Suppose \(R\) is not totally bounded, then there is an \(\epsilon_0 > 0\) such that \(R\) has no finite \(\epsilon_0\)​-net. Choose any point \(a_1 \in R\). Then \(R\) contains at least one point, say \(a_2\), such that \(\rho(a_1, a_2) > \epsilon_0\), since otherwise \(a_1\) would be an \(\epsilon_0\)​-net for \(R\). Moreover, \(R\) contains a point \(a_3\) such that \(\rho(a_1, a_3) > \epsilon_0\), \(\rho(a_2, a_3) > \epsilon_0\), since otherwise the pair \(a_1, a_2\) would be an \(\epsilon_0\)​-net for \(R\). More generally, once having found the points \(a_1, a_2, \ldots a_n\), we choose \(a_{n+1} \in R\) such that \(\rho(a_k, a_{n+1}) > \epsilon_0 \quad (k = 1, 2, \ldots, n)\). This construction gives an infinite sequence of distinct points \(a_1, a_2, \ldots, a_n, \ldots\) with no limit points, since \(\rho(a_j, a_k) > \epsilon_0\) if \(j \neq k\). But then \(R\) cannot be countably compact.

Since \(R\) is totally bounded, by Theorem 1, \(R\) has a finite \((1/n)\)​-net for every \(n=1, 2, \ldots\). The union of all these nets is then a countably everywhere dense subset of \(R\). It follows from Theorem 5 Topological Spaces: Basic Concepts that \(R\) has a countable base.

An immediate consequence of Corollary 1 and Theorem 9 Topological Spaces: Compactness.

To see that compactness of \(R\) implies completeness of \(R\), we need only note that if \(R\) has a Cauchy sequence \(\{x_n\}\) with no limit, then \(\{x_n\}\) has no limit point in \(R\). Hence if \(R\) is compact, \(R\) is complete. By Theorem 1, if \(R\) is compact, \(R\) is totally bounded.

Conversely, suppose \(R\) is totally bounded and complete, and let \(\{x_n\}\) be any infinite sequence of distinct points in \(R\). Let \(N_1\) be a finite \(1\)​-net for \(R\), and construct a closed sphere of radius 1 about every point of \(N_1\). Since these spheres cover \(R\) and there are finitely many of them, at least one of the spheres, say \(S_1\), contains an infinite subsequence \[x_1^{(1)}, \ldots, x_n^{(1)}, \ldots\] of the sequence \(\{x_n\}\). Let \(N_2\) be a finite \(1/2\)​-net for \(R\), and construct a closed sphere of \(1/2\) for every point of \(N_2\). Then at least one of these spheres, say \(S_2\), contains an infinite subsequence \[x_1^{(2)}, \ldots, x_n^{(2)}, \ldots\] of the sequence \(\{x_n^{(1)}\}\). Continue this construction indefinitely, where \(S_n\) has radius \(1/2^{n-1}\). Let \({S’}_n\) be the closed sphere with the same centre as \(S_n\) but with a radius \(r_n\) twice as large (i.e., equal to \(1/2^n\)). Then clearly \[{S’}_1 \subseteq {S’}_2 \subseteq \ldots \subseteq {S’}_n \subseteq \ldots,\] and moreover \(r_n \rightarrow 0\) as \(n \rightarrow \infty\). Since \(R\) is complete, it follows from the nested sphere theorem (Theorem 2, Complete Metric Spaces) that \[\cap_{n=1}^\infty {S’}_n \neq \emptyset.\] In fact, there is a point \(x_0 \in R\) such that \[\cap_{n=1}^\infty {S’}_n = \{x_0\}\] (recall Problem 3, Complete Metric Spaces). Clearly \(x_0\) is a limit point of the original sequence \(\{x_n\}\), since every neighbourhood of \(x_0\) contains some sphere \(S_k\) and hence some infinite subsequence \(\{x_n^{(k)}\}\). Therefore every infinite sequence \(\{x_n\}\) of distinct points of \(R\) has a limit point of \(R\). It follows that \(R\) is countably compact and hence compact, by Corollary 2.

An immediate consequence of Theorem 2 and the fact that a closed subset of a complete metric space is itself complete.

We first prove necessity and then sufficiency.

