# Dedekind completions

## Idea

The Dedekind completion of a linear order $L$ is a new linear order $\overline{L}$ that contains suprema for all inhabited bounded subsets, and such that a supremum in $L$ is still a supremum in $\overline{L}$.

While Dedekind completeness was traditionally described in the context of the real numbers, it can be stated for any linear order, although it really works best for dense? and unbounded (without top or bottom) linear orders. Intuitively, a linear order is Dedekind complete if Dedekind cuts don’t give any ‘new’ elements.

Any paragraph containing the string ‘duiq’ is original research (although lower duiqs at least are known in domain theory).

## Definitions

Let $S$ be a set with a dense? linear order $<$ without top or bottom elements.

###### Definition

A cut in $S$ is a pair of subsets $L,U\subset S$ of $S$ that satisfy the following eight properties:

1. $L$ is inhabited;
2. Dually, $U$ is inhabited;
3. If $x, then $x\in L$;
4. Dually, if $x>y\in U$, then $x\in U$;
5. If $x\in L$, then $x for some $y$;
6. Dually, $x\in U$, then $x>y\in U$ for some $y$;
7. If $x, then $x\in L$ or $y\in U$;
8. If $x\in L$ and $y\in U$, then $x.
###### Definition

The linearly ordered set $S$ is Dedekind complete if every cut $\left(L,U\right)$ is of the form

$L=\left\{x\phantom{\rule{thickmathspace}{0ex}}\mid \phantom{\rule{thickmathspace}{0ex}}xa\right\}$L = \{x \;| \;x \lt a\},\; U = \{x \;|\; x \gt a\}

for some unique $a\in S$.

###### Definition

If $T$ is also an unbounded dense linear order, $S$ is Dedekind complete, and we have a universal arrow $u:T\to S$ in the category of linear orders, then $S$ (equipped with $u$) is the Dedekind completion of $T$.

The set of Dedekind cuts of rational numbers –the set of real numbers– is Dedekind complete. In fact, starting with any unbounded dense linearly ordered set $S$, the set of Dedekind cuts is isomorphic to the reals as long as $S$ is a countably infinite set.

The operation of forming the set of Dedekind cuts is idempotent, so the Dedekind completion can be constructed as the set of Dedekind cuts. More precisely, the Dedekind-complete linear orders form a reflective subcategory of the category of dense unbounded linear orders, so that Dedekind completion is a kind of completion in the abstract categorial sense.

## Generalisations

One can generalise Dedekind completion from linear orders to quasiorders.

###### Definition

A duiq (dense unbounded inhabited quasiorder) is a quasiordered set $S$ such that, given finite (here always meaning Kuratowski-finite) subsets $F$ and $G$ of $S$ such that $x whenever $x\in F$ and $z\in G$, we have some $y$ in $S$ such that $x whenever $x\in F$ and $z\in G$.

Note that, for $S$ a linear order, $S$ is a duiq iff $S$ is dense, unbounded, and inhabited, hence the term ‘duiq’. (Using linearity, we may assume that $F$ and $G$ are subsingletons; then two singleton subsets is denseness, one singleton subset and one empty subset is unboundedness, and two empty subsets is inhabitedness.)

###### Definition

Given a duiq $S$, a cut is a pair $\left(L,U\right)$ of subsets such that:

1. $L$ is inhabited (which is a special case of (5) for $F$ the empty subset);
2. Dually, $U$ is inhabited (a special case of (6));
3. If $x, then $x\in L$;
4. Dually, if $x>y\in U$, then $x\in U$;
5. If $F$ is a finite subset of $L$ (which we may assume inhabited if we include (1)), then for some $x\in L$, every $y\in F$ satisfies $y;
6. Dually, if $F$ is a finite (inhabited) subset of $U$, then for some $x\in U$, every $y\in F$ satisfies $y>x$;
7. If $L and $L, then $x=y$;
8. If $x\in L$ and $y\in U$, then $x.

We then define Dedekind-complete duiqs and Dedekind completions of duiqs the same as for dense linear orders, using this notion of cut.

A good example of a duiq is the set of rational-valued functions on any set $X$; the Dedekind completion is the set of real-valued functions on $X$.

Sections 4.31–39 of HAF do things in even more generality, but I don't really understand it yet.

## One-sided Dedekind completions

At least in classical mathematics, considering only $L$ (for a lower cut) or $U$ (for an upper cut) doesn't really give us anything new for linear orders; we have only the technicality that $\infty ≔\left(S,\varnothing \right)$ or $-\infty ≔\left(\varnothing ,S\right)$ is a cut (depending on the side), and we can rule even these out by simply requiring that $L$ have an upper bound or that $U$ have a lower bound.

In constructive mathematics, one-sided cuts are more general; see one-sided real number for a discussion of the case where $S$ is the linear order of rational numbers.

Even in classical mathematics, one-sided cuts do give us something new for quasiorders. Here, we have first more general one-sided notions of duiqs: a lower duiq need only satisfy the condition of a duiq for $G$ a singleton, and an upper duiq need only satisfy the condition for $F$ a singleton. Then the lower Dedekind completion of a lower duiq is its set of lower cuts, and the upper Dedekind completion of an upper duiq is its set of upper cuts.

For example, let $X$ be a compactum and let $S$ be the quasiordered set of continuous real-valued functions on $X$. Then $S$ is a duiq, hence both a lower and upper duiq. Its lower Dedekind completion is the set of lower semicontinuous? functions on $X$ taking values in the lower reals (which classically are all either real or $\infty$); and dually on the upper side. Even working classically and ignoring the technicality of $\infty$, semicontinuous functions are much more general than continuous ones.

## References

• Page 249&250 of Continuous Lattices and Domains (Google book) covers lower Dedekind completions of duiqs (although not under that name) in the context of domain theory.

Revised on October 1, 2012 07:02:12 by Toby Bartels (98.23.147.236)