Bourbaki-Witt Fixed Point Theorem

Theorem
Let $X$ be a non-empty chain complete poset (that is, a poset in which every chain has a least upper bound). Let $f : X \to X$ be a function such that $f(x) \geq x$. For every $x \in X$ there exists $y \in X$ with $y \geq x$ such that $f(y) = y$.

Proof
Let $\gamma$ be the Hartog number of $X$. Define $g : \gamma \to X$ by transfinite induction as follows:


 * Let $g(0) = x$.
 * $g(\alpha + 1) = f(g(\alpha))$
 * $g(\alpha) = \sup \{ f(\beta) : \beta < \alpha \}$ when $\alpha$ is a limit ordinal.

If $f(x) > x$ for every $x \in X$ then $g$ is strictly increasing and thus an injection. but by construction there is no injection from $\gamma$ to $X$. Hence $f(y) = y$ for some $y$ in the image of $g$, and every element of the image of $g$ is $\geq x$. QED.