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I have been trying to write up some notes on completion of ordered fields, ideally in the general case (i.e., not just completing $\mathbb{Q}$ to get $\mathbb{R}$ but considering the completion via Cauchy sequences of any ordered field). I have found the technical details of this to be surprisingly thorny, especially compared to the relatively clean construction of the completion of a metric space (which takes $\mathbb{R}$ as given).

Finally I found a clean, concise treatment of the general case in $\S$ 8.7 of P.M. Cohn's text Basic Algebra: Groups, Rings and Fields. He does the general case and also includes the universal property of completion. Following his terminology, let's say an ordered field is complete if every Cauchy sequence in that field is convergent (I would say "sequentially complete" to differentiate both from Dedekind completeness and also the possibility of considering more general Cauchy nets).

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda: K \rightarrow \tilde{K}$ such that to each order-embedding $f: K \hookrightarrow L$ into a complete ordered field $L$ there is a unique order-embedding $f': \tilde{K} \rightarrow L$ such that $f = f' \circ \lambda$.

I am happy with the existence of $\lambda$. However, Cohn's proof of the universal property says:

"Finally, let $f: K \hookrightarrow L$ be an order-embedding in a complete field $L$. Any element $\alpha$ of $\tilde{K}$ is obtained as the limit of a Cauchy sequence $\{a_n\}$ in $K$; it is easily seen that $\{f(a_n)\}$ is a Cauchy sequence in $L$ and so has a limit..."

This sounds reasonable at first glance...but it's false, I think. Suppose $K = \mathbb{Q}$ and $L = \mathbb{R}((t))$, the latter given the unique ordering in which $t$ is positive and smaller than every positive real number. It is well-known that $L$ is complete (all I'm using about $L$ is that it is complete and non-Archimedean, and such fields certainly exist, so if you prefer just take any such $L$). Let $a_n = \frac{1}{n}$, so of course $f(a_n) = \frac{1}{n}$. But we have a problem: the sequence $\frac{1}{n}$ is not in fact Cauchy (or equivalently, convergent) in $L$ because $L$ is non-Archimedean: the open interval $(-t,t)$ about $0$ contains no terms of the sequence!

[In fact one can make the problem even less subtle. It is known that ultraproducts and such construct ordered fields which have uncountable cofinality (equivalently, are not first countable in the order topology). In such a field, the only Cauchy sequences are the eventually constant ones. In particular such fields are automatically complete. However, such a field contains subfields like $\mathbb{Q}$ which have plenty of not eventually constant convergent sequences. Clearly an order embedding cannot make a not-eventually-constant sequence eventually constant!]

Thus I claim the distressing fact that a homomorphism of ordered fields need not be even sequentially continuous with respect to the given order topologies: a fortiori it need not be continuous.

So the question is: am I making some silly mistake here? (Cohn is a gold-standard algebraist and his book is lovely and authoritative, even compared to many of the more familiar standard texts.) If not, is the result even true? I think it isn't: instead of $\mathbb{R}((t))$ we could take $\mathbb{Q}((t))$ which I believe is still complete -- a sequence is Cauchy iff for each $n$ the sequence of $n$th Laurent series coefficients is eventually constant; hence Cauchy sequences are convergent -- and then the embedding from $\mathbb{Q}$ will not extend to $\mathbb{R}$. If the result is false, I wonder:

What is the appropriate universal property of the completion of an ordered field? (Note that I asked a similar question about metric spaces here a while back.) We could require for instance the maps to be "sequentially Cauchy" (i.e., to preserve Cauchy sequences) and then the proof goes through. But that seems a little weak...

Added: One fix is to impose the condition that $f$ too be a dense embedding (i.e., such that in between any two distinct elements of $L$ lies an element of the image of $f$). This condition is sufficient for Cauchy sequences to map to Cauchy sequences (I think it is not necessary: imagine embedding a field of uncountable cofinality into another field of bigger cofinality). If we put this in then essentially we get a rather wordy way of saying that the completion is unique, but that is indeed the main application of this result anyway, so far as I know.


