Inverse of Increasing Bijection need not be Increasing

Theorem

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.

Let $\phi: S \to T$ be a bijection which is increasing.

Then $\phi^{-1}: T \to S$ is not necessarily also increasing.

Let $S := \powerset {\set {a, b} }$.

Let $T := \set {1, 2, 3, 4}$.

Clearly so is $\struct {T, \le}$ (although its ordering is in fact total, it is still technically an ordered set).

Let $\phi: S \to T$ be defined as:

$\ds \map \phi \O$

$=$

$\ds 1$

$\ds \map \phi {\set a}$

$=$

$\ds 2$

$\ds \map \phi {\set b}$

$=$

$\ds 3$

$\ds \map \phi {\set {a, b} }$

$=$

$\ds 4$

By inspection it can be seen that $\phi$ is a bijection.

Also by inspection it can be seen that $\phi$ is increasing.

But note that while $2 \le 3$, it is not the case that $\set a \subseteq \set b$.

That is, $\phi^{-1}$ is not increasing.

$\blacksquare$