Completions and the Archimedean property August 28, 2009
Posted by David Pierce in Uncategorized.trackback
In an ordered Abelian group, one positive element can be described as infinite with respect to another if the former exceeds every integral multiple of the other. If there are no such elements, the group can be called Archimedean.
Non-Archimedean ordered Abelian groups exist: for example, the group of ordered pairs (x,y) of integers, with the left-lexicographic ordering, so that (x,y)<(a,b) if and only if either x<a, or else x=a and y<b.
The main point of this article is to observe that an ordered
Abelian group is Archimedean if and only if it has a completion.
I do not know of a reference for this result, though I can well imagine that a reference exists.
The observation about completeness arises from considering that the field of real numbers is complete in two ways:
- It is complete as an ordered field, because every nonempty
subset with an upper bound has a least upper bound. - It is complete as a valued field, because every Cauchy
sequence (with respect to the absolute value function) converges.
In sense (2), the field of real numbers is just one example of a
complete field. The complex numbers compose another such field,
as do the p-adic completions of the field of rational numbers.
Every valued field has a completion.
In sense (1) however, the field of real numbers is unique.
I have encountered at least one mathematician who seemed not to
be aware of this, or to have forgotten it, having apparently
confused completeness of valued fields with completeness of
ordered fields.
Possibly the distinction between ordered fields and valued fields
is like that between induction and completion: a distinction that
may be overlooked in one’s early education and then never
returned to.
At the end of his book Calculus, Michael Spivak constructs
the field of real numbers and proves its uniqueness (up to
isomorphism). It was from Spivak’s book that, as a student, I
first learned of the uniqueness of the real field. Spivak
praises the “one truly first-rate idea” in its construction:
Dedekind’s notion of a cut. Yet Spivak is disparaging of
the “drudgery” of going through the details of the construction.
I revisited the construction recently, in a course on
non-standard analysis at the Nesin Mathematics Village. If the
construction of the real numbers was going to be drudgery, I
wanted to see what more general results could be found in the
process.
The Dedekind cut construction gives a completion to every
ordered set (that is, totally ordered set). Indeed, let A be
such a set, and if x is in A, let (x) be the set of elements of A
that are less than or equal to x. Such sets compose a basis for
a topology on A. A cut of A can be defined as a nonempty closed
set in this topology, except A itself, unless this has a greatest
element. Let c(A) be the set of cuts of A. Then:
-
The set c(A) is ordered by inclusion
and is complete with
respect to this ordering. - The ordered set A embeds in c(A) under the map taking x to (x).
Again, an ordered Abelian group is Archimedean if, for any two
positive elements a and b, some multiple of a exceeds b. In
other words, a and b have a ratio in the sense of
Definition 4 of Book V of Euclid’s Elements. Indeed, the
positive part of an Archimedean ordered Abelian group would seem
to be just the set of magnitudes that have a ratio to some given
magnitude: the set is closed under addition, and under
subtraction of a lesser magnitude from a greater.
For any ordered Abelian group A, Archimedean or not, one can take
the completion of the underlying ordered set, and then extend the
definition of addition to the completion. One way to do this is
to define the sum of non-empty proper closed subsets X and Y of A
as the closure of the set of sums x+y, where x is in X and y is
in Y. Then c(A) becomes an Abelian monoid.
However, if A is a not Archimedean, then A cannot be complete,
since if the set of integral multiples of a positive element a is
bounded above by b, then b-a is also an upper bound. In c(A),
the set of multiples of a does have a supremum, c; but then c+a =
c.
Among Archimedean ordered Abelian groups, the group of integers
is the only discrete example, and this is complete. If A is a
dense Archimedean ordered Abelian group, then c(A) is the same.
If A and B are complete dense ordered Abelian groups, with
positive elements a and b respectively, then there is a unique
isomorphism from A to B taking a to b. The idea is that the
ordered field of rational numbers embeds in A under the map
taking 1 to a; then this map extends uniquely to the completion
of the rational field, which is the real field.
Consequently, for every real number a greater than 1, the
additive group of real numbers is isomorphic to the
multiplicative group of positive real numbers under a map taking
1 to a: this is the map commonly denoted by x |—> ax. One
shows that multiplication distributes over addition on the
positive reals, then on all reals, and so the reals compose a
complete ordered field, which must be unique, because the
underlying group is unique as a dense complete ordered Abelian
group.
Alexandre Borovik
dcorfield
David Pierce
A Dialogue on Infinity 
This is true, that completions in the two senses are distinct. However, if your absolute value comes from an order, then I believe the two are the same.
