A progressive map with no fixed point, on a poset that is not chain-complete

Refuted claim: the chain-completeness hypothesis of the Bourbaki–Witt fixed point theorem is decoration, that is every progressive map $f : P \to P$ on a poset has a fixed point (Bourbaki–Witt fixed point theorem, Chain-complete poset).

The witness is $(\mathbb{N}, \le)$ (Order on the natural numbers) with the successor map $f(n) = \sigma(n) = n + 1$. It is progressive, and it has no fixed point at all. There is no conflict with Bourbaki–Witt, because $(\mathbb{N}, \le)$ is not chain-complete: $\mathbb{N}$ is a chain (Chain in a poset) with no upper bound in $\mathbb{N}$ ($(\mathbb{N}, \le)$ has no maximal element: Zorn's chain hypothesis fails), hence with no least upper bound (Upper bound, least upper bound, and strict upper bound).

• The failure is sharp, and it is a missing supremum. Adjoin one element $\infty$ above every natural number. The enlarged poset is chain-complete: a subset containing $\infty$ has supremum $\infty$, a subset of $\mathbb{N}$ with an upper bound in $\mathbb{N}$ has a least one by The well-ordering principle, a subset of $\mathbb{N}$ with none has supremum $\infty$, and the empty chain has supremum $0$. Extending $f$ by $f(\infty) = \infty$ keeps it progressive, and the fixed point Bourbaki-Witt promises is $\infty$, precisely the supremum that was missing. Any progressive map on the enlarged poset must fix $\infty$, since $\infty$ is greatest.

• Monotonicity is not the issue. The map $f(n) = \sigma(n)$ is order preserving as well as progressive: $\sigma(m) = m + \sigma(0)$ and $\sigma(n) = n + \sigma(0)$ by the Given, and adding a fixed natural number preserves $\le$ in both directions (Order is compatible with addition), so $m \le n$ gives $\sigma(m) \le \sigma(n)$. So this is not a case of a badly behaved map defeating the theorem; a perfectly well behaved map is defeated by the poset. Conversely Bourbaki–Witt fixed point theorem assumes no monotonicity at all, which is what lets Zorn's lemma apply it to a map built from an arbitrary choice function.

• No iteration argument could have worked. Starting at $0$ and iterating $f$ walks up $\mathbb{N}$ forever without converging, and the fixed point in Bourbaki-Witt is not reached by iterating: it is the supremum of the smallest set closed under $f$ and under suprema of its chains. When that supremum does not exist there is nothing to reach.

• This is the same defect as in $(\mathbb{N}, \le)$ has no maximal element: Zorn's chain hypothesis fails, read one notch higher up the scale of bounds. There the chain $\mathbb{N}$ had no upper bound, which is what Zorn's lemma asks of every chain; here the same chain has no least upper bound, which is what Bourbaki–Witt fixed point theorem asks of every chain through chain-completeness. A least upper bound is in particular an upper bound, so the first failure implies the second, and adjoining one top element repairs both at once.