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 $\left({G, *}\right)$

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

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

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

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

The existence of $x \mathbin{\tilde *} y$ is justified by Characterization of Dedekind Completion.

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

From Supremum of Subset, it follows that $\tilde \preceq$ is compatible with $\tilde *$.

It remains to show that $\bigl({\tilde G, \tilde *}\bigr)$ is a group.

We prove that $\tilde *$ is associative.

Let $x, y, z \in \tilde G$.

Define:
 * $A = \bigl\{{\phi \left({a * b * c}\right): \phi \left({a}\right) \mathrel{\tilde \preceq} x, \, \phi \left({b}\right) \mathrel{\tilde \preceq} y, \, \phi \left({c}\right) \mathrel{\tilde \preceq} z}\bigr\}$

It suffices to show that:
 * $\bigl({x \mathbin{\tilde *} y}\bigr) \mathbin{\tilde *} z = \sup A = x \mathbin{\tilde *} \bigl({y \mathbin{\tilde *} z}\bigr)$

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

We have that $\bigl({x \mathbin{\tilde *} y}\bigr) \mathbin{\tilde *} z$ is an upper bound for $A$.

Let $u \in \tilde G$ be an upper bound for $A$.

It is to be shown that $\bigl({x \mathbin{\tilde *} y}\bigr) \mathbin{\tilde *} z \mathrel{\tilde \preceq} u$.

By Characterization of Dedekind Completion, we have that:
 * $u = \inf {\bigl\{{\phi \left({g}\right): \phi \left({g}\right) \mathrel{\tilde \succeq} u}\bigr\}}$

Therefore, it suffices to show that:
 * $\forall g \in G: \phi \left({g}\right) \mathrel{\tilde \succeq} u \implies \phi \left({g}\right) \mathrel{\tilde \succeq} \bigl({x \mathbin{\tilde *} y}\bigr) \mathbin{\tilde *} z$

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

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

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

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

Hence:
 * $\bigl({x \mathbin{\tilde *} y}\bigr) \mathbin{\tilde *} z \mathrel{\tilde \preceq} \phi \left({g}\right)$

It follows that $\tilde *$ is associative.

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

By Characterization of Dedekind Completion, it follows that $\phi \left({e}\right)$ is the identity of $\bigl({\tilde G, \tilde *}\bigr)$.

We prove the existence of inverses in $\bigl({\tilde G, \tilde *}\bigr)$.

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

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

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

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

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

It follows that $x \mathbin{\tilde *} y \mathrel{\tilde \preceq} \phi \left({e}\right)$.

We have that:
 * $\forall k \in G: \phi \left({k}\right) \mathrel{\tilde \succeq} x \implies \bigl({\forall g \in G: \phi \left({g}\right) \mathrel{\tilde \preceq} x \implies \phi \bigl({k^{-1}}\bigr) \mathrel{\tilde \preceq} \phi \bigl({g^{-1}}\bigr)}\bigr)$

It follows that:
 * $\forall k \in G: \phi \bigl({k^{-1}}\bigr) \mathrel{\tilde \preceq} y$

Suppose that $p \in G$, $\phi \left({p}\right) \mathrel{\tilde \succeq} x \mathbin{\tilde *} y$.

Then:
 * $\forall k \in G: \phi \left({k}\right) \mathrel{\tilde \succeq} x \implies x \mathbin{\tilde *} \phi \bigl({k^{-1}}\bigr) \mathrel{\tilde \preceq} x \mathbin{\tilde *} y \mathrel{\tilde \preceq} \phi \left({p}\right)$

By the associativity of $\tilde *$, it follows that:
 * $\forall k \in G: \phi \left({k}\right) \mathrel{\tilde \succeq} x \implies \phi \left({p * k}\right) \mathrel{\tilde \succeq} x$

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

It follows that:
 * $\forall n \in \N: \phi \left({p^n * b}\right) \mathrel{\tilde \succeq} x \mathrel{\tilde \succeq} \phi \left({a}\right)$

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

We have shown that:
 * $\forall p \in G: \phi \left({p}\right) \mathrel{\tilde \succeq} x \mathbin{\tilde *} y \implies p \succeq e$

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

Hence, $x \mathbin{\tilde *} y \mathrel{\tilde \succeq} \phi \left({e}\right)$.

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