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The recursive definitions of ordinal addition and ordinal multiplication arewhere the limit clauses apply to nonzero limit ordinals. Ordinal exponentiation has ${\alpha }^0=1$, ${\alpha }^{\beta +1}={\alpha }^{\beta }\cdot \alpha$, and for $\alpha >0$, ${\alpha }^{\lambda }={\sup }_{\beta <\lambda }{\alpha }^{\beta }$. Complete the zero-base convention by $0^{\beta }=0$ for $\beta >0$, with $0^0=1$. Stating the zero-base exception avoids including the value $0^0$ in a naive supremum at a positive limit.

For finite $n$, induction gives $1+n=n+1$. Now prove commutativity for finite $m,n$ by induction on $m$. The case $m=0$ is immediate. Associativity, the finite identity just obtained, and the induction hypothesis giveThus $\boxed{m+n=n+m\text{ for finite ordinals}}$.

It fails for general ordinals: $\boxed{1+\omega =\omega <\omega +1}$. The equality follows from the limit clause, since the finite sums $1+n$ are cofinal in $\omega$.

For nonzero $\alpha$, consider the exponents $\eta$ with ${\omega }^{\eta }\le \alpha$. The set is nonempty because it contains zero, and bounded because ${\omega }^{\eta }\ge \eta$ by transfinite induction. Let $\delta$ be its supremum. If this supremum is a successor it is attained; if it is a limit, continuity of ordinal exponentiation gives ${\omega }^{\delta }\le \alpha$. The zero case is immediate. Thus $\delta$ itself belongs to the set and is its maximum. Maximality givesPut $\rho ={\omega }^{\delta }>0$. Since ${\omega }^{\delta +1}=\rho \cdot \omega ={\sup }_{N<\omega }\rho N$, there is a least positive natural number $N$ with $\alpha <\rho N$. We have $N\ge 2$ because $\rho \le \alpha$. Set $n=N-1$. ThenFor ordinals $\xi \le \alpha$, the final segment of $\alpha$ after its initial segment $\xi$ has an order type $\zeta$ satisfying $\alpha =\xi +\zeta$. Apply this with $\xi =\rho n$. If $\zeta \ge \rho$, monotonicity would give $\alpha \ge \rho n+\rho =\rho (n+1)$, a contradiction. ThereforeThese intervals also make the leading exponent and coefficient unique. The remainder is unique because addition by a fixed left prefix is strictly increasing in its right operand.

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.

The recursive ordinal addition definitions are $a+0=a$, $a+(b+1)=(a+b)+1$, and $a+\lambda ={\sup }_{b<\lambda }(a+b)$ for a nonzero limit ordinal. Ordinal multiplication is $a\cdot 0=0$, $a\cdot (b+1)=a\cdot b+a$, and $a\cdot \lambda ={\sup }_{b<\lambda }a\cdot b$. For $a>0$, ordinal exponentiation is $a^0=1$, $a^{b+1}=a^b\cdot a$, and $a^{\lambda }={\sup }_{b<\lambda }{a}^b$; the exceptional zero base has $0^0=1$ and $0^b=0$ for $b>0$.

Transfinite induction gives ${\omega }^a\ge a$. At zero this is immediate; at a successor, ${\omega }^{b+1}={\omega }^b\omega \ge {\omega }^b+1\ge b+1$. At a limit, continuity and the induction hypothesis give ${\sup }_{b<a}{\omega }^b\ge {\sup }_{b<a}b=a$. Also $a\mapsto {\omega }^a$ is strictly increasing: its successor values strictly increase, and continuity preserves the strict inequality between any two distinct indices.

For $\alpha >0$, ${\omega }^{\alpha +1}>{\omega }^{\alpha }\ge \alpha$, so there is a least $\delta$ with ${\omega }^{\delta }>\alpha$. It cannot be zero and cannot be a limit: at a limit, a supremum greater than $\alpha$ has a term already greater than $\alpha$, contradicting minimality. Thus $\delta ={\alpha }_0+1$, andStrict increase proves uniqueness, and ${\alpha }_0\le {\omega }^{{\alpha }_0}\le \alpha$ proves the requested index bound.

