Sometimes there are multiple equivalent ways of defining the same structure; for example, topological spaces are determined by their open sets, but also by their closed sets. I'm looking for a way of defining new structures that avoids committing to any one particular definition.

Let me elaborate a bit.

Sometimes "non-commitment" occurs without needing special attention. For instance, suppose we've already defined the terms antisymmetric and irreflexive. Now consider two cases. We could go on to define asymmetric as "antisymmetric+irreflexive." Alternatively we could go on to define asymmetric as "Not both $x<y$ and $y<x$." But whichever approach we choose, the same endpoint is attained. For example, in both approaches it holds that:

  1. A relation is asymmetric iff it is both antisymmetric and irreflexive, and
  2. A relation $<$ is asymmetric iff it never holds that both $x<y$ and $y<x$.

More generally, the sentences that are "true" (also known as theorems) are identical between the different approaches. So we eventually arrive at the same place, albeit via a different routes. If we think of each approach as a path, then the end result is "path-independent."

Unfortunately, definitions of structures (e.g. groups, ordered sets, topological spaces) tend to be path-dependent.

For instance, we could define a topological space as an ordered pair $(X,\mathcal{O})$, where $\mathcal{O}$ is viewed as the collection of open sets. On the other hand, we could define a topological space as an ordered pair $(X,\mathcal{K})$, where $\mathcal{K}$ is viewed as the collection of closed sets.

Notice that, unlike in the previous example, there exist sentences whose truthvalues are dependent on the particular approach taken. For instance, the sentence, "For all topological spaces $(X,\mathcal{O})$, it holds that $\mathcal{O}$ is closed with respect to arbitrary unions," is true in the first approach but false in the second. In some sense, we've "committed" to a particular definition of a topological space; and, in doing so, made a decision that actually effects the end result.

This is bad, because we tend to write/talk about math in approach-independent ways. For instance, we're more likely to write, "Let $X$ denote a topological space and $\mathcal{K}$ denote the set of all closed sets of $X$," as compared to, "Let $(X,\mathcal{K})$ denote a topological space." Even though the latter is more terse, nonetheless the former is clearly better, because its meaning is independent of the details of how a topological space was defined.

So here's what I'm looking for.

Ideally, a full solution would consist of a better way of defining structures, such that we can avoid "committing" to any particular definition of a topological space, ordered set, group etc. We would continue making statements like, "A topological space is determined by its open sets," but would never make a statement like, "A topological space is an ordered pair $(X,\mathcal{O})$ such that..." Thus, the path wherein you define the term "topological space" via open sets would eventually meet up with the path wherein you define the term "topological space" via closed sets; and all sentences that are true in the open-set approach would be true in the closed-set approach, and vice versa.

Less ambitiously, a partial solution would show that, "It doesn't matter which way you do it, the resulting theory is essentially the same." I'll happily accept a category-theoretic solution to the problem, as long as I'm still able to do math in the classical way. For instance, concepts like "the underlying set of a topological space" and "the set of all clopen sets" need to continue making sense. In particular, defining things up to isomorphism (i.e. renaming of the underlying set) is not an option.

  • $\begingroup$ Your notion of path-dependence seems ill defined. For example you could have taken your sentence to be "For all topological spaces $(X, \mathcal O)$, it holds that the class of open sets is closed with respect to arbitrary unions". I also am not sure what you mean when you say that there is no theorem we can prove that would have the two approaches coincide, because those are absolutely equivalent definitions of a topology. $\endgroup$ – Jim Feb 14 '13 at 5:10
  • $\begingroup$ Yes but that would be a different sentence... the point is, there exists a sentence (eg, the one I wrote) that has a different truthvalue in one approach versus the other. $\endgroup$ – goblin Feb 14 '13 at 5:19
  • $\begingroup$ Your example is misleading. It just so happens that the two axiomatisations of topological spaces you chose have compatible signatures, in the sense that they are both of the form (set, subset of power set). Of course things go pear-shaped if you interpret "open" in the first axiomatisation to mean "closed" in the second axiomatisation. So I say the problem you bring up is spurious. $\endgroup$ – Zhen Lin Feb 14 '13 at 22:49
  • $\begingroup$ Even if they had different signatures, the problem persists. For instance, groups can be defined as ordered pairs $(G, *)$. Alternatively, as ordered triples $(G,*,e)$. The sentence "Every group is an ordered triple," is false in the first approach but true in the second. What I want is to be able to make statements like, "A group is determined by its underlying set and binary operation," without having to commit to any particular notion of what a group "is". $\endgroup$ – goblin Feb 14 '13 at 23:18
  • $\begingroup$ For some reason my comment above isnt displaying correctly; so just let me know if i need to clarify. $\endgroup$ – goblin Feb 14 '13 at 23:25

Notice that what you are really asking for is comparing two classes of structures and not two particular instances of these structures (judging from the example you gave regarding different definitions of topological spaces).

