Why the set of all countable ordinals is not chain complete

Question

Why the set of all countable ordinals (under their usual ordering) is $\omega$-complete, and why it is not a chain complete. Any reference to this question is welcome.

Answer 1

The set of all countable ordinals is a chain. Hence chain completeness would imply that this chain has an upper bound $\alpha$ in the set - but then $\alpha+1$ is also a countable ordinal.

Let $\alpha_0\le \alpha_1\le\ldots$ be an $\omega$-chain. Then $\bigcup_{n\in\omega}\alpha_n$ is a countable ordinal that is $\ge \alpha_n$ for each $n$.