Choose the largest positive integer $a_0$ with ${\omega }^{{\alpha }_0}{a}_0\le \alpha$. It exists because $\alpha <{\omega }^{{\alpha }_0}\omega ={\sup }_{m<\omega }{\omega }^{{\alpha }_0}m$. The remaining tail of the ordinal interval has an order type $\rho$ with $\alpha ={\omega }^{{\alpha }_0}{a}_0+\rho$; maximality implies $\rho <{\omega }^{{\alpha }_0}$. If $\rho >0$, repeat the construction on it. The leading exponents strictly decrease, and there is no infinite strictly decreasing sequence of ordinals: its set of values would have a least member followed by a smaller one. Hence the process terminates, yielding the Cantor normal formThe first exponent is determined by the preceding band, the first coefficient by maximality, and the tail by the ordinal interval left after that initial segment. Recursion therefore proves uniqueness of every exponent and coefficient. The PDF's second coefficient is printed ${\alpha }_1$ rather than $a_1$; the stated natural-number coefficients and the construction identify the intended expression.

For the final assertion, powers of $\omega$ are additively indecomposable: if $\tau <{\omega }^{\eta }$, then $\tau +{\omega }^{\eta }={\omega }^{\eta }$ for $\eta >0$. One proves this by induction on $\eta$. At a successor, bound $\tau$ by a finite multiple of ${\omega }^{\eta -1}$ and use the supremum over all such multiples; at a limit, bound $\tau$ by a smaller power and use continuity. For $\eta =0$ the only possible $\tau$ is zero. Thus a smaller leading exponent is absorbed when a larger-leading ordinal is added on its right. If $\beta ,\gamma$ have unequal leading exponents, either $\beta +\gamma =\gamma$ or $\gamma +\beta =\beta$. If their leading exponents coincide, the leading coefficient of either sum is the sum of their two positive leading coefficients; it exceeds each coefficient, so neither absorption equality holds. Zero is never commensurable with an ordinal. Consequently commensurability is exactly equality of the leading exponent, equivalently membership of both positive ordinals in one of the stated bands.

The recursive definition of ordinal addition is $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ for a nonzero limit ordinal $\lambda$. The synthetic definition is the order type of the ordered disjoint union of a copy of $\alpha$ followed by a copy of $\beta$.

For the synthetic operation, adding the empty order does nothing, adding a successor appends one final point, and concatenation with a limit order is the increasing union of the initial concatenations with all its proper initial segments. Their order types have supremum ${\sup }_{\beta <\lambda }(\alpha +\beta )$. Thus the synthetic operation satisfies the recursive equations. Transfinite induction gives uniqueness of the recursively defined operation, proving equivalence.

The recursive definitions of ordinal multiplication and ordinal exponentiation areThe last equation in each line is for nonzero limit $\lambda$. In particular the convention $0^0=1$ belongs to this recursive definition.

No such ordinal exists. For $\alpha >1$, ordinal multiplication gives ${\alpha }^2=\alpha \alpha \ge \alpha \cdot 2=\alpha +\alpha >\alpha$. Since ordinal exponentiation is nondecreasing in the exponent for base $\alpha >1$, ${\alpha }^{\omega }\ge {\alpha }^2>\alpha$.

No. The definition of ordinal exponentiation gives $2^{\omega }={\sup }_n{2}^n=\omega ={\sup }_n{3}^n=3^{\omega }$, although $2<3$.

