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From: Toby Bartels <categories@tobybartels.name>
To: categories@mta.ca
Subject: Re: Uniform locales in Shv(X)
Date: Thu, 28 Aug 2014 10:10:13 -0700	[thread overview]
Message-ID: <E1XNA80-0006AW-1v@mlist.mta.ca> (raw)
In-Reply-To: <E1XMeQW-0007GO-Kx@mlist.mta.ca>

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


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  reply	other threads:[~2014-08-28 17:10 UTC|newest]

Thread overview: 9+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
     [not found] <d8a1028f315c96d7f569314dddff0e83.squirrel@www.normalesup.org>
2014-08-22  2:24 ` Jeff Egger
2014-08-22 15:15   ` Toby Bartels
2014-08-25 15:40     ` Jeff Egger
2014-08-25 20:24       ` henry
2014-08-27  5:05       ` Toby Bartels
2014-08-28 17:10         ` Toby Bartels [this message]
2014-08-29 15:29 Giovanni Curi
  -- strict thread matches above, loose matches on Subject: below --
2014-08-20 23:25 Jeff Egger
2014-08-21 14:51 ` henry

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