Could someone show me how to do the following proof?
Consider the set A such that A={p/(2^n) where p is a natural number, n is a natural number, and p/(2^n) is between 0 and 1 (not inclusive).
Let f(x)=0 if x is in A.
Let f(x)=1 if x is in [0,1] - A.
Prove that limit (as x approaches a) of f(x) does not exist when a is in [0,1].
Any help here would be appreciated.
The basic idea here is that every neighbourhood of 0 contain a root of
f(x)=1, and also a root of f(x)=0.
In which case there is no k in R such that given epsilon >0, there is
beta>0, such that when 0<x<beta
|f(x)-k|< epsilon
because every inteval [0,beta) contains points at which f(x)=1 and f(x)=0.
RonL
Easy is in the eye of the beholder. This can be hard not knowing the theorems you have to work with. There is one basic lemma that is important to this problem: Given any two numbers in [0,1] there is a dyadic rational number between them. Do you need to prove that? Also between any two numbers there is an irrational number. If and then the neighborhood of a, contains both a dyadic rational number and an irrational number in . Therefore, the values of at those numbers will differ be 1. So there can be no limit at .