Further Added: The following recent note seems to face some issues on the topology of ordered fields more head-on than what I've seen otherwise in the literature. Here is the MathSciNet review:

Tanaka, Yoshio Topology on ordered fields. Comment. Math. Univ. Carolin. 53 (2012), no. 1, 139–147. An ordered field is a field $(K,+,∗)$ equipped with a linear order $<$ and the open interval topology $λ(<)$ where the algebra and order are related as follows: if $a,b,c \in K$ and $a<b$, then $a+c<b+c$; and if $c>0$ then $a∗c<b∗c$. Consequently, any ordered field contains an algebraic copy of $\mathbb{Q}$, the usual field of rational numbers. A key issue in this paper is that for a subset $A \subset K$, the topology $λ(<)|A$ that $A$ inherits as a subspace of $(K,λ(<))$ might, or might not, be the same as the open interval topology defined on $A$ by the restricted order $<|A$. The author discusses the Archimedean property of an ordered field, showing, for example, that K is Archimedean if and only if Q is dense in K, and that for any ordered field K, either Q is dense in K or is closed discrete in K. Other results concern Dedekind completeness of K and another result shows that if K is an Archimedean ordered field and if f:K→K′ is a surjective algebraic homomorphism, then the following are equivalent: f is continuous; f is a homeomorphism; and f is order preserving. The paper ends with examples of various ordered fields and homomorphisms.

I haven't been able to get my hands on a copy, but if someone can and it seems relevant, I'd take it as a favor if they'd let me know.

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  • $\begingroup$ At the end of the quotation of Cohn's theorem 8.7.1, should "$f = \lambda \circ f'$" instead be "$f = f'\circ\lambda$"? $\endgroup$ – John Bentin Aug 28 '13 at 7:37
  • $\begingroup$ @Pete: Is it true when $L$ is Archimedean? For example Dedekind-complete ordered fields are Archimedean. $\endgroup$ – Martin Brandenburg Aug 28 '13 at 8:20
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    $\begingroup$ @Martin: Yes, that's a special case of my suggested fix, because Q is order-dense in K iff K is Archimedean. It follows that every homomorphism into an Archimedean field is order-dense. $\endgroup$ – Pete L. Clark Aug 28 '13 at 9:27
  • $\begingroup$ Is there a reason why Cohn only considers sequences? $\endgroup$ – François G. Dorais Aug 28 '13 at 19:19
  • $\begingroup$ @Francois: Well, the title of the section is "The Field of Real Numbers", and the chapter goes on to discuss quadratic forms over fields, formally real and otherwise. Neither nets nor filters appear anywhere in the text, so far as I can see. I think he's really trying to go one up on the standard (and kind of boring) construction of $\mathbb{R}$ (as am I in my notes)...and I'm catching him out as going a little too far. $\endgroup$ – Pete L. Clark Aug 28 '13 at 19:39
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The statement can be corrected by adding one word:

Theorem 8.7.1: Let $K$ be an ordered field. Then there is a complete ordered field $\tilde{K}$ and a dense order-embedding $\lambda:K \to \tilde{K}$ such that to each cofinal order-embedding $f:K \to L$ into a complete ordered field $L$ there is a unique order-embedding $f′:\tilde{K} \to L$ such that $f=f′\circ \lambda$.

Note that there are only two possibilities for a subfield $K$ of an ordered field $L$: either $K$ is cofinal in $L$ or $K$ is discrete in $L$. [If $K \cap (0,\varepsilon) = \varnothing$ then $K \cap (x-\varepsilon,x+\varepsilon) = \{x\}$ for each $x \in K$.] Similarly an embedding $f:K \to L$ is either cofinal (and uniformly continuous) or discrete (and wildly discontinuous).

There cannot be a theorem that allows discrete embeddings $f:K \to L$. The reason is that every ordered field $K$ is discretely embeddable into the complete ordered field $K((T))$, where $T$ is infinitesimal with respect to $K$, and then there is no copy of $\tilde{K}$ in $K((T))$ in which $K$ is cofinal unless $K$ is already complete.

