From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/1313 Path: news.gmane.org!not-for-mail From: Peter Freyd Newsgroups: gmane.science.mathematics.categories Subject: Pre-calculus Date: Fri, 31 Dec 1999 17:54:56 -0500 (EST) Message-ID: <199912312254.RAA21550@saul.cis.upenn.edu> NNTP-Posting-Host: main.gmane.org X-Trace: ger.gmane.org 1241017744 30679 80.91.229.2 (29 Apr 2009 15:09:04 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:09:04 +0000 (UTC) To: categories@mta.ca Original-X-From: rrosebru@mta.ca Fri Dec 31 21:57:31 1999 -0400 Original-Received: (from Majordom@localhost) by mailserv.mta.ca (8.9.3/8.9.3) id UAA31833 for categories-list; Fri, 31 Dec 1999 20:14:46 -0400 (AST) X-Authentication-Warning: mailserv.mta.ca: Majordom set sender to cat-dist@mta.ca using -f Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 22 Original-Lines: 61 Xref: news.gmane.org gmane.science.mathematics.categories:1313 Archived-At: Starting with the closed interval, regarded as an ordered midpoint algebra with duality, what's the best way to get the reals? My previous answer was first to reflect the midpoint algebra in the subcategory of abelian groups and then define the multiplication via monotonic endomorphisms. I think there's a more natural way. First, a re-cap of the definitions. A midpoint operation is a binary operation, with values here denoted as x|y, satisfying: idempotence, x|x = x; commutativity, x|y = y|x; middle-two-interchange, (u|v)|(x|y) = (u|x)|(v|y); and cancelation, x|y = x|z => y = z. The closed interval is a totally ordered midpoint algebra with anti-involution, with values here denoted as -x. It follows that there's a unique element of the form (-x)|x. We'll call it the center. The middle-two-interchange law is just what is needed to see that the monoid of midpoint-endomorphisms is itself a midpoint algebra (where the midpoint of f and g is defined pointwise: (f|g)(x) = (fx)|(gx)). If f and g are both order-preserving or both order- reversing then it's easy to see that so is f|g. But we'll want all the _monotonic_ endomorphisms (the OED defines "monotonic" as "varying in such a way that it either never increases or never decreases") and there's no apparent reason to expect that monotonicity is preserved by midpointing. The wonderful fact, though, is: Midpoint-preserving functions between intervals are monotonic. (The axiom of choice yields 2^{2^N} counterexamples if the interval is replaced by the reals.) It follows that a midpoint-preserving function is determined by its values on any two points. The monoid of midpoint- and duality- preserving endo-functions on a closed interval is naturally isomorphic (via evaluation at Top) to the closed interval. For the entire reals use the stalk at the center of the germs of such functions. PS For a proof suppose that f:[-1,1] --> [-K,K] preserves midpoints and duality. For any x:[-1,1] and dyadic rational q such that qx:[-1,1], we may uniquely identify qx using just midpoints and duality, hence f(qx) = qf(x). Suppose that f is not monotonic. We wish to reach a contradiction. Let u < v and x < y in [-1,1] be such that fu < fv and fx > fy. Define a = (-u)|v and b = (-x)|y to obtain a,b > 0 and fb < 0 < fa. Let q > 0 be a dyadic rational such that a/b - (fa)/(bK) < q < a/b and define c = (-qb)|a. Let r be a dyadic rational such that (2K)/(fa) < r < 1/c. Then rc is in [-1,1] but f(rc) can not be in [-K,K] (because fc > (fa)/2). The theorem I stated doesn't say anything about preserving duality. So let g be a midpoint-preserving function between intervals. Let a be the dual of the value of g at the center. Then the function \x.a|(gx) preserves the center and continues to preserve midpoints. Since -x may be uniquely identified by (-x)|x = 0 the preservation of the center implies preservation of the duality and \x.a|(gx) is monotonic. Since \y.a|y not only preserves but reflects the order, we know that g is monotonic.