[proofplan]
Both claims are proved by transfinite induction, splitting on the three cases of the ordinal hierarchy (zero, successor, limit). Strict monotonicity (1) extends the local successor-increase condition — built into the definition of a normal function — to arbitrary comparisons by induction on the upper ordinal, with the continuity condition handling limit stages. Domination of the identity (2) follows by induction on $\alpha$: at successors we use (1) to jump up, and at limits we use continuity to pass the inductive hypothesis through the supremum. The two claims are linked — monotonicity is needed to lift the successor bound to every smaller ordinal in the limit case of (2).
[/proofplan]
[step:Fix the two defining properties of a normal function]
Recall that a function $f$ on the ordinals is **normal** if it is:
- **Strictly increasing at successors:** $f(\gamma) < f(\gamma^+)$ for every ordinal $\gamma$, equivalently $f(\gamma^+) \geq f(\gamma)^+ > f(\gamma)$;
- **Continuous at limits:** for every limit ordinal $\lambda$,
\begin{align*}
f(\lambda) = \sup\{f(\gamma) : \gamma < \lambda\}.
\end{align*}
Throughout the proof these two properties are the only facts about $f$ we use.
[/step]
[step:Prove strict monotonicity by induction on the larger ordinal $\beta$]
We prove claim (1): for every ordinal $\alpha$,
\begin{align*}
\alpha < \beta \implies f(\alpha) < f(\beta).
\end{align*}
Fix $\alpha$. We proceed by transfinite induction on $\beta$, over ordinals $\beta > \alpha$.
**Base case: $\beta = \alpha^+$.** The successor property of normal functions gives $f(\alpha) < f(\alpha^+) = f(\beta)$ directly.
**Successor step: $\beta = \gamma^+$ with $\gamma > \alpha$.** The inductive hypothesis (applied to $\gamma$, which satisfies $\alpha < \gamma < \beta$) gives $f(\alpha) < f(\gamma)$. The successor property of $f$ gives $f(\gamma) < f(\gamma^+) = f(\beta)$. Chaining these two strict inequalities:
\begin{align*}
f(\alpha) < f(\gamma) < f(\beta).
\end{align*}
**Limit step: $\beta$ is a limit ordinal with $\alpha < \beta$.** Since $\alpha < \beta$ and $\beta$ is a limit, the successor $\alpha^+$ also satisfies $\alpha^+ < \beta$ (a limit ordinal contains the successor of every ordinal strictly below it). By the continuity of $f$ at the limit $\beta$,
\begin{align*}
f(\beta) = \sup\{f(\gamma) : \gamma < \beta\}.
\end{align*}
Since $\alpha^+ < \beta$, the ordinal $f(\alpha^+)$ is among the terms of this supremum, so $f(\alpha^+) \leq f(\beta)$. Combined with $f(\alpha) < f(\alpha^+)$ from the base case,
\begin{align*}
f(\alpha) < f(\alpha^+) \leq f(\beta).
\end{align*}
This exhausts the three cases, so claim (1) holds.
[guided]
We want to prove $f(\alpha) < f(\beta)$ whenever $\alpha < \beta$. The definition of "normal" only directly tells us about successor steps: $f(\gamma) < f(\gamma^+)$. How do we leverage a single-step increase into a strict increase across arbitrary ordinal gaps? Transfinite induction on $\beta$ is the natural tool — at each $\beta$ we relate $f(\beta)$ to $f$ at some strictly smaller ordinal, which either is $\alpha$ itself or lies strictly between $\alpha$ and $\beta$.
Fix $\alpha$ and induct on $\beta > \alpha$.
**Base case: $\beta = \alpha^+$.** This is exactly the successor-step property of a normal function: $f(\alpha) < f(\alpha^+)$.
**Successor step: $\beta = \gamma^+$ with $\gamma > \alpha$.** We must show $f(\alpha) < f(\gamma^+)$. Since $\alpha < \gamma < \beta$, the inductive hypothesis at $\gamma$ gives $f(\alpha) < f(\gamma)$. The successor property at $\gamma$ gives $f(\gamma) < f(\gamma^+) = f(\beta)$. Together:
\begin{align*}
f(\alpha) < f(\gamma) < f(\beta).
\end{align*}
**Limit step: $\beta$ is a limit ordinal with $\alpha < \beta$.** Here is where continuity enters. By definition of a limit ordinal, if $\alpha < \beta$ then $\alpha^+ < \beta$ as well — limits are closed under successor. Continuity says
\begin{align*}
f(\beta) = \sup\{f(\gamma) : \gamma < \beta\},
\end{align*}
and since $\alpha^+ < \beta$, the value $f(\alpha^+)$ is one of the values whose supremum is taken. Therefore $f(\alpha^+) \leq f(\beta)$. Combined with the base-case inequality $f(\alpha) < f(\alpha^+)$:
\begin{align*}
f(\alpha) < f(\alpha^+) \leq f(\beta).
