Proving Dirichelet's Function Converges to 0: A Mathematical Proof

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SUMMARY

The discussion focuses on proving that Dirichlet's function converges to 0 as x approaches any point in the interval (0,1). The function is defined as f(x) = 0 for irrational x and f(x) = 1/q for rational x = p/q, where p and q are coprime integers. The key argument presented involves selecting a natural number n such that 1/n ≤ ε, which leads to the conclusion that only a finite number of rational numbers can exceed ε, thus demonstrating that the limit of f(x) as x approaches any point in (0,1) is indeed 0.

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Homework Statement


Consider the Dirichelet's function defined on (0,1) by

f(x)= 0 is x is irrational
and
f(x)= 1/q if x=p/q

where p and q are positive integers with no common factors. Show that lim f(x) for any x in (0,1) is 0

The Attempt at a Solution



Here is the first line of the proof the professor gave us.

"Let eps>0 and n so large that (1/n)<=eps, where n is a natural number. The only numbers x for which f(x) can > eps are 1/2, 1/3, 2/3, ..., 1/n, ...(n-1)/n. "

I don't understand how the any p/q with n as q can be greater than epsilon is n is chosen so large that 1/n <=eps.
 
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barksdalemc said:
I don't understand how the any p/q with n as q can be greater than epsilon is n is chosen so large that 1/n <=eps.
had trouble reading this last sentence

do you mean you don't understand how it's possible that p/n > eps given that 1/n <= eps?
 
Last edited:
yes that is exactly where I am confused, but he is saying that f(p/n) > eps. f(p/n) given 1/n.
 
i'd list all those numbers to exclude them when I'm choosing delta, then any f(x)<1/n<=eps
 
The point is that, within any finite distance of a number a, there can be only a finite number of numerators for a fraction with denominator n: and so only a finite number of fractions with denominator less than or equal to n.
 

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