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Orderings & Lattices · step 8 of 13

Well-orderings & the least-element principle

A total order is a well-order if every non-empty subset has a least element. ℕ is the canonical example; ℤ and the non-negative rationals are not, since ℤ has no least element at all and {x ∈ ℚ : x > 0} has bounded but unattained infima. Well-ordering is equivalent to induction, and it is the engine behind minimal-counterexample proofs.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

  1. Check for a least element in every subsetTo refute, exhibit one non-empty subset with none — usually an infinite strictly decreasing sequence.
  2. Use minimal counterexamplesTo prove ∀n P(n), assume the set of counterexamples is non-empty, take its least element, and derive a contradiction.
  3. Derive a smaller counterexampleThe contradiction almost always comes from producing a counterexample below the least one, which is impossible.
  4. Note the equivalence with inductionWell-ordering of ℕ, weak induction and strong induction are interderivable, so use whichever makes the argument shortest.

Worked example

Use the well-ordering principle to prove every integer n ≥ 2 has a prime divisor.

  1. Suppose not, and let S be the set of integers ≥ 2 with no prime divisor. Assume S ≠ ∅.
  2. By well-ordering, S has a least element m.
  3. m is not prime (a prime divides itself), so m = ab with 1 < a < m.
  4. a ≥ 2 and a < m, so a ∉ S, meaning a has a prime divisor p. But p ∣ a and a ∣ m give p ∣ m — contradiction.

Answer. S must be empty, so every integer n ≥ 2 has a prime divisor.

Where marks get dropped

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Well-orderings & the least-element principle — frequently asked questions

Is the well-ordering principle the same as the well-ordering theorem?

No. The principle is the statement that ℕ is well-ordered. The theorem — that EVERY set can be well-ordered — is equivalent to the axiom of choice and far stronger.

How does it relate to induction?

They are equivalent over the usual axioms for ℕ. A minimal-counterexample proof can always be rewritten as strong induction and vice versa.

Can an infinite set be well-ordered?

Yes — ℕ is. What well-ordering forbids is an infinite strictly decreasing sequence, which is exactly why it justifies termination arguments.

The rest of Orderings & Lattices

Partial orders, Hasse diagrams, bounds, lattices. Each subtopic below has its own method, worked example and mark-losing traps.

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