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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 interesting.

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, whether 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 infinitesimal operations?

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    $\begingroup$ 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 $\mathbb Q$. $\endgroup$ – Git Gud Jun 19 '14 at 17:31
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    $\begingroup$ 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. $\endgroup$ – Mark Bennet Jun 19 '14 at 17:34
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    $\begingroup$ @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.) $\endgroup$ – Qiaochu Yuan Jun 19 '14 at 17:41
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    $\begingroup$ 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". $\endgroup$ – Jack M Jun 19 '14 at 17:42
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    $\begingroup$ This does not answer the question, but you might be interested in the paper hereL tac.mta.ca/tac/volumes/20/10/20-10.pdf It studies the closed interval from a categorical POV, not the entire real line. $\endgroup$ – Steven Gubkin Jun 20 '14 at 13:30
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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.)

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    $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Jun 19 '14 at 18:26
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    $\begingroup$ 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 $K$. $\endgroup$ – Martin Brandenburg Jun 19 '14 at 18:36
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    $\begingroup$ @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). $\endgroup$ – Qiaochu Yuan Jun 19 '14 at 18:55
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    $\begingroup$ Very good, thanks. $\endgroup$ – MJD Jun 19 '14 at 18:56
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    $\begingroup$ @QiaochuYuan Thinking aloud here: part of the reason that that's so is because there are 'enough' rotations to make the group non-discrete. On the surface this is because sin and cos are 'real-closed' but not closed over smaller structures; a rotation of e.g. $\mathbb{Q}^n$ almost certainly can't be superimposed back onto that lattice, and so we lose many members of the rotation group. (cont) $\endgroup$ – Steven Stadnicki Jun 19 '14 at 19:02
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An interesting alternative approach to defining the reals was discussed on the category theory mailing list many years ago: http://comments.gmane.org/gmane.science.mathematics.categories/1319

Let $S$ be the ring of functions $s: \mathbb{N}\rightarrow \mathbb{N}$ such that the function of two variables $(m,n) \mapsto s(m+n) - s(m) - s(n)$ 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 $\mathbb{N}$.

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    $\begingroup$ 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? $\endgroup$ – user14972 Jun 19 '14 at 21:00
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    $\begingroup$ 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... $\endgroup$ – Dan Piponi Jun 19 '14 at 21:08
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    $\begingroup$ 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 $n \mapsto \lfloor n r \rfloor$. $\endgroup$ – Martin Brandenburg Jun 20 '14 at 8:03
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    $\begingroup$ a) arxiv.org/pdf/math/0301015v1.pdf $\endgroup$ – Martin Brandenburg Jun 20 '14 at 8:22
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    $\begingroup$ Also maths.mq.edu.au/~street/EffR.pdf $\endgroup$ – Dan Piponi Jun 20 '14 at 14:01
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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.

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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – goblin Feb 6 '16 at 16:30
  • $\begingroup$ @goblin I'm not quite sure what you are asking. $\endgroup$ – Ittay Weiss Feb 8 '16 at 9:01
  • $\begingroup$ Goblin asks: If you have a commutative ring with a compatible uniform structure, is the uniform completion also a commutative ring in a compatible way? (s.t. this generalizes the situation "$\mathbb Q$ to $\mathbb R$") $\endgroup$ – Stefan Perko Jul 7 '16 at 7:23
  • $\begingroup$ @goblin yes, if the ring structure maps are uniformly continuous, then they will extend to the completion. $\endgroup$ – Ittay Weiss Jul 7 '16 at 10:45
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There is certainly a broad consensus that the real numbers are extremely useful and moreover categorical. However, they are not "the optimal formalization of our intuitions of geometry and infinitesimal operations" as you put it. The hyperreal numbers are a better candidate for this task since they literally contain infinitesimal numbers, and moreover allow functions to be naturally extended.

In more detail, the hyperreals are the only theory within ZFC that give a theory of infinitesimals useful in analysis. Other theories don't even provide natural extensions of functions like the sine to the extended domain. Furthermore, a hyperreal field is definable in ZF. There are two main competitors: the surreals and the Smooth Infinitesimal Analysis. The former cannot even extend the sine function. The latter requires intuitionistic logic and does not work in ZFC.

Keisler has a unique characterisation of a hyperreal line in terms of a natural system of axioms. In a different direction, the group around Di Nasso has also provided such characterisations.

Since the OP mentioned both the real numbers and infinitesimals in his question, it is worth mentioning that Edward Nelson developed a more versatile set-theoretic foundation called internal set theory (which is however a conservative extension of ZFC) where numbers behaving like infinitesimals can be found in the real number field itself. These are still the usual real numbers so in particular they enjoy all the properties of categoricity that we all rave so much about.

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    $\begingroup$ The hyperreals are the only theory within ZFC that give a theory of infinitesimals useful in analysis. Other theories don't even provide natural extensions of functions like the sine to the extended domain. @StefanPerko $\endgroup$ – Mikhail Katz Jul 6 '16 at 13:48
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    $\begingroup$ "The hyperreals are the only theory within ZFC that give a theory of infinitesimals useful in analysis" - I am not sure I believe this without backup. At any rate, this is besides the point: This question asks for a category theoretic description; can you provide a universal property for the hyperreals? Furthermore, I didn't necessarily speak about theories within ZFC, I was thinking a little bit more generally. $\endgroup$ – Stefan Perko Jul 6 '16 at 13:52
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    $\begingroup$ I don't think your interpretation of "universal property" coincides with mine. - Okay, I can see, that you provide an answer to the second part of the question. That's certainly okay. - Don't know, what makes you claim that "hyperreals are the only theory within ZFC that give a theory of infinitesimals useful in analysis"? $\endgroup$ – Stefan Perko Jul 6 '16 at 14:00
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    $\begingroup$ @Stefan, I am familiar with two main competitors: the surreals and the Smooth Infinitesimal Analysis. The former cannot even extend the sine function. The latter requires intuitionistic logic and does not work in ZFC. Do you have any other candidates? $\endgroup$ – Mikhail Katz Jul 6 '16 at 14:06
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    $\begingroup$ My points are basically: 1) It is not necessary that the concept can be formalized in ZFC. 2) Just because we (specifically you and me) don't know about other approaches to infinitesimals doesn't mean there is nothing "better" out here; and most importantly: 3) From the category-theoretic POV a concept becomes way more interesting, if it can be somehow expressed by a universal property (en.wikipedia.org/wiki/Universal_property). $\endgroup$ – Stefan Perko Jul 6 '16 at 14:14
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$\mathbb{C}$ 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.

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    $\begingroup$ Algebraic closures stop well short of $\mathbb{C}$, though - this doesn't really pick out $\mathbb{R}$ from the real part of $\mathbb{\bar{Q}}$... $\endgroup$ – Steven Stadnicki Jun 20 '14 at 6:22
  • $\begingroup$ @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. $\endgroup$ – Eric Towers Jun 20 '14 at 15:55

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