## 6.6. A Function Primer

### Countable Discontinuities

*g*is continuous precisely at the irrational numbers, and discontinuous at all rational numbers.

### 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

*via*

**Q***f(r) = 1 / q*. Then the limit of

*f(r)*as

*r*approaches any real number is zero.

### Proof of Lemma:

Take any sequence*{ r*of rational numbers converging to a fixed number

_{n}}*a*(which could be rational or irrational). Since the sequence converges, it is bounded. For simplicity assume that all rational numbers

*r*are in the interval

_{n}*[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*r*in the interval_{n}*[0, N]*can have denominator equal to*1* - at most
*2 K*of the rational numbers*r*in the interval_{n}*[0, N]*can have denominator equal to*2* - ...
- at most
*M * K*of the rational numbers*r*in the interval_{n}*[0, N]*can have denominator equal to*M*

*r*can have a denominator less than or equal to

_{n}*K*. That means, however, that there exists an integer

*N*such the denominator of

*r*is bigger than

_{n}*M*for all

*n > N*. But then

Since| f(r_{n}) | < 1 / M < for all n > N

*> 0*was arbitrary, that means that the limit of

*f(r*must be zero, as needed. Our assumption that the numbers

_{n})*r*were all positive can easily be dropped, and a similar proof would work again.

_{n}

Using this lemma it should not be too hard to prove the original assertion.