Identity Mapping is Order Isomorphism/Proof 1

Theorem

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

The identity mapping $I_S$ is an order isomorphism from $\struct {S, \preceq}$ to itself.

Proof

By definition:

$\forall x \in S: \map {I_S} x = x$

So:

$x \preceq y \implies \map {I_S} x \preceq \map {I_S} y$

As $I_S$ is a bijection, we also have:

$\map {I_S^{-1} } x = x$

So:

$x \preceq y \implies \map {I_S^{-1} } x \preceq \map {I_S^{-1} } y$

$\blacksquare$