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:
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
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′.