Re: naive question from a non-mathematician

On Sun, 28 May 2006 19:21:22 +0200, Denis Feldmann
<denis.feldmann.asupprimer@xxxxxxxxxxxxxxxx> wrote:

David C. Ullrich a écrit :
On 27 May 2006 20:42:18 -0700, "Gene Ward Smith"
<genewardsmith@xxxxxxxxx> wrote:

Stephen Montgomery-Smith wrote:

Like I said in a different post, it depends upon your framework. Most
modern mathematicians define everything in terms of sets and set theory.
Most modern mathematicians define things up to isomorphism.

Thus the natural numbers are, more or less, defined as the set of
finite ordinals, the integers are pairs of natural numbers quotiented
out by the equivalence relation (a,b)~(c,d) iff a+d=b+c, the rationals
are pairs of integers (the second being non-zero) quotiented out by
another approprate equivalence relation, the reals are constructed from
the rationals usually either by Dedekind sections, or by some quotient
of the cauchy sequences, and the complex numbers are pairs of real numbers.
If you like, you can do things this way. Nobody forces you to, and you
could do it other ways.

Surely there are many ways of setting up all these things. But:

You could, for instance, define the reals
axiomatically,

For example the reals are a complete ordered field. This raises the
question of how we know that there exists a complete ordered field;
if there is no such thing then the theory is a little boring.

and recover the rationals and integers from that.

There are just so many ways of defining these objects in set theory, and
so to say R is a subset of C just doesn't cut it if you are going to be
nit-picking.
Let's say I define the complex numbers C as an algebraically closed
field of characteristic zero and cardinality the continuum with a
distinguished automorphism conj(z) of degree two.

Same question here.

Moreover, its not a univoque definition : take a non-standard extension
of C with same cardinal (for instance, the quotient of C^N by the
classical* relation induced by a (non-trivial) ultrafilter on N)



* This is (z_k)~(z'k) <=> {k/ z_k=z'k} \in U , U being the ultrafilter.



Now I define the real
numbers R as the subextension fixed by this automorphism, which I now
dub "complex conjugation".


This is R* (the same non-standard extension of R)


Now I define an archimedian absolute value
by |z| = sqrt(z conj(z)).

Why archimedian????


Now C and R are topological fields under the
topology defined by this, and conj is continuous. Now I define Q as the
intersection of all subfields of C (or R.)

There is a much simpler way : Q is the field generated by 1.


Z is the ring of integers of
Q. I've defined things so that, by definition, Q is a subfield of C. Of
course doing it your way I have a Q in C which is uniquely isomorphic
to the original Q which was constructed, and so forth blah blah.

Serious question: If you start with a structure satisfying your
definition of C and then do all this, does it follow that the
R you get is complete (in the sense that every nonempty set
bounded above has a least upper bound?) I bet it doesn't.

Sure : it is not even archimedian :-)

That's what I thought.


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

David C. Ullrich
.

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