Table of contents |
2 Proof 3 Notes |
Let f : Ord → Ord be a normal function. Then for every α ∈ Ord, there exists a β ∈ Ord such that β ≥ α and f(β) = β.
First of all, it is clear that for any α ∈ Ord, f(α) ≥ &alpha. If this was not the case, we could choose a minimal α with f(α) < α; then, since f is normal and thus monotone, f(f(α)) < f(α), which is a contradiction to α being minimal.
We now declare a sequence <αn> (n < ω) by setting α0 = α, and αn + 1 = f(αn) for n < ω, and define β = sup <αn>. There are three possible cases:
It is easily checked that the function f : Ord → Ord, f(α) = אα is normal; thus, there exists an ordinal Θ such that Θ = אΘ. In fact, the above lemma shows that there are infinitely many such Θ.Formal version
Proof
Notes