0
$\begingroup$

Let $U$ be a subset of $\mathbb{R}$.

$U$ inherits both a metric structure and a total order from its embedding in $\mathbb{R}$, which is both a metric space and a totally ordered set.

Thus one can ask if $U$ is metrically complete (every Cauchy sequence converges) and independently whether it is order-complete (every nonempty, bounded-above subset has a least upper bound).

The answers are not always the same. For example, an open interval (not the whole line) is not metrically complete (consider a Cauchy sequence converging to one of the missing endpoints) but it is order-complete as it is order-isomorphic to the whole real line (e.g. by a function like $\tan^{-1}$).

One can always take the metric completion $\bar U$ (the set of Cauchy sequences modulo an appropriate equivalence relation) of $U$. Since $U\subset \mathbb{R}$ and $\mathbb{R}$ is a complete metric space, $\bar U$ is naturally embedded in $\mathbb{R}$, so it is totally ordered in a natural way, and $U$ will have a natural embedding in $\bar U$.

Meanwhile, I believe one can always take the order-completion $\tilde U$ of $U$: the set of nonempty, downward-closed subsets of $U$, modulo an appropriate equivalence relation, ordered by inclusion. (TBH, I'm not sure of the equivalence relation in full generality. I know how this works in the classical case of building $\mathbb{R}$ from $\mathbb{Q}$, but for an arbitrary total order I'm not sure.) If things work as I believe they do, then $\tilde U$ will be an order-complete totally ordered set. $U$ will have a natural embedding in $\tilde U$ respecting the order.

My preliminary question is: are the things I just said about order-completion correct?

If so, here is my real question:

Is there a natural condition characterizing those $U$ such that $\bar U$ and $\tilde U$ are order-isomorphic, with an isomorphism respecting the embedding of $U$ in each?

Context/backstory (you can ignore):

I was writing up for my students a proof that order-completeness as defined above is equivalent to Dedekind's original definition, which is that for any partition $U= A\dot\cup B$ such that $a<b$ for all $a\in A, b\in B$, there exists a $c\in U$ such that $c\geq a, \forall a\in A$ and $c\leq b, \forall b\in B$. The proof turned out to be much subtler than I intended for it to be. In particular, if you assume Dedekind's original formulation holds for $U$, then $c$ is not necessarily uniquely determined by $A,B$: in particular, if $A$ has a greatest element and $B$ has a least element (for example if $U=(-\infty,0]\cup [1,\infty)$), there will be two possible $c$'s. In this situation, if you want to verify the standard definition of order-completeness for $U$, you need to make sure you choose the right $c$ in the proof.

This made the whole argument way trickier than I planned, so I was thinking about how I might make it more transparent. I was considering spending some time working through some examples of possible $U$'s other than $\mathbb{Q}$ or $\mathbb{R}$ to illustrate the issues that come up. I started to write about $[0,1]\cup (2,3]$, when I realized... wait. This guy is actually order-isomorphic to $[0,2]$! So it is order-complete (which is the only kind of completeness I have discussed with my students)! This is horrible. It will not be consistent with the intuition I've tried to develop with them about what completeness is all about. It is obviously "missing" the point at $x=2$... but only in the sense of metric completion, not order-completion.

For another interesting example, two disjoint open intervals $(0,1)\cup (2,3)$ are missing four points from the persective of metric completion, but only one point (a least upper bound for $(0,1)$) from the perspective of order-completion.

On the other hand, metric and order completion famously coincide for $\mathbb{Q}$. (With due respect to Cantor and Dedekind, respectively.)

So, is there a nice theorem characterizing which $U$'s have them coincide?

$\endgroup$
3
  • $\begingroup$ You wrote a lot, trying to follow it. Would it be easier if you think that you obtain the order-completion by inserting the gaps, where a gap is thought of as "the new element that should be inserted" or "the missing element" between a downward-closed non-empty A and upward-closed non-empty B that form a partition of U, and such that A has no largest element, and B has no smallest element? This is not really different than using Dedekind's completion, might be a different point of view easier to grasp, I have yet to read in detail what you describe, and what you ask. $\endgroup$
    – Mirko
    Apr 20 '18 at 1:00
  • $\begingroup$ I assumed there was some standard construction of order-completion. (I couldn't find it with a quick search though.) After writing, I thought it was probably "the set of downward-closed, nonempty sets modulo the equivalence $A\sim A'$ if $A,A'$ differ by one element which is the greatest in one and not the other." I bet your construction is equivalent. $\endgroup$ Apr 20 '18 at 1:10
  • 1
    $\begingroup$ here is a related question on MSE math.stackexchange.com/questions/455911/… $\endgroup$
    – Mirko
    Apr 20 '18 at 1:22
1
$\begingroup$

There is a standard way to define order-completion, but I cant seem to find a good reference online (there is too much for too general cases and partial orders, but not for what I was looking for).

Call a pair $(A,B)$ a cut if it partitions $U$, both $A$ and $B$ are non-empty, and $A$ is downward-closed (if $a\in A$ and $c\in U$ with $c<a$ then $c\in A$) and $B$ is upward-closed. Call a cut a gap if $A$ has no largest element and $B$ has no smallest element. We obtain the order-completion of $U$ by inserting an element to serve both $\sup A$ and $\inf B$ for each gap $(A,B)$. We need not worry about cuts that are not gaps. (Sometimes people would add smallest and last elements to $U$ if these were not already present, the topology resulting from this type of completion is compact, have seen it in the context of Suslin lines, but cannot find a reference.)

So, here are some necessary conditions for $\bar U$ and $\tilde U$ to be order-isomorphic. If $(A,B)$ is a gap, then (working in $\mathbb R$) we must have $\sup A=\inf B$. Otherwise the metric completion would add two elements, $\sup A\not=\inf B$, while the order completion would add only one element.

Another necessary condition, if for some bounded above (in $\mathbb R$) $C\subseteq U$ we have that $\sup C=c\not\in U,$ then we must have that (a): $\emptyset\not=(c,\infty)\cap U$, and (b): $(c,\infty)\cap U$ does not have a least element. To prove (a) assume to the contrary that $U\subseteq (-\infty,c)$. The metric completion must add $c$, while the order-completion won't (as we only aim to build a completion that is bounded-complete https://en.wikipedia.org/wiki/Bounded_complete_poset ). To prove (b), if $(c,\infty)\cap U$ did have a least element $d$ then the order-completion need not add any element between $(-\infty,c)$ and $[d,\infty)$, while the metric completion needs to add $c$. Similar necessary conditions holds for bounded below $C\subseteq U$.

In particular, either $\sup U=\infty$, or else $\sup U\in U$. Similarly, either $\inf U=-\infty$, or else $\inf U\in U$.

The above implies that whenever $x\in \bar U\setminus U$ then we must have that $\Bigl((-\infty,x)\cap U,(x,\infty)\cap U)\Bigr)$ is a gap, in particular both $(-\infty,x)\cap U$ and $(x,\infty)\cap U$ are non-empty, and moreover $x=\inf ((x,\infty)\cap U) =\sup ((-\infty,x)\cap U)$ (where $\inf$ and $\sup$ are, again, taken in $\mathbb R$). I think it is easy to verify that this condition is also sufficient to imply that $\bar U=\tilde U$ (and the isomorphism would be the identity).

$\endgroup$
1
  • $\begingroup$ I take "and the isomorphism would be the identity" to mean that in this situation, the order completion has a unique embedding in $\mathbb{R}$ that extends the embedding of $U$, and this embedding coincides with the metric completion. $\endgroup$ Aug 21 '18 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.