Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 13I / b

The first assertion is false: $1<2$, but $1+\omega =2+\omega =\omega$. The second is true. In the concatenation definition of ordinal addition, the copy of $\alpha$ is a proper initial segment of the copy of $\beta$ when $\alpha <\beta$, so $\gamma +\alpha$ is a proper initial segment of $\gamma +\beta$.

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

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 $\alpha >1$, ${\alpha }^{\omega }\ge {\alpha }^2>\alpha$.

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 ${\epsilon }_0<{\omega }_1$.