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From: pupa@entropy.UUCP (Marek Rychlik)
Newsgroups: net.math
Subject: Re: weird functions
Message-ID: <154@entropy.UUCP>
Date: Mon, 11-Mar-85 23:51:38 EST
Article-I.D.: entropy.154
Posted: Mon Mar 11 23:51:38 1985
Date-Received: Wed, 13-Mar-85 00:44:51 EST
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Organization: UW MathStat, Seattle
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> A slight modification of the function I posted:        
> 
> 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

	This function is continuous at all irrationals and discontinuous
at the rationals.
Proof. Fix eps>0 and an irrational x0. 
The set of those rational x that g(x)<1/eps is finite.
Therefore, there is delta>0 such that for every rational x with the property
|x-x0|<delta we have g(x)>1/eps. This implies f(x)<eps and f is continuous at
x0. 
	It is discontinuous at x0 rational, since 
1) f(x0)>0 
2) x0 is a limit of irrationals, where f has value 0.