Identity Mapping is Order Isomorphism

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 1

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:

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

$\blacksquare$


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.

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.1: \ 1^\circ$
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.6$
  • 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 6.30$
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