# Why Do Structured Sets Often Get Referred to Only by the Set?

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?

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Looks like "not a real question" (but a lament) to me... – Grigory M Jun 14 '11 at 19:03
@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. – Jim Belk Jun 14 '11 at 19:08
You 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! – Mariano Suárez-Alvarez Jun 14 '11 at 20:11
@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. – amWhy Jun 14 '11 at 20:20
@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". – Mariano Suárez-Alvarez Jun 14 '11 at 20:30

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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And 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. – Arturo Magidin Jun 14 '11 at 19:03
@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. – Arturo Magidin Jun 14 '11 at 19:18
@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. – Arturo Magidin Jun 14 '11 at 20:03
@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. – Arturo Magidin Jun 14 '11 at 20:37
@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. – Pete L. Clark Jun 14 '11 at 22:23

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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If 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. – Doug Spoonwood Jun 14 '11 at 20:19
@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. – Zhen Lin Jun 14 '11 at 20:29
@Zhen Thomas didn't just say proofs, he said "proofs and writing in mathematics". – Doug Spoonwood Jun 14 '11 at 20:32

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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In 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. – Doug Spoonwood Jun 14 '11 at 20:13