Re: [HM] Infinitesimals

Gordon Fisher (gfisher@shentel.net)
Sat, 14 Nov 1998 14:24:21

At 01:55 PM 11/14/98 -0500, Roger Cooke wrote:
> In what sense is "complete" being used here? A linearly ordered field
> that is complete in the order topology need not be Archimedean, if memory
> serves me right. (I'll have to think about the counterexample.)

From *Algebra* by Birkhoff and MacLane, 1967, p 175-177

"The real field R is distinguished from other ordered fields by being
*complete*. Completeness can be defined in various ways, for example by
the condition (converse of Lemma 5) that every Cauchy sequence has a limit.
We prefer to adopt another, equivalent definition which involves the
concept of lower bound. ..... [some definitions and a proposition omitted
here] .....
We now define an ordered domain D to be *complete* when every non-empty set
S of positive elements of D has a g.l.b., and in D. Our basic axiom on the
real field, whose existence we assume (cf. Exercise 13 below), is the
following.
Axiom. The real numbers R form a complete ordered field. .....
[another proposition omitted here; "ordered" here means linearly ordered]
.....
Proposition 13 (Archimedean law). Given a > 0 and b > 0 in R, there
exists a natural number n such that na > b." The proof of this proposition
uses only this kind of completeness, which I take it is the standard one,
and linear order in a [an integral] domain.

B & M remark: "Note that Proposition 13 does not hold in the (incomplete)
ordered field R[[t]] of real formal power series, since t > 0 and 1 > 0, yet
nt < 1 for all integers n."

This theorem is often proved in real variable texts, often (at least a
while back) without assuming existence of the real field, i.e. some
"construction" is given, usually using Dedekind cuts or Cauchy sequences.
There are many equivalent formulations of completeness. I seem to recall a
book -- was it by Olmstead? -- which contained a whole bunch of equivalent
formulations.

Gordon Fisher gfisher@shentel.net