Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The familiar number sets $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$ all have "natural constructions", which indicate, why they are mathematically interesting.

For example, equipping $\mathbb{N}$ with the usual sucessor function and the constant $0$, it can be described as the initial $(0,1)$-Algebra. Or, if we want it to be an additive monoid, it is the free monoid on some one-point set. Since monoids are something very elementary and can be described purely categorically in many equivalent ways, this gives an idea, why considering the natural numbers might be interisting.

Now the forgetful functor $\mathbf{Groups}\longrightarrow\mathbf{Monoids}$ has a left adjoint sending $\mathbb{N}$ on the additive group of integers $\mathbb{Z}$. Alternatively, $\mathbb{Z}$ is the initial ring. If one questions that considering rings is interesting, we could reply that they are just monoids in the "natural" category of abelian groups with the "natural" tensor product making it a monoidal category.

From $\mathbb{Z}$ to $\mathbb{Q}$ it's not far, since the rationals are the image of $\mathbb{Z}$ under the left adjoint of the embedding $\mathbf{Fields}\longrightarrow\mathbf{Domains}$.

(One could go one step further and move on to the algebraic numbers as the field-theoretic completion of the rationals.)

My question now is: Why do we consider the reals $\mathbb{R}$, from a structural point of view? It is clear to me (or at least I don't feel I have the right to question it) that considering real numbers in physics, finance, etc. is necessary, because it provides good models for our reality.

I rather wonder, wether there are mathematical reasons making the reals interesting. The only description of the real numbers by some universal property that I am aware of is that they are the Cauchy-Completion of the rationals as a metric space - but since the definition of a metric space already depends on some notion of the reals, this is just a cheap trick. (One could of course define a metric space as a set $X$ with a set-function into some archimedian ordered field, satisfying the usual axioms, but this is a little artificial, I think.) Also note that it is almost impossible to do Algebra, Set Theory or Graph Theory without knowing what the natural numbers are, whereas one can prove many results of Algebra and Topology without ever coming across the reals.

I hope you can provide some ideas, showing why the reals numbers considered as a topological space/a field/a group/an ordered set are interesting in conceptual (i.e. category-based) mathematics. Of course, if one questions that the real numbers are interesting, one also has to question complex numbers $\mathbb{C}$ and other concepts based on these notions (like almost all of Analysis, Differential topology, etc.), so I am well aware of the fact, that I should not refuse the real numbers as old-fashioned even in case, I don't receive many answers.

Edit: Of course, my question is implicitly based on my strong belief that mathematical interestingness and categorical interestingness are equivalent concepts. As suggested by some of the comments, I would like to rephrase my question: How can I deduce from the mathematical properties of the real numbers that they are mathematically interesting, and thus that they are the optimal formalization of our intuitions of geometry and infenitesimal operations?

share|cite|improve this question
One would like $\sup \left(\left\{x\in \mathbb Q\colon x^2<2\right\}\right)$ to exist, it's natural. Well, it doesn't exist in $\mathbb Q$. – Git Gud Jun 19 '14 at 17:31
The reals allow you to take limits without having to worry too much about whether they exist. Note also that $\mathbb Q$ is a metric space which does not need the reals in its definition or any artificiality - the reals are not essential to the definition of a metric space. – Mark Bennet Jun 19 '14 at 17:34
@Stahl: I disagree! The universal properties encode precisely why they are interesting. $\mathbb{N}$ counts things; more precisely, it counts repetitions of things, and this is encoded in the fact that $\mathbb{N}$ equipped with the successor operation is the universal dynamical system. (One can get all of the other universal properties of $\mathbb{N}$ from this one, e.g. as a monoid or as a semiring.) – Qiaochu Yuan Jun 19 '14 at 17:41
Your question body seems to be at odds with the title. The title seems to be asking for practical or philosophical reasons why we should study the real numbers, but the question body seems to be asking "How do the real numbers fit into category theory?". In particular, I think when you say "from a mathematician's point of view" you mean "from a category theorist's point of view". – Jack M Jun 19 '14 at 17:42
@QiaochuYuan Maybe I came off a bit strong - I don't really mean that universal objects aren't interesting, I'm suggesting that these things are interesting from a mathematician's point of view because we have experience with them and a use for them. A lot of mathematics has been done because we have some experience, or because there is necessity for a construction (like a limit). Later on, we realize they fit into a bigger picture, but they're also interesting simply because they're useful or occur "naturally." tl;dr (I feel) something isn't uninteresting because it's not universal. – Stahl Jun 19 '14 at 17:46

I think you are not asking the question you mean to ask. The question you mean to ask is something like "what kind of universal properties does $\mathbb{R}$ satisfy?" which is very different from "why should mathematicians care about $\mathbb{R}$?" Of course the answer to that question is to model lots of phenomena of obvious mathematical interest, e.g. differential equations and manifolds.

