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I understand that a set is like a list of things, except that the order doesn't matter and that you can't have any duplicates in a set. For example: $\{3, 1, 4, 2\}$ is the same set as $\{1, 2, 3, 4\}$; and that $\{2, 3, 2, 2, 3\}$ doesn't make sense as a set because it has elements that appear more than once..

I also understand that relations are sets of ordered pairs, and that functions are a subset of relations. Like, I feel comfortable with basic discrete math.

But what's "structured" about a set? I picture an ASCII-art cathedral spire composed of curly braces and numbers and commas.

Zev Chonoles
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Josh Tilles
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1 Answers1

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Structure is an additional information on the set. It means that there are some relations, constants and operations associated to that set.

For example $\Bbb N$ is a set, but we can give is structure like addition or order, or both. Then the elements of $\Bbb N$ may have importance relative to the structure. If we added the order, then $0$ is a minimum; if we added multiplication then $1$ is a unit; and so on.

It is important to remember that while addition and so on are very natural for us on $\Bbb N$ we can also take relations and operations which make no sense for us, and it will be mathematically valid to do. We can of course give structure to strange sets, and it may not make much sense at first but it is still a valid thing that we can do.

To be able and use the structure mathematically we have a language which allows us to express wanted properties of the structure, or its elements, or subsets, and so on. This language contains symbols for our constants, for any relations we need, as well function symbols.

For example, if we consider $\Bbb Z$ with only the [usual] order for structure then we might as well be talking about it in the language which contain a single relation symbol which we will interpret as $\leq$. If we decide to talk about $\Bbb Z$ with addition and multiplication as well, this structure is called a ring, and we use the language of rings which includes symbols for $+$ and $\cdot$ (and sometimes symbols for the constants $0$ and $1$).

It is important to point out that there is a big difference between language and structure, but they are tied together. Often when we work in mathematics we implicitly assume that we have a language which contains the needed symbols and we interpret it in such way that we give a compatible structure to a particular set.

The basics of this idea can be found in many introductory logic (or model theory) books, where we study language and structure, also Wikipedia.

Asaf Karagila
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  • This can be the curriculum for an entire course, or at least a good and heavy 4-hours a week course for over a month. I tried to give some idea what's the idea behind structure but there is no substitute for an actual course or a book. – Asaf Karagila Feb 23 '13 at 00:01
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    Great answer! Worked a lot for me. – Integral Feb 23 '13 at 00:08
  • I tried to add a bit about the issue of language, because structures are essentially tied to languages. I'm still convinced that there is a lot more to say, and that a "complete" answer would require more or less a small books to be published as lecture notes later. – Asaf Karagila Feb 23 '13 at 13:41
  • @AsafKaragila, is structure synonymous with: system of relations and operators? – alancalvitti Feb 23 '13 at 15:40
  • @alan: Is drinking synonymous with beer? – Asaf Karagila Feb 23 '13 at 15:44
  • I read (I think in Cohn's book Universal Algebra) that 1st order language is a system of relations and operators. So are these a "1st-order" representation of structure? What structure can't be represented in terms of relations and operators? – alancalvitti Feb 23 '13 at 15:52
  • @alan: I can't come up with a counterexample, and I doubt there exists one. One can always take power sets (or iterate power sets and add predicates for each iteration if needed) to allow higher-order quantification. So I'm not even sure if such structure can be specified really. Everything I can think of can be made into a first-order structure, or a higher-order structure if one prefers to think of it that way. – Asaf Karagila Feb 23 '13 at 15:58
  • I'd like to ask why Tarski's program on relational algebra as foundations was stopped. I don't think he was interested in closure systems for example. Operators are as necessary as relations. – alancalvitti Feb 23 '13 at 16:04
  • @alan: I don't know why. Do note that every operator is a relation, and you can simply add an axiom saying that it is an operator (in addition to a relation). Similarly constants are just unary relations that have a unique element satisfying them. It is easier to work with operations and constants, yes. But those can be easily subsumed into relations if one does not mind to pay the price of additional axioms. – Asaf Karagila Feb 23 '13 at 16:08
  • Yes there is a Galois connection between relations and operators expressed in terms of their invariants and polars, but I think this was extraneous to Tarski's program. I think Tarski was primarily trying to remove variables. In any case both constants can variables are expressible as monomorphisms in cat. theory. Math is full of circular connections. But would like to minimize the axioms. – alancalvitti Feb 23 '13 at 16:13
  • @alan: I really don't know anything about Tarski's program that you mention. So I can't help you there. If you prefer minimizing the axioms then operators are better than relations, that much is obvious. – Asaf Karagila Feb 23 '13 at 16:17
  • I think it's like algebra and geometry - better to consider them both and connect them than to focus on just one or the other. – alancalvitti Feb 23 '13 at 17:18
  • @alan: Your comment doesn't make much sense in this context because you completely ignore the context of why someone might want to work with a language that only has relations, or why someone might be willing to accept a lot of new axioms for this convenience. Model theoretically it can be much easier when all your terms are variables, which means that any induction on a structure of formula does not need to include an induction on the structure of terms. – Asaf Karagila Feb 23 '13 at 17:37
  • Why define 1st order languages in terms of relations and operators? – alancalvitti Feb 23 '13 at 20:35
  • @alan: How else would you define it? – Asaf Karagila Feb 23 '13 at 21:26
  • As per your suggestion, in terms of relations alone – alancalvitti Feb 23 '13 at 21:37
  • @alan: Again, this is a matter of convenience and preference. Next you will insist that defining the real numbers as quotient of Cauchy sequences is better than defining it as Dedekind cuts - both methods have merits and both have shortcomings with respect to the other one. Allowing the language to contain operations and constants makes flexible and easily writable terms, but it can make syntactical analysis difficult. On the other hand, adding axioms may be problematic for one reason or another. There are many reasons and you should study model- and proof- theory to understand them better – Asaf Karagila Feb 23 '13 at 21:44
  • I would if it led to understanding (let alone new results) of other branches of maths, as opposed to more model- and more proof- theory. Do you promise if I learn some it will lead to "non-esoteric" knowledge? – alancalvitti Feb 23 '13 at 21:52
  • @alan: The only promise I can make is that you will die at the end. I can't even tell you when the end is, though. I will point out that Hrushovski is actively researching all sort of things which may actually have real life applications (e.g. a student of his told me something about studying structures which might be groups, that is there is an $x%$ chance the structure is a group, and you want to say meaningful things regardless). – Asaf Karagila Feb 23 '13 at 21:55