For every limit ordinalγ (i.e. γ is neither zero nor a successor), it is the case that f(γ) = sup{f(ν): ν < γ}.
For all ordinals α < β, it is the case that f(α) < f(β).
Examples
A simple normal function is given by f(α) = 1 + α (see ordinal arithmetic). But f(α) = α + 1 is not normal because it is not continuous at any limit ordinal (for example, ). If β is a fixed ordinal, then the functions f(α) = β + α, f(α) = β × α (for β ≥ 1), and f(α) = βα (for β ≥ 2) are all normal.
Proof: If not, choose γ minimal such that f(γ) < γ. Since f is strictly monotonically increasing, f(f(γ)) < f(γ), contradicting minimality of γ.
Furthermore, for any non-empty set S of ordinals, we have
f(sup S) = sup f(S).
Proof: "≥" follows from the monotonicity of f and the definition of the supremum. For "≤", consider three cases:
if sup S = 0, then S = {0} and sup f(S) = f(0) = f(sup S);
if sup S = ν + 1 is a successor, then sup S is in S, so f(sup S) is in f(S), i.e. f(sup S) ≤ sup f(S);
if sup S is a nonzero limit, then for any ν < sup S there exists an s in S such that ν < s, i.e. f(ν) < f(s) ≤ sup f(S), yielding f(sup S) = sup {f(ν): ν < sup S} ≤ sup f(S).
Every normal function f has arbitrarily large fixed points; see the fixed-point lemma for normal functions for a proof. One can create a normal function f′: Ord → Ord, called the derivative of f, such that f′(α) is the α-th fixed point of f.[2] For a hierarchy of normal functions, see Veblen functions.