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 | 7 Comments

Cuts and Sequences are Equivalent

Sorry to not get this posted until so late, but the end of the semester has been a bit hectic.

We’ve used Dedekind cuts to “complete” the order on the rational numbers — to make sure that every nonempty set of numbers with an upper bound has a least upper bound. We’ve also used Cauchy sequences to “complete” the uniform structure on the rational numbers — to make sure that every Cauchy sequence converges. But do we actually get the same thing in each case?

If we take a real number $x$ represented by a Cauchy sequence $x_n$ it’s easy to come up with a cut. Given a rational number $q$ we use the constant sequence $q_n=q$ and compare it to $x_n$. If $x_n-q_n$ is eventually nonnegative then $q$ is less than $x$, and should go into the left set $X_L$. On the other hand, if it’s eventually nonpositive then $q$ is greater than $x$ and should go into the right set $X_R$. It’s straightforward to show that this function from $\mathbb{R}$ to the set of cuts preserves the order.

Now let’s start with the cut $(X_L,X_R)$ and write down a Cauchy sequence. Pick some $x_L\in X_L$ and $x_R\in X_R$, and construct the sequence as follows. First write down $x_0=x_L$ and $x_1=x_R$. Now set $x_2=\frac{x_L+x_R}{2}$. This value will either be in $X_L$ or it won’t. If it is, replace $x_L$ by $x_2$, and otherwise replace $x_R$ by $x_2$. Then define $x_3$ as the midpoint between our two left and right points, and again replace either the left or the right point. Keep going, and we see that all future numbers in the sequence are closer to each other than the current $x_L$ and $x_R$ are to each other. And these two always keep moving closer and closer to each other, halving their distance at each step. So the sequence has to be Cauchy. If we picked a different $x_L$ and $x_R$ to start with, we’d get an equivalent sequence. I’ll leave this to you to show.

Notice here that the points in the sequence that lie in $X_L$ are moving steadily upwards towards the cut, and those in $X_R$ are moving steadily downwards towards it. Eventually, the sequence will rise above any point in $X_L$ and fall below any point in $X_R$, and so if we take this sequence and build a cut from it we will get back the exact same cut we started with. Also, if we build a cut from a Cauchy sequence, and then a sequence from it, we get back an equivalent sequence. Thus we have set up a bijection between the set of cuts and the set of equivalence classes of Cauchy sequences, and we’ve already seen that it preserves the order structure.

Now let’s look at the map from sequences to cuts and verify that it preserves addition and multiplication of positive numbers. This will make the map into an isomorphism of ordered fields, and so both constructions are describing essentially the same thing. So if we have Cauchy sequences $x_n$ and $y_n$, which give right sets $X_R$ and $Y_R$ of rational numbers, then what’s the right set of the sequence $x_n+y_n$? It’s the set of rational numbers $q$ so that $x_n+y_n-q$ is eventually nonnegative. But any such $q$ can be broken up as $q=q_x+q_y$, where $x_n-q_x$ and $y_n-q_y$ are both eventually nonnegative. That is, $(X+Y)_R$ is the set of sums of elements of $X_R$ and $Y_R$, and so addition is preserved. The proof for multiplication is essentially the same.

So both methods of extending the real numbers give us essentially the same ordered field, which is thus both complete as a uniform space and Dedekind complete.

December 7, 2007 Posted by | Fundamentals, Numbers | 1 Comment