# The Unapologetic Mathematician

## The Topological Field of Real Numbers

We’ve defined the topological space we call the real number line $\mathbb{R}$ as the completion of the rational numbers $\mathbb{Q}$ as a uniform space. But we want to be able to do things like arithmetic on it. That is, we want to put the structure of a field on this set. And because we’ve also got the structure of a topological space, we want the field operations to be continuous maps. Then we’ll have a topological field, or a “field object” (analogous to a group object) in the category $\mathbf{Top}$ of topological spaces.

Not only do we want the field operations to be continuous, we want them to agree with those on the rational numbers. And since $\mathbb{Q}$ is dense in $\mathbb{R}$ (and similarly $\mathbb{Q}\times\mathbb{Q}$ is dense in $\mathbb{R}\times\mathbb{R}$), we will get unique continuous maps to extend our field operations. In fact the uniqueness is the easy part, due to the following general property of dense subsets.

Consider a topological space $X$ with a dense subset $D\subseteq X$. Then every point $x\in X$ has a sequence $x_n\in D$ with $\lim x_n=x$. Now if $f:X\rightarrow Y$ and $g:X\rightarrow Y$ are two continuous functions which agree for every point in $D$, then they agree for all points in $X$. Indeed, picking a sequence in $D$ converging to $x$ we have
$f(x)=f(\lim x_n)=\lim f(x_n)=\lim g(x_n)=g(\lim x_n)=g(x)$.

So if we can show the existence of a continuous extension of, say, addition of rational numbers to all real numbers, then the extension is unique. In fact, the continuity will be enough to tell us what the extension should look like. Let’s take real numbers $x$ and $y$, and sequences of rational numbers $x_n$ and $y_n$ converging to $x$ and $y$, respectively. We should have
$s(x,y)=s(\lim x_n,\lim y_n)=s(\lim(x_n,y_n))=\lim x_n+y_n$
but how do we know that the limit on the right exists? Well if we can show that the sequence $x_n+y_n$ is a Cauchy sequence of rational numbers, then it must converge because $\mathbb{R}$ is complete.

Given a rational number $r$ we must show that there exists a natural number $N$ so that $\left|(x_m+y_m)-(x_n+y_n)\right| for all $m,n\geq N$. But we know that there’s a number $N_x$ so that $\left|x_m-x_n\right|<\frac{r}{2}$ for $m,n\geq N_x$, and a number $N_y$ so that $\left|y_m-y_n\right|<\frac{r}{2}$ for $m,n\geq N_y$. Then we can choose $N$ to be the larger of $N_x$ and $N_y$ and find
$\left|(x_m-x_n)+(y_m-y_n)\right|\leq\left|x_m-x_n\right|+\left|y_m-y_n\right|<\frac{r}{2}+\frac{r}{2}=r$
So the sequence of sums is Cauchy, and thus converges.

What if we chose different sequences $x'_n$ and $y'_n$ converging to $x$ and $y$? Then we get another Cauchy sequence $x'_n+y'_n$ of rational numbers. To show that addition of real numbers is well-defined, we need to show that it’s equivalent to the sequence $x_n+y_n$. So given a rational number $r$ does there exist an $N$ so that $\left|(x_n+y_n)-(x'_n+y'_n)\right| for all $n\geq N$? This is almost exactly the same as the above argument that each sequence is Cauchy! As such, I’ll leave it to you.

So we’ve got a continuous function taking two real numbers and giving back another one, and which agrees with addition of rational numbers. Does it define an Abelian group? The uniqueness property for functions defined on dense subspaces will come to our rescue! We can write down two functions from $\mathbb{R}\times\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$ defined by $s(s(x,y),z)$ and $s(x,s(y,z))$. Since $s$ agrees with addition on rational numbers, and since triples of rational numbers are dense in the set of triples of real numbers, these two functions agree on a dense subset of their domains, and so must be equal. If we take the ${0}$ from $\mathbb{Q}$ as the additive identity we can also verify that it acts as an identity real number addition. We can also find the negative of a real number $x$ by negating each term of a Cauchy sequence converging to $x$, and verify that this behaves as an additive inverse, and we can show this addition to be commutative, all using the same techniques as above. From here we’ll just write $x+y$ for the sum of real numbers $x$ and $y$.

What about the multiplication? Again, we’ll want to choose rational sequences $x_n$ and $y_n$ converging to $x$ and $y$, and define our function by
$m(x,y)=m(\lim x_n,\lim y_n)=m(\lim(x_n,y_n))=\lim x_ny_n$
so it will be continuous and agree with rational number multiplication. Now we must show that for every rational number $r$ there is an $N$ so that $\left|x_my_m-x_ny_n\right| for all $m,n\geq N$. This will be a bit clearer if we start by noting that for each rational $r_x$ there is an $N_x$ so that $\left|x_m-x_n\right| for all $m,n\geq N_x$. In particular, for sufficiently large $n$ we have $\left|x_n\right|<\left|x_N\right|+r_x$, so the sequence $x_n$ is bounded above by some $b_x$. Similarly, given $r_y$ we can pick $N_y$ so that $\left|y_m-y_n\right| for $m,n\geq N_y$ and get an upper bound $b_y\geq y_n$ for all $n$. Then choosing $N$ to be the larger of $N_x$ and $N_y$ we will have
$\left|x_my_m-x_ny_n\right|=\left|(x_m-x_n)y_m+x_n(y_m-y_n)\right|\leq r_xb_y+b_xr_y$
for $m,n\geq N$. Now given a rational $r$ we can (with a little work) find $r_x$ and $r_y$ so that the expression on the right will be less than $r$, and so the sequence is Cauchy, as desired.

Then, as for addition, it turns out that a similar proof will show that this definition doesn’t depend on the choice of sequences converging to $x$ and $y$, so we get a multiplication. Again, we can use the density of the rational numbers to show that it’s associative and commutative, that $1\in\mathbb{Q}$ serves as its unit, and that multiplication distributes over addition. We’ll just write $xy$ for the product of real numbers $x$ and $y$ from here on.

To show that $\mathbb{R}$ is a field we need a multiplicative inverse for each nonzero real number. That is, for each Cauchy sequence of rational numbers $x_n$ that doesn’t converge to ${0}$, we would like to consider the sequence $\frac{1}{x_n}$, but some of the $x_n$ might equal zero and thus throw us off. However, there can only be a finite number of zeroes in the sequence or else ${0}$ would be an accumulation point of the sequence and it would either converge to ${0}$ or fail to be Cauchy. So we can just change each of those to some nonzero rational number without breaking the Cauchy property or changing the real number it converges to. Then another argument similar to that for multiplication shows that this defines a function from the nonzero reals to themselves which acts as a multiplicative inverse.

December 3, 2007