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Lemma 2: ("Real Induction") Let A be a subset of real numbers.
If
(0) α ∈ A
(1) ∀x ∈ A. ∃y. [x,y) ⊆ A
(2) ∀xy. [x, y) ⊆ A ⇒ y ∈ A
then [α,∞) ⊆ A.
Proof:
Suppose towards a contradiction that [α,∞) ∖ A is nonempty. Set β = inf ([α,∞) ∖ A).
(1) means we can't have β ∈ A, because it would collide with the infimum.
So β > α and β ∉ A. But then [α,β) ⊆ A, and so β ∉ A, a contradiction.
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