Revision as of 07:36, 29 December 2009 edit 71.119.248.102 (talk) →Background and formal statement ← Previous edit		Revision as of 15:23, 29 December 2009 edit undo JRSpriggs (talk \| contribs) Autopatrolled, Extended confirmed users, Pending changes reviewers 18,932 edits →Background and formal statement: sup {} = 0, but f(0) may not be 0 Next edit →
Line 5: * ''f'' is '''strictly increasing''': ''f''(α) < f(β) whenever α < β. * ''f'' is '''continuous''': for every limit ordinal λ, ''f''(λ) = sup { f(α) : α < λ }. It can be shown that if ''f'' is normal then ''f'' commutes with [[supremum\|suprema]]; for any nonempty set ''A'' of ordinals, :''f''(sup ''A'') = sup {''f''(α) : α ∈ ''A'' }. Indeed, if sup ''A'' is ~~not~~ a ~~limit~~successor ordinal then sup ''A'' is an element of ''A'' and the equality follows from the increasing property of ''f''. If sup ''A'' is a limit ordinal then the equality follows from the continuous property of ''f''. A '''fixed point''' of a normal function is an ordinal β such that ''f''(β) = β.

Fixed-point lemma for normal functions: Difference between revisions