User:Abcxyz/Sandbox/Dedekind Completions of Ordered Sets

Feel free to comment. --abcxyz (talk) 16:44, 9 January 2013 (UTC)

Join-Dense
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $T \subseteq S$.

Then $T$ is join-dense in $S$ iff:
 * $\forall x \in S: x = \sup {\left\{{y \in T: y \preceq x}\right\}}$

The dual notion is meet-dense.

Dedekind Completion of Dual is Dual of Dedekind Completion
Let $S$ be an ordered set.

Let $\bigl({\tilde{S}, \phi}\bigr)$ be a Dedekind completion of $S$.

Then the dual of $\tilde{S}$, together with $\phi$, is a Dedekind completion of the dual of $S$.

Proof
Follows directly from Dedekind Completeness is Self-Dual.

Image of Order Embedding is Join-Dense in Dedekind Completion
Let $S$ be an ordered set.

Let $\bigl({\tilde{S}, \phi}\bigr)$ be a Dedekind completion of $S$.

Then the image $\phi \left({S}\right)$ is join-dense in $\tilde{S}$.

Proof
Define:
 * $T = \bigl\{{x \in \tilde{S}: x = \sup {\left\{{a \in \phi \left({S}\right): a \le x}\right\}}}\bigr\}$

Then, by definition, the image $\phi \left({S}\right)$ is join-dense in $T$.

Let $A \subseteq T$ be non-empty and bounded above.

Then $A$ admits a supremum in $\tilde{S}$, and, by Supremum of Suprema:
 * $\sup A = \sup {\left\{{a \in \phi \left({S}\right): \exists x \in A: a \le x}\right\}} \le \sup {\left\{{a \in \phi \left({S}\right): a \le \sup A}\right\}}$

It follows that $\sup A \in T$; hence, $T$ is Dedekind complete.

Let $\psi: S \to T$ denote the restriction of $\phi$ to $S \times T$.

By definition, there exists an order embedding $f: \tilde{S} \to T$ such that:
 * $f \circ \phi = \psi$

It follows that:
 * $\forall x \in \tilde{S}: f \left({x}\right) = \sup {\left\{{a \in \phi \left({S}\right): a \le f \left({x}\right)}\right\}} = \sup {\left\{{a \in \phi \left({S}\right): a \le f \left({f \left({x}\right)}\right)}\right\}} = f \left({f \left({x}\right)}\right)$

Since $f$ is an injection, it follows that $T = \tilde{S}$; the result follows.

Uniqueness of Dedekind Completion
Let $S$ be an ordered set.

Let $\left({X, f}\right)$ and $\left({Y, g}\right)$ be Dedekind completions of $S$.

Then there exists a unique order isomorphism $\psi: X \to Y$ such that:
 * $\psi \circ f = g$

Proof
By definition, there exist order embeddings $\tilde{f}: Y \to X$ and $\tilde{g}: X \to Y$ such that:
 * $\tilde{f} \circ g = f$
 * $\tilde{g} \circ f = g$

It follows that:
 * $\bigl({\tilde{f} \circ \tilde{g}}\bigr) \circ f = f$
 * $\bigl({\tilde{g} \circ \tilde{f}}\bigr) \circ g = g$

From Image of Order Embedding is Join-Dense in Dedekind Completion, the images $f \left({S}\right)$ and $g \left({S}\right)$ are join-dense in $X$ and $Y$, respectively.

It follows that:

Hence:
 * $\tilde{f} \circ \tilde{g} = \operatorname{id}_X$
 * $\tilde{g} \circ \tilde{f} = \operatorname{id}_Y$

Therefore, $\tilde{g}$ is an order isomorphism.

Suppose that $\psi: X \to Y$ is an order isomorphism such that:
 * $\psi \circ f = g$

Then:
 * $\bigl({\psi \circ \tilde{f}}\bigr) \circ g = \bigl({\tilde{g} \circ \tilde{f}}\bigr) \circ g = g$

It follows that:
 * $\psi \circ \tilde{f} = \operatorname{id}_Y$

Hence:
 * $\psi = \psi \circ \tilde{f} \circ \tilde{g} = \tilde{g}$

as desired.