Suppose $R$ is not totally bounded, then there is an ${ϵ}_0>0$ such that $R$ has no finite ${ϵ}_0$​-net. Choose any point $a_1\in R$. Then $R$ contains at least one point, say $a_2$, such that $\rho (a_1,a_2)>{ϵ}_0$, since otherwise $a_1$ would be an ${ϵ}_0$​-net for $R$. Moreover, $R$ contains a point $a_3$ such that $\rho (a_1,a_3)>{ϵ}_0$, $\rho (a_2,a_3)>{ϵ}_0$, since otherwise the pair $a_1,a_2$ would be an ${ϵ}_0$​-net for $R$. More generally, once having found the points $a_1,a_2,\dots a_n$, we choose $a_{n+1}\in R$ such that $\rho (a_k,a_{n+1})>{ϵ}_0(k=1,2,\dots ,n)$. This construction gives an infinite sequence of distinct points $a_1,a_2,\dots ,a_n,\dots$ with no limit points, since $\rho (a_j,a_k)>{ϵ}_0$ if $j\ne k$. But then $R$ cannot be countably compact.

Since $R$ is totally bounded, by Theorem 1, $R$ has a finite $(1/n)$​-net for every $n=1,2,\dots$. 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 $$x_1^{(1)},\dots ,x_n^{(1)},\dots$$ 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 $$x_1^{(2)},\dots ,x_n^{(2)},\dots$$ of the sequence $\{x_n^{(1)}\}$. Continue this construction indefinitely, where $S_n$ has radius $1/2^{n-1}$. Let ${S^{\prime }}_n$ 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 $${S^{\prime }}_1\subseteq {S^{\prime }}_2\subseteq \dots \subseteq {S^{\prime }}_n\subseteq \dots ,$$ and moreover $r_n\to 0$ as $n\to \mathrm{\infty }$. Since $R$ is complete, it follows from the nested sphere theorem (Theorem 2, Complete Metric Spaces) that $${\cap }_{n=1}^{\mathrm{\infty }}{S^{\prime }}_n\ne \mathrm{\varnothing }.$$ In fact, there is a point $x_0\in R$ such that $${\cap }_{n=1}^{\mathrm{\infty }}{S^{\prime }}_n=\{x_0\}$$ (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 $\mathrm{\Phi }$ is relatively compact in $C_{[a,b]}$. Then by Theorem 3, it is totally bounded, so given any $ϵ>0$, there is a finite $(ϵ/3)$​-net ${\varphi }_1,\dots ,{\varphi }_n$ in $\mathrm{\Phi }$ (see Problem 1). Being a continuous function defined on a closed interval, each ${\varphi }_i$ is bounded: $$|{\varphi }_i(x)|\le K(a\le x\le b).$$ Let $K=max\{K_1,\dots ,K_n\}+ϵ/3$. By the definition of an $(ϵ/3)$​-net, given any $\varphi \in \mathrm{\Phi }$, there is at least one ${\varphi }_i$ such that $$\rho (\varphi ,{\varphi }_i)=\underset{a\le x\le b}{max}|\varphi (x)={\varphi }_i(x)|\le ϵ/3.$$ Therefore $|\varphi (x)|\le |{\varphi }_i(x)|+ϵ/3\le K_i+ϵ/3\le K$, i.e. $\mathrm{\Phi }$ is uniformly bounded. Moreover, each function ${\varphi }_i$ in the $(ϵ/3)$​-net is continuous, and hence uniformly continuous, on $[a,b]$. Hence, given any $ϵ>0$, there is a ${\delta }_i$ such that $$|{\varphi }_i(x_1)-{\varphi }_i(x_2)|\le ϵ/3$$ whenever $|x_1-x_2|<{\delta }_i$. Let $\delta =min\{{\delta }_1,\dots ,{\delta }_n\}$. Then, given any $\varphi \in \mathrm{\Phi }$ and choosing ${\varphi }_i$ such that $\rho (\varphi ,{\varphi }_i)<ϵ/3$, we have $$\begin{array}{rl}|\varphi (x_1)-\varphi (x_2)|& \le |\varphi (x_1)-{\varphi }_i(x_1)+|{\varphi }_i(x_1)-{\varphi }_i(x_2)|+|{\varphi }_i(x_2)-\varphi (x_2)|\\ & <ϵ/3+ϵ/3+ϵ/3=ϵ\end{array}$$ whenever $|x_1-x_2|<\delta$. This proves the equicontinuity of $\mathrm{\Phi }$.

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

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

Since $$\begin{array}{rl}|\varphi (x_k)-\psi (x_k)|& \le ϵ/5,\\ |\varphi (x_{k+1})-\psi (x_{k+1})|& \le ϵ/5,\\ |\varphi (x_k)-\varphi (x_{k+1})|& \le ϵ/5,\end{array}$$ by construction, we have $$|\psi (x_k)-\psi (x_{k+1})|\le 3ϵ/5.$$ Moreover, $$|\psi (x_k)-\psi (x)|\le 3ϵ/5(x_k\le x\le x_{k+1}),$$ 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 $$|\varphi (x)-\psi (x)|\le |\varphi (x)-\varphi (x_k)|+|\varphi (x_k)-\psi (x_k)|+|\psi (x_k)-\psi (x)|<ϵ,$$ i.e. the set of polygonal lines $\psi (x)$ forms an $ϵ$​-net for $\mathrm{\Phi }$. But there are obviously only finitely many such lines. Therefore $\mathrm{\Phi }$ is totally bounded.

