Re: non-Archimedean models of Euclidean geometry?

On 9 maalis, 12:48, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Sun, 8 Mar 2009 08:50:39 -0700 (PDT), Gc <Gcut...@xxxxxxxxxxx>
wrote:



On 7 maalis, 06:31, Ben Crowell
<crowel...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
I've been trying to absorb a few ideas from Tarski's work leading up
to the proof that elementary Euclidean geometry is complete and
consistent. I'm guessing that the full proof is hopelessly huge
and technical, but I at least want to start by understanding
the interpretation of Tarski's axiomatization.

He has an axiom schema of continuity. It's basically the Dedekind
cut construction, but because he wants a first-order theory he defines
the partitions (A,B) not by quantifications over sets A and B but by
making an axiom schema in which A and B are defined by first-order
formulae. The axiom basically says that if every point in A is to the
left of every point in B, then there exists a point between A and B.

Now non-Archimedean models are generally going to violate this
axiom.

You've said in the past that your big ambition in life is to prove
me wrong about something. You should really be embarassed
at the way your attempts to do so all turn out to be
ridiculous - if I were you I'd be _very_ careful about
proving things I said in that regard.

I have no such ambitions anymore. I

There are no non-archimedian models for this theory.

The proof that there _are_ non-archimedian models
is very simple. Your refutations of that proof imply
some basic understanding, of what I'm not sure.


OK. Help me with this. How can you say that there are non-archimedian
models or even archimedian models, when you can`t define numbers in
that theory, eg you can`t extend the language so that it contains
numbersand those symbols actually mean something. Now consider models
of this theory. There are two relations R and Y, and then there is
universe A of a model, models are of the form omega = (A, R ,Y). Now
these models have nothing to do with numbers per se. If you wan`t to
understand even the original euclidean geometry you must understand
that you must in principle be able to construct everything without any
numbers in blank paper.

There is up to
isomorphism exactly one model for each infinite cardinal.

Supposing that were true, how would that show there
are no non-archimedian models?

It is not true, I am sorry. But Aatu knows this stuff and he gave an
example where the non isomorphic models have then same language. It is
no use to speak about "unique up to isomorphism" if the language are
not compatible.

That follows
from that the theory is complete.

What?????????????????????

This is the part you should be embarassed about. I gather
you're actually studying logic.

No I am not. I haven`t ever taken any logic classes and I haven`t read
anything about logic since last summer.

All I know about logic
is what I learned in a course thirty years ago, but _I_
know that "elementarily equivalent" is not the same
as "isomorphic". Jeez.

Yes you are right. I have complete forgot about that.

Here's something I'm pretty sure is a counterexample.
The language is FOL with equality, including exactly
one unary predicate P and no other predicates or
function symbols.

I thought we were speaking about the theory in OP. In your reply i
didn`t notice you saying that you are thinking about some other
theory.

The theory is "there exist infinitely
many x such that P(x) and there exist infinitely many
x such that ~P(x)". I can easily give a set of first-order
axioms that amounts to that, in case it's not obvious
how to do so.

It seems clear to me that this theory is complete,
although I haven't given a formal proof. By all
means explain why it's not.

What this has to do with the original theory?

The theory certain
has non-isomorphic models of the same cardinality.
If c is any uncountable cardinal there are at least
two non-isomoprhic models of cardinality c:
One where only countably many elements
satisfy (the interpretation of) P and one
where only countably many elements
satisfy ~P.

E.g., A could be the set of infinitesimals and B the positive
reals. However, I'm not clear on whether this still applies when A
and B have to be defined by first-order logical formulae. The first-
order language in which the formulae have to be constructed can't,
for example, define the set of all infinitesimals I used in the
counterexample above.

So I can tell that this axiom definitely rules out the rationals
as a model, but does it actually rule out the hyperreals or the
surreal numbers?

David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)

.

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