The continuum is intrinsically geometrical, so it is a shame to specify Peter's midpoint coalgebra on IxI with a mess of algebraic-looking equations as per my previous message when it has a relatively short and transparent geometric specification. (I must have got this idea from the paper on higher dimensional automata that I'm just finishing up.) Incidentally there were two ambiguities in my equations, arising from respectively equations (1) and (5). I was going to offer fixes for these, but one of the degeneracies that made these ambiguities possible in the first place (that with (5)) does not even arise in the geometric presentation, while the other, arising with (1), is just as readily dealt with geometrically as algebraically. In the following I'll refer to this coalgebra as Fm for Freyd's midpoint (unrelated to Scott's bottom). (For listeners who just tuned in to Fm, its role is to promote the final coalgebra I of Peter's fusion functor X v X from a linear order, and hence a topological space via the order topology, to a metric space understood as the real interval [-1,1]. It does this by furnishing IxI with a coalgebra, which determines a unique coalgebra homomorphism from IxI to I, I being the final coalgebra. This homomorphism is then taken to be (x+y)/2, while the endpoints of I are taken to be -1 and 1, thereby uniquely determining a continuous metric on I qua topological space.) For both the equational and geometric specifications, here's what the domain and codomain of Fm look like. (1,1) IxI ^ / \ / \ (1,1) (1,-1)< 1 >(-1,1) /\ \ / / \ Fm \ / (1,-1)< >(-1,1) -------------> 0 IxI v IxI \ / / \ \/ / \ (-1,-1) (1,-1)< -1 >(-1,1) \ / \ / V (-1,-1) So Fm maps pairs of the form (x,-x) to 0 (where the two copies of IxI are fused) and all other pairs (x,y) to a pair (h,(x',y')) whose head h specifies which copy of IxI to land in, 1 for upper and -1 for lower, and (x',y') gives the landing point within that copy. Here -1 <= x,y,x',y' <= 1, -x is defined by changing all the signs on the sequence representing x and h is unproblematically determined by whether (x,y) is in the upper or lower half of IxI oriented as shown. The tricky part is to specify x' and y' reasonably. (Although it is convenient to think of x and y as reals in [1,-1], at this stage they are merely infinite sequences over {-1,0,1} with the properties that once zero alway zero and infinite runs of -1 or 1 are only allowed in constant sequences.) The domain of Fm, namely I x I, can be blown up as (1,1) /\ / \ (1,0) (0,1) / \ / \ / \/ \ (1,-1)< (0,0) >(-1,1) \ /\ / \ / \ / (0,-1) (0,1) \ / \/ (-1,-1) In my previous mess-age, my five equations implicitly decomposed this domain into nine regions, namely the midpoint (0,0) (equation (2)), the four lines leading out of it (equation (3)), the upper and lower diamonds (equation (4)), and the left and right diamonds (equation (5)). For the geometric description of Fm, a simpler decomposition suffices, namely three regions: the upper and lower diamonds as one region, call it M for Middle, and the interior of each side diamond along with its outer sides, call them L and R. (So M is closed and L + R = IxI - M.) (1,1) /\ / \ . (1,0) (0,1) . / . \ / . \ / . \/ . \ (1,-1)< L . M (0,0) . R >(-1,1) \ . /\ . / \ . / \ . / . (0,-1) (0,1) . \ / \/ (-1,-1) In addition we need one more region, S for Shrunk, as a subregion of IxI v IxI, shown below. S is simply IxI v IxI shrunk in half (or pushed twice as far away). /\ / \ / \ / \ \ /\ / \/S \/ \ / \/ IxI v IxI /\ / \ /\ S/\ / \/ \ \ / \ / \ / \/ We now describe Fm as follows. Its restriction to M is the evident similarity between M and IxI v IxI. And its restriction to each of L and R lands in S, and in each case is the whole of Fm shrunk in half (the corecursion is essential, not even geometry can eliminate it). Note that these maps are defined on sequences, i.e. we are not assuming the metric we set out to construct. Fm as defined is not commutative, but can be made so by composing its restriction to L with commutativity of (x+y)/2, viz. reflection of L about its central vertical axis, thereby symmetrizing Fm. (This commutativity ambiguity also arose via equation (1) of my 5-equation approach.) This symmetrical Fm would appear to be the canonical choice for Freyd's midpoint coalgebra. (The apparent competitor obtained by reflecting R instead of L is slightly more discontinuous, if that's a reasonable tie breaker.) This coalgebra is of course not final, nor is it even injective, though it is surjective, witness subdomain M. It is also not continuous: points near (1,0) in L go to points near (1,(0,0)) in IxI v IxI, while points near (1,0) in M go to points near (1,(1,-1)) in IxI x IxI. On the other hand points near (1/2,0) in both M and L.M (the M of L) go to points near (1/2,0) in IxI v IxI, which has no counterpart in the above-mentioned competitor. Question. Is there a corecursively defined Freyd midpoint coalgebra that is continuous? (x,y) |--> ((x+y)/2,(x+y)/2) (projection onto the center axis) is a continuous midpoint coalgebra, but that assumes the metric before we've defined it. What corecursion would define this? Vaughan