Paul Taylor wrote:
I observed that N is overt discrete Hausdorff not compact 2^N is compact Hausdorff not discrete overt which Toby attributed to the fact that N is the free algebra for +1 whereas 2^N is the cofree coalgebra for a functor that is not directly analogous.
As Toby noted, the functor is x2. However it is more than merely analogous, it is the dual of +1 when 2 is taken to be the dual of 1 (as with the original Stone duality for Boolean algebras, more generally distributive lattices, more generally yet Chu(Set,2)), product of course being the dual of sum.
ASD might make things clearer here. Its 1-level theory, like that of an elementary topos, is not self-dual.
By "1-level" do you mean first order? *-autonomous categories constitute a natural elementary notion of abstract Stone duality that is self-dual. What advantage of the elementary fragment of ASD over *-autonomous categories justifies its failure of duality? And how does the *-autonomous abstraction of Stone duality relate to ASD's abstraction of it? Toby Bartels wrote:
Perhaps the asymmetry is simply between initial algebras and final colagebras. One is a colimit and the other is a limit; there are already several asymmetries between these, such as that products distribute over sums but not vice versa. Indeed, if final coalgebras preserve "some"s, this might be more than just a bad pun.
In general 2 is not the dual of 1. In order to exhibit a duality between N and 2^N, they should be defined in a self-dual category that does make 2 the dual of 1. The most general such that I'm aware of is Chu(Set,2) --- Chu(Set,3) embeds Chu(Set,2) (in 3x2 = 6 ways) but it makes 3 the dual of 1. If you know of another category that makes 2 the dual of 1 that does not embed in Chu(Set,2) I'm all ears. In situations where 2 is not the dual of 1 it's not clear to me why one should expect 2^N to turn out to be dual to N in a way where all properties of one have their dual counterpart of the other; in Chu(Set,K) the natural functor for a final coalgebra dual to N is FX = XxK. In the context of duality the way to think about X in a Chu space (A,X,r) is that it is the dual of A, with the relation r characterizing the particular duality: the dual of (A,X,r) is (X,A,r^) where r^ is the converse of r. The first component constitutes the underlying set while the second can be thought of as a generalized frame of open sets, generalized by dispensing with the traditional frame operations of finite meets and arbitrary sups, with r simply the binary relation of membership of points in open sets. For this reason, absent alternatives to Chu(Set,2) there is no alternative to taking 1 to be the Chu space (1,2) if you want 2 to turn out to be the dual of 1. Note that in doing so 2 becomes the Chu space (2,1), the coarsest possible consistent topology on 2 (thinking of (2,0) as inconsistent). It should be distinguished from discrete 2 = 1+1, which is (2,4). Taking 1 in Chu(Set,2) to be the discrete Chu space (1,2), the initial algebra for +1 is (1,2) + (1,2) + ... = (N,2^N), 2^N being the power set of N, a complete atomic Boolean algebra. The final coalgebra for x2 when 2 = (2,1), as the dual of the initial algebra for +1, is the dual of (N,2^N), namely (2^N,N). Back to Paul:
The symmetry between => T /\ = some overt discrete free algebra and <= F \/ != all compact Hausdorff cofree coalgebra is very strong in this, but not perfect, because N is overt discrete Hausdorff not compact 2^N is compact Hausdorff not discrete overt I have not managed to isolate convincingly the precise point where the symmetry breaks down.
If ASD makes 2 the dual of 1 then it's suspiciously non-abstract. The only framework I know of for analyzing this duality is Chu(Set,K) for K = 2. Any larger K doesn't work, the concreteness of Chu(Set,3) etc. notwithstanding, unless you set things up so that the final coalgebra of the functor XxK is independent of the cardinality of K (which can be done as Bill Lawvere noted long ago, making 3^N equivalent to 2^N, but I'd be very impressed if ASD incorporated Bill's machinery). In Chu(Set,2), if one takes X in (A,X,r) to be the open sets, with r(a,x) membership of a in x, then (N,2^N,r) is discrete (the discrete Chu spaces over 2 being those of the form (A,2^A,r)), Hausdorff (by discreteness), not compact (because infinite and discrete), and overt (because spatial, Elephant C3.1.16). There are two ways of computing which of these four properties (2^N,N,r^) has, depending on whether one applies the elementary definitions of the properties to the "chupology" (taking r^ to be the converse of membership, i.e. a subset of N can "belong" to an element of N), or organizes 2^N as a topological Boolean algebra with the (Stone) topology serving to make it a CABA so that the continuous ultrafilters are all and only the elements of N, in order for the duality to work. Here are the two ways, alongside what Paul claims for comparison. Chupology not compact not Hausdorff not discrete ?~?overt? Stone compact Hausdorff not discrete overt Paul compact Hausdorff not discrete overt (The chupology is not compact because the whole space is not open, in fact {} is not "in" any natural number. It is not Hausdorff because no two opens are disjoint to begin with. And it's obviously not discrete. However unless there's a definition of "overt" meaningful for all chupologies I don't know what "overt" would mean for chupologies whose open sets don't form frames---they do in (N,2^N) but not in (2^N,N).) Interestingly we have perfect agreement between the usual Stone topology one imposes on a Boolean algebra to make it a CABA (to make the duality work) and the topology Paul has in mind for 2^N. Paul, is this because you have in mind the Stone topology, with 2^N obviously being spatial hence overt, or for some other reason? If the former then that is the "precise point where the symmetry breaks down:" you don't take the precise dual of this when dualizing back to (N,2^N), which you leave instead as the original Chu space (N,2^N). This is a perfectly fine discrete, Hausdorff, overt, but not compact space, again in perfect agreement with you. However comparing it with the topology of a topological Boolean algebra is apples to oranges. If the latter however then I have no explanation and the ball is back in your court. Vaughan Pratt