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 $X$.

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.

Thus by Strictly Monotone Mapping is Injective, $g$ is an injection.

But by construction there is no injection from $\gamma$ to $X$.

Hence $f(y) = y$ for some $y$ in the image of $g$.

So every element of the image of $g$ is $\geq x$.