From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/4807 Path: news.gmane.org!not-for-mail From: Michael Barr Newsgroups: gmane.science.mathematics.categories Subject: Re: terminology in definitions of limits Date: Wed, 21 Jan 2009 20:47:04 -0500 (EST) Message-ID: Reply-To: Michael Barr NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed X-Trace: ger.gmane.org 1241020185 14924 80.91.229.2 (29 Apr 2009 15:49:45 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:49:45 +0000 (UTC) To: Charles Wells , catbb Original-X-From: rrosebru@mta.ca Thu Jan 22 22:27:34 2009 -0400 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Thu, 22 Jan 2009 22:27:34 -0400 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1LQBes-0002K0-5Z for categories-list@mta.ca; Thu, 22 Jan 2009 22:20:34 -0400 Original-Sender: categories@mta.ca Precedence: bulk X-Keywords: X-UID: 28 Original-Lines: 27 Xref: news.gmane.org gmane.science.mathematics.categories:4807 Archived-At: This is getting peripheral to the main point. AS far as I recall, I thought of T as a test object. As for epsilon-delta, Bishop required that delta be prescribed as a constructible function of epsilon in order that a function be continuous. He required that the convergence be uniform on every closed interval, so that this function on a closed interval was independent of the points in the interval. Michael On Wed, 21 Jan 2009, Charles Wells wrote: > Calculus teachers do something similar when they make an epsilon-delta proof > into a game: The opponent picks an epsilon (the test object) and you have > to come up with a delta. > There is one big difference between epsilon-delta proofs and limits. To > show that something is a limit you have to find, for each test object, the > unique arrow specified by the definition of limit. Thus you are producing a > function (indeed, a bijection). The delta for a given epsilon is not unique, > and so there is no natural function giving a delta for each epsilon. I am > pretty sure this makes epsilon-delta proofs harder for non-talented students > than proving something is a limit. I know some calculus teachers talk about > there being a function that takes epsilon to delta, but I suspect it is a > mistake to bring that up. > > Charles Wells >