Why do structured sets, like (N, +) often get referred to just by their set? Under this way of speaking, where N denotes the natural numbers, + addition, and * multiplication, (N, +, *) and (N, +) both can get referred to as N. But, due to our ancestors we can readily talk about (N, +) via Presburger Arithmetic, and (N, +, *) via Peano Arithmetic, which readily makes these structures different, since equalities in (N, +) can get decided algorithmically, but they can't for (N, +, *). But, the way of referring to these structures by the set N masks all of this. So, why even bother referring to a structured set by its set in the first place?
Why Do Structured Sets Often Get Referred to Only by the Set?
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2Looks like "not a real question" (but a lament) to me... – 2011-06-14
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2@Grigory: It seems like a perfectly reasonable question to me. This would only qualify as a "lament" if it were clear that everyone agreed with the OP, and that nothing can be done about it. – 2011-06-14
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7You have probably never worked with cobraided topological Hopf algebra in the category of relative Yetter-Drinfeld modules over a Hopf algebra $H$ and $H$-comodule algebra $A$... What you propose is simply unworkable except in the most simple cases! – 2011-06-14
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0@Mariano: with respect to *this* site, questions often arise with respect to "most simple cases", when students are first learning, e.g. abstract algebra and definitions of a group, ring, etc., in which case great emphasis is placed (and I believe, appropriately) on being clear about a structure's operation(s), under which it qualifies as a group, e.g., etc. It may very well be a contextual issue, but I'd rather error on the side of fuller specification rather than lesser, at least here, when answering (or asking) questions. – 2011-06-14
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0@Mariano et.al.: I realize the question asked was more general in scope than just about the practices here on Math.SE, but I believe it's a good question for users of this site to consider when answering questions. – 2011-06-14
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5@amWhy: the question makes a difference between $(N,+)$ and $(N,+,*)$ based on the algorithmic decidability of two standard first order theories attached to those structres... It is not a great leap of the imagination to assume the question is asked from a context distinct from "first learning the definition of a group". – 2011-06-14
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0@Mariano After reading The Ignorance of Bourbaki http://www.dpmms.cam.ac.uk/~ardm/bourbaki.pdf earlier today, I think it meaningful to ask such a question from plenty of contexts. – 2011-06-14
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0@Mariano: Nothing I wrote suggested "this" question was asked from someone first learning the definition of a group. I simply stated that the point brought up by this user is well worth considering...Perhaps it wasn't the intended point of the OP...but I believe it has relevance here (on this site). – 2011-06-14
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0@Doug: I don't understand what relation there is between that article and your question on whether one should include all possibly relevant information in the notation one uses... – 2011-06-14
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0Note: It appears that the source of this question is the long [comment thread here.](http://math.stackexchange.com/questions/37648/subjects-studied-in-number-theory/37699#37699) – 2011-08-20
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0@Bill Though, as I recall, that thread did originate my curiosity here, so far as I can tell, there does exist more to it. Someone around here recently asked about the metric space R. I could only guess the person wanted to suggest the reals under the absolute value function. As perhaps a better example than the OP, the metric spaces (R, absolute value function), (R, d) where d indicates the function d(x, y)=0 if x=y, d(x, y)=1 otherwise, aren't the same, even though they have the same underlying set. – 2011-08-20
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0@Doug The purpose of the above comment was primarily to serve as a link between the two related threads. When you post a question based on a prior discussion here it is best for all to include a link to such prior discussion(s). You'll get more informed answers and folks won't waste time repeating what has already been said, etc. – 2011-08-20
3 Answers
Because people are lazy there is value in lossy compression for the sake of communication. I agree that this can be a bad convention in the sense that it can create confusion, but 1) usually by context you can tell what structure is assumed, and 2) sometimes people want to consider multiple compatible structures without explicitly listing them, which are again usually clear from context.
