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 16:09:55 -0500

In <cvjo7r$9m$1@gander.coarse.univie.ac.at>, on 02/25/2005
   at 03:10 AM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:

>So, I'll assume you meant "<_R" and "<_Q":

Those were the only ones that were clear. I meant pred_ in the
construction pred_(<_R)(x).

>field(<_R)

I'm not sure what that means either.

>A model of R is a set of the cardinality of the powerset of N
>together with some "<_R" defining the topology on R

That's half of it. A model of R has all of: a set of elements, two
binary operations and an order, satisfying the axioms for a complete
Archimedean field. You don't need to make the cardinality part of the
definition, because you can prove it from the other properties.

>the question is what other properties a model of R is supposed to
>fulfill (other than the existance of an injective f:Q->R under which
>Q is dense in R,

That's a consequence of the other properties.

>the question is if "<_R" does need to be defined as a linear order.

Yes, because otherwise it's not a complete Archimedean field and hence
not a model of R.

>that's the question! maybe it is true for dedekind-cuts, but other
>models merely define some injective f:Q->R with "<" only defined on
>Q to be an ordered field, nothing whatsoever is said about the
>existance of a linear order on R!

No; it's not a model of R unless it's true.

>maybe you could create a well-ordering on R according to AC,

The order for R is not a well ordering.

>but how do you prove that some linear order on R does exist for
>all models of R?

How do you prove that all pink grapefruits are pink? If it isn't a
complete complete Archimedean field then it isn't a model of R. Now,
there might be a literature for Analysis over non-Archimedean fields,
or even over fields with only partial orders, but such literature
would deal with something other than Real Analysis.

>since "<_R'" is transitive and Q is dense in R' under "<_R'", so
>naturally a map of Q->R' could preserve the orders of elements of
>R'\Q in some sense.

How does a function preserve a property of elements that are not in
its domain? To say that f preseres order means that x<y implies
f(x)<f(y).

>so you actually meant "homomorphism of fields"?

The topic under discussion was fields, so what else could I have
meant?

>then how do you prove that those properties you mentioned already
>are sufficient for showing that they preserve the order of Q and are
>injective?

First you prove that the additive subgroup generated by 1 is
isomorphic to Z. Then you prove that the subfield generated by that
subgroup is isomorphic to Q. Since we are dealing with an ordered
field, we are talking about order-preserving isomorphisms, but that
doesn't add much complexity.

>again, I think you are talking about some kind of "isomorphism of
>fields", while in set-theory "structure-preserving isomorphism" was
>explicitely defined as a structure-preserving bijective function
>where its inverse is structure-preserving too.

An isomorphism if Field Theory *is* a "structure-preserving bijective
function where its inverse is structure-preserving too"; the structure
in question is (+,*) or (+,*,<), depending on whether you are talking
about a filed or an ordered filed.

>"tree" is a notion of graph-theory,

Then why not use standard nomenclature for trees?

>no, "DIRECTED" is a consequence from the fact that removing a node
>does result in 2 seperate trees iff that node had more than one
>neighbours,

No.

>apart from all finite paths between the root and some rational
>numbers I also allow for paths of infinite length

Before you "allow" it define what you mean by it.

>additionally, for each node x I put a "direction" on the set of
>neighbours of x which are seperated from the root by the given node
>x (those neighbours are what you called
>>>children).

No. The nodes that I call children of x are the nodes that have a link
from x to them.

>OK, then please show me that dedekind-cuts can be proven to fulfill
>these axioms without first proving that only equivalence-classes of
>size 2 are required once for each rational number. you can't!

Huh? How could you have an equivalence class of size greater than 2?

>actually you still did not prove that dedekind-cuts *do* have no
>other ambiguity-issues than "whether to use (-oo,q] and (q,oo) or to
>use (-oo,q) and [q,oo)"!

Then consult the literature; I gave a proof for Cauchy sequences
already.

>I suspect

When the proof has been available for a century, "suspect" is no
substitute for reading.

>with some nodes of the tree I constructed

Then don't use the "tree" that you constructed; you still haven't
defined what it is and it has no relevance to either of the two simple
and well known models. 

-- 
Shmuel (Seymour J.) Metz, SysProg and JOAT  <http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action.  I reserve the
right to publicly post or ridicule any abusive E-mail.  Reply to
domain Patriot dot net user shmuel+news to contact me.  Do not
reply to spamtrap@library.lspace.org