Here is one: $\mathbb{R}$ is the terminal archimedean field. (But unlike the example of $\mathbb{N}$ I don't consider this the last word on why the real numbers are interesting. This doesn't really explain why we use the real numbers to model Euclidean space, for example.)

share|cite|improve this answer
I don't feel like arguing about this, but I want to leave my opinion here. I don't think 'the' answer to ""why should mathematicians care about $\mathbb R$?" is what you say, but rather we want the supremum of any non-empty bounded above set to exist. – Git Gud Jun 19 '14 at 18:04
@Git: that also doesn't explain why we use the real numbers to model Euclidean space, and in particular doesn't explain why e.g. $\mathbb{R}^n$ has such a large symmetry group, larger than expected from just being an $n$-fold product of some order that we care about because it's order-complete. – Qiaochu Yuan Jun 19 '14 at 18:26
Here are some details for the proof that $\mathbb{R}$ is the terminal archimedean ordered field (which is omitted at the nlab): Let $K$ be an archimedian ordered field and $u \in K$. Choose $n \in \mathbb{N}$ such that $u \leq n$. Then $\{q \in \mathbb{Q} : q \leq u\}$ is bounded in $\mathbb{R}$, namely by $n$. Hence, it has a supremum $f(u)$. Then one checks that $f : K \to \mathbb{R}$ is a homomorphism of ordered fields. It is unique, because $\mathbb{Q}$ is dense in $K$. – Martin Brandenburg Jun 19 '14 at 18:36
In what sense does $\Bbb R^n$ have a large symmetry group? How large would we expect it to be? – MJD Jun 19 '14 at 18:44
@MJD: once we equip it with the Euclidean metric, it has symmetry group $\mathbb{R}^n \rtimes \text{O}(n)$. This is much larger than one expects for $X^n$ where $X$ is some generic object in a category with finite products; generically we only expect something like the wreath product $\text{Aut}(X) \wr S_n$. So there's really something special about $\mathbb{R}^n$ with the Euclidean metric, which is of clear importance in mathematics and whose properties don't seem to obviously follow from Git Gud's desired characterization of $\mathbb{R}$ (or mine, for that matter). – Qiaochu Yuan Jun 19 '14 at 18:55

An interesting alternative approach to defining the reals was discussed on the category theory mailing list many years ago:

Let $S$ be the ring of functions $s: \mathbb{N}\rightarrow \mathbb{N}$ such that the function of two variables $(m,n) \rightarrow s(m+n) - s(m) - s(n)$ is bounded. Addition is element-wise and multiplication is functional composition. Let $I$ consist of the bounded sequences. Then $\mathbb{R} = S/I$.

So $S$ includes these things that are almost but not quite homomorphisms. This is a little surprising as defining approximate things (ie. actually approximate by a finite amount rather than infinitesimally) and composing them often leads to things getting badly behaved. So there's something special about $\mathbb{R}$ and its relationship with $\mathbb{N}$.

share|cite|improve this answer
An interesting construction. I think the number $r$ corresponds to the map $n \mapsto \lfloor rn \rfloor$? In some sense, this is still an infinitesimal approximation, because your numbers are $\Theta(n)$ and your errors are $o(n)$. Does this approach produce the correct real number object in a topos? – Hurkyl Jun 19 '14 at 21:00
I don't know much about toposes. But curiously the people who've written about this construction seem to work with toposes a lot. I feel like in some sense this construction gets to the very heart of what real numbers are about. It seems like a generalization of reasonable (in some sense) strategies for sharing $M$ objects between $N$ people. I haven't worked out the details of that yet though... – Dan Piponi Jun 19 '14 at 21:08
I've tried to verify this description. a) Is there any literature about it? b) We have to replace $\mathbb{N}$ by $\mathbb{Z}$, right? Otherwise we have no chance to get an additive group. c) $S$ is not a ring, but just a near-ring, right? The distributive law $f \circ (g+h)=f \circ g + f \circ h$ doesn't hold. However the difference is bounded so that the near-ring quotient $S/I$ is a ring. d) The isomorphism is given by mapping $s : \mathbb{Z} \to \mathbb{Z}$ to $\lim_{n \to \infty} \frac{s(n)}{n} \in \mathbb{R}$, and by mapping $r \in \mathbb{R}$ to $n \mapsto \lfloor n r \rfloor$. – Martin Brandenburg Jun 20 '14 at 8:03
Also – Dan Piponi Jun 20 '14 at 14:01

