Set of well-orderings of a given ordinal

Question

Let $\alpha$ be an ordinal. Consider the set $$S = \{ \operatorname{type}(\alpha, R) : R \text{ well orders } \alpha \}$$ Clearly $\alpha \in S$. Can anything be said about the relation of an arbitrary $\beta \in S$ to $\alpha$?

Answer 1

$S$ has cardinality $\alpha^+$ and in fact $[\alpha, \alpha^+) \subseteq S \subseteq \alpha^+$.

To see this, let $\beta \in [\alpha, \alpha^+)$. Fix a bijection $$ f \colon \beta \to \alpha. $$ Define $R \subseteq \alpha \times \alpha$ via $$ R = \{ (f(i), f(j)) \mid i < j < \beta \}. $$ It's easy to see that $R$ is a wellorder of $\alpha$ and that $\mathrm{otp}(R) = \beta \in S$.

I'll leave it to you to prove that $S \subseteq \alpha^+$.

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$\sup S$ is what is known as Hartogs' number (of $\alpha$). If $\alpha$ is an ordinal or can be wellordered, this is just the cardinal successor of $\alpha$.

In general, given any set $X$ (which may not have a wellorder), we let $$ H(X) := \sup \{ \alpha \in \mathrm{On} \mid \exists f \colon \alpha \to X \text{ injective} \}. $$ Let us define $S(X) := \{ \alpha \in \mathrm{On} \mid \exists f \colon \alpha \to X \text{ injective} \}$. If $\alpha \in S(X)$, witnessed by $f \colon \alpha \to X$, then we get a well-order $R$ of $\mathrm{img}(f)$ given by $$ R := \{ (f(i), f(j)) \mid i < j < \alpha \}. $$ Hence $S(X)$ consists of those ordertypes that are realizable as wellorders on (some subset of) $X$. In this sence, $S(X)$ is a generalization of the $S$ in your original post.