Questions: Limit Ordinals and Omega

Question 1

Which of the following is a limit ordinal?

• A. 5 (the ordinal {0, 1, 2, 3, 4})
• B. ω + 3 (three successor steps past ω)
• C. ω · 2 (also written ω + ω)
• D. ω + 1 (one successor step past ω)

Answer: (C)

A limit ordinal has no immediate predecessor — it cannot be written as β + 1 for any β. ω · 2 = ω + ω is the second limit ordinal: it is the union of all ordinals ω + n (for n finite), and there is no single ordinal whose successor it is. Options A (5), B (ω + 3), and D (ω + 1) are all successor ordinals — each equals some β + 1.

Question 2

What distinguishes ω from every finite ordinal?

• A. ω is the largest finite ordinal — it bounds all others from above
• B. ω has no immediate predecessor — it cannot be reached by adding 1 to any single smaller ordinal
• C. ω contains more elements than every finite ordinal, making it uncountable
• D. ω can only be constructed via transfinite induction, not direct set-theoretic definition

Answer: (B)

The defining property of ω as a limit ordinal is that no ordinal n satisfies ω = n + 1. For any finite n, n + 1 is still finite and strictly less than ω. ω is 'approached from below' by the sequence 0, 1, 2, … but never reached by a single successor step. Option A is wrong: ω is infinite, not finite. Option C is wrong: ω is countably infinite (it is the order type of ℕ).

Question 3

Nearly every ordinal greater than 0 is a successor ordinal.

• T. True
• F. False

Answer: False

The classification theorem states every ordinal is exactly one of: zero (0), a successor ordinal (β + 1 for some β), or a limit ordinal. Limit ordinals — ω, ω + ω, ω², and infinitely many others — are not successors of any single predecessor. Ignoring this third category is the central misconception about ordinal structure.

Question 4

ω equals the set of all finite ordinals: {0, 1, 2, 3, …}.

• T. True
• F. False

Answer: True

In the von Neumann construction, ω is defined as the union of all finite ordinals. Since each finite ordinal n = {0, 1, …, n − 1}, their union is {0, 1, 2, 3, …} = ω. This set is itself an ordinal: its elements are exactly the ordinals strictly less than it, and they are well-ordered. ω is simultaneously a set, an ordinal, and the order type of the natural numbers.

Question 5

Why does transfinite induction require a separate 'limit case' in addition to the base case and the successor case?

Model answer: Standard induction has two cases: base (0) and successor (n → n + 1), which together cover all finite ordinals. But a limit ordinal like ω has no immediate predecessor — there is no n with n + 1 = ω, so the successor step can never 'reach' it. The limit case bridges this gap: it shows that if a property holds at every ordinal below λ, it holds at the limit ordinal λ itself. Without this third case, inductive proofs could not cross the infinite gaps separating limit ordinals from all successor ordinals below them.

This three-way structure mirrors the three-way ordinal classification: 0, successors, limits. Every inductive argument over the ordinals must explicitly handle all three. The limit case is often the most subtle — it requires showing the property 'survives' taking a supremum over an infinite initial segment, not just a single successor step.