# 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.

## Proof

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

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

From Subset Relation on Power Set is Partial Ordering, $\struct {S, \subseteq}$ is an ordered set.

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.

$\blacksquare$

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 7$