Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site faron.UUCP Path: utzoo!watmath!clyde!burl!ulysses!allegra!bellcore!decvax!wanginst!vaxine!encore!linus!faron!meister From: meister@faron.UUCP (Philip W. Servita) Newsgroups: net.math Subject: Re: weird functions Message-ID: <244@faron.UUCP> Date: Fri, 1-Mar-85 09:07:10 EST Article-I.D.: faron.244 Posted: Fri Mar 1 09:07:10 1985 Date-Received: Mon, 4-Mar-85 08:23:05 EST References: <445@spp2.UUCP> Reply-To: meister@faron.UUCP (Philip W. Servita) Distribution: net Organization: The MITRE Coporation, Bedford, MA Lines: 42 >Let g be a one-to-one map of the rationals to the positive integers. >Let f(x) = 1/g(x) if x is rational, f(x) = 0 otherwise. >Where is f continuous? > > gross (Howard Gross) {decvax,hplabs,ihnp4,sdcrdcf}!trwrb!trwspp!spp2 i dont think this changes the original problem at all,(the original problem was f(x) = 1/q if x rational, lowest form p/q, f(x) = 0 for x irrational) if we take a sequence X(i) of rationals s.t. lim X(i) = m (m irrational) i->inf define F(x) = f(x) x irrational, 1/f(x) x rational then since g(x) is one-to-one we are guaranteed that for each n in N(natural numbers) there exists k in N s.t. F(X(h)) > n for all h > k (h in N) (this step left to the reader) now by the archimedean principle, we can move to: for each d > 0 there exists k in N st f(X(h)) < d for all h > k, h in N so by the cauchy criterion f is continuous at each irrational #. NOW: let Z(i) be a sequence of irrationals s.t. lim Z(i) = m (m rational,= p/q) i->inf it is obvious that lim f(Z(i)) = 0 as each of the Z(i) is irrational. i->inf hence this limit is not = 1/q and f is discontinous at each rational #. this argument is practically the same as the one used for the original function, and should hold even if g(x) is an n-to-1 function, as long as n is some predetermined finite #. -the venn buddhist -- --------------------------------------------------------------------- is anything really trash before you throw it away? ---------------------------------------------------------------------