User:Abcxyz/Sandbox/Dedekind Completions of Archimedean Ordered Groups

Yes, I know to put singular nouns in the theorem title when this is put up. (The title of this page refers to its subject.) It's just that I haven't spent the time to come up with (or discuss) a suitable name for this theorem. Just be patient; I'll get to the matter in due course. --abcxyz (talk) 17:45, 20 January 2013 (UTC)

Definition:Archimedean Ordered Group
An ordered group $\left({G, *, \preceq}\right)$ is said to be Archimedean iff:
 * $\forall g \in G: \left({\exists h \in G: \forall n \in \N: g^n \preceq h}\right) \implies g \preceq e$

where:
 * $e$ denotes the identity element of $\left({G, *}\right)$
 * $g^n$ denotes the $n$th power of $g$ in $\left({G, *}\right)$

In other words, if $g \in G$, $g \npreceq e$, then $\left\{{g^n: n \in \N}\right\}$ is unbounded above in $\left({G, \preceq}\right)$.

Theorem
Let $\left({G, *, \le}\right)$ be an Archimedean ordered group.

Let $\bigl({\bigl({\tilde G, \preceq}\bigr), \phi}\bigr)$ be a Dedekind completion of $\left({G, \le}\right)$.

Then there exists a unique binary operation $\circledast$ on $\tilde G$ such that:
 * $({1}): \quad \bigl({\tilde G, \circledast, \preceq}\bigr)$ is an ordered group
 * $({2}): \quad \phi$ is a group homomorphism from $\left({G, *}\right)$ to $\bigl({\tilde G, \circledast}\bigr)$

Proof of Existence
We will show that we can define an operation on $\tilde G$ thus:

$\circledast: \tilde G \times \tilde G \to \tilde G$ be the binary operation on $\tilde G$ defined as:
 * $\forall x, y \in \tilde G: x \circledast y = \sup {\left\{{\phi \left({g * h}\right): \phi \left({g}\right) \preceq x, \, \phi \left({h}\right) \preceq y}\right\}}$

We will show that $(\tilde G, \circledast)$ is a group.

We will show that $\phi$ is a homomorphism from $\left({G, *}\right)$ to $\bigl({\tilde G, \circledast}\bigr)$.

We will also show that: From Supremum of Subset, it follows that $\preceq$ is compatible with $\circledast$.

It remains to verify the group axioms for $\bigl({\tilde G, \circledast}\bigr)$.

$({G0}):$ $\circledast$ is Well-Defined and $G$ is Closed Under It
Let $x, y \in \tilde G$.

We must show that the set
 * $T = \left\{{\phi \left({g * h}\right):g, h \in G, \, \phi \left({g}\right) \preceq x, \, \phi \left({h}\right) \preceq y}\right\}$

is non-empty and bounded above in $\tilde G$.

By Characterization of Dedekind Completion, there are elements $a_x,b_x\in \phi(G)$ such that $a_x \preceq x \preceq b_x$.

Thus there are elements $a_x',b_x' \in G$ such that $\phi(a_x') \preceq x \preceq \phi(b_x')$

Thus: $S_x = \left\{{g \in G: \phi\left({g}\right) \preceq x}\right\}$ is non-empty.

Defining $a_y, b_y, a_y', b_y'$ similarly, we can similarly prove that: $S_y = \left\{{h \in G: \phi\left({h}\right) \preceq y}\right\}$ is non-empty.

Thus $T$ is non-empty.

By their constructions, $\phi(S_x)$ and $\phi(S_y)$ are bounded above by $x$ and $y$, respectively.

Then $\phi(S_x)$ is bounded above by $\phi(b_x')$ and $\phi(S_y)$ is bounded above by $\phi(b_y')$.

Since $\phi$ is an order embedding:
 * $S_x$ is bounded above by $b_x'$.
 * $S_y$ is bounded above by $b_y'$.

Since $\le$ is compatible with $*$, User:Dfeuer/Operating on Ordered Group Relationships shows that
 * $S = \left\{{\left({g * h}\right):g \in S_x, \, h \in S_y}\right\}$

is bounded above by $b_x' * b_y'$

But $T = \phi(S)$ and $\phi$ is an order embedding, so $T$ is bounded above by $\phi(b_x'*b_y')$.

Since $\tilde G$ is Dedekind-complete and $T$ is non-empty and bounded above, $T$ has a supremum in $\tilde G$, so

$\circledast$ is well-defined and its values are elements of $\tilde G$.

$({G1}):$ Associativity
Let $x, y, z \in \tilde G$, and define:
 * $A = \left\{{\phi \left({a * b * c}\right): a,b,c\in G,\, \phi \left({a}\right) \preceq x, \, \phi \left({b}\right) \preceq y, \, \phi \left({c}\right) \preceq z}\right\}$

It suffices to show that:
 * $\left({x \circledast y}\right) \circledast z = \sup A = x \circledast \left({y \circledast z}\right)$

We only prove the first equality; the second follows similarly.

For $a,b,c \in G$:

We have that $\sup A \preceq \left({x \circledast y}\right) \circledast z$.

We now show that $\sup A \succeq \left({x \circledast y}\right) \circledast z$.

