Prove Theorem 46.7. Let \(X\) be a space; let \((Y, d)\) be a metric space. For the function space \(Y^X\), one has the following inclusions of topologies:
\[ (\text{uniform}) \supset (\text{compact convergence}) \supset (\text{pointwise convergence}). \]
If \(X\) is compact, the first two coincide, and if \(X\) is discrete, the second two coincide.
Solution
For \(f \in Y^X\), \(\varepsilon \in (0, 1]\), and \(C \subseteq X\), put \[B_C(f, \varepsilon) = \{g \in Y^X : \sup_{x \in C} d(f(x), g(x)) < \varepsilon\}, \qquad B(f, \varepsilon) = \{g \in Y^X : \sup_{x \in X} \bar{d}(f(x), g(x)) < \varepsilon\},\] where \(\bar{d} = \min(d, 1)\). The sets \(B_C(f, \varepsilon)\) with \(C\) compact form a basis for the topology of compact convergence, the sets \(B(f, \varepsilon)\) form a basis for the uniform topology, and the sets \(B_{\{x\}}(f, \varepsilon)\) form a subbasis for the topology of pointwise convergence. If \(g \in B_C(f, \varepsilon)\) and \(r = \sup_{x \in C} d(f(x), g(x))\), then \(B_C(g, \varepsilon - r) \subseteq B_C(f, \varepsilon)\) by the triangle inequality, and similarly for \(B(f, \varepsilon)\). So to show that a set \(B_C(f, \varepsilon)\) is open in a topology, it suffices to find, for every \(f\) and \(\varepsilon\), a neighborhood of \(f\) in that topology inside \(B_C(f, \varepsilon)\).
1. (pointwise) \(\subseteq\) (compact convergence).
For \(x \in X\), the singleton \(\{x\}\) is compact, so the subbasis element \(B_{\{x\}}(f, \varepsilon)\) of the pointwise topology is a basis element of the topology of compact convergence.
2. (compact convergence) \(\subseteq\) (uniform).
Let \(C \subseteq X\) be compact and \(\varepsilon \in (0, 1]\). If \(g \in B(f, \varepsilon)\), then \(\bar{d}(f(x), g(x)) < \varepsilon \le 1\) for all \(x\), so \(d(f(x), g(x)) = \bar{d}(f(x), g(x))\) and \(\sup_{x \in C} d(f(x), g(x)) < \varepsilon\). Hence \(B(f, \varepsilon) \subseteq B_C(f, \varepsilon)\).
3. If \(X\) is compact, the uniform topology equals the topology of compact convergence.
Take \(C = X\). For \(\varepsilon \in (0, 1]\), the argument of step 2 shows \(B_X(f, \varepsilon) = B(f, \varepsilon)\). So every uniform basis element is a basis element of the topology of compact convergence, and step 2 gives the reverse inclusion.
4. If \(X\) is discrete, the topology of compact convergence equals the topology of pointwise convergence.
A compact subset \(C\) of a discrete space is finite, say \(C = \{x_1, \ldots, x_k\}\). Then \[B_C(f, \varepsilon) = \bigcap_{i=1}^k B_{\{x_i\}}(f, \varepsilon)\] is open in the topology of pointwise convergence. Step 1 gives the reverse inclusion.
5. Q.E.D.
Steps 1 and 2 give the inclusions, and steps 3 and 4 the two coincidences.