For every countable ordinal $\beta$, ${\omega }^{\beta }$ is countable: successor stages use a countable product, and countable limit stages a countable union. Hence ${\omega }^{{\omega }_1}={\sup }_{\beta <{\omega }_1}{\omega }^{\beta }\le {\omega }_1$. Conversely ${\omega }^{\beta }\ge \beta$ by transfinite induction, so these values are cofinal in ${\omega }_1$. Therefore ${\omega }^{{\omega }_1}={\omega }_1$.

It is not the least fixed point. Set $a_0=1$, $a_{n+1}={\omega }^{a_n}$ and define epsilon zero as ${\epsilon }_0={\sup }_n{a}_n$. This is countable, and continuity of ordinal exponentiation at limits gives ${\omega }^{{\epsilon }_0}={\sup }_n{\omega }^{a_n}={\epsilon }_0$. Any fixed point of $\alpha \mapsto {\omega }^{\alpha }$ dominates every $a_n$ by induction, hence dominates ${\epsilon }_0$. Thus the least fixed point is ${\epsilon }_0<{\omega }_1$.

Ordinal exponentiation is defined by transfinite recursion:when $\lambda$ is a nonzero limit ordinal. If $\alpha \ge 1$, each successor step and each supremum show that $\beta \mapsto {\alpha }^{\beta }$ is nondecreasing. If $\alpha \ge 2$, transfinite induction strengthens this to strict increase: ${\alpha }^{\beta }<{\alpha }^{\beta +1}$, and the value at a limit exceeds every earlier value.

Strict increase can disappear when the base changes. For example, with $\omega <\omega +1<\omega +2$,because for every positive finite $n$, both finite powers have leading term ${\omega }^n$, and taking the supremum gives ${\omega }^{\omega }$.

We next proveApply transfinite induction to $\gamma$. The zero and successor cases follow directly from the recursive definitions and the associativity of ordinal multiplication. At a limit $\lambda$, continuity of ordinal multiplication in its right argument givesThe superficially similar identity $(\alpha \beta {)}^{\gamma }={\alpha }^{\gamma }{\beta }^{\gamma }$ is false: for $\alpha =\omega$, $\beta =2$, and $\gamma =2$,

Finally,For the first equivalence, bases zero and one fail, while strict increase embeds every countable ordinal into the powers of any base $\alpha \ge 2$. For the second, $\alpha \ge {\omega }_2$ immediately gives ${\alpha }^{{\omega }_1}\ge \alpha$. Conversely, if $\alpha <{\omega }_2$, then $|\alpha |\le {\aleph }_1$. Every exponent below the first uncountable ordinal ${\omega }_1$ is countable, so each ${\alpha }^{\beta }$ has cardinality at most ${\aleph }_1$; the supremum of ${\aleph }_1$ such ordinals still has cardinality at most ${\aleph }_1$ and is therefore below the second uncountable ordinal ${\omega }_2$.

Ordinal exponentiation is defined byfor limit $\lambda$. Transfinite induction gives ${\omega }^{\alpha }\ge \alpha$. Since ${\omega }^{\alpha +1}>\alpha$, there is a least ${\alpha }^{\ast }$ with ${\omega }^{{\alpha }^{\ast }}>\alpha$. If nonzero ${\alpha }^{\ast }$ were a limit, the defining supremum would imply ${\omega }^{\beta }>\alpha$ for some $\beta <{\alpha }^{\ast }$, a contradiction. Hence ${\alpha }^{\ast }$ is a successor.

For $\alpha >0$, write ${\alpha }^{\ast }=\beta +1$. Then ${\omega }^{\beta }\le \alpha <{\omega }^{\beta +1}$. There is a largest positive integer $n$ with ${\omega }^{\beta }n\le \alpha$, and ordinal division givesRepeating on $\gamma$ terminates because there is no infinite strictly decreasing sequence of ordinals, producing the Cantor normal form

For two monomials,For general $\alpha +{\alpha }^{\prime }$, compare the leading exponent ${\beta }_1^{\prime }$ with the exponents of $\alpha$: discard every trailing term of $\alpha$ having exponent below ${\beta }_1^{\prime }$, combine coefficients if the last retained exponent equals ${\beta }_1^{\prime }$, and then append all remaining terms of ${\alpha }^{\prime }$. This is its Cantor normal form.

