Let P(α): 1+α=α. Evaluate P(0), P(1) and P(ω). What can be concluded?

P does not hold for every ordinal, since it fails at 0 and 1 though it holds at ω

P holds at 0 vacuously, so the induction goes through without trouble

P holds only at ω, and that is enough for all limits

P holds for every ordinal, since the general step works

Exercise · ZW-TRANSFINITE-INDUCTION-010 · ★★★ Advanced

Let \(P(\alpha): 1+\alpha=\alpha\). Evaluate \(P(0)\), \(P(1)\) and \(P(\omega)\). What can be concluded?

Choose one option.

Show answer

D. \(P\) does not hold for every ordinal, since it fails at 0 and 1 though it holds at \(\omega\)

Explanation

\(P(0)\) and \(P(1)\) are false, since \(1+0=1\) and \(1+1=2\), while \(1+\omega=\omega\) holds. Because the base case fails, induction cannot conclude that \(P\) holds for every ordinal.

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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 \(\alpha\).

\[\forall\alpha[\forall\beta<\alpha P(\beta)\to P(\alpha)]\to\forall\alpha P(\alpha)\]

Let \(P(\alpha): 1+\alpha=\alpha\). Evaluate \(P(0)\), \(P(1)\) and \(P(\omega)\). What can be concluded?

Explanation

Review the concept: Transfinite induction

More practice problems: transfinite induction