Smallest ordinal closed under $\kappa$ sized suprema

Question

Let $\kappa$ be a cardinal. Is the smallest ordinal that is closed under suprema of sets with cardinality $\leq \kappa$ equal to $\kappa+1$?

In particular, is the smallest ordinal that is closed under countable suprema equal to $\aleph_1=\omega_1$, the first uncountable ordinal?

Answer 1

For general ordinals, $0$ is closed under suprema, since it has no elements. Any successor ordinal $\alpha+1$ is closed under suprema, since $\mathrm{sup} (\alpha+1)=\max(\alpha+1)=\alpha\in \alpha+1$. So that makes the question a bit trivial unless we only look at limit ordinals.

A limit ordinal $\alpha$ is closed under suprema of subsets $X\subseteq\alpha$ with cardinality $|X|\leq \kappa$ if and only if $\mathrm{cf}(\alpha)>\kappa$, that is, the cofinality of $\alpha$ is larger than $\kappa$.

Considering that every limit ordinal of cardinality $\leq \kappa$ has cofinality $\leq \kappa$, the least limit ordinal closed under suprema of sets of cardinality $\leq \kappa$ must have at least cardinality $\kappa^+$. Since successor cardinals are regular, $\mathrm{cf}(\kappa^+)=\kappa^+$, so this satisfies our property.

Therefore, $\omega_1$ is indeed the least limit ordinal closed under countable suprema.

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Note that there is a difference between $\kappa^+$, the least cardinal larger than $\kappa$, and $\kappa+1$. The latter is either:

• the least ordinal larger than $\kappa$ when $+$ is interpreted as ordinal addition, or
• $\kappa$ itself, when $+$ is interpreted as cardinal addition.