Let \(X\) be a topological space; let \(Y\) be a complete metric space. Show that \(\mathcal{C}(X, Y)\) is a Baire space in the fine topology (see E-0GM3H). [Hint: Given basis elements \(B(f_i, \delta_i)\) such that \(\delta_1 \leq 1\) and \(\delta_{i+1} \leq \delta_i/3\) and \(f_{i+1} \in B(f_i, \delta_i/3)\), show that
\[ \bigcap B(f_i, \delta_i) \neq \varnothing.] \]
Solution
The fine topology has as a basis the sets \(B(f, \delta) = \{g \in \mathcal{C}(X, Y) : d(f(x), g(x)) < \delta(x) \text{ for all } x \in X\}\), with \(f \in \mathcal{C}(X, Y)\) and \(\delta: X \to (0, \infty)\) continuous. A space is a Baire space if and only if, for every nonempty open set \(U_0\) and every sequence of dense open sets \(D_1, D_2, \ldots\), the set \(U_0 \cap \bigcap_n D_n\) is nonempty. Fix such \(U_0\) and \(D_n\).
1. There are \(f_n \in \mathcal{C}(X, Y)\) and continuous \(\delta_n: X \to (0, \infty)\) with \(\delta_1 \le 1\), \(B(f_1, \delta_1) \subseteq U_0 \cap D_1\), and, for \(n \ge 1\), \(\delta_{n+1} \le \delta_n/3\) and \(B(f_{n+1}, \delta_{n+1}) \subseteq B(f_n, \delta_n/3) \cap D_{n+1}\).
Since \(D_1\) is dense and open, \(U_0 \cap D_1\) is a nonempty open set; it contains a basis element \(B(f_1, \delta)\), and \(\delta_1 = \min(\delta, 1)\) is continuous and positive. Given \(f_n\) and \(\delta_n\), the set \(B(f_n, \delta_n/3) \cap D_{n+1}\) is open and nonempty because \(D_{n+1}\) is dense; it contains a basis element \(B(f_{n+1}, \delta)\), and \(\delta_{n+1} = \min(\delta, \delta_n/3)\) is continuous and positive.
2. For each \(x \in X\), \(f_n(x)\) converges to a point \(f(x) \in Y\), and \(d(f(x), f_n(x)) \le \delta_n(x)/2\) for all \(n\).
By step 1, \(f_{k+1} \in B(f_k, \delta_k/3)\) and \(\delta_k \le \delta_n/3^{k-n}\) for \(k \ge n\). So for \(m > n\), \[d(f_m(x), f_n(x)) \le \sum_{k=n}^{m-1} \frac{\delta_k(x)}{3} \le \delta_n(x) \sum_{j=1}^\infty 3^{-j} = \frac{\delta_n(x)}{2}.\] Since \(\delta_n(x) \le 3^{-(n-1)}\), the sequence \((f_n(x))\) is Cauchy, and it converges because \(Y\) is complete. Letting \(m \to \infty\) gives the bound.
3. \(f\) is continuous.
Let \(x_0 \in X\) and \(\varepsilon > 0\). Since \(\delta_N \le 3^{-(N-1)}\), choose \(N\) with \(\delta_N < \varepsilon/3\) everywhere. By continuity of \(f_N\), there is a neighborhood \(W\) of \(x_0\) with \(d(f_N(x), f_N(x_0)) < \varepsilon/3\) for \(x \in W\). For \(x \in W\), step 2 gives \[d(f(x), f(x_0)) \le d(f(x), f_N(x)) + d(f_N(x), f_N(x_0)) + d(f_N(x_0), f(x_0)) < \frac{\varepsilon}{6} + \frac{\varepsilon}{3} + \frac{\varepsilon}{6} < \varepsilon.\]
4. Q.E.D.
By steps 2 and 3, \(f \in \mathcal{C}(X, Y)\) and \(d(f(x), f_n(x)) \le \delta_n(x)/2 < \delta_n(x)\), so \(f \in B(f_n, \delta_n)\) for every \(n\). By step 1, \(f \in U_0 \cap \bigcap_n D_n\).