6.6. A Function Primer

Countable Discontinuities

The function g is continuous precisely at the irrational numbers, and discontinuous at all rational numbers.
Incidentally, it is impossible to have a function that is continuous only at the rationals, which will be proved in the section on Metric spaces and Baire categories.


The actual proof is left as an exercise. However, you may want to make use of the following fact:


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: In total, at most finitely many of the rn can have a denominator less than or equal to K. That means, however, that there exists an integer N such the denominator of rn is bigger than M for all n > N. But then
| f(rn) | < 1 / M < for all n > N
Since > 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.

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