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 order embedding $\phi: S \to \tilde S$ such that:


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


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

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


 * $\operatorname U \left({I}\right) = \left\{{x \in S: x}\right.$ is an upper bound for $\left.{I}\right\}$


 * $\operatorname L \left({I}\right) = \left\{{x \in S: x}\right.$ is a lower bound for $\left.{I}\right\}$

Note that, for all $I, J \subseteq S$:


 * $I \subseteq \operatorname {LU} \left({I}\right)$


 * $I \subseteq J \implies \operatorname {LU} \left({I}\right) \subseteq \operatorname {LU} \left({J}\right)$


 * $\operatorname {ULU} \left({I}\right) = \operatorname U \left({I}\right)$

All of the above follow directly from the relevant definitions.

Let:
 * $\tilde S = \left\{{I \subseteq S: I = \operatorname {LU} \left({I}\right)}\right.$, $I$ 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 is Smallest Superset: General Result, $\displaystyle \bigcup A$ is non-empty and bounded above.

It follows that:
 * $\displaystyle \operatorname{LU} \left({\bigcup A}\right) \in \tilde S$

From Subset of Union: General Result, it follows that $\displaystyle \operatorname{LU} \left({\bigcup A}\right)$ is an upper bound for $A$.

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

It follows that:
 * $\displaystyle \operatorname{LU} \left({\bigcup A}\right) \subseteq \operatorname{LU} \left({I}\right) = I$

Hence:
 * $\displaystyle \operatorname{LU} \left({\bigcup A}\right) = \sup A$

By definition, $\bigl({\tilde S, \subseteq}\bigr)$ is Dedekind complete.

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

where ${\bar\downarrow} \left({x}\right)$ 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 embedding.

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

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

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

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

Then $\forall x \in I: \forall u \in \operatorname U \left({J}\right): f \left({x}\right) \preceq \tilde f \left({J}\right) \preceq f \left({u}\right)$.

Since $f$ is an order embedding, it follows that $I \subseteq \operatorname {LU} \left({J}\right) = J$.

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

Also see

 * Dedekind Completion is Unique up to Isomorphism