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# β = 0. Then we have α<sub>''n''</sub> = 0 for all ''n'', and thus ''f''(β) = 0.
# β = δ + 1 for an ordinal number δ. Then there exists ''m'' < ω such that for all ''n'' ≥ ''m'', α<sub>''n''</sub> = δ + 1. It follows that ''f''(δ + 1) = ''f''(α<sub>''m''</sub>) = α<sub>''m'' + 1</sub> = δ + 1, and thus ''f''(β) = β.
# β is a [[limit ordinal]]. We first observe that sup <''f''(ν) : ν < β> = sup <''f''(α<sub>''n''</sub>) : ''n'' < ω>. "≥" is trivial; for "≤", we choose ν < β, then find an ''n'' with α<sub>''n''</sub> > ν, and since ''f'' is monotone, we have ''f''(α<sub>''n''</sub>) > ''f''(ν). Now we have ''f''(β) = sup <''f''(ν) : ν < β> (since ''f'' is continuous), and thus ''f''(β) = sup <''f''(α<sub>''n''</sub>) : ''n'' < ω> = sup < α<sub>''n''</sub> : ''n'' < ω > = β.
== Notes ==
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