From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/4638 Path: news.gmane.org!not-for-mail From: Martin Escardo Newsgroups: gmane.science.mathematics.categories Subject: Re: Group and abelian group objects in the category of Kelley spaces Date: Mon, 29 Sep 2008 00:10:56 +0100 Message-ID: NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit X-Trace: ger.gmane.org 1241020073 14151 80.91.229.2 (29 Apr 2009 15:47:53 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:47:53 +0000 (UTC) To: Michael Barr , categories@mta.ca Original-X-From: rrosebru@mta.ca Sun Sep 28 20:49:13 2008 -0300 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Sun, 28 Sep 2008 20:49:13 -0300 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1Kk5ua-0006Sg-Iw for categories-list@mta.ca; Sun, 28 Sep 2008 20:42:48 -0300 Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 134 Original-Lines: 56 Xref: news.gmane.org gmane.science.mathematics.categories:4638 Archived-At: It is interesting that an answer to this question, before it was asked here in the categories list, came up two weeks ago in a discussion I had with Jimmie Lawson and Matthias Schroeder. Schroeder showed recently that N^(N^N), where N is discrete and the exponential is calculated in k-spaces, is not regular, and hence not zero-dimensional either, which was an open problem (notice that the compact-open topology on N^(N^N) is easily seen to be zero-dimensional). Lawson observed that this is isomorphic to Z^(N^N), which, with the pointwise operations, is an abelian group in the category of k-spaces. This gives your counter-example. Lawson also said that counter-examples to complete regularity of k-groups where previously known among the experts in the subject, but were more complicated and/or artificial. (I don't know references.) I hope this helps. Martin Escardo Michael Barr writes: > I have not thought deeply on this, but it strikes me that the basic > problem is that such a group might not have a uniform topology. Such a > group will have, I think, a separately continuous multiplication and > hence, if U is a neighborhood of the identity {xU|x \in G} will be a > cover, but there would seem no obvious reason for it to have a > *-refinement. A continuous homomorphism would be uniformly continuous for > those covers, if they do form a uniformity, it seems to me. > > If only John Isbell were still around to answer this kind of question, a > wish I have wished many times since and well before his demise. But have > you looked in his uniform spaces book? That is the sort of thing he might > well have considered. If I were around the math library, I would look. > > Michael > > On Thu, 25 Sep 2008, Bill Rowan wrote: > > > Hi all, > > > > Does anyone know of a good place where someone has written down the basic > > properties of such objects? As an example, if we have an (abelian, say) > > topological group, there is a natural uniform topology on the group such > > that the operations are uniformly continuous. Does the same hold for > > abelian group objects in the category of Kelley spaces? But anything > > would be helpful. > > > > Bill Rowan > > > > > >