Bourbaki-Witt Fixed Point Theorem

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Theorem

Let $\struct {X, \le}$ be a non-empty chain complete poset (that is, an ordered set in which every chain has a supremum).

Let $f: X \to X$ be an inflationary mapping, that is, so that $\map f x \ge x$.


Then for every $x \in X$ there exists $y \in X$ where $y \ge x$ such that $\map f y = y$.


Proof

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

Suppose that $x \in X$.


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

$\map g 0 = x$
$\map g {\alpha + 1} = \map f {\map g \alpha}$
$\map g \alpha = \sup \set {\map g \beta: \beta < \alpha}$ when $\alpha$ is a limit ordinal.


That $f$ is inflationary guarantees both that:

$(1) \quad \set {\map g \beta: \beta < \alpha}$ is a chain for each $\alpha < \gamma$
$(2) \quad g$ is increasing.



If $g$ is strictly increasing, then $g$ is strictly monotone and, by Strictly Monotone Mapping with Totally Ordered Domain is Injective, an injection.

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

Therefore there must be some $\alpha < \beta < \gamma$ with $\map g \alpha = \map g \beta$.

Then:

$\alpha + 1 \le \beta$

so:

$\map g \alpha \le \map g {\alpha + 1} \le \map g \beta = \map g \alpha$

whence:

$\map f {\map g \alpha} = \map g \alpha$

Because $x = \map g 0$:

$\exists y \in \map g \gamma \subseteq \set {y \in X : x \le y}: \map f y = y$

$\blacksquare$


Source of Name

This entry was named for Nicolas Bourbaki and Ernst Witt.


Sources