(Necessity.) Suppose \(\Phi\) is relatively compact in \(C_{[a, b]}\). Then by Theorem 3, it is totally bounded, so given any \(\epsilon > 0\), there is a finite \((\epsilon/3)\)​-net \(\phi_1, \ldots, \phi_n\) in \(\Phi\) (see Problem 1). Being a continuous function defined on a closed interval, each \(\phi_i\) is bounded: \[|\phi_i(x)| \leq K \quad (a \leq x \leq b).\] Let \(K = \max\{K_1, \ldots, K_n\} + \epsilon/3\). By the definition of an \((\epsilon/3)\)​-net, given any \(\phi \in \Phi\), there is at least one \(\phi_i\) such that \[\rho(\phi, \phi_i) = \max_{a \leq x \leq b} |\phi(x) = \phi_i(x)| \leq \epsilon/3.\] Therefore \(|\phi(x)| \leq |\phi_i(x)| + \epsilon/3 \leq K_i + \epsilon/3 \leq K\), i.e. \(\Phi\) is uniformly bounded. Moreover, each function \(\phi_i\) in the \((\epsilon/3)\)​-net is continuous, and hence uniformly continuous, on \([a, b]\). Hence, given any \(\epsilon > 0\), there is a \(\delta_i\) such that \[|\phi_i(x_1) - \phi_i(x_2)| \leq \epsilon/3\] whenever \(|x_1-x_2| < \delta_i\). Let \(\delta = \min\{\delta_1, \ldots, \delta_n\}\). Then, given any \(\phi \in \Phi\) and choosing \(\phi_i\) such that \(\rho(\phi, \phi_i) < \epsilon/3\), we have \[\begin{equation} \begin{split} |\phi(x_1) - \phi(x_2)| &\leq |\phi(x_1) - \phi_i(x_1) + |\phi_i(x_1) - \phi_i(x_2)| + |\phi_i(x_2) - \phi(x_2)| \\ & < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon \end{split}\end{equation}\] whenever \(|x_1 - x_2| < \delta\). This proves the equicontinuity of \(\Phi\).

(Sufficiency.) Suppose \(\Phi\) is uniformly bounded and equicontinuous. Accordin to Theorem 3, to prove that \(\Phi\) is relatively compact in \(C_{[a, b]}\), we need only show that \(\Phi\) is totally bounded, i.e., that given any \(\epsilon > 0\), there exists a finite \(\epsilon\)​-net for \(\Phi\) in \(C_{[a, b]}\). Since \(\Phi\) is uniformly bounded, \(|\phi(x)| \leq K\) for all \(\phi \in \Phi\), and by equicontinuity, let \(\delta > 0\) be such that \[|\phi(x_1) - \phi(x_2)| < \epsilon/5\] for all \(\phi \in \Phi\) whenever \(|x_1-x_2|< \delta\). Divide the interval \(a \leq x \leq b\) along the x-axis into subintervals of length less than \(\delta\), by introducing points of subdivision \(x_0, x_1, \ldots, x_n\) such that \[a = x_0 < x_1 < \ldots < x_n = b\] and then draw a vertical line through each of these points. Similarly, divide the interval \(-K \leq y \leq K\) along the y-axis into subintervals of length less than \(\epsilon/5\), by introducing points of subdivision \(y_0, y_1, \ldots, y_m\) such that \[-K = y_0 < y_1 < \ldots < y_m= K\] and then draw a horizontal line through each of these points. In this way, the rectangle \(a < x < b\), \(-K < y < K\) is divided into \(nm\) cells of horizontal side length less than \(\delta\) and vertical side length less than \(\epsilon/5\).

We now associate with each function \(\phi \in \Phi\) a polygonal line \(y = \psi(x)\) which has vertices at points of the form \((x_k, y_l)\) and differs from the function \(\phi\) by less than \(\epsilon/5\) at every point \(x_k\) (the reader should draw a figure and convince themself on the existence of such a function).

Since \[\begin{equation} \begin{split} |\phi(x_k) - \psi(x_k)| &\leq \epsilon/5, \\ ​|\phi(x_{k+1}) - \psi(x_{k+1})| &\leq \epsilon/5, \\ ​|\phi(x_k) - \phi(x_{k+1})| &\leq \epsilon/5, \end{split}\end{equation}\] by construction, we have \[|\psi(x_k) - \psi(x_{k+1})| \leq 3\epsilon/5.\] Moreover, \[|\psi(x_k) - \psi(x)| \leq 3\epsilon/5 \quad (x_k \leq x \leq x_{k+1}),\] since \(\psi(x)\) is linear between the point \(x_k\) and \(x_{k+1}\). Let \(x\) be any point in \([a, b]\) and \(x_k\) the point of subdivision nearest to \(x\) on the left. Then \[|\phi(x)-\psi(x)| \leq |\phi(x)-\phi(x_k)| + |\phi(x_k)-\psi(x_k)| + |\psi(x_k)-\psi(x)| < \epsilon,\] i.e. the set of polygonal lines \(\psi(x)\) forms an \(\epsilon\)​-net for \(\Phi\). But there are obviously only finitely many such lines. Therefore \(\Phi\) is totally bounded.

By the continuity of \(f\), we have \(|f(x, y)| \leq K\) in some domain \(G’ \subseteq G\) containing the point \((x_0, y_0)\). Draw the lines with slopes \(K\) and \(-K\) through the point \((x_0, y_0)\). Then draw vertical lines \(x = a\) and \(x=b \quad (a < x_0 < b)\) which together with the first two lines form two isosceles triangles contained in \(G’\) with common vertex \((x_0, y_0)\). This gives a closed interval \([a, b]\), which will figure in the rest of the proof.

