[proofplan] We prove both directions. For the forward direction (total boundedness implies uniform boundedness and equicontinuity), we extract uniform boundedness from a $1$-net by the triangle inequality, and equicontinuity by contradiction using total boundedness. For the reverse direction (uniform boundedness and equicontinuity imply total boundedness), we discretise both the domain $K$ (using a finite $\delta$-net from the compactness of $K$) and the range $[-M, M]$ (using a grid of mesh $\varepsilon$), then show that each function in $\mathcal{F}$ is $\varepsilon$-close to one of finitely many "approximate profiles." [/proofplan] [step:Forward direction: extract uniform boundedness from a finite $1$-net] Assume $\mathcal{F}$ is totally bounded in $(C(K), \|\cdot\|_\infty)$. Applying the definition with $\varepsilon = 1$, there exists a finite set $\{g_1, \ldots, g_N\} \subset C(K)$ with $\mathcal{F} \subset \bigcup_{i=1}^N B_\infty(g_i, 1)$, where $B_\infty(g, r) = \{h \in C(K) : \|h - g\|_\infty < r\}$. For any $f \in \mathcal{F}$, there exists $i$ with $\|f - g_i\|_\infty < 1$, so \begin{align*} \|f\|_\infty \le \|f - g_i\|_\infty + \|g_i\|_\infty < 1 + \max_{1 \le j \le N} \|g_j\|_\infty. \end{align*} Setting $M = 1 + \max_{1 \le j \le N} \|g_j\|_\infty$, we have $\|f\|_\infty \le M$ for all $f \in \mathcal{F}$. Thus $\mathcal{F}$ is uniformly bounded. [/step] [step:Forward direction: extract equicontinuity from a finite $\varepsilon/3$-net] Fix $\varepsilon > 0$. Since $\mathcal{F}$ is totally bounded, there exists a finite set $\{g_1, \ldots, g_N\} \subset C(K)$ with $\mathcal{F} \subset \bigcup_{i=1}^N B_\infty(g_i, \varepsilon/3)$. Each $g_i$ is continuous on the compact metric space $K$, hence uniformly continuous by the [Heine-Cantor Theorem](/theorems/954). For each $i \in \{1, \ldots, N\}$, there exists $\delta_i > 0$ such that $d_K(x, y) < \delta_i$ implies $|g_i(x) - g_i(y)| < \varepsilon/3$. Set $\delta = \min(\delta_1, \ldots, \delta_N) > 0$ (the minimum of finitely many positive numbers is positive). Let $f \in \mathcal{F}$ and $x, y \in K$ with $d_K(x, y) < \delta$. Choose $i$ with $\|f - g_i\|_\infty < \varepsilon/3$. Then \begin{align*} |f(x) - f(y)| &\le |f(x) - g_i(x)| + |g_i(x) - g_i(y)| + |g_i(y) - f(y)| \\ &\le \|f - g_i\|_\infty + |g_i(x) - g_i(y)| + \|f - g_i\|_\infty \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \end{align*} Since $\delta$ is independent of $f$, the family $\mathcal{F}$ is equicontinuous. [guided] The idea is that a finite $\varepsilon/3$-net lets us "reduce the equicontinuity of the infinite family $\mathcal{F}$ to the uniform continuity of finitely many functions." Each $f \in \mathcal{F}$ is uniformly close to some $g_i$ in the net (within $\varepsilon/3$ in sup-norm), and $g_i$ is uniformly continuous on the compact space $K$. The $\varepsilon/3$-argument decomposes the oscillation of $f$ into three pieces: deviation from $g_i$ at $x$, oscillation of $g_i$ from $x$ to $y$, and deviation from $g_i$ at $y$. Fix $\varepsilon > 0$ and choose a finite $\varepsilon/3$-net $\{g_1, \ldots, g_N\}$ for $\mathcal{F}$. Each $g_i: K \to \mathbb{R}$ is continuous. Since $K$ is compact, the [Heine-Cantor Theorem](/theorems/954) guarantees that each $g_i$ is uniformly continuous: for each $i$, there exists $\delta_i > 0$ such that $d_K(x, y) < \delta_i$ implies $|g_i(x) - g_i(y)| < \varepsilon/3$. The critical observation is that we can take a single $\delta$ for all $i$: set $\delta = \min_{1 \le i \le N} \delta_i$. This minimum is positive because it is the minimum of finitely many positive numbers. Now for any $f \in \mathcal{F}$ and any $x, y \in K$ with $d_K(x, y) < \delta$: 1. Choose $i$ with $\|f - g_i\|_\infty < \varepsilon/3$. This gives $|f(x) - g_i(x)| < \varepsilon/3$ and $|f(y) - g_i(y)| < \varepsilon/3$. 