\end{align*}
Note the small subtlety: the continuity equation gives $f(\beta)$ as a *supremum*, not a *strict* upper bound, so we only get $f(\alpha^+) \leq f(\beta)$. The strictness of the final inequality comes from the successor step $f(\alpha) < f(\alpha^+)$, not from the limit step. This is why we cannot skip the intermediate ordinal $\alpha^+$.
[/guided]
[/step]
[step:Prove $f(\alpha) \geq \alpha$ by transfinite induction on $\alpha$]
We prove claim (2): $f(\alpha) \geq \alpha$ for every ordinal $\alpha$. We induct on $\alpha$.
**Base case: $\alpha = 0$.** Since $0$ is the least ordinal, $f(0) \geq 0$ holds automatically.
**Successor step: $\alpha = \delta^+$.** The inductive hypothesis at $\delta$ gives $f(\delta) \geq \delta$. By claim (1) already proved, $f(\delta) < f(\delta^+)$, i.e. $f(\delta^+) \geq f(\delta)^+$. Chaining,
\begin{align*}
f(\alpha) = f(\delta^+) \geq f(\delta)^+ \geq \delta^+ = \alpha,
\end{align*}
where the middle inequality uses that the successor operation is monotone: $\mu \geq \nu$ implies $\mu^+ \geq \nu^+$, applied with $\mu = f(\delta)$ and $\nu = \delta$.
**Limit step: $\alpha = \lambda$ is a limit ordinal.** By the continuity of $f$,
\begin{align*}
f(\lambda) = \sup\{f(\gamma) : \gamma < \lambda\}.
\end{align*}
For each $\gamma < \lambda$, the inductive hypothesis gives $f(\gamma) \geq \gamma$. Taking the supremum over all such $\gamma$ preserves the inequality term by term:
\begin{align*}
f(\lambda) = \sup\{f(\gamma) : \gamma < \lambda\} \geq \sup\{\gamma : \gamma < \lambda\} = \lambda,
\end{align*}
where the last equality is the standard characterisation of a limit ordinal as the supremum of the ordinals strictly below it.
This exhausts the three cases, completing the proof of claim (2).
[guided]
Claim (2) says $f$ dominates the identity. The strategy is again transfinite induction on $\alpha$, with three cases.
**Base case: $\alpha = 0$.** Every ordinal is $\geq 0$, so $f(0) \geq 0$ needs no further justification.
**Successor step: $\alpha = \delta^+$.** The inductive hypothesis at $\delta$ gives $f(\delta) \geq \delta$. We want $f(\delta^+) \geq \delta^+$. Here claim (1), already proved in the previous step, does the work: $f(\delta) < f(\delta^+)$ means $f(\delta)^+ \leq f(\delta^+)$. Since the successor operation on ordinals is monotone — if $\mu \geq \nu$ then $\mu^+ \geq \nu^+$ — the hypothesis $f(\delta) \geq \delta$ lifts to $f(\delta)^+ \geq \delta^+$. Chaining:
\begin{align*}
f(\delta^+) \geq f(\delta)^+ \geq \delta^+.
\end{align*}
Why does this step need claim (1)? Because the definition of "normal" only guarantees $f(\delta) < f(\delta^+)$; we had to prove strict monotonicity first in order to bootstrap from $f(\delta) \geq \delta$ to $f(\delta^+) \geq \delta^+$. This is why claim (1) was proved before claim (2).
**Limit step: $\alpha = \lambda$ is a limit ordinal.** By continuity of $f$:
\begin{align*}
f(\lambda) = \sup\{f(\gamma) : \gamma < \lambda\}.
\end{align*}
The inductive hypothesis says $f(\gamma) \geq \gamma$ for every $\gamma < \lambda$. Suprema respect termwise inequalities: if $a_i \leq b_i$ for every $i$, then $\sup_i a_i \leq \sup_i b_i$. Applying this with $a_\gamma = \gamma$ and $b_\gamma = f(\gamma)$:
\begin{align*}
\sup\{f(\gamma) : \gamma < \lambda\} \geq \sup\{\gamma : \gamma < \lambda\}.
\end{align*}
Finally, for a limit ordinal $\lambda$, the defining property is $\lambda = \sup\{\gamma : \gamma < \lambda\}$. Combining:
\begin{align*}
f(\lambda) = \sup\{f(\gamma) : \gamma < \lambda\} \geq \sup\{\gamma : \gamma < \lambda\} = \lambda.
\end{align*}
Both claims hold across all three ordinal cases, so the proof by transfinite induction is complete.
[/guided]
[/step]
