Existence of Dedekind Completion

Theorem
Let $S$ be an ordered set.

Then there exists a Dedekind completion of $S$.

That is, there exists a Dedekind complete ordered set $\tilde{S}$ and an increasing mapping $\phi: S \to \tilde{S}$ such that:


 * For all Dedekind complete ordered sets $X$, and for all order monomorphisms $f: S \to X$, there exists an order monomorphism $\tilde{f}: \tilde{S} \to X$ such that:


 * $\tilde{f} \circ \phi = f$

Proof
For all subsets $L \subseteq S$, define:


 * $L_+ = \left\{{x \in S: x}\right.$ is an upper bound for $\left.{L}\right\}$


 * $L_- = \left\{{x \in S: x}\right.$ is a lower bound for $\left.{L}\right\}$


 * $\underline{L} = \left({L_+}\right)_-$

Note that:


 * $L \subseteq \underline{L}$


 * $L \subseteq M \implies L_{\pm} \supseteq M_{\pm}$; it follows that $L \subseteq M \implies \underline{L} \subseteq \underline{M}$


 * $\underline{L}_+ = L_+$; it follows that $\underline{\underline{L}} = \underline{L}$

All of the above follow directly from the relevant definitions.

Let:
 * $\tilde{S} = \left\{{L \subseteq S:}\right.$ $L = \underline{L}$, $L$ is non-empty and bounded above$\left.{}\right\}$

We have that $\bigl({\tilde{S}, \subseteq}\bigr)$ is an ordered set.

Let $A \subseteq \tilde{S}$ be non-empty and bounded above.

By Union Smallest: General Result, $\displaystyle \bigcup A$ is non-empty and bounded above.

It follows that $\displaystyle \underline{\bigcup A} \in \tilde{S}$.

From Subset of Union: General Result, it follows that $\displaystyle \underline{\bigcup A}$ is an upper bound for $A$.

If $U \in \tilde{S}$ is an upper bound for $A$, then, by Union Smallest: General Result:
 * $\displaystyle \bigcup A \subseteq U$

It follows that:
 * $\displaystyle \underline{\bigcup A} \subseteq \underline{U} = U$

Hence, $\displaystyle \underline{\bigcup A} = \sup A$.

It has been established that $\bigl({\tilde{S}, \subseteq}\bigr)$ is Dedekind complete.

Let $\phi: S \to \tilde{S}$ be the increasing mapping defined as:
 * $\forall x \in S: \phi \left({x}\right) = {\overline{\downarrow}} \left({x}\right)$

where ${\overline{\downarrow}} \left({x}\right)$ denotes the weak lower closure of $x$ (in $S$).

Let $\left({X, \preceq}\right)$ be a Dedekind complete ordered set, and let $f: S \to X$ be an order monomorphism.

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

where $f \left({L}\right)$ denotes the image of $L$ under $f$.

Then $\tilde{f} \circ \phi = f$.

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

Let $x \in L$, and choose any $u \in M_+$.

Then $f \left({x}\right) \preceq \tilde{f} \left({M}\right) \preceq f \left({u}\right)$.

Since $f$ is an order monomorphism, it follows that $x \in \underline{M} = M$.

Hence, $L \subseteq M$, and so $\tilde{f}$ is an order monomorphism.