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== Proof ==
The first step of the proof is to verify that <math>f(\gamma)\ge\gamma</math> for all ordinals <math>\gamma</math> and that <math>f</math> commutes with suprema. Given these results, inductively define an increasing sequence <math>\langle\alpha_n\rangle_{n<\omega}</math> by setting <math>\alpha_0 = \alpha</math>, and <math>\alpha_{n+1} = f(\alpha_n)</math> for <math>n\in\omega</math>. Let <math>\beta = \sup_{n<\omega} \alpha_n</math>, so <math>\beta\ge\alpha</math>. Moreover, because <math>f</math> commutes with suprema,
:<math>f(\beta) = f(\sup_{n<\omega} \alpha_n)</math>
:<math>\qquad = \sup_{n<\omega} f(\alpha_n)</math>
:<math>\qquad = \sup_{n<\omega} \alpha_{n+1}</math>
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== Example application ==
The function ''f'' : Ord → Ord, ''f''(''α'') = ω<sub>''α''</sub> is normal (see [[initial ordinal]]). Thus, there exists an ordinal ''θ'' such that ''θ'' = ω<sub>''θ''</sub>. In fact, the lemma shows that there is a closed, unbounded class of such ''θ''.
==References==
{{refbegin}}
* {{cite book
| author = Levy, A.
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| year = 1908
| pages = 280–292
| doi= 10.2307/1988605
| issue = 3
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| issn= 0002-9947| doi-access = free
}}
{{refend}}
[[Category:Ordinal numbers]]
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