Let ''f'' : [[ordinal number|Ord]] →→ Ord be a [[normal function]]. Then for every αα ∈∈ Ord, there exists a ββ ∈∈ Ord such that ββ ≥≥ αα and ''f''(ββ) = ββ.
== Proof ==
We know that ''f''(γγ) ≥≥ γγ for all ordinals γγ. We now declare an increasing sequence <αα<sub>''n''</sub>> (''n'' < ωω) by setting αα<sub>0</sub> = αα, and αα<sub>''n''+1</sub> = ''f''(αα<sub>''n''</sub>) for ''n'' < ωω, and define ββ = sup <αα<sub>''n''</sub>>. Clearly, ββ ≥≥ αα. Since ''f'' commutes with [[supremum|suprema]], we have
(The last step uses the fact that the sequence <αα<sub>''n''</sub>> increases).
== Example application ==
It is easily checked that the function ''f'' : Ord →→ Ord, ''f''(αα) = אא<sub>αα</sub> is normal (see [[aleph number]]); thus, there exists an ordinal ΘΘ such that ΘΘ = אא<sub>ΘΘ</sub>. In fact, the above lemma shows that there are infinitely many such ΘΘ.
