Normal function

In axiomatic set theory, a function $f : Ord → Ord$ is called normal (or a normal function) if it is continuous (with respect to the order topology) and strictly monotonically increasing. This is equivalent to the following two conditions:


 * 1) For every limit ordinal $γ$ (i.e. $γ$ is neither zero nor a successor), it is the case that $f&hairsp;(γ) = sup\{f&hairsp;(ν) : ν < γ\}$.
 * 2) For all ordinals $α < β$, it is the case that $f&hairsp;(α) < f&hairsp;(β)$.

Examples
A simple normal function is given by $f&hairsp;(α) = 1 + α$ (see ordinal arithmetic). But $f&hairsp;(α) = α + 1$ 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 $f&hairsp;(α) = β + α$, $f&hairsp;(α) = β × α$ (for $β ≥ 1$), and $f&hairsp;(α) = β^{α}$ (for $β ≥ 2$) are all normal.

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

Properties
If $f$ is normal, then for any ordinal $α$,

Proof: If not, choose $γ$ minimal such that $f&hairsp;(α) ≥ α$. Since $f$ is strictly monotonically increasing, $f&hairsp;(γ) < γ$, contradicting minimality of $γ$.

Furthermore, for any non-empty set $S$ of ordinals, we have

Proof: "≥" follows from the monotonicity of $f$ and the definition of the supremum. For "$f&hairsp;(f&hairsp;(γ)) < f&hairsp;(γ)$", set $f&hairsp;(sup S) = sup f&hairsp;(S)$ and consider three cases:
 * if $≤$, then $δ = sup S$ and $δ = 0$;
 * if $S = \{0\}$ is a successor, then there exists $s$ in $S$ with $sup f&hairsp;(S) = f&hairsp;(0)$, so that $δ = ν + 1$. Therefore, $ν < s$, which implies $δ ≤ s$;
 * if $δ$ is a nonzero limit, pick any $f&hairsp;(δ) ≤ f&hairsp;(s)$, and an $s$ in $S$ such that $f&hairsp;(δ) ≤ sup f&hairsp;(S)$ (possible since $ν < δ$). Therefore, $ν < s$ so that $δ = sup S$, yielding $f&hairsp;(ν) < f&hairsp;(s)$, 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 $f&hairsp;(ν) < sup f&hairsp;(S)$, called the derivative of $f$, such that $f&hairsp;(δ) = sup \{f&hairsp;(ν) : ν < δ\} ≤ sup f&hairsp;(S)$ is the $α$-th fixed point of $f$. For a hierarchy of normal functions, see Veblen functions.