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

CHAPTER 1. Ω-CALCULUS
6. For all r ∈ R, Fr = {λ ∈ Pfin(Ω) : r ∈ λ} ∈ Q;
7. ∅ ∈ Q and Pfin(Ω) ∈ Q.
Proof. The first part of this Proposition follows from Proposition 1.4.4.
To prove part 2, we will show that, if A = Qϕ,ψ and B = Qη,γ, then A∩B = Qδ,φ,
and δ and φ have the same Ω-limit. If A ⊆ B or B ⊆ A, the thesis follows from
part 1. Let’s suppose that this is not the case. Let’s call ξ = limλ↑Ω ϕ(λ) and
ζ = limλ↑Ω η(λ). If ξ = ζ, Axiom 1 ensures that there is a function ν such as
limλ↑Ω ν(λ) = ξ −ζ. If we define η ′ = η +ν and γ ′ = γ +ν, we have that Qη ′ ,γ ′ = Qη,γ
and
lim η
λ↑Ω ′ = lim γ
λ↑Ω ′ = ζ + ξ − ζ = ξ
We have shown that, without loss of generality, we can suppose
Now, it’s clear that
and
lim ϕ(λ) = lim ψ(λ) = lim η(λ) = lim γ(λ)
λ↑Ω λ↑Ω λ↑Ω λ↑Ω
Qϕ−η,ψ−γ = Qϕ,ψ ∩ Qη,γ
lim(ϕ(λ)
− η(λ)) = lim(ψ(λ)
− γ(λ)) = 0
λ↑Ω λ↑Ω
and that’s what we wanted to prove.
The proof of part 3 is by converse. Let’s suppose A ∈ Q and B ∈ Q. From
the definition we have that, for all functions ϕ and ψ, the hypothesis Qϕ,ψ = A or
Qϕ,ψ = B implies limλ↑Ω ϕ(λ) = limλ↑Ω ψ(λ). Since the implication holds if both the
premises are true, we have that, if for ϕ and ψ the equality Qϕ,ψ = A ∪ B holds,
then limλ↑Ω ϕ(λ) = limλ↑Ω ψ(λ), and so we conclude A ∪ B ∈ Q, as we wanted.
To prove the first implication of part 4, let’s suppose A ∈ Q, so that there exist
ϕ, ψ such as A = Qϕ,ψ and limλ↑Ω ϕ = limλ↑Ω ψ. Let’s define the functions:
and
η(λ) =
γ(λ) =
ϕ(λ) if λ ∈ A
1 if λ ∈ A c
ϕ + 1 if λ ∈ A
1 if λ ∈ A c
Clearly, Qη,γ = A c and, by part 3 of Proposition 1.4.4, limλ↑Ω η(λ) = limλ↑Ω ϕ(λ),
while limλ↑Ω γ(λ) = limλ↑Ω ϕ(λ)+1. This shows that A c ∈ Q. Conversely, if A c ∈ Q,
we can apply the same argument to show that A = (A c ) c ∈ Q.
To prove part 5, we can consider the function c0 and the function
c
0 if λ ∈ F
ϕ(λ) =
1 if λ ∈ F
By definition, we have Qc0,ϕ = F c , so ϕ is eventually constant (take λ0 = n
i=1 λi)
and, by Axiom 2, limλ↑Ω ϕ(λ) = 0. This implies F c ∈ Q and, by part 4, F ∈ Q.
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