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Re: Proving archimedian property from "dedekind complete + orderer field"
Posted:
Jul 28, 2013 11:29 AM


On Sun, 28 Jul 2013 03:16:24 0700 (PDT), Lax Clarke <lax.clarke@gmail.com> wrote:
>I'm confused as to why the archimedian property (unbounded natural numbers in reals) needs to be proven from {least upper bound property + ordered field}" > >Aren't the naturals "obviously" infinite (meaning provable without all the other machinery)? Because n+1 is bigger than n which are different natural numbers (you can prove n+1 > n > 0 using the order properties of reals). > >Is it my understanding that this proof is intended to exclude any case where the naturals could wrap around on each other like 1+1+1...+1=0 in a finite field?
Decided to answer this in a separate post so as not to disrupt the flow of the other one.
No, that's really not it. "Ordered field" rules out 1 + ... + 1 = 0, but "ordered field" is not enough to show that the naturals are not bounded. There _are_ ordered fields in which the naturals _are_ bounded. (Of course they cannot be leastupperbound complete, by the theorem we're discussing.)
I suspect that people will mention the "nonstandard reals", or "hyperreals" as an example. There are much simpler examples.
For example, let F be the set of all rational functions: quotients of (real) polynomials. Given a rational function f, define f > 0 to mean that
there exists a real number x such that f(t) > 0 for all t > x.
It's not hard to show that F is now an ordered field. Define f(t) = t. Then f is an uppper bound for the naturals in F.
(The multiplicative identity in F is the constant function 1, so the "naturals" are the elements of the form 1 + ... + 1, that is, the constant functions that take a natural number for their value.)
>Why can't we prove that from the totality properly of the order < ? > >How would one prove that the naturals inside the rationals (without the sup property which is used in the proof I talk about above) is unbounded above? > >Thanks for the help.