By the continuity of $f$, we have $|f(x,y)|\le K$ in some domain $G^{\prime }\subseteq G$ 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 $x=b(a<x_0<b)$ which together with the first two lines form two isosceles triangles contained in $G^{\prime }$ 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 $L_1,L_2,\dots ,L_n,\dots$ 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 $n\to \mathrm{\infty }$. Let ${\varphi }_n$ be the function with graph $L_n$. Then this gives a family of functions ${\varphi }_1,{\varphi }_2,\dots ,{\varphi }_n,\dots$, 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 $\{{\varphi }_n\}$ is relatively compact and therefore contains a uniformly convergent subsequence $${\varphi }^{(1)},{\varphi }^{(2)},\dots ,{\varphi }^{(n)},\dots .$$ Let $\varphi (x)=\underset{n\to \mathrm{\infty }}{lim}{\varphi }^{(n)}(x)$. Then clearly $\varphi (x_0)=y_0$, so that the curve $y=\varphi (x)$ passes through the point $(x_0,y_0)$. We now show that the $y=\varphi (x)$ satisfies the differential equation $\frac{dy}{dx}=f(x,y)$ in the open interval $(a,b)$. This means showing that, given any $ϵ>0$ and any points $x^{\prime },x^{″}\in (a,b)$, we have $$|\frac{\varphi (x^{″})-\varphi (x^{\prime })}{x^{″}-x^{\prime }}-f(x^{\prime },\varphi (x^{\prime }))|<ϵ$$ whenever $|x^{″}-x^{\prime }|$ is sufficiently small, or equivalently that $$|\frac{{\varphi }^{(n)}(x^{″})-{\varphi }^{(n)}(x^{\prime })}{x^{″}-x^{\prime }}-f(x^{\prime },{\varphi }^{(n)}(x^{\prime }))|<ϵ$$ whenever $n$ is sufficiently large and $|x^{″}-x^{\prime }|$ is sufficiently small. Let $y^{\prime }=\varphi (x^{\prime })$. Then, by continuity of $f$, given any $ϵ>0$, there is a number $\eta >0$ such that $$f(x^{\prime },y^{\prime })-ϵ<f(x,y)<f(x^{\prime },y^{\prime })+ϵ$$ whenever $|x-x^{\prime }|<2\eta$, $|y-y^{\prime }|<4K\eta$. 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 $$|\varphi (x)-{\varphi }^{(n)}(x)|<K\eta .$$ Then all the Euler lines $L_n$ with $n>N$ live inside the rectange $Q$ (why?). Suppose $L_n$ has vertices $(a_0,b_0),(a_1,b_1),\dots ,(a_{k+1},b_{k+1})$, where $a\le x^{\prime }<a_1<a_2<\dots <a_k<x^{″}\le a_{k+1}$. [Note we assume that $x^{″}>x^{\prime }$. The case $x^{″}<x^{\prime }$ is treated similarly.] Then $$\begin{array}{rl}{\varphi }^{(n)}(a_1)-{\varphi }^{(n)}(x^{\prime })& =f(a_0,b_0)(a_1-x^{\prime }),\\ {\varphi }^{(n)}(a_{i+1})-{\varphi }^{(n)}(a_i)& =f(a_i,b_i)(a_{i+1}-a_i)(i=1,2,\dots ,k-1),\\ {\varphi }^{(n)}(x^{″})-{\varphi }^{(n)}(a_k)& =f(a_k,b_k)(a^{″}-a_k).\end{array}$$ Hence, if $|x^{″}-x^{\prime }|<\eta$, $$\begin{array}{rl}[f(x^{\prime },y^{\prime })-ϵ](a_1-x^{\prime })& <{\varphi }^{(n)}(a_1)-{\varphi }^{(n)}(x^{\prime })<[f(x^{\prime },y^{\prime })+ϵ](a_1-x^{\prime }),\\ [f(x^{\prime },y^{\prime })-ϵ](a_{i+1}-a_i)& <{\varphi }^{(n)}(a_{i+1})-{\varphi }^{(n)}(a_i)\\ & <[f(x^{\prime },y^{\prime })+ϵ](a_{i+1}-a_i)(i=1,2,\dots ,k-1),\\ [f(x^{\prime },y^{\prime })-ϵ](x^{″}-a_k)& <{\varphi }^{(n)}(x^{″})-{\varphi }^{(n)}(a_k)<[f(x^{\prime },y^{\prime })+ϵ](x^{″}-a_k).\end{array}$$ Adding these inequalities, we get $$[f(x^{\prime },y^{\prime })-ϵ](x^{″}-x^{\prime })<{\varphi }^{(n)}(x^{″})-{\varphi }^{(n)}(x^{\prime })<[f(x^{\prime },y^{\prime })+ϵ](x^{″}-x^{\prime })$$ if $|x^{″}-x^{\prime }|<\eta$, which is equivalent to $$|\frac{{\varphi }^{(n)}(x^{″})-{\varphi }^{(n)}(x^{\prime })}{x^{″}-x^{\prime }}-f(x^{\prime },{\varphi }^{(n)}(x^{\prime }))|<ϵ.$$