We’ve defined the topological space we call the real number line as the completion of the rational numbers 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 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 is dense in (and similarly is dense in ), 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 with a dense subset . Then every point has a sequence with . Now if and are two continuous functions which agree for every point in , then they agree for all points in . Indeed, picking a sequence in converging to we have
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 and , and sequences of rational numbers and converging to and , respectively. We should have
but how do we know that the limit on the right exists? Well if we can show that the sequence is a Cauchy sequence of rational numbers, then it must converge because is complete.
Given a rational number we must show that there exists a natural number so that for all . But we know that there’s a number so that for , and a number so that for . Then we can choose to be the larger of and and find
So the sequence of sums is Cauchy, and thus converges.
What if we chose different sequences and converging to and ? Then we get another Cauchy sequence of rational numbers. To show that addition of real numbers is well-defined, we need to show that it’s equivalent to the sequence . So given a rational number does there exist an so that for all ? 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 to defined by and . Since 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 from 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 by negating each term of a Cauchy sequence converging to , 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 for the sum of real numbers and .
What about the multiplication? Again, we’ll want to choose rational sequences and converging to and , and define our function by
so it will be continuous and agree with rational number multiplication. Now we must show that for every rational number there is an so that for all . This will be a bit clearer if we start by noting that for each rational there is an so that for all . In particular, for sufficiently large we have , so the sequence is bounded above by some . Similarly, given we can pick so that for and get an upper bound for all . Then choosing to be the larger of and we will have
for . Now given a rational we can (with a little work) find and so that the expression on the right will be less than , 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 and , so we get a multiplication. Again, we can use the density of the rational numbers to show that it’s associative and commutative, that serves as its unit, and that multiplication distributes over addition. We’ll just write for the product of real numbers and from here on.
To show that is a field we need a multiplicative inverse for each nonzero real number. That is, for each Cauchy sequence of rational numbers that doesn’t converge to , we would like to consider the sequence , but some of the might equal zero and thus throw us off. However, there can only be a finite number of zeroes in the sequence or else would be an accumulation point of the sequence and it would either converge to 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.