# The Unapologetic Mathematician

## The Order on the Real Numbers

We’ve defined the real numbers $\mathbb{R}$ as a topological field by completing the rational numbers $\mathbb{Q}$ as a uniform space, and then extending the field operations to the new points by continuity. Now we extend the order on the rational numbers to make $\mathbb{R}$ into an ordered field.

First off, we can simplify our work greatly by recognizing that we just need to determine the subset $\mathbb{R}^+$ of positive real numbers — those $x\in\mathbb{R}$ with $x\geq0$. Then we can say $x\geq y$ if $x-y\geq0$. Now, each real number is represented by a Cauchy sequence of rational numbers, and so we say $x\geq0$ if $x$ has a representative sequence $x_n$ with each point $x_n\geq 0$.

What we need to check is that the positive numbers are closed under both addition and multiplication. But clearly if we pick $x_n$ and $y_n$ to be nonnegative Cauchy sequences representing $x$ and $y$, respectively, then $x+y$ is represented by $x_n+y_n$ and $xy$ is represented by $x_ny_n$, and these will be nonnegative since $\mathbb{Q}$ is an ordered field.

Now for each $x$, $x-x=0\geq0$, so $x\geq x$. Also, if $x\geq y$ and $y\geq z$, then $x-y\geq0$ and $y-z\geq0$, so $x-z=(x-y)+(y-z)\geq0$, and so $x\geq z$. These show that $\geq$ defines a preorder on $\mathbb{R}$, since it is reflexive and transitive. Further, if $x\geq y$ and $y\geq x$ then $x-y\geq0$ and $y-x\geq0$, so $x-y=0$ and thus $x=y$. This shows that $\geq$ is a partial order. Clearly this order is total because any real number either has a nonnegative representative or it doesn’t.

One thing is a little hazy here. We asserted that if a number and its negative are both greater than or equal to zero, then it must be zero itself. Why is this? Well if $x_n$ is a nonnegative Cauchy sequence representing $x$ then $-x_n$ represents $-x$. Now can we find a nonnegative Cauchy sequence $y_n$ equivalent to $-x_n$? The lowest rational number that $y_n$ can be is, of course, zero, and so $\left|y_n-(-x_n)\right|\geq x_n$. But for $-x_n$ and $y_n$ to be equivalent we must have for each positive rational $r$ an $N$ so that $r\geq\left|y_n-(-x_n)\right|\geq x_n$ for $n\geq N$. But this just says that $x_n$ converges to ${0}$!

So $\mathbb{R}$ is an ordered field, so what does this tell us? First off, we get an absolute value $\left|x\right|$ just like we did for the rationals. Secondly, we’ll get a uniform structure as we do for any ordered group. This uniform topology has a subbase consisting of all the half-infinite intervals $(x,\infty)$ and $(-\infty,x)$ for all real $x$. But this is also a subbase for the metric we got from completing the rationals, and so the two topologies coincide!

One more very important thing holds for all ordered fields. As a field $\mathbb{F}$ is a kind of a ring with unit, and like any ring with unit there is a unique ring homomorphism $\mathbb{Z}\rightarrow\mathbb{F}$. Now since $1gt;0$ in any ordered field, we have $2=1+1>0$, and $3=2+1>0$, and so on, to show that no nonzero integer can become zero under this map. Since we have an injective homomorphism of rings, the universal property of the field of fractions gives us a unique field homomorphism $\mathbb{Q}\rightarrow\mathbb{F}$ extending the ring homomorphism from the integers.

Now if $\mathbb{F}$ is complete in the uniform structure defined by its order, this homomorphism will be uniformly complete. Therefore by the universal property of uniform completions, we will find a unique extension $\mathbb{R}\rightarrow\mathbb{F}$. That is, given any (uniformly) complete ordered field there is a unique uniformly continuous homomorphism of fields from the real numbers to the field in question. Thus $\mathbb{R}$ is the universal such field, which characterizes it uniquely up to isomorphism!

So we can unambiguously speak of “the” real numbers, even if we use a different method of constructing them, or even no method at all. We can work out the rest of the theory of real numbers from these properties (though for the first few we might fall back on our construction) just as we could work out the theory of natural numbers from the Peano axioms.

December 4, 2007 -