Newsgroups: sci.math
From: Erland.Gadde@sm.luth.se (Erland Gadde )
Subject: Re: Metrization (Was: Re: Non-standard analysis)
Date: Mon, 20 Feb 1995 16:46:27 GMT

I wrote:

> To me, the field of rational functions over R doesn't seem to have
> a countable basis. I can't prove that right now, but I dare to make the
> following stronger conjecture:
> 
> Let F be an ordered field, with the topology which has the family of
> all open intervals as a basis. 
> If F has a countable basis, then F can be isomorphically and homeomorphically
> embedded in R.
> 
> In fact, I suspect that this true even if F is assumed to be metrizable.
> 
> 
> Does anyone know if I'm right or wrong?


Well, I was right about the rational functions over R having no countable 
basis, but my conjecture was wrong.


Theorem 1.
Let F denote the field of rational functions over R,
with the order relation defined as p<q (p,q in F) if there is a real M
such that for all x>M, p(x) and q(x) are both defined and p(x)<q(x),
as numbers. With the topology which has the family of all open intervals 
in F as a basis, F is not second countable.

Proof:
For each n>0, let e_n, denote the rational function given by e_n(x)=1/x^n,
for all x in R. For each constant rational function k, the intervals I_n=
(k-e_n,k+e_n) is a strictly decreasing local basis at k. If B is a basis
for F, then, if we put i_1=1, there must exist a set U_i_1 in B such that
k is in U_i_1 and U_i_1 subset I_i_1. Then, there is a integer i_2>i_1
such that I_i_2 subset U_i_1. Then, there is an U_i_2 in B such that
k is in  U_i_2 and U_i_2 subset I_i_2. Continuing in this manner, we ontain
a sequence {U_i_j} (j>0) of sets around k from B. There is such a sequence
of sets from B around every constant k, and no two such sets around different
k:s intersect each other. Since there are uncountably many constant functions 
in F, B is uncountable. But B was an arbitrary basis for B, so F has no
countable basis Q.E.D. 


Theorem 2.
Let F be as in Theorem 1, but with Q substituted for R everywhere.
Then F is metrizable.

Proof:
The family of all open intervals over in F is a countable basis for F,
and F is clearly regular. Hence, (by the theorem quoted by T. Murphy) F is
metrizable Q.E.D.


Erland Gadde.