Ernie Manes wrote:
I don't think anything goes wrong if AC fails as far as establishing a distributive law is concerned. No doubt much further discussion would be needed to understand well what happens, precisely, but note that many AA (e.g. principal filters) have non-empty C(AA). In any case, a sub-example obtains replacing P with the finite subsets monad, in which case no AC is needed.
Ernie
Dear Ernie, Here is why I think the distributive law goes wrong if AC fails. The original definition was this:
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In a paper under preparation on distributive laws, joint with Philip Mulry, we show that the families monad arises by a distributive law of the power set monad P over itself. To describe that law, first establish some notation as follows. For AA in PPX a family of subsets of X, let C(AA) be the cartesian product of AA, that is, the set of all choice families x = (x_A : A in AA) with x_A in A. Let I(x) be the image of x, i.e. {x_A : A in AA}. The distributive law in question, PPA -> PPA, maps AA to {I(x) : x \in C(AA)}. <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< One of the conditions of a distributivity is that the following diagram commutes: PPPX ---- P mu X ------> PPX | | sigma PX | | | v | PPPX sigma | | P sigma X | | | v v PPPX ---- mu P X ------> PPX where mu is the multiplication and sigma is the distributivity. (Hope the format was clear. Down the left we have three copies of PPPX, with vertical morphisms involving sigma and P. Down the left are two copies of PPX, with vertical morphism sigma. Across are two horizontal morphisms involving mu and P.) Start from AAA at top left. Going across and down gives {I(x) | x in C{U(AA)) | AA in AAA}} (1) where U is union (i.e. mu). Going down and across gives {I(z) | (exists y in C(AAA)) z in C(I(y))} (2) Suppose we have such an x in C{U(AA)) | AA in AAA}. We want to find corresponding y and z to demonstrate that I(x) is in (2). Now for each AA in AAA, x chooses an element x_U(AA) in U(AA) and so there is some A in AA with x_U(AA) in A. We'd like our y in C(AAA) to have y_AA as such an A, but unfortunately A is not uniquely defined and so we need the axiom of choice to find such a y. Similar uses of choice appear in other places, so that is why it appears to me that the distributive law depends on AC. If we replace P by a finite powerset F, then we have a separate question: How do we know, if AA is in FFX, that sigma(AA) is finite? This depends on the constructive notion of finiteness. In my work I was using Kuratowski finiteness (for which the finite powerset monad is the free semilattice monad), and for that we find that C(AA) is not finite in general - which is why I replaced it with Ch(AA), using total choice relations instead of choice functions. But that definition of sigma does not give a distributivity even classically. All the best, Steve. 17-Feb-2005 15:44:28 -0400,1085;000000000000-00000000