Universal property characterizing $\Bbb R$ Is it possible to characterize the field of real numbers in a natural way with the language of category theory?
For example, $\Bbb Q$ is the initial object in the category of ordered fields and $\Bbb Z$ is the initial object in Ring. I'm hoping for something similar with $\Bbb R$.
EDIT: Of course any object in the (quite boring) category of complete ordered fields will do, but completeness is a very strong property (in particular it's second-order logic), so I was hoping that one can express this aspect of the real numbers by only using arrows, and that the category with which you deal is given by simple properties, favourably those which can be stated in first-order language.
 A: This question leads directly to the notion of a real closed field. These are ordered fields which "believe that they are $\mathbb{R}$". More precisely, they satisfy the same first order sentences as $\mathbb{R}$. It turns out that already two sentence schemes suffice: Every positive element is a square, and every polynomial of odd degree has a root. There are several other interesting characterizations, which can be found in the Wikipedia article. See also the references given there.
Every ordered field has a real closed algebraic extension, the so called real closure. This gives lots of examples of real closed fields, not isomorphic to $\mathbb{R}$. For example, the real closure of $\mathbb{Q}$ is $\mathbb{R} \cap \overline{\mathbb{Q}}$, i.e. the field of real algebraic numbers.
This strongly suggests that there isn't any first-order universal property of $\mathbb{R}$ within the category of ordered fields, since solutions for universal problems are always isomorphic.
At least we can observe the following: If $K$ is an ordered field, then there is at most one homomorphism of fields $\mathbb{R} \to K$. For a proof, let $f : \mathbb{R} \to K$ be such a homomorphism. Then it has to be monotonic: If $x \leq y$, then $y-x$ is a square, hence also $f(y-x)=f(y)-f(x)$ is a square, which implies $f(x) \leq f(y)$. When $K$ is equipped with the order topology, then $f$ becomes continuous. Since $f|_{\mathbb{Q}}$ is unique and $\mathbb{Q}$ is dense in $\mathbb{R}$, it follows that $f$ is unique.
Observe that $f$ exists if and only if all rational Cauchy sequences have a limit in $K$.
