Fixed Point of Composition of Inflationary Mappings

Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $f, g: S \to S$ be inflationary mappings.

Let $x \in S$.

Then:

$x$ is a fixed point of $f \circ g$

$x$ is a fixed point of both $f$ and $g$.

$\Box$

Let $h = f \circ g$.

Let $x$ be a fixed point of $h$.

Then by definition of composition:

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

Since $f$ is inflationary:

$x \preceq \map g x$

Aiming for a contradiction, suppose $x \ne \map g x$.

Then $x \prec \map g x$.

Since $f$ is also inflationary:

$\map g x \preceq \map f {\map g x}$

$x \prec \map f {\map g x}$

But this contradicts the assumption that $x$ is a fixed point of $f \circ g$.

Therefore, $x = \map g x$, and $x$ is a fixed point of $g$.

Aiming for a contradiction, suppose $\map f x \ne x$.

Then $x \prec \map f x$.

As we have shown that $x = \map g x$, it follows that:

$x \prec \map f {\map g x}$

But this contradicts assumption that $x$ is a fixed point of $f \circ g$.

Hence, $x$ is also a fixed point of $f$.

$\blacksquare$