User:Abcxyz/Sandbox/Dedekind Completions of Ordered Sets

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

Definition: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.

Theorem
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.

Theorem
Let $S$ be an ordered set.

Let $\bigl({\tilde S, \le}\bigr)$ be a Dedekind complete ordered set.

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

Then $\bigl({\tilde S, \phi}\bigr)$ is a Dedekind completion of $S$ iff:
 * $({1}): \quad$ The image $\phi \left({S}\right)$ is join-dense and meet-dense in $\tilde S$.


 * $({2}): \quad \forall x \in \tilde S: \exists a, b \in \phi \left({S}\right): a \le x \le b$

Necessary Condition
Suppose that $\bigl({\tilde S, \phi}\bigr)$ is a Dedekind completion of $S$.

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

Then, by definition:
 * $({1}): \quad$ The image $\phi \left({S}\right)$ is join-dense in $T$.


 * $({2}): \quad \forall x \in T: \exists b \in \phi \left({S}\right): x \le b$

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

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 $\epsilon: \tilde S \to T$ such that:


 * $\epsilon \circ \phi = \psi$

It follows that:


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

Since $\epsilon$ is an injection, it follows that $T = \tilde S$.

The result follows from Dual of Dedekind Completion is Dedekind Completion of Dual.

Sufficient Condition
Suppose that:
 * $({1}): \quad$ The image $\phi \left({S}\right)$ is join-dense and meet-dense in $\tilde S$.


 * $({2}): \quad \forall x \in \tilde S: \exists a, b \in \phi \left({S}\right): a \le x \le b$

Let $\left({X, \preceq}\right)$ be a Dedekind complete ordered set.

Let $f: S \to X$ be an order embedding.

Let $\tilde f: \tilde S \to X$ be the increasing mapping defined as:
 * $\forall x \in \tilde S: \tilde f \left({x}\right) = \sup {\left\{{f \left({s}\right): \phi \left({s}\right) \le x}\right\}}$

The existence of $\tilde f$ is justified by hypothesis $({2})$.

We have that:
 * $\tilde f \circ \phi = f$

Let $x, y \in \tilde S$ such that $\tilde f \left({x}\right) \preceq \tilde f \left({y}\right)$.

Then:

Hence, $\tilde f$ is an order embedding.

Theorem
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 Equivalent Conditions for 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:
 * $\bigl({\psi \circ \tilde f}\bigr) \circ g = g$

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

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

as desired.