From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/2121 Path: news.gmane.org!not-for-mail From: Dusko Pavlovic Newsgroups: gmane.science.mathematics.categories Subject: Re: Cauchy completeness of Cauchy reals Date: Fri, 24 Jan 2003 08:56:29 -0800 Message-ID: <3E31703D.9030801@kestrel.edu> References: NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii; format=flowed Content-Transfer-Encoding: 7bit X-Trace: ger.gmane.org 1241018424 2579 80.91.229.2 (29 Apr 2009 15:20:24 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:20:24 +0000 (UTC) To: CATEGORIES LIST Original-X-From: rrosebru@mta.ca Fri Jan 24 14:00:54 2003 -0400 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Fri, 24 Jan 2003 14:00:54 -0400 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.10) id 18c87R-0005Kt-00 for categories-list@mta.ca; Fri, 24 Jan 2003 13:59:57 -0400 User-Agent: Mozilla/5.0 (X11; U; Linux ppc; en-US; rv:0.9.9) Gecko/20020604 X-Accept-Language: en-us, en Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 46 Original-Lines: 61 Xref: news.gmane.org gmane.science.mathematics.categories:2121 Archived-At: Andrej Bauer wrote: >There seems to be an open question in regard to this, advertised by >Alex Simpson and Martin Escardo: find a topos in which Cauchy reals >are not Cauchy complete (i.e., not every Cauchy sequence of reals has >a limit). For extra credit, make it so that the Cauchy completion of >Cauchy reals is strictly smaller than the Dedekind reals. > this will give me negative credits one way or another, but here it goes: let a = (a_i) be a cauchy sequence of rationals between 0 and 1. (cauchy means |a_m - a_n| < 1/m + 1/n, as in andrej's message.) let b = (b_i) be the subsequence b_i = a_{2^i+3}. b and a are equivalent in the sense from the message, because |a_m - b_n| < 1/m + 1/(2^n+3) < 2/m + 2/n note that |b_i - b_{i+k}| < 1/(2^i+3) + 1/(2^(i+k)+3) < 1/(2^i+2) now define x_i to be the simplest dyadic that falls in the interval between b_i - 1/2^(i+2) and b_i + 1/2^(i+2). in other words, to get x_i, begin adding 1/2 + 1/4 + 1/8... until you overshoot b_i - 1/2^(1+2). if you also overshoot b_i + 1/2^(1+2), and the last summand was 1/2^k, skip it, and try adding 1/2^(k+1), etc. one of them must fall in between. a less childish way to say this is that x_i is the shortest irredundant binary (no infinite sequences of 1) such that |b_i - x_i| < 1/2^(i+2) so x = (x_i) is a cauchy sequence equivalent to b, with |x_i - x_{i+k}| <= |x_i-b_i| + | b_i - b_{i+k}| + |b_{i+k} - x_{i+k}| < < 1/2^(i+2) + 1/2^(i+2) + 1/2^(i+k+2) < < 1/2^i this means that the first i digits of x_i and x_{i+k} coincide. now let X be the binary number such that its first i digits are the same as in x_i, for every i. (if it ends on an infinte sequence of 1s, replace it by the corresponding irredundant representative.) this is clearly a constant cauchy sequence equivalent to x, so it is its limit. since x is equivalent to b |b_m - x_n| <= |b_m - b_n| + |b_n - x_n| < 1/2^(m+3) + 1/2^(n+3) + 1/2^(n+2) < 2/m + 2/n and b is equivalent to a, X is also the limit of a. is there an error in the above reasoning? i can't find it. on the other hand, i printed out the paper by alex and martin, and the conjecture is stated rather strongly, so i guess i must be missing something. -- dusko