Let \((Y, d)\) be a metric space; let \(X\) be a space. Define a topology on \(\mathcal{C}(X, Y)\) as follows. Given \(f \in \mathcal{C}(X, Y)\), and given a positive continuous function \(\delta: X \to \mathbb{R}_+\) on \(X\), let
\[ B(f, \delta) = \ts{g \mid d(f(x), g(x)) < \delta(x) \text{ for all } x \in X}. \]
Show that the sets \(B(f, \delta)\) form a basis for a topology on \(\mathcal{C}(X, Y)\). We call it the fine topology.
Show that the fine topology contains the uniform topology.
Show that if \(X\) is compact, the fine and uniform topologies agree.
Show that if \(X\) is discrete, then \(\mathcal{C}(X, Y) = Y^X\) and the fine and box topologies agree.
Solution
1. Part (a): The sets \(\mathcal{B} = \{B(f, \delta) : f \in \mathcal{C}(X, Y), \delta \in \mathcal{C}(X, \mathbb{R}_+)\}\) form a basis.Proof:
1.1. Covering: For any \(f \in \mathcal{C}(X, Y)\), the constant function \(\delta_1(x) = 1\) is continuous and positive, and \(d(f(x), f(x)) = 0 < 1\) for all \(x\), so \(f \in B(f, \delta_1)\).
1.2. Intersection: Let \(g \in B(f_1, \delta_1) \cap B(f_2, \delta_2)\).
1.3. For each \(x \in X\), \(d(f_1(x), g(x)) < \delta_1(x)\) and \(d(f_2(x), g(x)) < \delta_2(x)\).
1.4. Define \(\delta_3: X \to \mathbb{R}_+\) by:\[\delta_3(x) = \min\{\delta_1(x) - d(f_1(x), g(x)), \; \delta_2(x) - d(f_2(x), g(x))\}.\]
1.5. Since \(f_1, f_2, g, \delta_1, \delta_2\) and the metric \(d\) are continuous, the map \(x \mapsto d(f_i(x), g(x))\) is continuous on \(X\). Hence \(\delta_3\) is continuous and strictly positive everywhere (\(\delta_3(x) > 0\)).
1.6. For any \(h \in B(g, \delta_3)\) and each \(x \in X\), the triangle inequality gives:\[d(f_1(x), h(x)) \le d(f_1(x), g(x)) + d(g(x), h(x)) < d(f_1(x), g(x)) + (\delta_1(x) - d(f_1(x), g(x))) = \delta_1(x).\]
1.7. Thus \(h \in B(f_1, \delta_1)\). Similarly \(h \in B(f_2, \delta_2)\).
1.8. Hence \(g \in B(g, \delta_3) \subseteq B(f_1, \delta_1) \cap B(f_2, \delta_2)\), proving that \(\mathcal{B}\) is a basis.
2. Part (b): The fine topology contains the uniform topology.Proof:
2.1. The uniform topology has as a basis the balls \(B_{\bar{\rho}}(f, \varepsilon) = \{g \in \mathcal{C}(X, Y) : \sup_{x \in X} \bar{d}(f(x), g(x)) < \varepsilon\}\) with \(0 < \varepsilon \le 1\), where \(\bar{d} = \min(d, 1)\); for such \(\varepsilon\), \(\bar{d}(f(x), g(x)) < \varepsilon\) if and only if \(d(f(x), g(x)) < \varepsilon\).
2.2. Let \(g \in B_{\bar{\rho}}(f, \varepsilon)\) and put \(r = \bar{\rho}(f, g) < \varepsilon\). Let \(\delta\) be the constant function \((\varepsilon - r)/2\), which is continuous and positive. For \(h \in B(g, \delta)\) and \(x \in X\), \(d(f(x), h(x)) < r + (\varepsilon - r)/2\), so \(\bar{\rho}(f, h) \le r + (\varepsilon - r)/2 < \varepsilon\). Hence \(g \in B(g, \delta) \subseteq B_{\bar{\rho}}(f, \varepsilon)\).
2.3. Thus every uniform ball is open in the fine topology, so the fine topology contains the uniform topology.
3. Part (c): If \(X\) is compact, the fine and uniform topologies coincide.Proof:
3.1. By 2, the fine topology is at least as fine as the uniform topology.
3.2. Conversely, let \(B(f, \delta)\) be a basic open set in the fine topology and let \(g \in B(f, \delta)\). By step 1, there is a continuous positive \(\delta_3\) with \(g \in B(g, \delta_3) \subseteq B(f, \delta)\).
3.3. Since \(X\) is compact and \(\delta_3: X \to (0, \infty)\) is continuous, the extreme value theorem gives\[\varepsilon := \min\left(\min_{x \in X} \delta_3(x), 1\right) > 0.\]
3.4. If \(h \in B_{\bar{\rho}}(g, \varepsilon)\), then for all \(x \in X\), \(d(g(x), h(x)) < \varepsilon \le \delta_3(x)\), so \(h \in B(g, \delta_3)\).
3.5. Thus \(g \in B_{\bar{\rho}}(g, \varepsilon) \subseteq B(f, \delta)\). Every point of \(B(f, \delta)\) has a uniform ball inside it, so every fine-open set is uniform-open.
4. Part (d): If \(X\) is discrete, \(\mathcal{C}(X, Y) = Y^X\) and the fine topology equals the box topology.Proof:
4.1. When \(X\) is discrete, every map \(f: X \to Y\) is continuous, so \(\mathcal{C}(X, Y) = Y^X\).
4.2. Furthermore, every function \(\delta: X \to \mathbb{R}_+\) is continuous.
4.3. A basic open set in the box topology on \(\prod_{x \in X} Y\) containing \(f\) is \(\prod_{x \in X} B_d(f(x), \varepsilon_x)\) for an arbitrary collection of positive numbers \((\varepsilon_x)_{x \in X}\).
4.4. Defining \(\delta(x) = \varepsilon_x\) for each \(x \in X\), \(\delta\) is a continuous positive function on the discrete space \(X\), and \(B(f, \delta) = \prod_{x \in X} B_d(f(x), \varepsilon_x)\).
4.5. Hence the bases for the fine topology and the box topology are identical. Q.E.D.