Last time we talked about ordered fields, we discovered that among our favorite examples:

- The real and rational numbers are ordered.
- Finite fields are not ordered.
- Neither are the complex numbers

There’s one more property of fields left to discuss, what’s called *completeness*. Intuitively, we say that a field is complete when there are no holes in it. For an example of a hole, let’s consider the rational numbers whose square isn’t two (that is, all of them.) When we square a positive rational number we can get arbitrarily close to 2 from below, by squaring 1, 1.4, 1.41, 1.414, 1.4142, etc, or from above, by squaring 2, 1.5, 1.42, 1.415, 1.4143, etc, but we can never get exactly two. We know where a square root of 2 would go (a little past 1.4142135623730951), but where it should be there’s a hole. For ordered fields there’s a simple, standard definition of completeness:

C1: Any non-empty subset of the field that is bounded above has a smallest upper bound.

In other words, create a subset S, say all the positive numbers whose square is less than 2. It’s bounded above, because all of them are less than 2. But in the rationals we can show that there is no smallest upper bound. Take any number U which is greater than all the members of S. U*U is greater than 2, and there is going to be some positive integer N large enough that U*U – 2*U/N is also greater than 2. Then (U-1/N)*(U-1/N) = U*U – 2*U/N + 1/N*N is also greater than 2, so U-1/N is an upper bound for S than is smaller than U. Thus, for any upper bound to S we can find a smaller one.

In the real numbers, there is no problem finding the smallest upper bound: it’s the square root of 2 (call it SQ, for short). It’s easy to see that all members of S are smaller than SQ, because their squares are smaller than SQ*SQ, and anything smaller than SQ would have a square smaller than 2 and thus not be an upper bound for S.

This is the real difference between the rationals and the reals, that C1 is false for the former and true for the latter. Thus, the reals are called a *complete ordered field*, and, in fact, are in a sense the only complete ordered field. If we found another one, we’d find that it’s just the real numbers under a different name. It would have a o, and a 1, and a fifth root of 12, and a pi, etc, etc, and even if they had different names they would act exactly the same way.

Now, C1 depends on the field being ordered, since without that there’s no such thing as an upper bound. That makes C1 useless for asking whether non-ordered fields are complete. There’s another way to define completeness, but we need to develop some machinery for it. The first concept we need is that of a sequence, which is simple enough. It’s an infinite list of members of a field. Here are some examples:

- 0,0,0,0,0,0,…
- 1,2,3,4,5,6,…
- 1,-1,1,-1,1,-1,1,-1,…
- 1,.1,.01,.001,.0001,…
- 1,1.4,1.41,1.414,1.4142,…

In each case, it should be clear where the sequence goes after the terms that are shown. One of the obvious question to ask about a sequence is whether it has a *limit*, that is, gets arbitrarily close to some number as it goes on. We make this precise by saying that

A sequence S has a limit a if and only if for any positive number e, no matter how small, if you go far enough out, all the members of the sequence are no further away from a than e.

Sequences that have a limit are said to *converge*. Obviously, 2 and 3 don’t converge (infinity is not a number, at least for these purposes.) 1 and 4 converge to 0. And 5? If it’s a sequence of reals, it has the limit SQ. If it’s a sequence of rationals, it has no limit, because there is no rational number equal to SQ. So it does or doesn’t converge, depending.

But it’s unsatisfying to say that sometimes 5 is like 2 and 3, because obviously it has a lot in common with 1 and 4. For one thing, its members get closer and closer together, even if they don’t eventually go somewhere specific. So let’s find a way to make that precise:

A sequence is called a Cauchy (pronounced *Coh-she*) sequence if and only if for any positive number e, no matter how small, if you go far enough out, all the members of the sequence are no further away from *each other* than e. That’s very similar to our definition of convergence. The only real difference is that it doesn’t mention a limit.

Obviously, any sequence that converges is Cauchy. Choose an e: if you go far enough out, the members are no further away from the limit than e/2, so they’re no further away from each other than e. 2 and 3 aren’t Cauchy, since as far as you go out, there are member of 3 that are 2 units apart and members of 2 that are as far apart as you like. 5 is Cauchy even if it’s a sequence of rationals.

When you think about it, “numbers can get closer and closer together but not go to some limit” is a pretty good definition of a hole. In fact, the alternate definition of completeness we’ve been working towards is:

C1′: All Cauchy sequences converge.

And this applies to all fields, because it doesn’t mention ordering. So we can ask about our other favorite fields.

