Bourbaki–Witt theorem

Remarks

The map $f$ need not be monotone; the hypothesis $x \le f(x)$ says that $f$ is inflationary. Proof idea: fix $a \in P$ and let $T$ be the smallest subset of $P$ that contains $a$, is closed under $f$, and contains the supremum of each of its nonempty chains. A careful induction shows that $T$ is a chain, so it has a supremum $s$, and $s \in T$. Then $f(s) \in T$, so $f(s) \le s \le f(s)$ and $f(s) = s$. The proof uses no form of the axiom of choice. Combined with the axiom of choice, the theorem shows that such a poset has a maximal element. Otherwise a choice of some $f(x) > x$ for each $x$ would be an inflationary map with no fixed point. This is a standard step in the proof of Zorn's lemma from the axiom of choice. The hypothesis $x \le f(x)$ is needed: on the chain $\set{0, 1}$, where every chain has a supremum, the map swapping $0$ and $1$ has no fixed point. The completeness hypothesis is needed too: on $\N$, the inflationary map $n \mapsto n + 1$ has no fixed point, and the chain $\N$ has no supremum.

Going deeper

For a student

The theorem is a choice-free fixed-point principle for chain-complete posets. It is a standard route to Zorn's lemma from the axiom of choice, and it is used to build objects by transfinite recursion without naming ordinals. Unlike the Knaster–Tarski theorem, it does not need monotonicity, but it says nothing about least fixed points.