Normal function

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; that is, the inverse image of the one-point open set {λ + 1} is the set {λ}, which is not open when λ is a limit ordinal. If β is a fixed ordinal, then the functions f(α) = β + α, f(α) = β × α (for β ≥ 1), and f(α) = β^α (for β ≥ 2) are all normal.

More important examples of normal functions are given by the aleph numbers ${\displaystyle f(\alpha )=\aleph _{\alpha }}$, which connect ordinal and cardinal numbers, and by the beth numbers ${\displaystyle f(\alpha )=\beth _{\alpha }}$.

If f is normal, then for any ordinal α,

f(α) ≥ α.[1]

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 "≤", set δ = sup S and consider three cases:

• if δ = 0, then S = {0} and sup f(S) = f(0);
• if δ = ν + 1 is a successor, then there exists s in S with ν < s, so that δ ≤ s. Therefore, f(δ) ≤ f(s), which implies f(δ) ≤ sup f(S);
• if δ is a nonzero limit, pick any ν < δ, and an s in S such that ν < s (possible since δ = sup S). Therefore, f(ν) < f(s) so that f(ν) < sup f(S), yielding f(δ) = sup {f(ν) : ν < δ} ≤ sup f(S), as desired.

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.

• Johnstone, Peter (1987), Notes on Logic and Set Theory, Cambridge University Press, ISBN 978-0-521-33692-5.