Is the supremum of a set of ordinal numbers always the supremum of a countable subset?

Question

Let $\lambda$ be a limit ordinal, that is $\lambda = \sup\{\beta \mid \beta<\lambda\}$. Can we always find a countable subset $M \subseteq \{\beta \mid \beta < \lambda\}$ such that $\lambda = \sup M$?

Answer 1

No, we can't always find such a countable subset. Take, for instance, the first uncountable ordinal $\omega_1$. Any countable subset of $\omega_1=\{\lambda\mid \lambda<\omega_1\}$ will have a countable supremum. That's because the union of countably many countable sets is countable, and the supremum of any set of ordinals is either their union, or the successor of that union.

Answer 2

No. The least cardinality of such a subset is $\operatorname{cof}\lambda$, the cofinality of $\lambda$. In general, $\operatorname{cof}\lambda\le \lvert\lambda\rvert$, but for instance $\operatorname{cof}\aleph_1=\aleph_1$. For infinite initial ordinals (i.e. the sequence of cardinals $\aleph_\alpha$), $$\operatorname{cof}\aleph_\alpha=\begin{cases}\aleph_\alpha&\text{if }\alpha=0\lor \alpha\text{ is a successor ordinal}\\ \operatorname{cof}\alpha&\text{if }\alpha\text{ is a limit ordinal}\end{cases}$$