set in one
maps into an open set in another.

>thanks for stating the obvious.

It may be obvious but you still don't get it.

>But as I said, my question is if the fact that this model is defined
>to have some Q being dense in it,

What does that mean? Given a complete archimedean field, there exists
a homomorphism of Q unto a dense subfield. It doesn't matter how you
define the model, just that you prove it to be a complete archimedean
field.

>if this property is already capable of proofing that each open set
>(defined as F(x) for each cauchy-filter F and each element x)

That's not the standard definintion of open sets in R. If you want to
prove that it is equivalent, go ahead, but the relevant properties of
the standard topology of R are trivial in terms of the standard
definition.

>and that it actually is an archimedean field.

That what is? Some construction using filters? Well, if it isn't an
archimedean field then it isn't a model of R.

>nice theorem, unfortunately the set-theoretical approach doesn't
>follow this path,

What gives you that idea? There are several pedagogical approaches,
but a common one is to define a model of R and then to prove that the
properties don't depend on the model. You have the same type of
situation in other parts of Mathematics, e.g., Homology Theory.

>neither the model with cauchy-filters nor the model with
>dedekind-cuts are proven to fulfill this axiom, as far as I could
>see.

Consult any Analysis book. The proof for Dedekind cuts is no more
complicated than the one for Cauchy Sequences.

>you attempted to prove by relying on the property that (a + (b-a)/3,
>b - (b-a)/3) is an open set, which I still can not believe.

What don't you believe? That a + (b-a)/3 < b - (b-a)/3 when a < b? The
definition[2] of open set in R is a union of open intervals, and
that's clearly an open interval.

>so, again:

You need to collate all of your text, fix the typos and present it
cohesively in one place.

[1] Note that Robinson's NSA is not a model in the same sense.

[2] For the standard topology. 

-- 
Shmuel (Seymour J.) Metz, SysProg and JOAT  <http://patriot.net/~shmuel>
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