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: 02/20/05 

Date: Sat, 19 Feb 2005 19:31:51 -0500

In <cv1jfb$cl$1@gander.coarse.univie.ac.at>, on 02/17/2005
   at 05:45 PM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:

>again I forgot to explicitely write down the dependencies:

>1) for-all r1 and r2 non-equal in R there exists q in Q such that
>either r1 <_R q <_R r2 or r2 <_R q <_R r1 with <_R being transitive
>and anti-reflexive.

>2) for-all r1 and r2 in R there exists q in Q such that from r1 <_R
>r2 follows that r1 <_R q <_R r2, again with <_R being transitive and
>anti-reflexive.

Why not the simpler for all r1 and r2 in R with r1 <_R r2, there
exists q in Q with r1 <_R q <_R r2?

>the difference is that 1) does enforce anti-symetry, linearity and
>extensionality from <_Q (since pred_(<_R)(x) non-equal pred_(<_R)(y)
>for x<y is a result of some q between x and y) onto <_R restricted
>to QxQ. the point is that if <_R is not a linear order, then 2)
>could still hold!

Please clarify your nomenclature.

>a strict quasi-order is defined to be only transitive and
>anti-reflexive, this does imply that some pair of elements could
>exist such that neither "<" nor ">" nor "=" is true. the members of
>such a pair are defined as "incompareble", written as x _|_ y, and
>otherwise they are compareable.

R is an ordered field, so all elements are comparable in that sense.
With only a partial order it wouldn't be R. But even if you were
talking about some R1 that has only a partial order, a map of Q->R'
could only preserve the orders of elements of Q, not of R'.

>define R as a set fulfilling some properties among which also the
>existance of a quasi-order and the related quasi-denseness are
>found. the question is if this set does exist and if it is in some
>way unique.

The standard models of R show that a complete Archimedean field does
exist. If you relax the properties to a field with a partial order
then you will lose uniqueness and you will get structures that are not
models of R.

>as I said, f2 is "quasi-order-preserving" since any quasi-order on R
>will have corresponding elements in Q wherever compareable.

Q is ordered, not just partially ordered. The proof that f1=f2 if both
are order-preserving homomorphisms is trivial.

>I don't know how you define homomorphism,

Then maybe you should do some reading. A homomorphism of fields is a
function that preserves zero, unity, addition and multiplication.

>but an isomorphism is a bijection, and it is called
>structure-preserving when both directions (i.e. also the inverse
>function) are structure-preserving,

No. A bijection that is not structure preserving is, by definition,
not an isomorphism.

>I am using the language from Graph-theory since I do not know enough
>about the vocabulary on trees.

Then why did you write "tree" and "root"?

>a tree is a graph without circles,

ITYM acyclic DIRECTED graph.

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

If they're separated then they're not neighbors.

>tell me, which elements of Q are not represented?

1/3

>if you create a tree with elements of R as its nodes, then there
>are.

Recipe for tiger stew: first catch the tiger. You've sort of kind of
defined a structure that does not include all elements of R, or even
all elements of Q. Whether you could construct other structures that
included one of the missing elements is irrelevant.

>what I was asking is if your construction of "<" is independant of
>the choice of representatives, and how to prove that.

Well, I'm neither Cauchy nor Dedekind, but I already gave you a proof
for Cauchy sequences.

>the point is that apart from ultra-filters you do use
>equivalence-classes to describe R,

For Cauchy sequences; not for Dedekind cuts.

>and the existance of ultra-filters again is questionable.

Are you arguing for or against?

>if you do have equivalence-classes then you must walk the long way
>through proofing that properties are independant of the
>representants used,

So? That involves less work than dealing with ultrafilters.

>and you need to say something about the size and number of
>equivalence-classes.

No I don't; I need only show that the construction yields a complete
Archimedean field.

>but you still need to prove that equivalence-classes are seperate
>and that this property does not depend on the representants used.

What equivalence classes? The only ambiguity is whether to use (-oo,q]
and (q,oo) or to use (-oo,q) and [q,oo), and that can be eliminated by
an appropriate change to the definition of a cut. 

-- 
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