Exercise · ZW-TRANSFINITE-INDUCTION-003 · ★★☆ Intermediate
How is the successor case \(\alpha=\beta+1\) handled in the general version?
Choose one option.
Show answer
- C. It is covered by the general step, since \(\beta<\beta+1\)
Explanation
In the general version the successor is a special case: \(\beta<\beta+1\) lets us use the hypothesis.
Review the concept: Transfinite induction
Principle that proves a property holds for all ordinal numbers \(\mathrm{On}\) by showing that if it holds for all ordinals below a given \(\alpha\), it also holds for \(\alpha\).
\[\forall\alpha[\forall\beta<\alpha P(\beta)\to P(\alpha)]\to\forall\alpha P(\alpha)\]
More practice problems: transfinite induction
- ZW-TRANSFINITE-INDUCTION-001 What does the general step of transfinite induction require?
- ZW-TRANSFINITE-INDUCTION-002 If P(α) follows whenever P(β) holds for all β<α, what do we conclude?
- ZW-TRANSFINITE-INDUCTION-004 Which of these ordinals is a limit ordinal?
- ZW-TRANSFINITE-INDUCTION-005 On which kind of ordered sets can transfinite induction be applied?
- ZW-TRANSFINITE-INDUCTION-006 Let P(α): α∉α. The general step assumes α∈α. What follows?
- ZW-TRANSFINITE-INDUCTION-007 If 0+α=α for every ordinal α, what is 0+(ω+1)?
- ZW-TRANSFINITE-INDUCTION-008 If 0+β=β for all β<λ and λ is a limit ordinal, what is 0+λ?
- ZW-TRANSFINITE-INDUCTION-009 Is it valid to conclude that P holds for every ordinal if P(0) holds and P(β) implies P(β+1) for ev…
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