Find a function continuous at a single point of an interval

Find a function continuous at a single point of an interval, but discontinuous at all other points of the interval.

Consider the interval $(-a,a)$ for any $a \in \mathbb{R}$ , and define

$f(x) = \begin{cases} x & \text{if } x \text{ is rational},\\ 0 & \text{if } x \text{ is irrational}. \end{cases}$

for all $x \in (-a,a)$ .

We claim $f$ is continuous at 0, but discontinuous at every other point of the interval.

Proof. First, we show $f$ is continuous at 0. To do this we must show that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $|f(x)| < \varepsilon$ whenever $0 < |x| < \delta$ . So, let $\varepsilon > 0$ be given. Then, choose $\delta = \varepsilon$ . Then we have,

$|x| < \delta \quad \implies \quad |x| < \varepsilon \quad \implies \quad |f(x)| < \varepsilon$

since $|f(x)| \leq |x|$ from the definition of $f(x)$ . Thus, $f$ is continuous at 0.

Now, we must show $f$ is discontinuous at every other point of the interval. Let $p \neq 0$ be any nonzero point of the interval.

We consider two cases:
Case #1: $p$ is rational. Then $f(p) = p$ . Since $p \neq 0$ by assumption, we consider $\varepsilon = |p| > 0$ . Since the irrationals are dense in the reals (see Exercise #9 of Section I.3.12 here) we know that for any $\delta > 0$ there is some irrational number $x$ such that $|p-x| < \delta$ . But then, $|f(p) - f(x)| = |p - 0| = |p| \not < \varepsilon$ . This means $f$ is not continuous at $p$ since we have found an $\varepsilon > 0$ such that there is no $\delta > 0$ we can choose such that $|f(p) - f(x)| < \varepsilon$ whenever $|p - x| < \delta$ .

Case #2: $p$ is irrational. Since $p$ is irrational we know $f(p) = 0$ . Consider $\varepsilon = |p| > 0$ . Since the rationals are dense in the reals (see Exercise #6 of Section I.3.12 here) we know that for any $\delta > 0$ there exists some rational number $r$ such that $|p| < |r| < |p| + \delta$ . But then, $|f(p) - f(r)| = |0-r| = |r| \not < \varepsilon$ . This means $f$ is not continuous at $p$ . (Again, because we have shown that there exists an $\varepsilon > 0$ such that for every $\delta > 0$ there is some number $r$ such that $|f(p) - f(r)| \not < \varepsilon$ with $|p-r| < \delta.) \qquad \blacksquare$

(Note: There are much nicer ways to do this using sequences, but Apostol doesn’t develop those techniques, so we have to work straight from the epsilon-delta definitions.)

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Find a function continuous at a single point of an interval

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