Content deleted Content added
mNo edit summary |
mNo edit summary |
||
Line 9:
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 <α<sub>''n''</sub>> (''n'' < ω) by setting α<sub>0</sub> = α, and α<sub>''n'' + 1</sub> = ''f''(α<sub>''n''</sub>) for ''n'' < ω, and define β = sup <α<sub>''n''</sub>>. There are three possible cases:
# β = 0. Then we have α<sub>''n''</sub> = 0 for all ''n'', and thus ''f''(β) = 0.
| |||