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== Formal version ==
Let ''f'' : [[ordinal number|Ord]] → Ord be a [[normal function]]. Then for every α ∈ Ord, there exists a β ∈ Ord such that β ≥ α and ''f''(β) = β.
== Proof ==
We know that ''f''(γ) ≥ γ for all ordinals γ. We now declare an increasing 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>>. Clearly, β ≥ α. Since ''f'' commutes with [[supremum|suprema]], we have
:''f''(β) = ''f''(sup {α<sub>''n''</sub> : ''n'' < ω})
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== Example application ==
It is easily checked that the function ''f'' : Ord → Ord, ''f''(α) = א<sub>α</sub> is normal (see [[aleph number]]); thus, there exists an ordinal Θ such that Θ = א<sub>Θ</sub>. In fact, the above lemma shows that there are infinitely many such Θ.
==References==
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