Identity Mapping is Order Isomorphism

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

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

Proof 1

By definition:

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


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

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

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


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


Proof 2

An ordered set is a relational structure where order isomorphism is a special case of relation isomorphism.

The result follows directly from Identity Mapping is Relation Isomorphism.