Countable Discontinuities
Proof:
The actual proof is left as an exercise. However, you may want to make use of the following fact:Lemma
If r = p / q is a rational number (in lowest terms) define a function with domain Q via f(r) = 1 / q. Then the limit of f(r) as r approaches any real number is zero.Proof of Lemma:
Take any sequence { rn } of rational numbers converging to a fixed number a (which could be rational or irrational). Since the sequence converges, it is bounded. For simplicity assume that all rational numbers rn are in the interval [0, K] for some integer K. Now take any > 0 and pick an integer M such that 1 / M < . Because each rational number is the quotient of two positive integers, we have:- at most K of the rational numbers rn in the interval [0, N] can have denominator equal to 1
- at most 2 K of the rational numbers rn in the interval [0, N] can have denominator equal to 2
- ...
- at most M * K of the rational numbers rn in the interval [0, N] can have denominator equal to M
| f(rn) | < 1 / M < for all n > NSince > 0 was arbitrary, that means that the limit of f(rn) must be zero, as needed. Our assumption that the numbers rn were all positive can easily be dropped, and a similar proof would work again.
Using this lemma it should not be too hard to prove the original assertion.