There is a characterisation of the reals purely in terms of its order structure which explains its ubiquity as a model in mathematics and physics: a totally ordered, Dedekind complete set which has a countable order dense subset but no largest or smallest element. This is the basis of many results which show that natural orderings which arise there are induced by a real- valued function. This is one of the central problems of the theory of measurement and explains how one can pull the reals out of a hat from a system of axioms which do not contain the concept of number explicitly. Examples are: entropy (from the ordering "adiabatically accessible" on the states of a thermodynamical substance), temperature (hotter than), price in economics from the relationship "worth more than" (under the generic name of "utility function") and of course in the synthetic approach to axiomatic euclidean geometry (using the ordering induced by the "in between" axiom). The important point is that these concepts in the physical world can be experimentally verified in the ordinal sense without recourse to a numerical scale.

share|cite|improve this answer

There are various topological characterizations of the topological space $\mathbb{R}$. See MO/76134. For example it is the unique connected, locally connected separable regular space, such that deleting any point gives two components. To some extent this is exactly what we want the continuum to be.

But the question seems to be aimed more at algebraic characterizations. In this case, it would be nice to give a characterization of $\mathbb{R}$ within the category of fields. This fits nicely into my question SE/634010 if the category of fields is rigid (still unsolved), which would give a categorical characterization of any field. At least, we have a characterization of those fields which are elementary equivalent to (i.e. satisfy the same sentences as) $\mathbb{R}$. These are the real closed fields, which have various purely algebraic characterizations. A very short one is that $F$ is real closed if $F^{alg} / F$ is a proper finite extension.

share|cite|improve this answer

Following Bourbaki, the reals are a completion of $\mathbb Q$, not as metric spaces but as uniform spaces. In more detail, every metric space is a uniform space, but not vice-versa. The familiar construction of the reals and of metric completions by means of Cauchy sequences (i.e., Cantor's construction) is more of a metric construction then a uniform construction. In particular, it does not work for arbitrary uniform spaces. However, there are general uniform completions that turn any uniform space into a complete one. These constructions agree with the metric completions of course.

Now, forgetting any notion of metric, the space $\mathbb Q$ with its standard topology is a uniform space. Thus, just like any other space, it has a completion. That completion is $\mathbb R$. Viewed this way, $\mathbb R$ is a completion just like any other completion of any other uniform space. To me this makes the construction of the reals seem more uniform, and perhaps this point-of-view relates to your question.

share|cite|improve this answer
Can you write a little more about this? For example, would it be fair to say that: 0) essentially what's going on is that we're viewing $\mathbb{Q}$ as a commutative ring object in the category of uniform spaces, and 1) that every algebraic structure internal to the category of uniform spaces has a completion in the same signature? – goblin yesterday

$\mathbb{C}$ and $\mathbb{F}_{p^n}$ give you algebraic closures, polynomials being the obvious generalization of the linear equations that $\mathbb{Q}$ is sufficient to solve. Once one has made it to $\mathbb{C}$, it's easy to pick out $\mathbb{R}$ -- it's the maximal Archimedean subfield fixed under and the "polynomial operations" of addition, multiplication and positive integer powers.

(The Pythagoreans would have gladly stopped at $\mathbb{Q}$, but the irrational constructibles eventually won the argument.)

Edit 20140620: Remembered to add "Archimedean". Thanks to Qiaochu Yuan for his answer reminding me that I'd forgotten a condition.

share|cite|improve this answer
Algebraic closures stop well short of $\mathbb{C}$, though - this doesn't really pick out $\mathbb{R}$ from the real part of $\mathbb{\bar{Q}}$... – Steven Stadnicki Jun 20 '14 at 6:22
@Steven: Give me $\mathbb{Z}$ and I'll find the algebraics, though only in an artificial way, categorically. Then give me sequences and I'll find $\mathbb{C}$ and this is not categorically artificial. – Eric Towers Jun 20 '14 at 15:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.