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8And because often writing everything out can get in the way of understanding. Trying to write out even a basic group theory theorem unambiguously and technically correct by distinguishing between the groups, its underlying sets, its operations (making sure that different groups get operations named with different symbols) and the like would likely create a statement in which the technical issues overshadow the simple statement trying to be conveyed. Or, try to do calculus by explicitly describing each real as an equivalence class of rational Cauchy sequences, and see how far you can get. – 2011-06-14
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1@Qiaochu: I disagree. Referring to structured objects by their underlying sets is an excellent convention that makes mathematics easier to understand and easier to communicate. – 2011-06-14
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0@Jim: but you can't deny there are situations where it is genuinely confusing. For example, there was a confusing thread a month or so ago about Hilbert spaces where one point of confusion was that there were two spaces $A \subset B$ with $A, B$ both Hilbert spaces, but where the induced inner product on $A$ coming from $B$ wasn't the inner product given to $A$. The standard notation for Hilbert spaces does not easily allow us to express this subtlety, and if it did, the whole discussion would've been cleared up sooner. – 2011-06-14
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2@Qiaochu: As with any convention, the convention is adopted so long as it is useful; yes, there are situations when the convention could be confusing, at which point it is (or should be) abandoned in favor of non-confusing statements. (E.g., when we talk about "the additive group of rationals" instead of simply saying "$\mathbb{Q}$"). But just because there are exceptional situations in which a convention might prove a hindrance is no reason to denigrate the entire convention, or to scrap it entirely, especially if the lack of convention is likely to prove a hindrance far more often than not. – 2011-06-14
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1@Arturo: sure. I don't think I actually suggested that we scrap the convention entirely, and I also didn't say that being lazy was a bad thing. A math teacher of mine used to say that laziness was my greatest asset as a mathematician. – 2011-06-14
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1@Qiaochu: Fair enough, but keep in mind that "lazy" usually carries negative connotations. It is the OP that argues for abolishing the convention; see, e.g., [recent comments here](http://math.stackexchange.com/questions/37648/subjects-studied-in-number-theory/37699#37699) which I suspect is what sparked this query. – 2011-06-14
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1@Qiaochu: It's certainly true that this convention causes confusion on occasion, but the gains from efficiency of communication far outweigh the loss of precision. – 2011-06-14
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1@Arturo I don't see the point about doing calculus by equivalence classes of rational Cauchy sequences, since that's not the only way to construct the reals. I haven't argued that the convention never get used. I would hope that such conventions get recognized as having potential problems, and since if it seems the convention gets overly estimated in terms of value, there exists sufficient ground to denigrate it. After all, how else would one point out problems with such a convention other than by saying something which seems derogatory towards it? – 2011-06-14
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3@Doug: Your toolbox does seem to be rather lacking, if the only way you can think of to point out a (potential) problem with something is by denigrating it. – 2011-06-14
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1@Arturo Pointing out a problem with something, in my opinion, denigrates that something by pointing out a problem. If pointing out a problem isn't by nature a denigration, then there exists no denigration necessarily in the original post here. – 2011-06-14
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2@Doug: **Criticize**: tr. v. 1. To evaluate the merits and demerits and judge accordingly; evaluate. 2. to find fault with; point out the faults of. **Denigrate**: tr. v. 1. to attack the reputation of; defame. 2. to deny the importance or validity of; belittle. (**Deny:** tr. v. 1. To declare untrue; 2. to refuse to admit or acknowledge: disavow. 3. a. To give a negative answer to; b. to refuse to grant; c. to restrain (oneself) from gratification of desires. 4. *archaic* decline. 5. to refuse the accept the existence, truth, or validity of). Webster's online dictionary. – 2011-06-14
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1@Arturo I have no idea why you've quoted a dictionary here, since dictionaries like Webster's don't dictate the meaning of words, and generally get written by comparing the usage of terms in several works of literature (or by comparing several usages of terms in speech). – 2011-06-14
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0@Doug: I've quoted it to point out that, as seems to be often the case, your usage of words and your view of the world is at odds with common usage. No doubt you will demure, all evidence to the contrary, again as usual, so I will only add *adieu*. – 2011-06-14
