Bourbaki-Witt Fixed Point Theorem

Theorem
Let $\left({X, \le}\right)$ 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 mapping such that $f \left({x}\right) \geq x$.

Then for every $x \in X$ there exists $y \in X$ where $y \geq x$ such that $f \left({y}\right) = y$.

Proof
Let $\gamma$ be the Hartogs number of $X$.

Suppose that $x \in X$.

Define $g : \gamma \to X$ by transfinite recursion as follows:


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

Suppose, with a view to obtaining a contradiction, there is no $y \in g \left({\gamma}\right)$ such that $f \left({y}\right) = y$.

Then:
 * $\forall y \in X: f \left({y}\right) > y$

That is, $f$ is strictly increasing.

If $f$ is strictly increasing, then $g$ is also strictly increasing, and so by definition strictly monotone.

By Strictly Monotone Mapping is Injective, $g$ is an injection.

But $\gamma$ is, by definition, the least ordinal such that there is no injection from $\gamma$ to $X$.

From this contradiction it follows that our supposition that there is no $y \in g \left({\gamma}\right)$ such that $f \left({y}\right) = y$ must have been false.

Thus:
 * $\exists y \in g \left({\gamma}\right) \subseteq \left\{{y \in X : x \leq y}\right\}: f \left({y}\right) = y$

Hence the result.