Hi, My point about the entourage approach was that it does not require overtness at least for the basic definitions: For example, you can state the axiom "for all entourage a there exists an entourage c such that c.c <= a$ as" using the following form instead of the composition of entourage: pi_12 ^* c (intersection) pi_23^* c <= pi_13^*a where pi_12, pi_23 and pi_13 denotes the three projection from X^3 to X^2. the uniformly below relation can also be defined as: a is uniformly below b if there exist an entourage c such that pi_1^* a (intersection) c <= pi_2^* b and hence you can state the admissibility axiom without overtness. The absence of overtness still yields a lots of problems: for example the uniformly below relation is no longer interpolative. (and I agree that it is not clear at all that this is the good way of doing things) When focusing on overt space everything work properly and one can defines for example completeness and completion (it is actually a direct consequence of the results about localic metric space in my thesis) but at some point it yields other problems related to the fact that subspaces of uniform spaces are no longer uniform space and this gives examples of things that should be uniform spaces but which are not overt. If you are willing to restrict to open maps then using the entourage approach give a not to awful answer to your initial question: a relative uniform structure on an open map f:Y ->X, is the given by a collection of a family of open sublocales of Y \times_X Y which correspond to open sublocales of the form {y_1,y_2,x | x \in U, (y_1,y_2) \in S(x) } for U an open subset of X and S a section over U of the sheaf of entourage of Y in sh(X). You just need to write down all the axioms that these things has to satisfy but they are going to be relatively natural (and if you are only interested to a notion of "basis of entourage" they I think they are not going to be to complicated) Cheers, Simon

Hi Toby,?

Many thanks for your comments. ?

Over the weekend, I noticed a (probable) mistake in my previous posting: namely, the assertion that _only_ the admissibility axiom could require overtness of the locale in question. ?In fact, the natural way to define composition of entourages (qua relations) is to apply first the inverse image of? ? 1xDeltax1 : LxLxL ---> LxLxLxL,? and then the open image of? ? 1x!x1: LxLxL ---> LxL. ? Thus the intuitive way of writing down the square-refinement axiom also requires L to be overt. ?(Btw, when discussing that axiom in my previous post, I carelessly wrote "epi" where I obviously meant "cofinal".) ?Perhaps there is some other way of expressing that axiom, but I no longer care: I've reconciled myself to the idea that uniformity requires overtness.

I also made some progress on?

...

...

Perhaps, if we define a Weil pseudo-uniformity on L to be exactly as above minus the admissibility axiom, it is possible---at least in the case of a localic topos E=Shv(X)---to describe a second locale underlying the pseudouniformity? If so, then we don't really need formulate the admissibility axiom in the internal language of the topos.

Namely, I have an idea of how this might be done, but it requires a detour through gauge spaces (=uniformities defined via families of pseudometrics, for those not familiar with the terminology) which I'm not sure is constructively valid. ?In any case, I like this idea: in particular, if it works, then one could apply it to the case where L is a discrete locale and investigate to what extent the idea of a uniform space as a _set_ together with a family of entourages fails to capture the general idea. ?

Cheers, Jeff.

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I wrote in part:

I don't know what to do about the dependent choice, however.

Now I do! Generalize from gauge spaces to prometric spaces. There is a very precise analogy as follows: equivalence relation : uniformity :: pseudometric : prometric. So just as an entourage need not be transitive, but instead there is a transitivity condition on the entire family, so a distance map in a prometric need not satisfy the triangle inequality, but instead a triangle inequality is imposed on the entire family. A gauge generates a prometric, but not every prometric arises thus. Then it's straightforward to get the uniformity underlying a prometric, and also to get a universal gauge or universal prometric from a uniformity. The question is whether this gives you back your original uniformity, and this in turn hinges on whether there are enough distance maps (which in the case of the gauge are pseudometrics) in the prometric. Specifically, given an entourage U, we need a natural number n and a uniformly continuous d such that {x,y | d(x,y) < 1/n} is in U. An argument involving dependent choice and infima says yes, for the gauge. For the prometric, we have a simpler argument, using only infima. This argument works internally to any topos with a natural-numbers object as long as the distance maps in our prometrics are valued in upper reals. Specifically, given the entourage U, define d_U as follows: d_U(x,y) := inf {t | t = 0 and (x,y) in U}. That is, d_U(x,y) = 0 iff (x,y) in U, and d_U(x,y) = infinity iff (x,y) not in U. (If you don't want to allow infinite distances, use a finite upper bound; this gives a different prometric but one with the same uniformity. You can even get the same prometric by using an unbounded sequence of bounds.) This violates the triangle inequality whenever U is not transitive, but d_U(x,z) <= d_V(x,y) + d_V(y,z) when V . V <= U, so we have a prometric if we start with a uniformity. So anyway, if you use prometrics valued in upper real numbers, then every uniformity is prometrizable, even if not gauge-able. This is all for pointwise spaces, but the construction of d_U should also work for a uniform cover in a uniform locale. I have not checked this carefully, however. --Toby [For admin and other information see: http://www.mta.ca/~cat-dist/ ]