Do all sets of cardinals have a countable subset with the same supremum?

Question

Consider a set of cardinals S and let m be the supremum of S, that is, the smallest cardinal greater than all elements of S. Such a cardinal m must exist by the well ordering of cardinals. Must S have a countable subset whose supremum is also m? If so I'd like a proof and if not a counter example. Thanks for your time.

Answer 1

The answer is no, and the keyword to google is "cofinality".

For an explicit example you can take $S=\{\kappa\mid\kappa<\aleph_{\omega_1}\}$. Clearly $\sup S=\aleph_{\omega_1}$, but since $\mathrm{cof}(\aleph_{\omega_1})=\mathrm{cof}(\omega_1)=\omega_1$, any subset of $S$ with the same sup must have at least $\omega_1$ many elements.