Cuts have one major existential advantage: they require only two sets to define them. If you want to avoid invoking infinite sets to the extent that’s possible, it’s better to go from a model of the rationals to a model of the reals by cuts.
On the other hand, in a pedagogical context like Spivak’s they’re just terrible. There’s very little intuition for reconstructing the field properties on the set of cuts and showing that they properly extend the field properties on the set of rationals. Cauchy sequences, though, are algebraically clear to the student. They have the added advantage of being still useful in the sequel, when talking about sequential notions of convergence and continuity.
The completions of an ordered field with respect to an ordering and with respect to the associated absolute value function need not be the same, unless absolute values are required to be real numbers—which they usually are, but this is not required in the actual formation of the completion.
For example, if R is the field of real numbers, then by introducing a positive infinitesimal t we can form the ordered field R(t) of rational functions. This has a completion, both with respect to the absolute value function, and with respect to the t-adic valuation; if I’m not mistaken, in each case the completion is the same: R((t)), the field of formal Laurent series.
The ordered field R(t), simply as an ordered set, also has a completion; but again this completion is not even a group.
Regarding pedagogy: Now matter how you approach it, I think calculus is going to be hard. When methods of improving calculus instruction are proposed, I wonder if they don’t seem easy only to those people who have already spent a lot of effort to understand calculus.
Spivak’s Calculus (and the high school teacher named Donald J. Brown for whose course Spivak’s book was a reference) first taught me mathematics, and I revere Spivak’s book, despite the imperfections I find with it as I get older. I’m not sure I actually read Spivak’s chapter on the Dedekind-cut construction in high school; it may have been from Dedekind’s own book that I learned the construction first in college. In any case, I like the construction: it seems to follow naturally from the wish to fill in the gaps in the rational number line.
I don’t see it quite so naturally.
What do you mean by a “gap” in the number line? I, for one, mean that I can take a sequence of steps on the rational numbers that seem like they “should be” converging, but they don’t converge to any rational number. This is formalized in the notion of a Cauchy sequence, and the Cauchy completion just says that this sequence “is” the number you’re looking for. And you then note that many other sequences seem to indicate the same “gap”, and so you define the number to be the equivalence class of such sequences. The desired topological property (the nonexistence of such “gaps”) is immediate.
From there it’s a simple matter to show that all the rational numbers can also be derived in this matter, and that the obvious termwise operations (with some care in the case of division) define a totally ordered field structure on the new collection of equivalence classes, which is identical to the old structure when restricted to just the rational numbers.
For the Dedekind construction, you assert that a “gap” is a place where you can cut the rationals into two sets, but which doesn’t correspond to an existing rational number. So you define a real number to be all the ways of cutting up the rationals into two sets like that. But now how do you know that there aren’t the same sort of gaps in this new collection? How do you know you’re done? It’s possible to prove, but it’s not nearly so immediate.
In Dedekind’s original conception as I understand it (but I haven’t got his book in front of me), if one breaks a geometrical line in two pieces, then on expects to find a point at the end of one of the pieces. There may not be such a rational point; therefore there are gaps among the rationals. But the point that determines the break is itself determined by knowing which rationals fall on which side.
The original reason for posting on this topic was the observation that, if an ordered abelian group is completed with respect to the ordering, then the ordered-group structure extends to the completion if and only if the original group is Archimedean.
This has lead the the observation that, for ordered fields, completion-by-Dedekind-cut and completion-by-Cauchy-sequence are not equivalent constructions. They give the same result only for Archimedean ordered fields. So it may not be meaningful to claim that one construction is better than the other. Perhaps the Cauchy-sequence construction is of more general interest.
“The main point of this article is to observe that an ordered
Abelian group is Archimedean if and only if it has a completion.”
Great! This answers a question I had been struggling with for some time: what happens when one Dedekind completes a non-Archimedean ordered field? One cannot get a Dedekind complete ordered field, because then it would be Archimedean and so would be its subfield. But one has to get a Dedekind complete SOMETHING, because that’s what Dedekind completion does. So somehow Dedekind completion loses the “field” part. This post finally explains exactly how and why this occurs.
“I do not know of a reference for this result, though I can well imagine that a reference exists.”
I’m sure that some experts must know this result, but I haven’t found it in the literature either. It would be a service to the community to spend an afternoon or so really looking for a reference, and if you don’t find one, writing and publishing a short note with the content of this post. I could see it appearing as a note in the American Mathematical Monthly, for instance.