Composition of Inflationary Mappings is Inflationary
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be inflationary mappings.
Then $f \circ g$, the composition of $f$ and $g$, is also inflationary.
Proof
Let $x \in S$.
\(\text {(1)}: \quad\) | \(\ds x\) | \(\preceq\) | \(\ds \map g x\) | $g$ is inflationary | ||||||||||
\(\text {(2)}: \quad\) | \(\ds \map g x\) | \(\preceq\) | \(\ds \map f {\map g x}\) | $f$ is inflationary | ||||||||||
\(\ds x\) | \(\preceq\) | \(\ds \map f {\map g x}\) | $(1)$ and $(2)$ and $\preceq$ is an ordering and hence transitive | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\preceq\) | \(\ds \map {\paren {f \circ g} } x\) | Definition of Composition of Mappings |
Since this holds for all $x \in S$, $f \circ g$ is inflationary.
$\blacksquare$