# Find the smallest cardinal $\kappa$ such that $\omega + \omega$ is the supremum of $\kappa$ smaller ordinals

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.

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

$$\omega$$ 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, $$\kappa=\mathrm{card}(S)=\omega.$$