Order Completion Unique up to Isomorphism

From ProofWiki
Jump to navigation Jump to search


Let $\left({S, \preceq_S}\right)$ be an ordered set.

Suppose that both $\left({T, \preceq_T}\right)$ and $\left({T', \preceq_{T'}}\right)$ are order completions for $\left({S, \preceq_S}\right)$.

Then there exists a unique order isomorphism $\psi: T \to T'$.

In particular, $\left({T, \preceq_T}\right)$ and $\left({T', \preceq_{T'}}\right)$ are isomorphic.


Both $\left({T, \preceq_T}\right)$ and $\left({T', \preceq_{T'}}\right)$ are order completions for $\left({S, \preceq_S}\right)$.

Hence they both satisfy condition $(4)$ (and also $(1)$, $(2)$ and $(3)$).

Thus, applying condition $(4)$ to $\left({T, \preceq_T}\right)$ (with respect to $\left({T', \preceq_{T'}}\right)$), obtain a unique order-preserving mapping $\phi: T' \to T$.

Applying $(4)$ to $\left({T', \preceq_{T'}}\right)$ (with respect to $\left({T, \preceq_T}\right)$) gives also a unique order-preserving mapping $\psi: T \to T'$.

By Composite of Order-Preserving Mappings is Order-Preserving, their composites $\psi \circ \phi: T' \to T'$ and $\phi \circ \psi: T \to T$ are also order-preserving.

Now applying $(4)$ to $\left({T, \preceq_T}\right)$ (with respect to itself), it follows that $\phi \circ \psi$ is unique.

Now from Identity Mapping is Order Isomorphism, the identity mapping $\operatorname{id}_T: T \to T$ is also order-preserving.

Thus, uniqueness of $\phi \circ \psi$ implies that $\phi \circ \psi = \operatorname{id}_T$.

Similarly, it follows that $\psi \circ \phi = \operatorname{id}_{T'}$.

It follows that $\psi: T \to T'$ is a bijection from Bijection iff Left and Right Inverse.

Thus, $\psi$ is an order-preserving bijection whose inverse is also order-preserving.

That is, $\psi$ is an order isomorphism.

Its uniqueness was already remarked above.