Abian–Brown theorem

Remarks

Proof idea: define $x_0 = a$, $x_{\alpha + 1} = f(x_\alpha)$, and $x_\lambda = \sup_{\alpha < \lambda} x_\alpha$ at limit ordinals $\lambda$. By transfinite induction, $(x_\alpha)$ is increasing and each $x_\alpha$ satisfies $x_\alpha \le f(x_\alpha)$, so the suprema exist. The sequence cannot be strictly increasing for all ordinals, since $P$ is a set, so $x_{\alpha + 1} = x_\alpha$ for some $\alpha$, and $p = x_\alpha$ is a fixed point. If $q$ is a fixed point with $a \le q$, induction gives $x_\alpha \le q$ for all $\alpha$, so $p$ is the least such fixed point. The theorem contains the existence part of the Knaster–Tarski theorem: in a complete lattice, take $a$ to be the least element. The hypothesis $a \le f(a)$ is needed. On the two-element antichain $\set{a, b}$, every nonempty chain is a single point, and the swap of $a$ and $b$ is monotone and has no fixed point.

Going deeper

For a student

The theorem weakens the completeness hypothesis of the Knaster–Tarski theorem to chain-completeness, at the price of needing a starting point $a$ with $a \le f(a)$. In a chain-complete poset with a least element, every monotone map has a least fixed point. Compared with the Bourbaki–Witt theorem, it asks for monotonicity instead of $x \le f(x)$ for all $x$, and in return it gives a least fixed point above $a$.