[ Jargon | {(A::A)} from (set:function:set) and set as {members} | Set - a category ]
We can describe any {(A::A)} as an arrow world in (at least) two ways: one allows (f,g) to be composable precisely if (g|) is a subset of (|f); the other allows arbitrary (f,g) composable. Each has the same composite, f o g: the restricted composition (that of Set), given (g|) a subset of (|f), gives (|fog)=(|g); composition is only allowed when the composite will accept every input that g would.
The utterly liberal composition of Partial, on the other hand, merely has (|fog) a subset of (|g) with no guarantee of equality - indeed, many composable pairs have empty composite simply because the range of one does not even intersect the domain of the other. None the less, the empty set, {}, is a subset of every set, thus of every Cartesian product of sets: it is consequently a partial function (there is nothing else to check) between any pair of sets.
As defined elsewhere, one set, S, is described as a subset of another, T, precisely when every member of S is a member of T. I now pause to look at the structure induced on functions and partial functions by the subset relationship.
For any set A, consider the subsets of {(a,a): a in A}. Each of these is a partial function (A::A): we can think of them as the definitive embeddings of the subsets of A in A. The composite of any two is simply the definitive embedding of the intersection of the subsets they severally embed. Likewise, the union of any two is the definitive embedding of the union of the subsets severally embedded.
The strict embeddings (ie {f= (A: i->i :A): (|f) is a proper subset of (f|)}) form a strict partial ordering. Combined with the identities on subsets of A, they form a partial order category. One of the characteristics of Set is that any collection of subsets of some set has union and intersection among the subsets of the given set.
If we only consider some subsets of a set, their embeddings may still form a partial ordering (strict if their identities are omitted): we can ask for unions of valid subsets to be valid, or for their unions - but we need not require both together in all cases; we can restrict such discussion to finite unions or intersections (independently) or extend it arbitrarily. For reasons which I'll try to explain in discussions of science, I am particularly interested in collections of subsets which admit rules of kind `whenever two members of the collection have intersection also in the collection, so is their union' with a further constraint that the measures (probabilities associated with) the two thus-intersected collections sum to the sum of the measures of their union and intersection.
A topology on a set S is defined to be a collection, open(S), of subsets of S, with S and empty as members, which is closed under finite intersection and arbitrary union. A co-topology on a set S is defined to be a collection, compact(S), of subsets of S, with S and empty as members, which is closed under finite union and arbitrary intersection. Any open(S) supplies us with a collection of subsets, H, of S for which: for any subset T of open(S) whose union has H as a subset, the union of some finite subset of T also has H as a subset. With a little work, one can show that these form a co-topology of S. I suspect I want to shed empty from compact. I would sooner stipulate that every point-set (or range of a post-terminal arrow) is a member of compact(S), and that the union of all such is S, which need not be compact (it might not be a finite union, after all). In Set, the domain of any pre-initial arrow is empty, as opposed to the post-terminal arrows having `point' ranges, with union the entire set. So the topology, open(S), has every domain of a pre-initial arrow as a member: in Set, that repeatedly accepts {} as a member. So we have point-sets compact and empty open, with the union of all opens (which is necessarily open) being S and the intersection of all compacts being compact and, typically, empty. What gets interesting is when we deal with a world in which the intersection is undecidable - we could perhaps derive an absurdity from any putative member, but yet not be able to prove that it is empty - a question which arises naturally once we start abandoning reduction ad absurdum. Describe a subset of S as compact (with respect to this topology) precisely if: whenever the union of (:f:open(S)) has C as a subset, there is some finite H for which the union of (H:f:) has C as a subset. Thus open(S) implies the collection, compact(S), of compact subsets of S: this, in turn, has empty as a member and is closed under finite union and arbitrary intersection. One can, in fact, define topology in terms of compact, deriving open from it.
Written by Eddy.