Let \(g: \mathbb{Z}_+ \to \mathbb{Q}\) be a bijective function; let \(x_n = g(n)\). Define \(f: \mathbb{R} \to \mathbb{R}\) as follows:
\[ \begin{array}{ll} f(x_n) = 1/n & \text{for } x_n \in \mathbb{Q}, \\ f(x) = 0 & \text{for } x \notin \mathbb{Q}. \end{array} \]
Show that \(f\) is continuous at each irrational and discontinuous at each rational. Can you find a sequence of continuous functions \(f_n\) converging to \(f\)?
Solution
1. Continuity at every irrational point \(x_0 \in \mathbb{R} \setminus \mathbb{Q}\):Proof:
1.1. At an irrational point \(x_0\), \(f(x_0) = 0\).
1.2. Let \(\varepsilon > 0\). Choose an integer \(N \in \mathbb{Z}_+\) such that \(\frac{1}{N} < \varepsilon\).
1.3. The set \(S_N = \{x_1, x_2, \dots, x_N\}\) is finite, and since \(x_0\) is irrational, \(x_0 \notin S_N\).
1.4. Define \(\delta = \min_{1 \le n \le N} |x_0 - x_n| > 0\).
1.5. If \(|x - x_0| < \delta\), then \(x \notin S_N\).
1.6. Case 1: If \(x \notin \mathbb{Q}\), \(|f(x) - f(x_0)| = |0 - 0| = 0 < \varepsilon\).
1.7. Case 2: If \(x \in \mathbb{Q}\), then \(x = x_k\) for some \(k > N\), so \(|f(x) - f(x_0)| = \frac{1}{k} < \frac{1}{N} < \varepsilon\).
1.8. Thus \(|f(x) - f(x_0)| < \varepsilon\) for all \(x \in (x_0 - \delta, x_0 + \delta)\), so \(f\) is continuous at \(x_0\).
2. Discontinuity at every rational point \(x_m \in \mathbb{Q}\):Proof:
2.1. At \(x_m \in \mathbb{Q}\), \(f(x_m) = \frac{1}{m} > 0\).
2.2. Let \(\varepsilon_0 = \frac{1}{2m} > 0\).
2.3. For any \(\delta > 0\), the interval \((x_m - \delta, x_m + \delta)\) contains an irrational number \(y\) by density of \(\mathbb{R} \setminus \mathbb{Q}\) in \(\mathbb{R}\).
2.4. At this point \(y\), \(|f(y) - f(x_m)| = |0 - \frac{1}{m}| = \frac{1}{m} > \varepsilon_0\).
2.5. Hence \(f\) is discontinuous at \(x_m\).
3. Construction of continuous functions \(f_k \to f\) pointwise:Proof:
3.1. For each fixed \(k \in \mathbb{Z}_+\), consider the first \(k\) rational numbers \(\{x_1, \dots, x_k\}\).
3.2. For each \(n \in \{1, \dots, k\}\), choose \(\delta_{n, k} > 0\) such that the intervals \((x_n - \delta_{n, k}, x_n + \delta_{n, k})\) are pairwise disjoint for \(1 \le n \le k\) and \(\delta_{n, k} < \frac{1}{k}\).
3.3. Define a continuous “tent” function \(\phi_{n, k}: \mathbb{R} \to [0, \frac{1}{n}]\) supported on \([x_n - \delta_{n, k}, x_n + \delta_{n, k}]\) by:\[\phi_{n, k}(x) = \max\left\{0, \; \frac{1}{n}\left(1 - \frac{|x - x_n|}{\delta_{n, k}}\right)\right\}.\]
3.4. Define \(f_k: \mathbb{R} \to \mathbb{R}\) by \(f_k(x) = \sum_{n=1}^k \phi_{n, k}(x)\). Each \(f_k\) is continuous as a finite sum of continuous functions.
3.5. For any rational \(x_m\), for all \(k \ge m\) we have \(f_k(x_m) = \phi_{m, k}(x_m) = \frac{1}{m} = f(x_m)\), so \(\lim_{k \to \infty} f_k(x_m) = f(x_m)\).
3.6. Let \(x\) be irrational and \(\varepsilon > 0\). Choose \(N\) with \(1/N < \varepsilon\) and put \(\eta = \min_{1 \le n \le N} |x - x_n| > 0\). For \(k > \max(N, 1/\eta)\), \(\delta_{n, k} < 1/k < \eta\), so \(\phi_{n, k}(x) = 0\) for \(n \le N\). The supports of the tents \(\phi_{1,k}, \ldots, \phi_{k,k}\) are disjoint, so \(f_k(x) = \phi_{n, k}(x) \le 1/n < \varepsilon\) for at most one \(n > N\), and \(f_k(x) = 0\) otherwise. Hence \(\lim_{k \to \infty} f_k(x) = 0 = f(x)\).
3.7. Thus \(f_k \to f\) pointwise on \(\mathbb{R}\). Q.E.D.