Composition of Inflationary Mappings is Inflationary

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Theorem

Let $\left({S, \preceq}\right)$ 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 $x \in S$.

\((1):\quad\) \(\displaystyle x\) \(\preceq\) \(\displaystyle g \left({x}\right)\) $g$ is inflationary
\((2):\quad\) \(\displaystyle g \left({x}\right)\) \(\preceq\) \(\displaystyle f \left({g \left({x}\right)}\right)\) $f$ is inflationary
\(\displaystyle x\) \(\preceq\) \(\displaystyle f \left({g \left({x}\right)}\right)\) $(1)$ and $(2)$ and $\preceq$ is an ordering and hence transitive
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(\preceq\) \(\displaystyle \left({f \circ g}\right) \left({x}\right)\) Definition of composition

Since this holds for all $x \in S$, $f \circ g$ is inflationary.

$\blacksquare$