# Thread: Infinitesimals and Completeness Axiom

1. ## Infinitesimals and Completeness Axiom

Under the hyperreal number system, infinitesimal numbers are nonzero, so then how would it be consistent with R being complete?

If completeness axiom says that every bounded set has a least upper bound, then I know that the rationals don't have the completeness property since the open interval from 0 to pi doesn't have a least upper bound in Q. But under the hyperreal system, what about analogous interval from 0 to, say, some infinitesimal plus pi?

2. I don't quite understand what you are asking here. Yes, R is complete but R*, the hyperreals, are not. There is a proof that the reals, as a subset of the hyperreals, is complete here:
Math Forum: Ask Dr. Math: A Mathematical Essay

3. Basically I'm saying that if you can show (0, root 2) interval has no least upper bound that is a rational (because root 2 is irrational between two rationals and makes the rational set incomplete), why can't you say the analogous with (o, root 2 + infinitesimal)? What would the least upper bound real number be for that interval?

If R is complete, then anything between real numbers should also be real numbers. But if *R contains R, then where would the infinitesimal elements be located? If, say, root 2 + infinitesimal is between root 2 and [(root 2) + 1].

4. Originally Posted by anomaly
Basically I'm saying that if you can show (0, root 2) interval has no least upper bound that is a rational (because root 2 is irrational between two rationals and makes the rational set incomplete), why can't you say the analogous with (o, root 2 + infinitesimal)? What would the least upper bound real number be for that interval?

If R is complete, then anything between real numbers should also be real numbers.
No, this does not follow.

But if *R contains R, then where would the infinitesimal elements be located? If, say, root 2 + infinitesimal is between root 2 and [(root 2) + 1].

5. Then where does x + dx go if dx is infinitesimal? Is *R even ordered?