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I'm asked to find the smallest cardinal $\kappa$ such that $\omega + \omega$ is the supremum of $\kappa$ smaller ordinals.

By definition, $\omega + \omega = \sup \{\omega + n : n < \omega\}$. I need to find some set $S\subseteq \omega + \omega$ (a set of smaller ordinals) such that $\omega + \omega = \sup S$. Given the above, I'm tempted to take $S=\{\omega + n: n < \omega\}$; its cardinality is $\omega$, and then the answer would be $\omega$. But how do I know if its cardinality is the smallest? Maybe there is some other set $S'$ that is smaller in cardinality than $S$ such that $\omega + \omega = \sup S'$?

Also, after I posted the question I realized that I don't quite understand why my $S$ (i.e. $S=\{\omega + n: n < \omega\}$) is a subset of $\omega + \omega$, i.e. why $\{\omega + n: n < \omega\} \subseteq \sup\{\omega + n: n < \omega\} $. I'm tempted to say that that's because $\{\omega + n: n < \omega\} \leq\sup\{\omega + n: n < \omega\} $, but $\subseteq $ and $\leq$ is the same for ordinals, whereas $\{\omega + n: n < \omega\} $ is a set of ordinals, not an ordinal.

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  • $\begingroup$ $\sup{\{\omega + \omega\}}$? $\endgroup$ Mar 18, 2022 at 0:07

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$\omega$ is the only infinite cardinal smaller than the ordinal $\omega+\omega.$ So if there exists some $S\subseteq\omega+\omega$ such that $\sup(S)=\omega+\omega$ and $\mathrm{card}(S)\lt\omega,$ then $S$ is a finite set. Is there a finite set $S$ such that $\sup(S)=\omega+\omega$? No, and I think this is trivial to see. Therefore, $S$ must be infinite. Therefore, $\kappa=\mathrm{card}(S)=\omega.$

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