Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 16H / c / Solution

Every proper initial segment of ${\omega }_1\setminus \omega$ is countable, while the whole tail is uncountable. Its order type is therefore the least uncountable ordinal itself:Equivalently, part (b) gives $\omega +\gamma ={\omega }_1$, and the unique possible uncountable initial ordinal $\gamma$ is ${\omega }_1$.