If the two structures are given by theories $T_1,T_2$ over the same language then you can say that the theories are essentially the same if $T_i$ implies each sentence in $T_j$, for $i,j\in \{1,2\}$. If the two theories are over different languages, say $L_1, L_2$ respectively, then it's a bit more difficult. You might try to first find a common refinement of the two languages and proceed as above. For instance, if you axiomatize groups using the language $\{\cdot, e\}$ by letting $T_1$ be any usual axiomatization and by letting $T_2$ be that usual axiomatization augmented with infinitely many new sentences describing generalized associativity then you can quite easily prove the two theories are essentially the same.

You will run into a bit more difficulty if you try to compare heaps and groups and you will run into a hell of a lot of difficulty if you try to compare the theory of the homotopy of topological spaces and the theory of simplicial sets. These are all cases where the two structures are essentially the same but where working with a theory describing them is quite hopeless (especially the second example).

You seem not to want a categorical solution to the problem, though it is the most natural one. Quite often any definition of structure comes with a definition of morphism. Then collecting these together almost always results in a category (or a weak $\infty $-category). Now, you can say that the two structures are essentially the same if the categories are essentially the same. The latter can be taken to mean anything you want, depending on how much flexibility you want to allow. For instance, requiring the resulting categories to be isomorphic means that the structures are essentially the same in a much stronger sense then requiring the categories to just be equivalent.

In mathematics the relevant notion of categories being essentially the same is categorical equivalence. Given the (huge) class of all categories you can define an equivalence relation on categories by categorical equivalence. Now, you can consider each equivalence class to be the purest form of the definition of a mathematical concept. For instance, the equivalence class of the category $Grp$ of groups (defined in the usual way) includes all equivalent categories, no matter what axiomatization is used to give rise to them. To use the words of your own question, you can think of the particular representatives of an equivalence class to be the different paths that lead to the pure abstract concept, and the equivalence class as being the actual concept being defined. Then independence of paths holds and any particular axiomatization is just a way to describe (indirectly) an equivalence class of categories. A direct, constructive characterization of the equivalence of the resulting categories just in terms of the defining theories is quite hopeless.

  • $\begingroup$ I have no issue with a categorical solution, which is why I put "category-theory" in the tags. But your "categorical solution" is awfully vague, to the point where I don't know whether my problem is solved or not. For instance, suppose instead of axiomatizing the term "magma," I instead axiomatize the phrase, "magma category." A magma is then (a somewhat meaningless name for) an object of any such category. Now given any magma $M$ that is an object of a magma category $C$, is it possible to define, "The set of all left-cancellative elements of $M$"? Or, "The underlying set of $M$?" $\endgroup$ – goblin Feb 14 '13 at 6:12
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    $\begingroup$ Yes, you can express such properties of magmas purely within any representative of the equivalence class of the magama category. For instance, a magma with one object is characterized as a terminal magma. The underlying set of a magma is the set of morphisms from the terminal magma to the given magma. Existence of free magmas allow one to identify a choice of an element in a magma as a morphism from the free magma on one letter to the given magma. And so on ... $\endgroup$ – Ittay Weiss Feb 14 '13 at 8:00
  • $\begingroup$ Does this work for the class of structures satisfying any conceivable axiom system? For instance, I don't think it would work for groups.... $\endgroup$ – goblin Feb 14 '13 at 8:14
  • $\begingroup$ to be answered properly 'this' and 'conceivable axiom system' will have to be defined properly. $\endgroup$ – Ittay Weiss Feb 14 '13 at 8:17
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    $\begingroup$ That might actually solve the problem. Thanks for the reference. $\endgroup$ – goblin Feb 14 '13 at 8:35

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