I wrote:

More precisely, I claim that the natural way to build up the set of such maps uses 2 vertices (0-cells), 6 edges (1-cells), 18 faces (2-cells), 54 3-cells, etc.

I forgot to give the details of this. The (open) k-cells consist of all the polyhedral functions from (0,1) to {0,1} with exactly k points of discontinuity. There are only two polyhedral functions from (0,1) to {0,1} with 0 points of discontinuity, namely the constant function 0 and the constant function 1. These are the two 0-cells. The functions with 1 point of discontinuity are of the form { a if t < t_1 f(t) = { b if t = t_1 { c if t > t_1 where t_1 is some element of (0,1) and a,b,c are elements of {0,1} not all equal to one another. There are 2^3-2=6 ways to choose a, b, and c, and for each such choice, we get a 1-cell isomorphic to (0,1). So we have 6 1-cells. The functions with 2 points of discontinuity are of the form { a if t < t_1 { b if t = t_1 f(t) = { c if t_1 < t < t_2 { d if t = t_2 { e if t > t_2 where t_1 < t_2 are elements of (0,1) and a,b,c,d,e are elements of {0,1}, with a,b,c not all equal to one another and c,d,e not all equal to one another. There are 18 ways to choose a, b, c, d, and e, and for each such choice, we get a 2-cell isomorphic to the open triangle {(t_1,t_2): 0 < t_1 < t_2 < 1}. So we have 18 2-cells. Etc. I want to add that my proposed method of computing an Euler measure for infinite dimensional polytopes of this kind is not accompanied by any supporting geometrical ideas. It's just a system of formal noodling that points to the possibility of extending the definition of Euler measure beyond where it has gone before (and perhaps beyond where it ought to go). Jim Propp