Re: naive question from a non-mathematician

On 29 May 2006 14:24:36 -0700, "Gene Ward Smith"
<genewardsmith@xxxxxxxxx> wrote:


David C. Ullrich wrote:

There is a first order theory of algebraically closed fields of
characteristic zero; just add an axiom scheme for characteristic zero,
and another for algebraic closure, to the axioms for a field. It has a
model in the algebraic numbers, so it has models in every infinite
cardinality. It can be shown to be categorical in uncountable
cardinality. So, we do have an algebraically closed field of
cardinality C, without constructing it, from model theory; and hence we
could use the top-down method(s) I've outlined.

_How_ does it follow that the reals are least-upper-bound complete?
That's not a first-order property.

I'm afraid you need to add that assumption, which is what I did when I
said that the valuation was to a group which was height one *and*
order-complete.

I don't recall you mentioning the completeness when you added
that bit.

I'll confess that I don't know what a group of height one is.
But if we're talking about an order-complete group of height
one presumably it's an ordered group - I do know what that is.

How do you prove the _existence_ of a complete ordered group?

Or, more to the point: How do you prove that there exists a
complete ordered group by an argument that cannot be trivially
modified to prove the existence of a complete ordered field?

Not going to bother suggesting that this is getting a little
silly. Yes, when someone said something about _the_ foundations
that was inappropriate, surely there are many ways to set
things up. But possibly you should have had the example
better prepared before starting to present it? You say an
alternative is to just do the following, but the following
continues to get more complicated as people raise objections.

The slightly silly part is that so far I've seen applications
of the completeness theorem from logic, as well as of algebra
that strikes me as non-trivial (which proves very little
since I know no algebra). The idea that this is a _better_
way to construct the complex numbers and then the reals
seems a little far-fetched. Where "better" just means
that it has some property such that someone somewhere
would actually use it in an actual foundational "project"...


Elsewhere you say that it _does_ follow. I'm not sure what you
mean by "categorical in uncountable cardinality":

(i) If aleph is an uncountable cardinal then any two models
of cardinality aleph are elementarily equivalent

(ii) Any two uncountable models are elementarily equivalent.

I mean any two models of cardinality aleph are isomorphic.

Not that I have an actual proof, but if anything it seems
to me that this would tend to indicate that the
construction does _not_ imply the reals are lub-complete.

It doesn't; you need to assume that to get the reals out of all the
various real-closed fields of cardinality c.


************************

David C. Ullrich
.

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