Order Embedding is Injection

Theorem

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.

Let $\phi: S \to T$ be an order embedding.

That is:

$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

Then $\phi$ is an injection.

Proof

Suppose $\phi: S \to T$ is a mapping such that:

$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$

Then, for all $x, y \in S$:

\(\ds \map \phi x\)

\(=\)

\(\ds \map \phi y\)

\(\ds \leadsto \ \ \)

\(\ds \map \phi x \preceq_2 \map \phi y\)

\(\land\)

\(\ds \map \phi y \preceq_2 \map \phi x\)

Ordering $\preceq_2$ is Reflexive

\(\ds \leadsto \ \ \)

\(\ds x \preceq_1 y\)

\(\land\)

\(\ds y \preceq_1 x\)

Definition of Order Embedding

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds y\)

Ordering $\preceq_1$ is Antisymmetric

So $\phi$ is an injection.

$\blacksquare$

Sources

• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $19$