Define
See the graph of
Proof:
-
Discontinuity at Rational Points:
Let
$c = \frac{m}{k}$ be rational and in lowest terms. Take a sequence of irrational numbers$x_{n_{n=1}}^{\infty}$ such that$\lim_{n \to \infty} x_n = c$ . Then,$\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} 0 = 0,$ but
$f(c) = \frac{1}{k} \neq 0.$ So,
$f$ is discontinuous at$c$ . -
Continuity at Irrational Points:
Let
$c$ be irrational, so$f(c) = 0$ . Take a sequence$x_{n_{n=1}^{\infty}}$ in$(0, 1)$ such that$\lim_{n \to \infty} x_n = c$ . Given$\epsilon > 0$ , find$K \in \mathbb{N}$ such that$\frac{1}{K} < \epsilon$ by the Archimedean property.If
$\frac{m}{k} \in (0, 1)$ and$m, k \in \mathbb{N}$ , then$0 < m < k$ . So there are only finitely many rational numbers in$(0, 1)$ whose denominator$k$ in lowest terms is less than$K$ .As
$\lim_{n \to \infty} x_n = c$ , every number not equal to$c$ can appear at most finitely many times in$x_{n_{n=1}^{\infty}}$ .Hence, there is an
$M$ such that for$n \geq M$ , all the rational numbers$x_n$ have denominators larger than or equal to$K$ . Thus, for$n \geq M$ ,$f(x_n) < \epsilon.$ Therefore,
$f$ is continuous at irrational$c$ .
- Prof. Elisabeth Werner: For teaching the foundational concepts of Analysis in Math 321 at Case Western Reserve University.
- Basic Analysis I by Jiří Lebl: Reference material from Chapter 3. CONTINUOUS FUNCTIONS