Prove the following.
Theorem. If \(D\) is a countable dense subset of \(\mathbb{R}\), there is no function \(f: \mathbb{R} \to \mathbb{R}\) that is continuous precisely at the points of \(D\).
Show that if \(f: \mathbb{R} \to \mathbb{R}\), then the set \(C\) of points at which \(f\) is continuous is a \(G_\delta\) set in \(\mathbb{R}\). [Hint: Let \(U_n\) be the union of all open sets \(U\) of \(\mathbb{R}\) such that \(\operatorname{diam} f(U) < 1/n\). Show that \(C = \bigcap U_n\).]
Show that \(D\) is not a \(G_\delta\) set in \(\mathbb{R}\). [Hint: Suppose \(D = \bigcap W_n\), where \(W_n\) is open in \(\mathbb{R}\). For \(d \in D\), set \(V_d = \mathbb{R} - \ts{d}\). Show \(W_n\) and \(V_d\) are dense in \(\mathbb{R}\).]
Solution
For \(n \in \ZZ_+\) let \(U_n\) be the union of all open sets \(U \subseteq \RR\) with \(\operatorname{diam} f(U) < 1/n\). Each \(U_n\) is open, as a union of open sets.
1. (a) \(C = \bigcap_{n=1}^\infty U_n\), so \(C\) is a \(G_\delta\) set.
1.1. \(C \subseteq \bigcap_n U_n\).
Let \(x_0 \in C\) and \(n \in \ZZ_+\). By continuity at \(x_0\), there is an open interval \(U \ni x_0\) with \(f(U) \subseteq (f(x_0) - \frac{1}{3n}, f(x_0) + \frac{1}{3n})\). Then \(\operatorname{diam} f(U) \le \frac{2}{3n} < \frac1n\), so \(x_0 \in U \subseteq U_n\).
1.2. \(\bigcap_n U_n \subseteq C\).
Let \(x_0 \in \bigcap_n U_n\) and \(\varepsilon > 0\); choose \(n\) with \(1/n < \varepsilon\). Since \(x_0 \in U_n\), some open \(U \ni x_0\) has \(\operatorname{diam} f(U) < 1/n\). For \(y \in U\), \(\abs{f(y) - f(x_0)} \le \operatorname{diam} f(U) < \varepsilon\). So \(f\) is continuous at \(x_0\).
1.3. Q.E.D.
Steps 1 and 2 give \(C = \bigcap_n U_n\), a countable intersection of open sets.
2. (b) \(D\) is not a \(G_\delta\) set.
Write \(D = \{d_k : k \in \ZZ_+\}\) and suppose \(D = \bigcap_n W_n\) with each \(W_n\) open. Each \(W_n\) contains the dense set \(D\), so it is dense. Each \(V_k = \RR \setminus \{d_k\}\) is open, and it is dense because \(\RR\) has no isolated points. By the Baire category theorem for the complete metric space \(\RR\), the intersection of the countable family of dense open sets \(W_n\) and \(V_k\) is dense, hence nonempty. But \[\bigcap_n W_n \cap \bigcap_k V_k = D \cap (\RR \setminus D) = \varnothing.\]
3. Q.E.D.
Step 1 proves part (a) and step 2 proves part (b). If \(f\) were continuous precisely at the points of \(D\), then \(D = C\) would be a \(G_\delta\) set by step 1, contradicting step 2.