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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
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