Let ℕ = {0,1,2,…}. (If you start at 1, shift base cases accordingly.) This post proves that eight standard principles are all equivalent—different faces of the well-foundedness of (ℕ, ≤).
Table of Contents (click to expand)
-
Eight equivalent principles
- (1) Ordinary Induction (PMI)
- (2) Complete (Strong) Induction (CMI)
- (3) Least Number Principle (LNP) / Well-Ordering
- (4) Principle of Finite Descent (PFD)
- (5) Well-Founded Induction on (ℕ, <)
- (6) No Infinite Decreasing Sequence
- (7) Course-of-Values Recursion / Iteration
- (8) Well-Ordering Theorem for ℕ
- Implication roadmap
- Proofs
- Conclusion
Eight equivalent principles
(1) Ordinary Induction (PMI)
If S ⊆ ℕ has 0 ∈ S and ∀n (n ∈ S ⇒ n+1 ∈ S), then S = ℕ.
(2) Complete (Strong) Induction (CMI)
If S ⊆ ℕ has 0 ∈ S and ∀n (({0,…,n} ⊆ S) ⇒ n+1 ∈ S), then S = ℕ.
(3) Least Number Principle (LNP) / Well-Ordering
Every nonempty A ⊆ ℕ has a least element.
(4) Principle of Finite Descent (PFD)
No infinite strictly decreasing sequence exists in ℕ.
(5) Well-Founded Induction on (ℕ,<)
If S ⊆ ℕ satisfies (∀m<n, m ∈ S) ⇒ n ∈ S, then S = ℕ.
(6) No Infinite Decreasing Sequence
Same as (4), stated explicitly for sequences.
(7) Course-of-Values Recursion / Iteration
Given X, a ∈ X, and an operator Φ, there is a unique f:ℕ→X with f(0)=a and f(n+1)=Φ(n, f restricted to {0,…,n}).
(8) Well-Ordering Theorem for ℕ
Every nonempty subset of ℕ has a minimum (essentially (3)).
Implication roadmap
(1) ⇔ (2), (1) ⇔ (3), (3) ⇔ (4) ⇔ (6), (2) ⇔ (5), (3) ⇒ (7), (7) ⇒ (2), and (8) ≡ (3).
Proofs
(1) ⇒ (2) and (2) ⇒ (1)
(1) implies (2) by defining T = {n : {0,…,n} ⊆ S} and applying induction. Conversely, if S satisfies PMI, then from {0,…,n} ⊆ S we know n ∈ S, hence n+1 ∈ S, so CMI applies.
(1) ⇒ (3) and (3) ⇒ (1)
(1) ⇒ (3): Assume A ⊆ ℕ nonempty. Let S = {n : no element of A ≤ n}. If 0 ∉ S, then 0 is least. Otherwise, inductively either a least element appears or S = ℕ, contradicting A ≠ ∅.
(3) ⇒ (1): Suppose S ⊆ ℕ with 0 ∈ S and successor-closed. If S ≠ ℕ, then T = ℕ\S is nonempty; by LNP it has a least element t. Minimality forces t ∈ S, contradiction. Thus S = ℕ.
(3) ⇔ (4) ⇔ (6)
(3) ⇒ (4): If an infinite descent exists, its set has no least element. (4) ⇒ (3): If A has no least, we can build an infinite descent. (4) and (6) are two forms of the same principle.
(2) ⇔ (5)
Well-founded induction on (ℕ,<) is exactly strong induction. Each implies the other directly.
(3) ⇒ (7)
Using the least bad index argument: if recursion fails at some n, take least such n, contradiction. So recursion exists and is unique.
(7) ⇒ (2)
Define f:ℕ→{0,1} via recursion to encode “{0,…,n} ⊆ S”. From uniqueness we deduce f(n)=1 always, hence S = ℕ. So recursion yields complete induction.
Conclusion
All eight principles are equivalent. They are different but interchangeable ways of saying that (ℕ, ≤) is well-founded: no infinite descent, every subset has a least element, induction works, and recursion is legitimate. Each tradition (number theory, logic, computer science) prefers one formulation, but logically they all belong to the same equivalence class.