- The complex numbers are complete. An easy way to see this is that a Cauchy sequence has to be Cauchy in both its real and imaginary parts, so each of them will have a limit, and their sum will be the limit of the sequence as a whole.
- Finite fields are complete, for a pretty silly reason. The only way a sequence of numbers from a finite field can be Cauchy is if it’s eventually constant, repeating the same number over and over like
- 3,4,1,0,3,3,3,3,3,3,…

And, in that case, it has a limit: the constant value. Thus finite fields are, trivially, complete.

This completes (so to speak) our survey of fields. Next time, instead of just talking about them, we’ll see how to make one.

UPDATE (since the original version of the Exercise was incorrect)

Exercise: Prove that, in an ordered field, C1 implies C1′.

I thought I mostly understood the Order! post and flattered myself by coming up with a proof for ” if a > b and 0 > c, then b * c > a * c” in my head that worked enough for me without writing it down but when we start getting into

eterritory, my brain locks up.JaybirdQuote Link

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Limits are hard. I didn’t want to double the size of the post by explaining them in detail, but they might be a good topic for next week.

Mike SchillingQuote Link

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I was tutoring my niece in calculus. She’s going into engineering, so I explained limits as something that mathematicians invented so they can justify dividing by zero or adding up an infinite number of terms, but that engineers never have to use. Once we pass Calculus I and II, that is.

KenQuote Link

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The field of quaternions is also complete, then. Quaternions with a norm of unity are used to represent three-dimensional rotations. Another method of representing rotations is as a series of three Euler angles. Unlike with quaternions, for whichever type of Euler angle representation you choose (e.g. 3-1-3), there are singularities in the field of Euler angle rotation sets; some rotations can be represented by an infinite number of sets of Euler angles. Does this mean an Euler angle representation is not complete? Or is it not even a field? I think maybe it is a field but is not complete. What do you think?

BoegiboeQuote Link

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I hate both you and Mike, in the same way that I generally hate people who are smarter than I.

ChrisQuote Link

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Quaternions are certainly complete, by the same argument as complex numbers. (Since C1′ uses only subtraction and absolute value, for those purposes complex numbers are effectively R2 and quaternions effectively R4.)

I’m no expert on Euler angles, but they look to me like a group with a single operation (composition), not a field or a metric space, so I don’t see how completeness would apply.

Mike SchillingQuote Link

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What I find curious here is that any set of Euler angles can be converted into a quaternion by a generic operation–generic for a given order of rotations (e.g. 3-1-3). So, whatever operations can be one with quaternions, one could say the Euler-angle equivalent can be done by converting to a quaternion, doing the operation, and then converting back. But here’s where the problem comes in, because converting back is not unambiguous. So, my question is really about this ambiguity.

BoegiboeQuote Link

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Like Mike said, rotations are not a field, but a group, because there is only one single associative operation.

That said, there is a lot of structure to this group. Locally, it looks very much like a R3, since you can describe it using three parameters (e.g. Euler angles), which makes it an example of what is more generally known as a Lie group.

This group (generally called SO(3)) looks like R3 wherever you look at it, so it is complete, just like R3 itself.

lukasQuote Link

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So are you saying the set of quaternions is a field, but the set of unit-norm quaternions is not? Yes, I see that now. Thanks! A few folks here know this about me, but for the benefit of those who don’t, I am a spacecraft dynamicist specializing in attitude determination and control. SO3 is my life :-)

BoegiboeQuote Link

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Groups are way cooler than fields.

PatrickQuote Link

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SO3 is my lifeCareful you don’t get it wet.

Mike SchillingQuote Link

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If you add two unit quaternions, you generally get a non-unit quaternion, so they aren’t a field.

lukasQuote Link

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C1′ does not imply C1. There are ordered fields in which the only convergent sequences are eventually constant. Such fields are not isomorphic to the real numbers, and so cannot be complete in the sense of C1.

Ken SQuote Link

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Oops. What I meant to say is that there are ordered fields in which all Cauchy sequences are eventually constant, and so they all converge. These fields all satisfy C1′, but not C1.

Ken SQuote Link

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I can’t picture what those fields would look like (if all Cauchy sequences are eventually constant, it’s pretty discrete, which makes it seem like the greatest member of a set would also be its least upper bound) but I’ll take your word for it. I do think that I have a correct proof that C1 implies C1′.

Mike SchillingQuote Link

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The construction goes like this: Given any ordered field F, an extension of F in which F is bounded is the field F(x) of rational functions with coefficients in F, where a(x) > b(x) if a(r) > b(r) for all sufficiently large r in F. Start from any ordered field F_0 you like, and and iterate this extension process aleph_1 times, to obtain the ordered field G. Since every sequence in aleph_1 is bounded, every increasing sequence in G is bounded. By taking reciprocals, every decreasing sequence of positive members of G has a positive lower bound, and so every Cauchy sequence must be eventually constant.

Ken SQuote Link

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