MAT 300 Mathematical Structures
Well-ordering Principle
Well-Ordering Principle (WOP).
Every nonempty subset of the natural numbers has a smallest element.
Comment: If we assume the Principle of Mathematical Induction (PMI) (or
the equivalent Principle of Complete Induction (PCI)) then may consider the
WOP a theorem. However, one may also consider the WOP a basic axiom that
defines the natural numbers (together with the other Peano axioms), and then
prove the PMI / PCI as theorems! Indeed, they are all equivalent.
Equivalent statements of the WOP:
(S⊆IN) −→ ((S=∅)∨(Shas a smallest element) )
((S⊆IN) ∧(S6=∅)) −→ (Shas a smallest element)
((S⊆IN) ∧(Sdoes not have a smallest element)) −→ (S=∅)
Proof (PCI ⇒WOP).
Suppose S⊆IN has no smallest element. To show that S=∅consider the
complement T= IN \Sand prove by induction that n∈Tfor all n∈IN.
Clearly 1 ∈Tsince otherwise 1 6∈ Timplies 1 ∈S, and necessarily 1 would be
the smallest element of S, which contradicts ( |
/
↓) our hypothesis.
Thus suppose that [n]⊆S. Again, suppose that (n+ 1) was in S. Since
S∩[n] = ∅, (n+ 1) would be the smallest element of S|
/
↓. Thus if n∈Tthen
also (n+ 1) ∈T. By induction T= IN follows, and hence S=∅.
Proof (WOP ⇒PMI).
Suppose S⊆IN is such that 1 ∈Sand whenever n∈Sthen also (n+ 1) ∈S.
Assuming the well-ordering principle we shall prove that S= IN. Again consider
the complement T= IN \Sand assume that Thas a smallest element, call this
n. The number ncan’t be equal to 1 since 1 ∈S. Thus n≥2 and n−1 is
a natural number. Since nis assumed to be the smallest element of T, and
(n−1) < n, (n−1) cannot be in T, i.e. it is in S. But by hypothesis, if
(n−1) ∈Sthen also n= ((n−1) + 1) ∈Swhich contradicts n∈T. Thus T
cannot have a smallest element and by the WOP T=∅, or S= IN. .
Exercise: Show that PMI ⇒PCI.
As a reference, recall:
(PMI): If S⊆IN,1∈IN and ∀n(n∈S→(n+ 1) ∈S)then S= IN.
(PCI): If S⊆IN,1∈IN and ∀n([n]⊆S→(n+ 1) ∈S)then S= IN.