For a nonzero ordinal $\alpha$, letThe set is nonempty because $0\in S$, and it is an initial segment because ordinal exponentiation by $\omega$ is strictly increasing. Put $\beta =\sup S$.

If $\beta$ is a successor, the definition of the supremum forces $\beta \in S$. If $\beta$ is a limit ordinal, then continuity of ordinal exponentiation givesso again $\beta \in S$. Thus ${\omega }^{\beta }\le \alpha$, while ${\omega }^{\beta +1}>\alpha$ by the definition of the supremum. Hence $\beta$ is the greatest required exponent.

Apply division by an additively indecomposable ordinal with $\lambda ={\omega }^{\beta }$:Since $\alpha \ge {\omega }^{\beta }$, the quotient $q$ is nonzero. It must be finite: if $q\ge \omega$, thencontradicting the maximality of $\beta$. Writing $q=n$ givesThis is the leading-term decomposition.

The Hartogs theorem says that for every set $X$ there is an ordinal that does not inject into $X$. Indeed, the well-orderings of subsets of $X$ form a set, since each is a relation in $\mathcal{P}(X\times X)$. By replacement, their order types form a set of ordinals. Its supremum plus one cannot be the order type of a well-ordered subset of $X$, and hence cannot inject into $X$. The least such ordinal is the Hartogs ordinal $h(X)$.

Ordinal exponentiation is defined byfor nonzero limit $\lambda$. Fix $\alpha ,\beta$ and induct on $\gamma$. The zero case is immediate. If the identity holds at $\gamma$, thenAt a limit $\lambda$, continuity of exponentiation and right continuity of ordinal multiplication give

Put $\rho ={\omega }^{\alpha }$. The setis nonempty and bounded, so let $\beta =\sup B$. Continuity of multiplication in its right argument gives $\rho \beta \le \gamma$. There is a unique tail $\delta$ with $\gamma =\rho \beta +\delta$. If $\delta \ge \rho$, then
$\rho (\beta +1)=\rho \beta +\rho \le \gamma$, contrary to maximality. Hence $\delta <\rho$. If
$\rho \beta +\delta =\rho {\beta }^{\prime }+{\delta }^{\prime }$ with both remainders below $\rho$ and $\beta <{\beta }^{\prime }$, thena contradiction. This proves existence and uniqueness in the division by an additively indecomposable ordinal.

Now let $X$ be nonempty and well ordered. Its least element is not a limit point, so $X^{\prime }\ne X$. Every derivative is itself a well-ordered subset. If $X^{(\alpha )}$ were nonempty for every $\alpha <h(X)$, then each differencewould be nonempty. Choosing its least member would inject $h(X)$ into $X$, since the nested successive differences are disjoint. This contradicts Hartogs' lemma. Thus some derivative is empty, as asserted by the derived-set iteration of a well-order.

For an ordinal $\xi$, a point $\gamma <\xi$ is a limit point exactly when it is a nonzero limit ordinal. Division by $\omega$ writes $\gamma =\omega \beta +n$ with $n<\omega$; this is a nonzero limit exactly when $n=0$ and $\beta >0$. HenceWithin this derived set, $\omega \beta$ is a limit point exactly when $\beta$ is a nonzero limit ordinal, equivalently $\beta =\omega \eta$ for some $\eta >0$. Therefore the derived sets of an ordinal satisfyFinally ${\omega }^{\prime }=\varnothing$, so the index of $\omega$ is $1$, while
$({\omega }^2{)}^{\prime }=\{\omega ,2\omega ,3\omega ,\dots \}$ and
$({\omega }^2{)}^{\prime \prime }=\varnothing$, so the index of ${\omega }^2$ is $2$.