Remark 1: A discrete embedding $f:K \to L$ can never be "sequentially Cauchy" unless all Cauchy sequences in $K$ are eventually constant. In the case where $K$ doesn't have nontrivial Cauchy sequences, then all embeddings are sequentially Cauchy but such a $K$ is also trivially complete. Therefore the theorem with "cofinal" and "sequentially Cauchy" are precisely equivalent.

Remark 2: Cohn's completion is unusual because he only considers Cauchy sequences. However, it does coincide with the usual topological completion when the ordered field is metrizable. An ordered field is metrizable precisely if it has countable cofinality. When a field has uncountable cofinality, all Cauchy sequences are eventually constant and Cohn's completion trivializes.

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  • $\begingroup$ Yes, I said that in my answer. Do we regard that as the optimal fix? E.g. the density condition is sufficient but not necessary for the map to extend. $\endgroup$ – Pete L. Clark Aug 28 '13 at 18:41
  • $\begingroup$ @Pete: Hm... I should have read to the end! I added a more explanation that suggests that nothing more can be done. $\endgroup$ – François G. Dorais Aug 28 '13 at 18:44
  • $\begingroup$ As I said in my question, if you include such a strong hypothesis as "density", you are essentially just asserting the uniqueness of the completion, and phrasing in terms of a universal property seems a little highfalutin. But are you of the opinion that this is what Cohn (possibly) intended? $\endgroup$ – Pete L. Clark Aug 28 '13 at 18:46
  • $\begingroup$ @Pete: It's the natural generalization of "$\mathbb{R}$ is the maximal archimedean ordered field." $\endgroup$ – François G. Dorais Aug 28 '13 at 18:50
  • $\begingroup$ It's a generalization, sure, and it seems pretty natural. But it doesn't feel like a nontrivial example of a universal mapping property. Given that the completion of a metric space has a bona fide universal mapping property which is stronger than just the analogous uniqueness statement, are you so sure that there is no stronger analogue here? Why? $\endgroup$ – Pete L. Clark Aug 28 '13 at 18:52
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Sine Emil seems to be rather busy, here is an outline of the Dedekind-style approach to completing an ordered field.


$\newcommand{\cut}[1]{\langle#1\rangle}$A good cut in $K$ is a pair $\cut{L,R}$ such that:

  1. $L$ is a nonempty initial part of $K$ with no maximal element.
  2. $R$ is a nonempty final part of $K$ with no minimal element.
  3. $L$ and $R$ do not overlap.
  4. For every $\varepsilon \gt 0$ there are $x \in L$, $y \in R$ such that $y - x \lt \varepsilon$.

The only new requirement is the last one. When $K$ is Archimedean this is equivalent to saying that $L \cup R$ omits at most one point of $K$. However, when $K$ is not Archimedean, there are some cuts that are not good, such as the cut with lower part $(-\infty,1)\cup(-\infty,2)\cup\cdots$

The field operations for good cuts are defined as usual.

Fact. If $\cut{L_1,R_1}$ and $\cut{L_2,R_2}$ are good cuts then so is $$\cut{L_1,R_1} + \cut{L_2,R_2} = \cut{L_1 + L_2,R_1 + R_2}.$$

It's not difficult to check that the good cut $\cut{(-\infty,0),(0,+\infty)}$ is an identity for addition of good cuts. Additive inverses also exist:

Fact. If $\cut{L,R}$ is a good cut then so is $\cut{-R,-L}.$ Moreover, $$\cut{L,R}+\cut{-R,-L} = \cut{(-\infty,0),(0,+\infty)}.$$

Note that the additional requirement comes in handy here since it is equivalent to $R - L = (0,+\infty)$ (modulo the other three requirements).