By Characterization of Dedekind Completion, we have that:
 * $\sup A = \inf {\left\{{\phi \left({g}\right): \phi \left({g}\right) \succeq \sup A}\right\}}$

Suppose that $g \in G$, $\phi \left({g}\right) \succeq \sup A$.

It suffices to show that $\phi \left({g}\right) \succeq \left({x \circledast y}\right) \circledast z$.

Suppose that $a, b, c \in G$, $\phi \left({a}\right) \preceq x$, $\phi \left({b}\right) \preceq y$, $\phi \left({c}\right) \preceq z$.

Then:
 * $\phi \left({a * b}\right) = \phi \left({a * b * c}\right) \circledast \phi \bigl({c^{-1}}\bigr) \preceq \phi \left({g}\right) \circledast \phi \bigl({c^{-1}}\bigr) = \phi \bigl({g * c^{-1}}\bigr)$

where $c^{-1}$ denotes the inverse of $c$ in $\left({G, *}\right)$.

By the definition of $\circledast$, it follows that:
 * $x \circledast y \preceq \phi \bigl({g * c^{-1}}\bigr)$

Therefore:
 * $\left({x \circledast y}\right) \circledast \phi \left({c}\right) \preceq \phi \bigl({g * c^{-1}}\bigr) \circledast \phi \left({c}\right) = \phi \left({g}\right)$

Hence:
 * $\left({x \circledast y}\right) \circledast z \preceq \phi \left({g}\right)$

It follows that $\circledast$ is associative.

$({G2}):$ Identity
Let $e$ denote the identity of $\left({G, *}\right)$.

Let $x \in \tilde G$.

By Characterization of Dedekind Completion, we have that:
 * $x = \sup {\left\{{\phi \left({g}\right): \phi \left({g}\right) \preceq x}\right\}}$

It follows that:
 * $x \circledast \phi \left({e}\right) = x = \phi \left({e}\right) \circledast x$

Therefore, $\phi \left({e}\right)$ is the identity of $\bigl({\tilde G, \circledast}\bigr)$.

$({G3}):$ Inverse
Let $x \in \tilde G$, and define:
 * $y = \inf {\left\{{\phi \bigl({g^{-1}}\bigr): \phi \left({g}\right) \preceq x}\right\}}$

where $g^{-1}$ denotes the inverse of $g$ in $\left({G, *}\right)$.

By Characterization of Dedekind Completion, it follows that the set:
 * $\left\{{\phi \bigl({g^{-1}}\bigr): \phi \left({g}\right) \preceq x}\right\}$

is non-empty and bounded below.

Therefore, by Dedekind Completeness is Self-Dual, $y \in \tilde G$.

We show that $y$ is the inverse of $x$ in $\bigl({\tilde G, \circledast}\bigr)$.

We only prove that $x \circledast y = \phi \left({e}\right)$; the equation $y \circledast x = \phi \left({e}\right)$ follows similarly.

For any $g, h \in G$ such that $\phi \left({g}\right) \preceq x$ and $\phi \left({h}\right) \preceq y$, we have that:
 * $\phi \left({g * h}\right) \preceq \phi \left({g}\right) \circledast y \preceq \phi \left({g}\right) \circledast \phi \bigl({g^{-1}}\bigr) = \phi \left({e}\right)$

It follows that $x \circledast y \preceq \phi \left({e}\right)$.

We now show that $x \circledast y \succeq \phi \left({e}\right)$.

By Characterization of Dedekind Completion, we have:
 * $x \circledast y = \inf {\left\{{\phi \left({p}\right): \phi \left({p}\right) \succeq x \circledast y}\right\}}$

Suppose that $p \in G$, $\phi \left({p}\right) \succeq x \circledast y$.

It suffices to show that $p \ge e$.

If $k \in G$, $\phi \left({k}\right) \succeq x$, then:
 * $\forall g \in G: \phi \left({g}\right) \preceq x \implies \phi \bigl({k^{-1}}\bigr) \preceq \phi \bigl({g^{-1}}\bigr)$

Therefore, $\phi \bigl({k^{-1}}\bigr) \preceq y$, and so:
 * $x \circledast \phi \bigl({k^{-1}}\bigr) \preceq x \circledast y \preceq \phi \left({p}\right)$

By the associativity of $\circledast$, it follows that $\phi \left({p * k}\right) \succeq x$.

By Characterization of Dedekind Completion, we can choose $a, b \in G: \phi \left({a}\right) \preceq x \preceq \phi \left({b}\right)$.

Since the above holds for any $k \in G$ such that $\phi \left({k}\right) \succeq x$, it follows by mathematical induction that:
 * $\forall n \in \N: \phi \left({p^n * b}\right) \succeq x \succeq \phi \left({a}\right)$

Since $\phi$ is an order embedding and $\left({G, *, \le}\right)$ is an Archimedean ordered group, it follows that $p \ge e$.

It follows that $y$ is the inverse of $x$ in $\bigl({\tilde G, \circledast}\bigr)$.

Proof of Uniqueness
Let $\odot$ be a binary operation on $\tilde G$ such that:
 * $({1}): \quad \bigl({\tilde G, \odot, \preceq}\bigr)$ is an ordered group
 * $({2}): \quad \phi$ is a group homomorphism from $\left({G, *}\right)$ to $\bigl({\tilde G, \odot}\bigr)$

Then, for all $x, y \in \tilde G$, we have: