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A nonzero ordinal $\delta$ is additively closed when $\beta +\gamma <\delta$ for all $\beta ,\gamma <\delta$. These are exactly the ordinals ${\omega }^{\alpha }$: sufficiency follows from Cantor normal form, while necessity follows by splitting any leading coefficient or nonzero tail below $\delta$ into two smaller summands.

Nonzero ordinals commute under addition precisely when they are positive finite right multiples of a common ordinal. This follows by comparing their Cantor normal forms at the first exponent where they differ.

Every nonzero ordinal $\alpha$ has a greatest exponent $\beta$ such that ${\omega }^{\beta }\le \alpha$. It is the leading exponent in the Cantor normal form of $\alpha$, andfor a positive finite integer $n$ and $\gamma <{\omega }^{\beta }$.

Let nonzero ordinals $\xi ,\eta$ have leading exponents $\rho ,\sigma$. If $\eta$ is finite, then $\xi \eta$ has leading exponent $\rho$. If $\eta$ is infinite, so that $\sigma >0$, then $\xi \eta$ has leading exponent $\rho +\sigma$. This follows by multiplying the leading terms in Cantor normal form and using continuity of ordinal multiplication at limit ordinals.

Replacing every base occurrence by omega in a hereditary base representation gives a Cantor normal form below epsilon zero. Hereditary base change preserves this ordinal expression, and subtraction of one lowers it strictly. The ordinal is a termination measure even when the natural-number value increases.

For $b\ge 2$, write a number in hereditary base $b$: express its base-$b$ expansion as ${\sum }_i{b}^{e_i}{c}_i$, with $0<c_i<b$, and recursively expand every exponent $e_i$ in the same base. Let $B_{b,c}(m)$, for $c>b$, be the number obtained by replacing every occurrence of the base $b$ in this hereditary expression by $c$. Coefficients and the symbols for addition and exponentiation are retained.

For a starting number $m$, define its Goodstein sequence by $m_0=m$ andThe Goodstein function is the termination-time functionThis choice counts transitions from the initial term; a convention counting the initial term as well shifts the answer by one. For example the sequence starting at $3$ is $3,3,3,2,1,0$, so $\mathcal{G}(3)=5$. Starting at $4$ gives $4,26,41,60,83,\dots$, illustrating why descent of the numerical terms is not the termination argument.

Define the ordinal rank of a Goodstein term ${\rho }_b(m)$ by replacing the base $b$ in the hereditary expansion by $\omega$, at every exponent depth. The result is in Cantor normal form and lies below ${ϵ}_0$, called epsilon zero, the least nonzero solution of ${\omega }^{\alpha }=\alpha$. Finite hereditary expressions have finite nesting depth, so their ranks are bounded by a sufficiently high finite tower of powers of $\omega$.

For a fixed base, ${\rho }_b$ is strictly increasing. To see this, compare two ordinary base-$b$ expansions at their largest differing exponent. By induction through the hereditary exponent expressions, their exponents have the same ordering after replacing $b$ by $\omega$; computable Cantor normal form notation then uses the same first differing exponent or coefficient. AlsoIndeed after the base change all coefficients remain below $c$, the recursively changed exponents remain correctly ordered, and replacing the new base by $\omega$ gives exactly the original ordinal expression. These facts can be proved together by structural induction for primitive recursive functions on the hereditary expression.

Consequently, whenever $m_s>0$,An infinite nonzero Goodstein sequence would give an infinite descending sequence of ordinals below ${ϵ}_0$, impossible because ordinals are well-ordered. Thus the Goodstein function is total. Its steps are effective finite operations, so simulating the sequence until zero also proves that it is a total computable function.

The same ordinal notation system gives the requested decidable well-order. Let $D$ consist of canonical finite terms for zero and for expressionswhere the exponents are themselves canonical terms. Membership in $D$ is decidable by recursively checking the syntax, positive coefficients and decreasing exponents. Comparison is decidable by recursive computable Cantor normal form notation, first of exponents, then coefficients, and then the remaining summands. All recursive calls inspect proper subterms, so the algorithm terminates.

Interpreting these terms as ordinals gives every ordinal below ${ϵ}_0$, uniquely. One way to connect this directly to hereditary base representations is to choose, for a given finite ordinal term, a base larger than every coefficient anywhere in that term. Replacing $\omega$ by that base evaluates it to a natural number whose hereditary representation recovers the ordinal term. Thus all bases together supply the complete notation system; a single fixed base is not enough.

Choose an effective injective coding of finite terms by natural numbers, and enumerate the valid codes in increasing numerical order. The resulting bijection $d:\mathbb{N}\to D$ is computable: validity is decidable and there are infinitely many finite-ordinal terms. Its inverse is computable by counting valid codes below a given code. DefineThis is a decidable relation on all of $\mathbb{N}$. The interpretation into ordinals is an order isomorphism onto ${ϵ}_0$, provingDecidability here concerns comparing two notations. Well-foundedness is supplied by their ordinal interpretation, the same mathematical descent principle used to prove Goodstein's theorem.

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.

True. The hypothesis says that $\alpha +\alpha$ and $\beta$ commute under ordinal addition. By the commuting ordinal addition classification, comparison of their Cantor normal forms implies that $\alpha$ and $\beta$ themselves commute. Thereforeso both outer expressions equal the middle one.

First let $\delta ={\omega }^{\alpha }$. If $\beta ,\gamma <\delta$, every exponent appearing in their Cantor normal forms is strictly below $\alpha$. The rules for ordinal addition can delete lower terms but cannot introduce an exponent at least $\alpha$. Thereforeso $\delta$ is additively closed.

Conversely, suppose nonzero $\delta$ is additively closed. Part (a) givesIf $\rho >0$, then both ${\omega }^{\alpha }n$ and $\rho$ are smaller than $\delta$, but their sum is $\delta$, contradicting closure. Hence $\rho =0$. If now $n>1$, thenwhileanother contradiction. Thus $n=1$, andThis proves the classification of additively closed ordinals.

Suppose first that $\delta >2$ is multiplicatively closed. It is also additively closed. Indeed, for $\beta ,\gamma <\delta$, let $\mu =\max \{\beta ,\gamma \}$. Monotonicity of ordinal addition giveswhere the last inequality uses $\mu <\delta$, $2<\delta$, and multiplicative closure. Part (b) therefore givesfor some nonzero ordinal $\lambda$.

For any $\rho ,\sigma <\lambda$, strict monotonicity givesMultiplicative closure and the exponent law now implyhence $\rho +\sigma <\lambda$. Thus $\lambda$ is additively closed, and part (b) gives $\lambda ={\omega }^{\alpha }$. Consequently

Conversely, let $\delta ={\omega }^{\lambda }$ with $\lambda ={\omega }^{\alpha }$. By part (b), $\lambda$ is additively closed. Take nonzero $\beta ,\gamma <\delta$, with leading exponents $\rho ,\sigma <\lambda$ in Cantor normal form. If $\gamma$ is finite, the product $\beta \gamma$ has leading exponent $\rho <\lambda$. If $\gamma$ is infinite, the leading exponent of an ordinal product isby additive closure of $\lambda$. In either caseProducts involving zero are immediate, so $\delta$ is multiplicatively closed. This is the multiplicative closure criterion for a power of omega and completes both directions.

Write the nonzero ordinal in Cantor normal formwhere ${\delta }_1>\cdots >{\delta }_r$ and $0<n_i<\omega$. Taking $\delta ={\delta }_1$, $n=n_1$, andgives $\alpha ={\omega }^{\delta }n+\eta$ with $\eta <{\omega }^{\delta }$.