Admissible subset (Bourbaki–Witt)

Let $(P, \le)$ be a chain-complete poset and $f : P \to P$ a progressive map (Chain-complete poset). A subset $A \subseteq P$ is $f$-admissible (or simply admissible, when $f$ is fixed) if:

• (C1) closed under $f$: $f(x) \in A$ for every $x \in A$;
• (C2) closed under chain suprema: $\sup C \in A$ for every chain $C \subseteq A$.

• (C2) already forces $\bot \in A$. The empty set is a chain contained in $A$ (Chain in a poset), and $\sup \emptyset = \bot$, so every admissible set contains the least element of $P$. Presentations that list "$\bot \in A$" as a third condition are therefore stating a consequence, not an extra hypothesis. This is the first of four places on this page where insisting that the empty set counts as a chain removes a case rather than creating one; the others are The cut at an extremal element is closed under chain suprema, A supremum of extremal elements is extremal and Every element of $M$ is extremal.
• The suprema in (C2) are taken in $P$ and exist there by chain-completeness; the condition is that they land back inside $A$.
• $P$ itself is admissible, so admissible sets exist. The content of A smallest admissible set exists is that there is a smallest one, and its minimality is what carries the two decisive steps of the Bourbaki–Witt argument, Everything in $M$ is comparable to an extremal element and Every element of $M$ is extremal: to prove that all elements of the smallest admissible set $M$ have some property, one shows that the elements of $M$ with that property again form an admissible set. The remaining lemmas use only admissibility of $M$, progressivity of $f$, and the order axioms.
• Nothing here assumes $f$ is monotone, injective, or continuous in any sense. Only (C1), (C2) and progressivity are ever used.