To understand multiplication of cuts, it helps to think how intervals multiply. The product $[x_1, y_1] \cdot [x_2, y_2]$ has to be an interval and its endpoints must be among the four products $x_1x_2, x_1y_2, y_1x_2, y_1y_2.$ Since it must contain all four products, the result is $$[\min(x_1x_2, x_1y_2, y_1x_2, y_1y_2),\max(x_1x_2, x_1y_2, y_1x_2, y_1y_2)].$$

Fact. If $\cut{L_1,R_1}$ and $\cut{L_2,R_2}$ are good cuts then so is $\cut{L_1,R_1}\cdot\cut{L_2,R_2} = \cut{L,R}$, where $$\begin{aligned} L &= \{\min(x_1x_2, x_1y_2, y_1x_2, y_1y_2) : x_1 \in L_1, y_1 \in R_1, x_2 \in L_2, y_2 \in R_2\}, \\ R &= \{\max(x_1x_2, x_1y_2, y_1x_2, y_1y_2) : x_1 \in L_1, y_1 \in R_1, x_2 \in L_2, y_2 \in R_2\}. \end{aligned}$$

(The verification that this product is a good cut is actually straightforward arguably less tedious than that with the more common definition by cases.)

It's not difficult to check that the good cut $\cut{(-\infty,1),(1,+\infty)}$ is an identity for multiplication of good cuts. Multiplicative inverses also exist for positive good cuts:

Fact. If $\cut{L,R}$ is a good cut and $0 \in L$ then $\cut{L^{-1},R^{-1}}$ is also a good cut where $$\begin{aligned} L^{-1} &= (-\infty,0] \cup \{y^{-1} : y \in R\}, \\ R^{-1} &= \{x^{-1} : x \in L \cap (0,+\infty)\}. \end{aligned}$$ Moreover, $\cut{L,R}\cdot\cut{L^{-1},R^{-1}} = \cut{(-\infty,1),(1,\infty)}.$

As expected, this last fact is the most tedious of all four to verify due to the fact that $x^{-1}$ is not uniformly continuous on $(0,+\infty)$. Multiplicative inverses for good cuts $\cut{L,R}$ with $0 \in R$ can be defined similarly or by taking the negative of the inverse of $\cut{-R,-L}.$

After verifying that the operations behave as required, we obtain the following.

Embedding Theorem. If $K$ is an ordered field then the set of all good cuts of $K$ forms an ordered field $\overline{K}$ in such a way that $$x \in K \mapsto \cut{(-\infty,x),(x,+\infty)} \in \overline{K}$$ is a dense emebedding of $K$ into $\overline{K}.$ Furthermore, $\overline{K}$ is complete in the sense that every good cut in $\overline{K}$ is filled.

The completion $\overline{K}$ has two universal properties:

Universal Properties. Suppose $K$ is an ordered field and $h:K \to \overline{K}$ is the embedding into its field of good cuts.

  1. If $f:K \to L$ is a dense embedding then there is a unique embedding $g:L \to \overline{K}$ such that $h = g \circ f$.

  2. If $f:K \to M$ is a cofinal embedding and $M$ is complete then there is a unique embedding $g:\overline{K} \to M$ such that $f = g \circ h$.

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Note: This answer has been made redundant by the latest version of the other answer. I am leaving it in place only to save the discussions attached to it.


You seem to have fully answered almost all of your questions. All I can do is try to offer a perspective that makes your findings seem more natural.

The Cauchy completion of an ordered field is fundamentally nothing but the completion of a Hausdorff topological group (the additive group with the order topology) that "happens to" admit the extension of some extra structure (multiplication and order). For completions of Hausdorff topological groups we have the simple universal property

Every continuous homomorphism of a group into a complete group has a unique continuous extension to a homomorphism of its completion.

This property carries over nicely to ordered fields. It "just happens", as a consequence of the extra structure, that the image of a continuous homomorphism of ordered fields is dense.


Edit: In the light of some of the comments I should explain that it does not make a significant difference to the above whether "Cauchy completion" is interpreted in terms of sequences or filters. If an ordered field is first countable, then the two are the same; if it isn't, then the sequential completion equals the original field and the universal property applies trivially.


Edit: I wrote earlier that the image of a continuous homomorphism of ordered fields is dense. Looking again at my reasoning I found I could only prove that it is cofinal. In fact there seems to be a continuous endomorphism of $R((x))$ that maps $x$ to $x^2$ and has a nowhere dense image. This suggests that it may be hard to state a universal property in order-theoretic terms.

