TITLE: Characterization properties of number sets $\mathbb{N},\mathbb{ Z},\mathbb{Q},\mathbb{R},\mathbb{C}$
QUESTION [11 upvotes]: When people say that a structure is defined up to isomorphism means, accordingly, that they assume certain properties that make it completely determined under certain operations and relations. 

So, I'd like to know what the properties are that characterize the different systems of numbers ($\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$) up to isomorphism with the operations of $+,\cdot$, and $\leq$. 

To make my question clearer, for example I  guess the principle of induction would be part of characterizing $\mathbb{N}$, the least upper bound property would be part of characterizing $\mathbb{R}$, etc. 
The thing is that there are a lot of properties like these and it's not clear, at least for me, to decide what are the main ones and what of them can be deduced and are redundant, etc. 

In other words I'd like to ask which properties characterize each of the sets $\mathbb{N},\mathbb{ Z},\mathbb{Q},\mathbb{R},\mathbb{C}$ based on their operations and orderings.

REPLY [7 votes]: Most of these things are distinguished as being initial objects in some suitable category. I don't have much practice with all of the statements, so I apologize for missteps. The notion of being "initial" is mostly what "being the smallest" means, as in Asaf's answer.

When they are distinguished as initial objects of categories of ring-like objects with orders

I think $\Bbb N$ must be the initial object of the category of totally ordered semirings (with identity). (Morphisms would have to preserve both $0$ and $1$.)
$\Bbb Z$ is initial in the category of totally ordered rings. (We have given up well-ordering from $\Bbb N$.)
$\Bbb Q$ is initial in the category of totally ordered fields. (We have given up "discreteness" from $\Bbb Z$ in the following sense: given any point of the ordered set, there isn't a least point strictly above it or a greatest point strictly beneath it.)
$\Bbb R$ is initial in the category of totally ordered fields with the least upper bound property, but it turns out to be a let-down since it is the only such ring. (We have given up "holes" that existed in $\Bbb Q$. Every subset now has a least upper bound. )
When we get to $\Bbb C$ we have a problem because $\Bbb C$ isn't orderable. I'm not aware of any categorically reasonable way to continue describing $\Bbb C$ with the program of categories with ordered objects. $\Bbb R$ seems to have reached an apex of total-orderedness and continuity, and it looks like $\Bbb C$ has gone out of bounds.


What if we just pay attention to order alone?

$\Bbb N$ is initial in the category of sets with a distinguished point and a successor function. (This is an exercise in MacLane's CFTWM.)
I'm not certain, but I think $\Bbb Z$ is initial in the category of sets with a distinguished point and a successor function and a predecessor function. (I'll see if a variant of MacLane's exercise works out for this.) After this point we abandon successors and predecessors and switch to plain order.
We might hope that $\Bbb Q$ is initial in the category of totally ordered sets with a distinguished point; however, I'm not entirely convinced this is true. It feels like even if we had an order preserving and distinguished-point preserving map from $\Bbb Q$ to another set, maybe you can "scale" the map to get a different one. (When we are working with rings, the extra algebraic structure eliminates this problem.)
$\Bbb R$ is in a similar state as $\Bbb Q$, since it's not clear that it's initial in the category of complete totally ordered sets.
$\Bbb C$ even lacks a natural order, and we would still have problems similar to $\Bbb Q$ and $\Bbb R$.


What if we just pay attention to the ring-like structure?

$\Bbb N$ is initial in the category of semirings (with identity).
$\Bbb Z$ is initial in the category of rings (with identity).
$\Bbb Q$ is initial in the category of characteristic $0$ fields.
it doesn't seem likely that $\Bbb R$ has a purely algebraic characterization, considering its construction inherently references topological properties. 
Given $\Bbb R$, then $\Bbb C$ does have the algebraic distinction of being the only algebraic field extension of $\Bbb R$ other than $\Bbb R$ itself. There is indeed a category of field extensions of $\Bbb R$, but this time $\Bbb C$ is not initial. Actually it seems to be terminal in this category. Apparently then it is initial in the opposite category :)
Finally, given any commutative ring $K$ (like a field), $K$ is bound to be initial in the category of associative $K$ algebras with identity.