Questions: Transfinite Induction

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A mathematician proves P(0) holds, and proves that for all ordinals α, if P(α) holds then P(α+1) holds. They conclude P holds for every ordinal. What is the error?

AThe base case should be P(1), not P(0)
BThis only establishes P for successor ordinals reachable from 0 by finite steps — it fails at ω because ω has no immediate predecessor, so the successor step has nothing to hand off from
CThe proof is valid — any property provable for 0 and preserved by successors holds for all ordinals
DP(0) is unnecessary if the successor step is sufficiently strong
Question 2 Multiple Choice

In a transfinite induction proof, the limit step states: if P(β) holds for all β < λ, then P(λ). Why must this step assume P holds for ALL β < λ, rather than just for the ordinal immediately before λ?

AIt is a convention that makes proofs easier to write, not a logical necessity
BBecause limit ordinals like ω have no immediate predecessor — there is no 'ordinal immediately before' them from which to inherit
CTo avoid circular reasoning, since λ is defined in terms of its predecessors
DBecause limit ordinals are uncountable and therefore cannot have immediate predecessors
Question 3 True / False

Every ordinal is exactly one of three kinds: zero, a successor ordinal, or a limit ordinal — and a transfinite induction proof must handle all three cases separately to cover every ordinal.

TTrue
FFalse
Question 4 True / False

A proof by transfinite induction that establishes P(0) (base case) and the successor step (P(α) → P(α+1)) is sufficient to prove P holds for most ordinals, including ω and beyond.

TTrue
FFalse
Question 5 Short Answer

Why does ordinary mathematical induction fail to reach the ordinal ω, and what does the limit step of transfinite induction add to fix this?

Think about your answer, then reveal below.