# The Unapologetic Mathematician

## Archimedean Fields

Whether we use Dedekind cuts or Cauchy sequences to construct the ordered field of real numbers $\mathbb{R}$ (and it doesn’t matter which), we are taking the ordered field of rational numbers and enlarging it to be “complete” in some sense or another. But we also aren’t making it too much bigger. The universality property we got from completing the uniform structure already gives evidence of that, but there’s another property which we can show is true of $\mathbb{R}$, and which shows that the real numbers aren’t too unwieldy.

In The Sand Reckoner, the ancient Greek mathematician Archimedes once set about the problem of should the number of grains of sand in existence to be finite. He does this by determining a (very weak) upper bound: the number of grains of sand it would take to fill up the entire universe, as he understood the latter term. He writes:

There are some … who think that the number of the sand is infinite in multitude; and I mean by the sand not only that which exists about Syracuse and the rest of Sicily but also that which is found in every region whether inhabited or uninhabited. Again there are some who, without regarding it as infinite, yet think that no number has been named which is great enough to exceed its multitude. And it is clear that they who hold this view, if they imagined a mass made up of sand in other respects as large as the mass of the earth filled up to a height equal to that of the highest of the mountains, would be many times further still from recognizing that any number could be expressed which exceeded the multitude of the sand so taken. But I will try to show you by means of geometrical proofs, which you will be able to follow, that, of the numbers named by me … some exceed not only the number of the mass of sand equal in magnitude to the earth filled up in the way described, but also that of a mass equal in magnitude to the universe

The deep fact here is a fundamental realization about numbers: the set of natural numbers has no upper bound in the real number system. That is, no matter how huge a real number we pick there’s always a natural number bigger than it. Equivalently, given any positive real number $x$ — even as small as the volume of a grain of sand — and another positive real number $y$ — even as large as the volume of (the ancient Greek conception of) the universe — there’s some natural number $n$ so that $nx\geq y$. When this happens in a given ordered field we say that the field is “Archimedean”.

So let’s show that $\mathbb{R}$ is Archimedean. If there were positive real numbers $x$ and $y$ so that $nx\leq y$ for all natural numbers $n$, then $y$ would be an upper bound for the set of $nx$. Then Dedekind completeness gives us a least upper bound $\sup\{nx\}$, and we can just take $y$ to be this least upper bound. Now $nx\leq y$, and also $(n+1)x\leq y$, and so $nx\leq y-x$. That is, $y-x$ is another upper bound for the set of multiples of $x$. But since $x$ was chosen to be positive we see that $y-x, contradicting the assumption that $y$ was the least such upper bound. So such a pair of real numbers can’t exist.

In particular, we can take a positive real number $x$ and consider the set of natural numbers $n$ which are larger than it. Since the natural numbers are well-ordered, there is a least such number, and it can’t be ${0}$ because we assume $x>0$. Subtracting one from this number will then give the largest natural number that is still below $x$ in the real number order, and we denote this number by $\lfloor x\rfloor$. We can thus write any positive number uniquely as the sum $\lfloor x\rfloor+r$ of a natural number and some remainder with $0\leq r<1$.

It turns out that the real numbers are actually the largest Archimedean field. That is, if $\mathbb{F}$ is any ordered field satisfying the Archimedean property, there will be an monomorphism of ordered fields $\mathbb{F}\rightarrow\mathbb{R}$, making (the image of) $\mathbb{F}$ a subfield of $\mathbb{R}$. I won’t prove this here, but I will note one thing about the meaning of this result: the Archimedean property essentially limits the size of an ordered field. That is, an ordered field can’t get too big without breaking this property. Dually, an ordered field can’t get too small without breaking Dedekind completeness or uniform completeness. Completeness pulls the field one way, while the Archimedean property pulls the other way, and the two reach a sort of equilibrium in the real numbers, living both at the top of one world and the bottom of the other.

December 7, 2007 - Posted by | Fundamentals, Numbers

1. “Completeness pulls the field one way, while the Archimedean property pulls the other way, and the two reach a sort of equilibrium in the real numbers, living both at the top of one world and the bottom of the other.”

A very pretty kinaesthetic metaphor. Which, by the way, was how Einstein and Feynman did a lot of their best work.

Comment by Jonathan Vos Post | December 7, 2007 | Reply

2. Just wondering what you mean when you say don’t make the rationals too much bigger to get the reals. We’re adding uncountably many numbers to get the reals…so in what sense are the reals not much bigger than the rationals.

Also, I wonder if there’s a universal property we can use to characterize the reals from the rationals, in a categorical sense. Looks like we can, similiar to the Grothendieck completion of a monoid or semigroup, but it’s too early to draw the commutative diagrams to check.

Comment by Josh | December 10, 2007 | Reply

3. Josh, I go more into that at the bottom of the post. The real numbers aren’t so big they can’t be Archimedean, but they aren’t so small they can’t be complete.

As for the universal property, that’s exactly how I did characterize them, using the completion functor on the category of uniform spaces. This is the left adjoint to the functor which includes the full subcategory of complete uniform spaces into the category of all uniform spaces. More precisely, any uniformly continuous function $f:\mathbb{Q}\rightarrow X$ from the rational numbers to a complete uniform space $X$ extends uniquely to a uniformly continuous function $\bar{f}:\mathbb{R}\rightarrow X$.

Comment by John Armstrong | December 10, 2007 | Reply

4. Just to add a bit to John’s response to Josh: there are ordered field extensions of the rationals much, much, much bigger than $\mathbb{R}$ that people sometimes consider, for example, fields of “hyperreals” which crop up in non-standard analysis (these are discussed at length in the math blog Mort Aux Triangles!). A rather tame example of this sort of thing, which is a bit smaller than these hyperreal extensions but moderately big nonetheless, is the field of rational functions with real coefficients, ordered by their growth “at infinity”; the ordinary reals sit inside as the subfield of constant functions.

But none of these big fields is complete; you can’t even complete them in the way John is discussing! Which is a pretty interesting fact. So in this way, completeness exerts control over the size: not too big, not too small. The standard reals are uniquely “just right”.

Comment by Todd Trimble | December 10, 2007 | Reply

5. [...] Groups and the Largest Archimedean Field Okay, I’d promised to get back to the fact that the real numbers form the “largest” Archimedean field. [...]

Pingback by Archimedean Groups and the Largest Archimedean Field « The Unapologetic Mathematician | December 17, 2007 | Reply

6. [...] Archimedean Fields « The Unapologetic Mathematician [...]

Pingback by Archimedean field - tutorial aa4969 | July 14, 2009 | Reply

7. [...] point of . Indeed, consider a point in the square and some radius . Since the real numbers are Archimedean, we can pick some , and as many people on Nick’s Twitter experiment (remember to follow [...]

Pingback by Iterated Integrals III « The Unapologetic Mathematician | December 18, 2009 | Reply