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1@Arturo I don't see how there exists a single thing as "common usage" of terms or a "common view of the world" to begin with. It's not like we all have the same linguistic background or have the same brains. – 2011-06-14
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1@Doug Spoonwood: When we "construct" set-theoretically even something as concrete as $\mathbb{Z}$, the elements of $\mathbb{Z}$ are equivalence classes of ordered pairs. In particular, technically, no element of $\mathbb{N}$ is in $\mathbb{Z}$! What a nuisance it would be if we continued to insist on distinguishing between $\mathbb{N}$ and the natural copy of it in $\mathbb{Z}$. – 2011-06-14
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0@Qiaochu: I agree with Arturo. The word "lazy" in common usage has an undeniable pejorative connotation. If someone has praised your mathematical laziness, then they meant something other than what readers of your answer here will easily be able to discern. Ironically, it seems that you are trying to defend the (inevitable and largely beneficial) use of context-dependent language by using language that is too context-dependent. (Or, if you can stand more, your use of "lazy" seems a little lazy. Why not say what you actually mean?) – 2011-06-14
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3@Doug: there does not exist a *single* "common usage". This is why one has to think about one's intended audience when speaking and writing: what can be assumed to be part of the common context, and what cannot? It is a sound pedagogical tool to act as though the common context is a little less than what you actually think or hope it to be, but to assume no common context at all is a serious mistake. The reason that these mathematical synecdoches are used is because they have been found to be useful to people when communicating mathematics to each other: it's as simple as that. – 2011-06-14
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0@Pete: touché. I'll amend my wording. – 2011-06-14
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0My first thought was also 'beause mathematicians are lazy'. Here I define 'lazy' as 'economic' or 'efficient' :) – 2011-06-15
It's a question of brevity, for the most part. Brevity is different from laziness, because brevity has the goal of clarity. In theory, we could require that all our proofs and writing in mathematics be so rigorous that a computer can read it, but then it would be unreadable by humans.
So, a news article will refer to "Secretary Clinton," or even "Clinton," perhaps only once referring to "Secretary of State Hillary Clinton." The reason is that humans are very good at determining context and meaning, and they find redundancy leads to confusion in communication. (This is why we use the word "it" in place of nouns, too, and that can cause confusion when misused, as can referring to "Clinton" if the article contains information about both Bill and Hillary.)
So, if the context isn't clear, then a person should definitely write $(\mathbb{N},+)$, but it's not always obvious when the context is clear or not.
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0If this "In theory, we could require that all our proofs and writing in mathematics be so rigorous that a computer can read it, but then it would be unreadable by humans." is really true, then how does undecidability of formal mathematical theories exist? I don't claim to know how good humans are in general at determining context and meaning. And I certainly doubt redundancy leads to confusion. Saying that all groups satisfy x+0=0+x=x qualifies as redundant, but it actually seems easier than just saying x+0=x. Additionally, your comment in parentheses refers to *ambiguity*. – 2011-06-14
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1@Doug: There's a difference between a theory being undecidable its proofs being unverifiable. So long as the theory is recursively axiomatisable, proofs can be verified, even if the theory is undecidable. – 2011-06-14
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0@Zhen Thomas didn't just say proofs, he said "proofs and writing in mathematics". – 2011-06-14
It is a general convention in mathematics that almost any structured object is primarily a set, with the structure as a sort of decoration. For example, the sphere $S^3$ is primarily the set of points in the sphere, and the various other metric, topolgoical, and algebraic structures on the sphere are considered secondary.
In the specific cases you have mentioned, the phrases "Presburger arithmetic" and "Peano arithmetic" refer primarily to specific first-order theories within the context of logic. I'm not sure why referring to a structured set via its first-order theory would be any more or less natural than referring to a structured set via its underlying set.
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0In the example of the original post, referring to the structured sets via their underlying sets I would think would suggest them as equivalent. Both sets (this is a sophism) have representation N here, and N=N, so how do the structures really differ? The first-order theories of these structured sets don't come as equivalent in terms of their metalogical "properties" (decidability vs. undecidability, completeness also, not sure if "properties" comes as the right term here). Though the representations don't suggest this exactly, (N, +)=(N, +, *) gets thrown out more easily. – 2011-06-14