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    $\begingroup$ That's true but Cohn's "completion" is not the usual topological completion. For example, every field of uncountable cofinality is "complete" in Cohn's sense since every Cauchy sequence is eventually constant. $\endgroup$ – François G. Dorais Aug 29 '13 at 16:16
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    $\begingroup$ On the other hand, Cohn's completion is the usual one for metrizable fields and these are precisely the ones with countable cofinality... $\endgroup$ – François G. Dorais Aug 29 '13 at 16:30
  • $\begingroup$ @FrançoisG.Dorais: That’s very true. What puzzles me is why would one ever want to use the Cauchy completion in the first place. The topological completion seems much more natural. $\endgroup$ – Emil Jeřábek Aug 29 '13 at 16:41
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    $\begingroup$ @Niels: Thanks for your answer. I agree with what you say (and even though you are dealing with a slightly more general completion process in general, I agree with you that it would be more natural to consider the more general case). I also hadn't yet noticed a that a homomorphism of ordered fields is continuous for the order topology if and only if it is dense (I knew only the "if" part). That's a nice result. $\endgroup$ – Pete L. Clark Aug 29 '13 at 17:56
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    $\begingroup$ @PeteL.Clark: I think what Emil is getting at is a more technical version of what I wrote to you last year about the distinction between "pores" and "proper gaps". I would love to see a full account of that. $\endgroup$ – Niels J. Diepeveen Aug 31 '13 at 2:11
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I wrote up the material on sequential completion of an ordered field. It appears in Chapter 16 of these notes. My perspective was that of Francois Dorais's answer: namely, the correction of Cohn's Theorem with the cofinality hypothesis added to the homomorphism $f: K \rightarrow L$. I also included the following results:

Let $f: F \rightarrow F'$ be a homomorphism of ordered fields.

Proposition: Let $\{x_n\}$ be a sequence in $F$. If $\{f(x_n)\}$ is Cauchy, then so is $\{x_n\}$.

$ $

Proposition (Dorais Dichotomy): Exactly one of the following holds:
(i) $f(F)$ is discrete in $F'$.
(ii) $f(F)$ is cofinal in $F'$.

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Theorem: The following are equivalent:
(i) $f$ is uniformly continuous. [Thus $f$ preserves Cauchy sequences.]
(ii) $f$ is continuous.
(iii) $f$ is continuous at $0$.
(iv) $f(F)$ is cofinal in $F'$.

Has anyone seen these results in the extant literature? I would be interested to know.

Ideally I would like to also give the construction of $\mathbb{R}$ from $\mathbb{Q}$ using Dedekind cuts. But from what I know about Dedekind completion, it can be performed on any linearly ordered set and yields a Dedekind complete linearly ordered set. But if you want the Dedekind completion of an ordered commutative group $G$ to retain the structure of a commutative group, it is necessary and sufficient for $G$ to be Archimedean. This makes the prospect of Dedekind completion of an ordered field look more limited. However, the comments to Niels Diepeveen's answer suggest that it should be possible to find the sequential completion as a subset of the Dedekind completion in a natural way. Again, I would be very interested to see that in more detail.

(Added: I just learned that the construction Emil and Francois are describing is from a 1970 paper of R. Baer. I wish I could read German!)

