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

CHAPTER 1. Ω-CALCULUS
Example 1.12.2. We want to calculate n(N). To do so, we use that, in our model,
limλ↑N |λ| = limλn,θ↑N |λn,θ|. Since
and
we deduce that, by definition of α,
λn,θ ∩ N = {1, . . . , n}
n = |λn,θ ∩ N| = max(λn,θ)
n(N) = lim
λ↑N |λ| = lim
λn,θ↑R max(λn,θ) = α (1.4)
Remark 1.12.3. Equation 1.4 shows us another interesting property of the number
α:
α = lim
λ↑R max(λ) = lim
λ↑N |λ|
We explicitely remark that this equality is true in the model, but is not entailed by
any axiom of Ω-Calculus. In other words: this equality is generally false, unless we
require that it holds by adding an appropriate axiom or by choosing a particular
model of Ω-Calculus.
We continue by calculating the numerosities of the sets E and O of even and odd
numbers, and those of Z and Q.
Example 1.12.4. We have already observed that, since N = E ∪ O, then, by finite
additivity of numerosity, n(N) = n(E) + n(O). Now, we will determine n(E) and
n(O). As in the previous example, n(E) = limλn,θ↑E |λn,θ|. If n > 1, then
|λn,θ ∩ E| = {2, . . . , n}
from which we deduce |λn,θ ∩ E| = n/2 is eventually true. We conclude n(E) = α/2
and n(O) = n(N) − n(E) = α/2.
Example 1.12.5. We want to calculate n(Z). Since Z = N ∪ −N ∪ {0}, by using
Corollary 1.6.3, we conclude n(Z) = 2n(N) + 1 = 2α + 1.
Example 1.12.6. We want to calculate n(Q). As in Example 1.12.2, we have
limλ↑Q |λ| = limλn,θ↑R |λn,θ ∩ Q|. We recall that, by definition,
λn,θ ∩ Q = λn = −n, −n2 + 1
, . . . , −
n
1 1
, 0,
n n , . . . , n2
− 1
, n
n
so |λn| = 2n 2 + 1. Taking the Ω-limit, we obtain n(Q) = 2α 2 + 1.
In our model, we can also answer the open question of Example 1.8.8. In particular,
we will show that min((0, 1) ◦ ) < 1/α.
Proposition 1.12.7. min((0, 1) ◦ ) < 1/α
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