Let P(α): α∉α. The general step assumes α∈α. What follows?

A contradiction, since it would give α<α in a strict order

That every β<α satisfies β∈β

Nothing, because α∈α is consistent

Exercise · ZW-TRANSFINITE-INDUCTION-006 · ★★☆ Intermediate

Let \(P(\alpha): \alpha\notin\alpha\). The general step assumes \(\alpha\in\alpha\). What follows?

Choose one option.

Show answer

D. A contradiction, since it would give \(\alpha<\alpha\) in a strict order

Explanation

If \(\alpha\in\alpha\), we would have \(\alpha<\alpha\), which contradicts the strictness of the order.

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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 \(P(\alpha): \alpha\notin\alpha\). The general step assumes \(\alpha\in\alpha\). What follows?

Explanation

Review the concept: Transfinite induction

More practice problems: transfinite induction