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    $\begingroup$ As Emil explained, the Dedekind-style completion does not use all cuts, only cuts where the left part $L$ and the right part $R$ get arbitrarily close in the sense that for every $\varepsilon > 0$ there are $x \in L$, $y \in R$ with $y-x < \varepsilon$. (This requirement is always true in the archimedean case.) So you don't get a field which is complete in the purely order-theoretic sense but it is complete with respect to the uniform topology. $\endgroup$ – François G. Dorais Sep 1 '13 at 8:50
  • $\begingroup$ @François: OK, thanks: this is starting to sink in. Is it as straightforward to show that you get a complete ordered field from this construction as it is in the Archimedean case (which still has some annoying details to check, IIRC)? $\endgroup$ – Pete L. Clark Sep 1 '13 at 9:56
  • $\begingroup$ The equivalence of (i), (ii), and (iii) is essentially proposition 1 in nlab.mathforge.org/nlab/show/topological+group $\endgroup$ – Niels J. Diepeveen Sep 1 '13 at 11:02
  • $\begingroup$ @Niels: sure, that part is very standard. (i) $\implies$ (ii) $\implies$ (iii) is immediate, and (iii) $\implies$ (i) is essentially: "If $x$ is close to $y$, then $x-y$ is close to $0$." (Although in fact the proof of (iii) $\iff$ (iv) is also immediate! I count that as a point in favor of the Theorem.) $\endgroup$ – Pete L. Clark Sep 1 '13 at 11:24
  • $\begingroup$ Where might I find Baer's paper? (I do read German) $\endgroup$ – Niels J. Diepeveen Sep 1 '13 at 11:43
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François G. Dorais has already described in detail the construction of completion using cuts, but let me add another perspective that might look more algebraically-minded.

The following are equivalent for any ordered field $K$:

  1. $K$ has no proper ordered extension $L$ in which it is dense (in the sense that there is an element of $K$ between any pair of elements of $L$).

  2. Every good cut in $K$ is filled.

For $2\to1$, it suffices to observe that any $a\in L$ determines a good cut $\langle(-\infty,a)\cap K,(a,+\infty)\cap K\rangle$ (condition 4 holds because of density of $K$ in $L$), and the only element of $L$ that can fill it is $a$ itself (again due to density). Thus, $a\in K$.

For $1\to2$, Let $\langle A,B\rangle$ be a good cut that is not filled in $K$. For every polynomial $f\in K[x]$, exactly one of the following must happen (using the fact that a polynomial can change sign only finitely many times in an ordered field):

(+) There are $a\in A$, $b\in B$, and $0<\epsilon\in K$ such that $f(u)>\epsilon$ for every $u\in(a,b)$.

(−) There are $a\in A$, $b\in B$, and $0<\epsilon\in K$ such that $f(u)>-\epsilon$ for every $u\in(a,b)$.

(0) For every $0<\epsilon\in K$, there are $a\in A$ and $b\in B$ such that $-\epsilon<f(u)<\epsilon$ for every $u\in(a,b)$.

The set $I$ of all polynomials satisfying (0) is a prime ideal, and taking (+) as the positive cone induces an ordering on $K[x]/I$, which canonically extends to its fraction field $L$. Then $L$ is an extension of $K$ where $x+I$ fills the cut $\langle A,B\rangle$. One can show that $K$ is dense in $L$ using local uniform continuity of polynomials.

So, let us call an ordered field complete when it satisfies either condition. By Zorn’s lemma, every $K$ has a maximal ordered extension $\hat K$ in which it is dense. Then $\hat K$ is complete. As above, every element of $\hat K$ is uniquely determined by a good cut on $K$ (by density), and conversely every such cut is filled in $\hat K$ (by completeness), hence $\hat K$ can be identified with the set of all good cuts on $K$. One can easily check that it satisfies the appropriate universal conditions (e.g., if $K$ is dense in $L$, the mapping sending each element of $L$ to the cut on $K$ it determines is the unique ordered $K$-embedding of $L$ in $\hat K$), which also implies that $\hat K$ is unique up to isomorphism.

EDIT: As for references, the construction appears to be due to Dana Scott [On completing ordered fields, in: Applications of Model Theory to Algebra, Analysis, and Probability, New York, 1969, pp. 274–278], but unfortunately I don’t have access to this book.

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  • $\begingroup$ Thanks for this. I've continued to find early precedents for this type of "proper cut" construction: it may go back at least to Veronese in the 19th century! $\endgroup$ – Pete L. Clark Sep 3 '13 at 5:57
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The universal property of the completion of an ordered field as stated in Cohn's book (I mean, P.M. Cohn, Basic Algebra, Theorem 8.7.1) is true if we replace the assumption "each dense order embedding $f:K \rightarrow L$" instead of "each order embedding $f:K \rightarrow L$".

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