Ω-Theory: Mathematics with Infinite and Infinitesimal Numbers - SELP

CHAPTER 1. Ω-CALCULUS
Proposition 1.1.5. An ordered field is Archimedean if and only if its only infinitesimal
number is 0.
Before we give the proof of this proposition, we recall that an ordered field is
Archimedean if the following property holds:
∀x > 0 ∀y > 0 ∃n ∈ N : ny > x
or, in other words, the iterated sums of any positive element y surpass any given
number x.
Proof of Proposition 1.1.5. Let ξ be an infinitesimal number. Without loss of generality,
we can suppose ξ > 0 (otherwise, choose −ξ). Since we have, by definition,
ξ < 1/n for all n ∈ N, if we consider the inverses we have n < 1/ξ for all n ∈ N or,
in other words, the field is not Archimedean.
Now, suppose that the field is not Archimedean. Then, by definition, we have
an element ξ that is a counterexample for the Archimedean property, that is n < ξ
for all n ∈ N. Then 1/ξ is a nonzero infinitesimal number, since 0 < 1/ξ < 1/n for
all n ∈ N.
Proposition 1.1.5 gives a very interesting characterization of non-Archimedean
fields, and shows that fields containing infinitesimal numbers don’t satisfy some
common properties we are used to take for granted.
1.2 Superreal Fields
In the mathematical practice and especially in analysis, the most used extension of
the real field is given by the complex numbers C. Instead, we want to concentrate
on those extensions of R that are still ordered fields.
Definition 1.2.1. A superreal field is an ordered field F that properly extends R
(e.g. R ⊂ F, R + ⊂ F + , and the inclusions are strict).
It turns out that any superreal field contains infinitesimal and infinite numbers,
as shown in the next Theorem.
Theorem 1.2.2. Any superreal field F is non-Archimedean.
Proof. Thanks to Proposition 1.1.5, we only need to show that F contains an infinite
or infinitesimal number. Choose ξ ∈ F\R. Without loss of generality, we can suppose
ξ > 0. If ξ > n for all n ∈ N, there’s nothing to prove. Let’s suppose this is not the
case, and 0 < ξ < n for some n ∈ N. Now, consider the set
X = {x ∈ R : ξ < x}
of the real upper bounds of ξ. By hypothesis, X is nonempty and bounded below.
Using the completeness property of R, the element r = inf X is a well defined real
number. We claim that ζ = ξ − r is an infinitesimal number.
To prove our claim, let’s pick n ∈ N. By the properties of least upper bound, we
have that ξ ≥ r − 1/n, so that ζ ≥ −1/n, and ξ < r + 1/n, so that ζ < 1/n. Since
both inequalities hold for all n ∈ N, we conclude that ζ is a nonzero infinitesimal
number.
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