The next step is to construct a family of polygonal lines, called Euler lines, associated with the differential equation \(\frac{dy}{dx} = f(x, y)\). We begin by drawing the line with slope \(f(x_0, y_0)\) through the point \((x_0, y_0)\). Next, choosing a point \((x_1, y_1)\) on the first line, we draw the line \(f(x_1, y_1)\) through the point \((x_1, y_1)\). Then, choosing a point \((x_2, y_2)\) on the second line, we draw the line with slope \(f(x_2, y_2)\) through the point \((x_2, y_2)\), and so on indefinitely. Suppose we construct a whole sequence \(L_1, L_2, \ldots, L_n, \ldots\) of such Euler lines going through the point \((x_0, y_0)\), with the property that the length of the longest line segment making up \(L_n\) approaches \(0\) as \(n \rightarrow \infty\). Let \(\phi_n\) be the function with graph \(L_n\). Then this gives a family of functions \(\phi_1, \phi_2, \ldots, \phi_n, \ldots\), all defined on the interval \([a, b]\), which is easily seen to be uniformly bounded and equicontinuous (as they are bounded by \(K\) and are straight lines). It follows from Arzelà’s theorem that the sequence \(\{\phi_n\}\) is relatively compact and therefore contains a uniformly convergent subsequence \[\phi^{(1)}, \phi^{(2)}, \ldots, \phi^{(n)}, \ldots.\] Let \(\phi(x) = \lim_{n \rightarrow \infty} \phi^{(n)}(x)\). Then clearly \(\phi(x_0) = y_0\), so that the curve \(y = \phi(x)\) passes through the point \((x_0, y_0)\). We now show that the \(y = \phi(x)\) satisfies the differential equation \(\frac{dy}{dx} = f(x, y)\) in the open interval \((a, b)\). This means showing that, given any \(\epsilon > 0\) and any points \(x’, x’’ \in (a, b)\), we have \[|\frac{\phi(x’’) - \phi(x’)}{x’’ - x’} - f(x’, \phi(x’))| < \epsilon\] whenever \(|x’’ - x’|\) is sufficiently small, or equivalently that \[|\frac{\phi^{(n)}(x’’) - \phi^{(n)}(x’)}{x’’-x’} - f(x’, \phi^{(n)}(x’))| < \epsilon\] whenever \(n\) is sufficiently large and \(|x’’ - x’|\) is sufficiently small. Let \(y’ = \phi(x’)\). Then, by continuity of \(f\), given any \(\epsilon > 0\), there is a number \(\eta > 0\) such that \[f(x’, y’) - \epsilon < f(x, y) < f(x’, y’) + \epsilon\] whenever \(|x - x’| < 2\eta\), \(|y-y’| < 4K\eta\). The set of points \((x, y)\) satisfying these inequalities is a rectange which we denote by \(Q\). Let \(N\) be so large that for all \(n > N\), the length of the longest segment making up \(L_n\) is less than \(\eta\) and moreover \[|\phi(x) - \phi^{(n)}(x)| < K\eta.\] Then all the Euler lines \(L_n\) with \(n > N\) live inside the rectange \(Q\) (why?). Suppose \(L_n\) has vertices \((a_0, b_0), (a_1, b_1), \ldots, (a_{k+1}, b_{k+1})\), where \(a \leq x’ < a_1 < a_2 < \ldots < a_k < x’’ \leq a_{k+1}\). [Note we assume that \(x’’ > x’\). The case \(x’’ < x’\) is treated similarly.] Then \[\begin{equation} \begin{split} \phi^{(n)}(a_1) - \phi^{(n)}(x’) & = f(a_0, b_0)(a_1 - x’), \\ \phi^{(n)}(a_{i+1}) - \phi^{(n)}(a_i) & = f(a_i, b_i)(a_{i+1} - a_i) \quad (i = 1, 2, \ldots, k-1), \\ \phi^{(n)}(x’’) - \phi^{(n)}(a_k) & = f(a_k, b_k)(a’’ - a_k). \end{split}\end{equation}\] Hence, if \(|x’’ - x’| < \eta\), \[\begin{equation} \begin{split} [f(x’, y’) - \epsilon](a_1 - x’) & < \phi^{(n)}(a_1) - \phi^{(n)}(x’) < [f(x’, y’) + \epsilon](a_1 - x’), \\ [f(x’, y’) - \epsilon](a_{i+1} - a_i) & < \phi^{(n)}(a_{i+1}) - \phi^{(n)}(a_i) \\ & < [f(x’, y’) + \epsilon](a_{i+1} - a_i) \quad (i=1, 2, \ldots, k-1), \\ [f(x’, y’) - \epsilon](x’’ - a_k) & < \phi^{(n)}(x’’) - \phi^{(n)}(a_k) < [f(x’, y’) + \epsilon](x’’ - a_k). \end{split}\end{equation}\] Adding these inequalities, we get \[[f(x’, y’) - \epsilon](x’’ - x’) < \phi^{(n)}(x’’) - \phi^{(n)}(x’) < [f(x’, y’) + \epsilon](x’’ - x’)\] if \(|x’’ - x’| < \eta\), which is equivalent to \[|\frac{\phi^{(n)}(x’’) - \phi^{(n)}(x’)}{x’’-x’} - f(x’, \phi^{(n)}(x’))| < \epsilon.\]