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For every ordinal base $\rho >1$, a nonzero ordinal has a unique finite ordinal addition expansion with strictly decreasing exponents ${\beta }_i$ and coefficients $0<{\xi }_i<\rho$. Choose the largest power at most the remaining ordinal and apply the ordinal division algorithm. The next remainder has smaller leading exponent, so well-foundedness forces termination. Each lower tail is smaller than its preceding power; uniqueness of division consequently fixes the exponent, coefficient and tail successively. For $\rho =\omega$ this is Cantor normal form.

For Fodor's theorem, let $\kappa$ be regular uncountable, $S\subseteq \kappa$ stationary, and $f:S\to \kappa$ regressive: $f(\alpha )<\alpha$ for every nonzero $\alpha \in S$. A club set is unbounded in $\kappa$ and closed under limits below $\kappa$; a stationary set meets every club. Removing $0$ preserves stationarity. We prove that some fibre of $f$ is stationary, not merely that one fibre is large.

We first check the club facts used in the argument. An intersection of fewer than $\kappa$ club subsets of $\kappa$ is club. Closure is immediate. For unboundedness, given $\gamma$ and clubs $(C_i{)}_{i<\mu }$ with $\mu <\kappa$, choose an increasing sequence ${\gamma }_n$ starting above $\gamma$, such that ${\gamma }_{n+1}$ is above a point of every $C_i$ beyond ${\gamma }_n$. Regularity bounds the supremum of the fewer-than-$\kappa$ points at each step. Then $\delta ={\sup }_{n<\omega }{\gamma }_n<\kappa$, and the chosen points in each $C_i$ are cofinal in $\delta$; closure puts $\delta$ in their intersection.

For a sequence $(C_{\xi }{)}_{\xi <\kappa }$ of clubs its diagonal intersectionis also club. To obtain a point above $\gamma$, recursively choose ${\gamma }_{n+1}>{\gamma }_n$ in ${\bigcap }_{\xi \le {\gamma }_n}{C}_{\xi }$, possible by the preceding result, and take the supremum $\delta$ of the sequence. For each $\xi <\delta$, every sufficiently late ${\gamma }_n$ belongs to $C_{\xi }$, so $\delta \in C_{\xi }$. For closure, if diagonal-intersection points are cofinal in $\delta <\kappa$, then for each $\xi <\delta$ their sufficiently late points lie in $C_{\xi }$, and closure again puts $\delta$ there.

Suppose every fibre $S_{\xi }=\{\alpha \in S:f(\alpha )=\xi \}$ were nonstationary. Choose a club $C_{\xi }$ disjoint from each fibre. Stationarity supplies $\alpha \in S\cap {\Delta }_{\xi <\kappa }{C}_{\xi }$, with $\alpha >0$. Put $\xi =f(\alpha )<\alpha$. By diagonal membership $\alpha \in C_{\xi }$, but by its definition $\alpha \in S_{\xi }$, a contradiction. ThusThis is precisely the pressing-down assertion. Regularity and uncountability were used to keep the club-building suprema below $\kappa$.

For the Cantor normal form theorem, every nonzero ordinal has a unique expressionAll operations are ordinal operations, so the order of the terms matters. The ordinal zero has the empty expression. We give the greedy existence proof and then prove uniqueness.

The powers ${\omega }^{\beta }$ are strictly increasing and continuous at limit exponents, and satisfy ${\omega }^{\beta }\ge \beta$. Hence the set of exponents with ${\omega }^{\beta }\le \alpha$ is nonempty and bounded, and its supremum ${\beta }_0$ belongs to it by continuity (or is already its last successor member). Thus there is a largest such exponent andSince the last product is the supremum of the finite multiples, there is a unique positive finite $n_0$ satisfyingThe initial segment of $\alpha$ of length ${\omega }^{{\beta }_0}{n}_0$ leaves a tail of unique order type $\rho$, soThe bound follows from strict increase of ordinal addition in its right argument: a tail at least ${\omega }^{{\beta }_0}$ would contradict the preceding interval. This is the relevant instance of the ordinal division algorithm.

If $\rho >0$, apply the same construction to it. Its leading exponent must be below ${\beta }_0$. Continue with the remainder. An infinite continuation would produce an infinite strictly descending sequence of ordinals, impossible by well-foundedness. Thus the process terminates after finitely many steps and gives the required form.

To prove uniqueness, first observe that a finite decreasing-exponent suffix with leading exponent $\gamma$ is smaller than ${\omega }^{\gamma +1}$. This follows by induction on its number of terms: the later tail is below ${\omega }^{\gamma }$, so adding it to a finite multiple of ${\omega }^{\gamma }$ stays below the next finite multiple and hence below ${\omega }^{\gamma }\cdot \omega$. Consequently the tail after a term with exponent $\beta$ is below ${\omega }^{\beta }$.

Any proposed normal form therefore places $\alpha$ in exactly the intervalIts first exponent is recoverable as the largest exponent whose power does not exceed $\alpha$, and its first coefficient is recoverable as the largest finite multiple not exceeding $\alpha$. The right remainder is unique because ordinal addition is strictly increasing in its right argument. Apply the same reasoning to the remainder, successively fixing every exponent and coefficient. This proves uniqueness without treating ordinal addition as commutative or cancelling arbitrary left summands.

Fix an ordinal $\rho >1$. The Generalized Cantor normal form asserts that every nonzero ordinal $\alpha$ has a unique expressionProducts, sums and powers here are ordinal multiplication, ordinal addition and ordinal exponentiation; the coefficients need not be finite if $\rho$ is infinite. The case $\rho =\omega$ is Cantor normal form, and a finite base on finite ordinals gives ordinary positional notation. Zero has the empty expansion.

We first prove the ordinal division algorithm. For $\delta >0$, the function $q\mapsto \delta q$ is strictly increasing, continuous at limit ordinals and unbounded. Hence there is a greatest $q$ with $\delta q\le \alpha$: the first $q^{\prime }$ with $\delta q^{\prime }>\alpha$ cannot be a limit by continuity and must equal $q+1$. The interval after the initial segment $\delta q$ has a unique order type $r$, so $\alpha =\delta q+r$. Maximality of $q$ gives $r<\delta$. These facts also prove uniqueness of $q,r$.

Similarly $\beta \mapsto {\rho }^{\beta }$ is strictly increasing, continuous at limits and unbounded, so there is a unique largest ${\beta }_0$ with ${\rho }^{{\beta }_0}\le \alpha$. Divide by ${\rho }^{{\beta }_0}$ to obtainThe quotient is positive, and $\alpha <{\rho }^{{\beta }_0+1}={\rho }^{{\beta }_0}\rho$ makes ${\xi }_0<\rho$. If $r_0>0$, repeat on $r_0$; its leading exponent is strictly smaller than ${\beta }_0$. This process terminates, since an infinite run would give an infinite descending sequence of ordinals. It constructs the required form.

For uniqueness, a finite expression with largest exponent $\beta$ and coefficients below $\rho$ has value below ${\rho }^{\beta +1}$. Prove this from the end backwards: a tail with largest exponent $\gamma <\beta$ is below ${\rho }^{\gamma +1}\le {\rho }^{\beta }$, andThus the displayed leading exponent is the uniquely determined largest $\beta$ with ${\rho }^{\beta }\le \alpha$. The ordinal division algorithm uniquely determines its coefficient and the remaining tail, and recursion proves uniqueness of the whole expression. No commutative rearrangement of ordinal sums is used.