Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 16F / iv / Solution

This assertion is always true. Ordinal addition obeys $\alpha +\beta \ge \alpha$ and $\alpha +\beta \ge \beta$. Thus both summands are at most ${\omega }_1$. If $0<\beta <{\omega }_1$ and $\alpha <{\omega }_1$, both ordinals are countable, and their disjoint ordered union is countable, so their sum cannot be ${\omega }_1$. If instead $\alpha ={\omega }_1$ and $\beta >0$, appending a nonempty second block gives a sum strictly larger than ${\omega }_1$. These exhaust the possibilities with $0<\beta <{\omega }_1$; hence $\beta =0$ or $\beta ={\omega }_1$.