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.

  • 3
    $\begingroup$ You need to specify more precisely just what aspects of these sets you want to characterize. For example $\langle\Bbb Z,+\rangle$ is the unique non-trivial free Abelian group that is also a free group. On the other hand, $\langle\Bbb Z,\le\rangle$ is the unique linear order without endpoints with the property that every interval $[a,b]$ is finite. $\endgroup$ Jul 11, 2013 at 20:39
  • $\begingroup$ This question is too vague/broad. Please try to improve the question. For example, pick a space and pick an operation. People could write a book answering this question. $\endgroup$ Jul 11, 2013 at 20:40
  • $\begingroup$ @BrianM.Scott My idea is that when someone define an arbitrary set, such a set is defined giving certain propierties. It's like saying $\mathbb{Q}$ is the set that....and then the list of propierties (the axioms). $\endgroup$ Jul 11, 2013 at 20:46
  • 5
    $\begingroup$ The problem is that this can be done in many different ways. If one views $\Bbb Q$ as a linearly ordered set, it has one characterization: it’s the unique countable dense linear order without endpoints. If one views it as a field, it has other characterizations; see Asaf’s answer. If one views it as a topological space with the usual topology, it has yet other characterizations. $\endgroup$ Jul 11, 2013 at 20:48
  • 1
    $\begingroup$ I think 90% of this question was lucid and focused, but the last sentence asked for "an exhaustive list of properties," which fooled people into thinking it was too broad. I think the OP was extremely clear what they were taking into consideration: the algebraic operations and an ordering. I've made edits to draw that out and suppress the "exhaustive list" request, but Daniela, if you really don't like the edits feel free to revert them. I think this is an interesting question and a good candidate for re-opening. $\endgroup$
    – rschwieb
    Jul 12, 2013 at 14:29

2 Answers 2

  1. $\Bbb N$ is the unique linearly ordered set that is infinite, but every initial segment is finite; it is also the smallest set which contains $0,1$ and closed under addition that satisfies the axiom $x+y=0\iff x=y=0$.

  2. $\Bbb Z$ is the smallest linearly ordered set on which both successor and predecessor operations are defined everywhere, and every element is a successor.

  3. $\Bbb Q$ is the smallest field which satisfies that $1+1+\ldots+1\neq 0$. It is also the smallest field which can be ordered.

  4. $\Bbb R$ is the unique ordered field which is Dedekind complete as an ordered set.

  5. $\Bbb C$ is the unique field which is algebraically closed, contains $\Bbb Q$ and is equipotent with $\Bbb R$. It is also the unique algebraic closure of $\Bbb R$.

  • $\begingroup$ Pre-congratz on 100k, which you will undoubtedly surpass before I get back to my computer again :p $\endgroup$
    – rschwieb
    Jul 11, 2013 at 21:06
  • $\begingroup$ @rschwieb: Thanks, although I'm not sure. Some of this reputation is only temporary until the serial voting script is scheduled to run in about five hours. ;-) $\endgroup$
    – Asaf Karagila
    Jul 11, 2013 at 21:08
  • $\begingroup$ @AsafKaragila After the valuable comments I though my question was going to be closed, thanks so much for saving it. Now, your answer couldn't be better and summarize everything in just few words. Since I'm just a simple mortal and also a beginner I need to start breaking all the concepts down, so just another question what book do you recommend me to read so I can learn all these concepts? $\endgroup$ Jul 11, 2013 at 21:30
  • $\begingroup$ @Daniela: First of all, the question is likely to be closed. It has four of the needed five votes for that right now. Secondly, these concepts are not going to be found in one book. It's a result of learning a lot of mathematics over the years. The observations on $\Bbb Q$ are likely to be found in an introductory book on field theory, or maybe even Galois theory. Similarly the ones about $\Bbb R$ and $\Bbb C$ are to be found in books about fields theory, and probably model theory for the latter. The ones about $\Bbb N$ and $\Bbb Z$ I don't know where you can find, actually. $\endgroup$
    – Asaf Karagila
    Jul 11, 2013 at 21:47
  • $\begingroup$ Oh, I see. Thanks again. $\endgroup$ Jul 11, 2013 at 22:09

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

  1. 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 $1$.)

  2. $\Bbb Z$ is initial in the category of totally ordered rings. (We have given up well-ordering from $\Bbb N$.)

  3. $\Bbb Q$ 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.)

  4. $\Bbb R$ 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. )

  5. 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?

  1. $\Bbb N$ is initial in the category of sets with a distinguished point and a successor function. (This is an exercise in MacLane's CFTWM.)

  2. 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.

  3. 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.)

  4. $\Bbb R$ 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.

  5. $\Bbb C$ 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?

  1. $\Bbb N$ is initial in the category of semirings (with identity).

  2. $\Bbb Z$ is initial in the category of rings (with identity).

  3. $\Bbb Q$ is initial in the category of characteristic $0$ fields.

  4. it doesn't seem likely that $\Bbb R$ has a purely algebraic characterization, considering its construction inherently references topological properties.

  5. 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 :)

  6. Finally, given any commutative ring $K$ (like a field), $K$ is bound to be initial in the category of associative $K$ algebras with identity.

  • $\begingroup$ Isn't $\Bbb Z$ a totally ordered ring which embeds into $\Bbb Q$? Also, if $\Bbb R$ is the unique ordered field which is Dedekind-complete, isn't $\Bbb R$ terminal in that category as well? And all the arrows are the identity map? :-) $\endgroup$
    – Asaf Karagila
    Jul 11, 2013 at 21:08
  • $\begingroup$ @AsafKaragila yes :( This was a hasty bad pre-commute pre-dinner post with lots of mistakes! Thanks for taking a look: I hope it will be better the next time you read it. $\endgroup$
    – rschwieb
    Jul 11, 2013 at 22:04
  • $\begingroup$ Consider the category $\cal C$ such that $\operatorname{obj}(\cal C)=\{\Bbb C\}$ and $\hom_\cal C(\Bbb{C,C})=\{\rm id\}$. Then $\Bbb C$ is the initial object of this category! ;-) $\endgroup$
    – Asaf Karagila
    Jul 11, 2013 at 23:53
  • $\begingroup$ @AsafKaragila I hope the improvements help. In my journeys I ran through the wiki and now wonder if the reals are still initial in the category of ordered complete (with respect to Cauchy sequences) but possibly nonarchimedian rings. Do you know offhand? $\endgroup$
    – rschwieb
    Jul 11, 2013 at 23:55
  • $\begingroup$ @AsafKaragila If I wanted easy categories whose initial objects are just a field $F$, couldn't I use categories of $F$-algebras? I felt like the "order" ideas in the original post kept me from doing that. $\endgroup$
    – rschwieb
    Jul 11, 2013 at 23:56

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