From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/4339 Path: news.gmane.org!not-for-mail From: Colin McLarty Newsgroups: gmane.science.mathematics.categories Subject: Re: question to Colin about uniqueness in his Replacement axiom Date: Tue, 18 Mar 2008 18:28:28 -0400 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 1241019881 12819 80.91.229.2 (29 Apr 2009 15:44:41 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:44:41 +0000 (UTC) To: categories@mta.ca Original-X-From: rrosebru@mta.ca Wed Mar 19 15:18:54 2008 -0300 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Wed, 19 Mar 2008 15:18:54 -0300 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1Jc2ld-0000xc-JD for categories-list@mta.ca; Wed, 19 Mar 2008 15:12:01 -0300 Content-Language: en Content-Disposition: inline Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 103 Original-Lines: 31 Xref: news.gmane.org gmane.science.mathematics.categories:4339 Archived-At: Thomas Streicher Tuesday, March 18, 2008 6:09 pm Wrote: > In your Replacement axiom (p.48 of your "Philosophia" article) you > psotultathe existence of a map f : S -> A such that S_x \cong x^*f > for all x : 1->X. > Can you prove that this f is unique up to isomorphism, i.e. that > wellpointedness for maps entails wellpointedness for families? Sure. It takes the axiom of choice of course, since without choice the result may be false (even two countably infinite families of countably infinite sets need not be isomorphic). It is the obvious argument by Zorn's lemma, which follows from choice: Given two families S-->A and S'-->A with corresponding fibers isomorphic, consider the set of all pairs with U a subset of A, and i an isomorphism over U from the restriction of S to the restriction of S'. By Zorn at least one of these is maximal (for the obvious ordering by inclusion) so call it . Since well-pointedness implies Boolean, U has a complement in A--which must be empty or else we could extend the isomorphism. best, Colin