gross@spp2.UUCP (Howard E. Gross) (02/22/85)
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?
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gross (Howard Gross) {decvax,hplabs,ihnp4,sdcrdcf}!trwrb!trwspp!spp2meister@faron.UUCP (Philip W. Servita) (03/01/85)
>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? ---------------------------------------------------------------------
pupa@entropy.UUCP (Marek Rychlik) (03/12/85)
> 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.