Order-Preserving Identity Mapping between Ordered Structures not necessarily Isomorphism
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Theorem
Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$ such that:
- $\forall a, b \in A: a \mathrel \RR b \implies a \mathrel \SS b$
Let $I_A$ denote the identity mapping on $A$.
Then it is not necessarily the case that $I_A$ is an order isomorphism from the ordered structures $\struct {A, \RR}$ and $\struct {A, \SS}$.
Proof
Let $\RR: \N \to \N$ be the ordering on the natural numbers $\N$ defined as:
- $\forall a, b \in \N: a \mathrel \RR b \iff a \le b \text { and } \map P a = \map P b$
where $\map P x$ is the parity of $x$.
That is:
- $0 \mathrel \RR 2 \mathrel \RR 4 \mathrel \RR \cdots$
and:
- $1 \mathrel \RR 3 \mathrel \RR 5 \mathrel \RR \cdots$
and so on, but (for example):
- $\lnot \paren {0 \mathrel \RR 1}$
Let $\SS: \N \to \N$ be the ordering on the natural numbers $\N$ defined as:
- $\forall a, b \in \N: a \mathrel \SS b \iff a \le b$
We have that:
\(\ds \forall a, b \in \N: \, \) | \(\ds a\) | \(\RR\) | \(\ds b\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(\le\) | \(\ds b\) | Definition of $\RR$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(\SS\) | \(\ds b\) | Definition of $\SS$ |
But note that:
\(\ds 0\) | \(\le\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(\SS\) | \(\ds 1\) | Definition of $\SS$ |
while it is not the case that $0 \mathrel \RR 1$.
Hence while it is true that:
- $\forall a, b \in \N: a \mathrel \RR b \implies a \mathrel \SS b$
it is not the case that:
- $\forall a, b \in \N: a \mathrel \SS b \implies a \mathrel \RR b$
and so the identity mapping on $\NN$ is not an order isomorphism.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(d)}$