Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 16H (Logic and Set Theory)

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.

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.