2. Since $d_K(x, y) < \delta \le \delta_i$, the uniform continuity of $g_i$ gives $|g_i(x) - g_i(y)| < \varepsilon/3$. 3. The triangle inequality combines these: \begin{align*} |f(x) - f(y)| \le |f(x) - g_i(x)| + |g_i(x) - g_i(y)| + |g_i(y) - f(y)| < \varepsilon. \end{align*} The $\delta$ does not depend on $f$ (only on $\varepsilon$ and the net functions $g_i$), so the bound holds uniformly over all $f \in \mathcal{F}$. This is exactly equicontinuity. The argument would fail if the net were infinite, since $\min_{i \ge 1} \delta_i$ could be zero. [/guided] [/step] [step:Reverse direction: discretise $K$ into a finite $\delta$-net using compactness] Assume $\mathcal{F}$ is uniformly bounded (with bound $M > 0$: $\|f\|_\infty \le M$ for all $f \in \mathcal{F}$) and equicontinuous. Fix $\varepsilon > 0$. By equicontinuity, there exists $\delta > 0$ such that $d_K(x, y) < \delta$ implies $|f(x) - f(y)| < \varepsilon/3$ for all $f \in \mathcal{F}$ and all $x, y \in K$. Since $K$ is compact, it is totally bounded. There exists a finite $\delta$-net $\{t_1, \ldots, t_L\} \subset K$ with $K \subset \bigcup_{j=1}^L B_{d_K}(t_j, \delta)$. [guided] The reverse direction is the deeper implication. The strategy is to reduce the problem of approximating continuous functions on $K$ to the problem of approximating their values at finitely many sample points. Fix $\varepsilon > 0$. Equicontinuity gives $\delta > 0$ controlling the oscillation of all functions in $\mathcal{F}$ simultaneously. Compactness of $K$ gives a finite $\delta$-net $\{t_1, \ldots, t_L\} \subset K$. The key observation is that any $f \in \mathcal{F}$ is determined up to $\varepsilon/3$ by its "fingerprint" $(f(t_1), \ldots, f(t_L)) \in \mathbb{R}^L$: for any $x \in K$, there exists $t_j$ with $d_K(x, t_j) < \delta$, so $|f(x) - f(t_j)| < \varepsilon/3$. Thus $f$ is within $\varepsilon/3$ of any function that agrees with $f$ at the sample points. [/guided] [/step] [step:Discretise the range and count approximate profiles] Partition the interval $[-M, M]$ into subintervals of length at most $\varepsilon/3$, using $P = \lceil 6M/\varepsilon \rceil$ subintervals. Let $\{v_1, \ldots, v_{P+1}\}$ be the endpoints (or midpoints) of these subintervals, chosen so that for every $s \in [-M, M]$, there exists $v_q$ with $|s - v_q| \le \varepsilon/6$. For each $f \in \mathcal{F}$, evaluate $f$ at the net points $t_1, \ldots, t_L$. Each value $f(t_j) \in [-M, M]$ (by uniform boundedness), so there exists $q_j \in \{1, \ldots, P+1\}$ with $|f(t_j) - v_{q_j}| \le \varepsilon/6$. The tuple $(q_1, \ldots, q_L) \in \{1, \ldots, P+1\}^L$ is the **approximate profile** of $f$. The total number of approximate profiles is at most $(P + 1)^L$, which is finite. [/step] [step:Construct a finite $\varepsilon$-net for $\mathcal{F}$ from representative functions] For each approximate profile $(q_1, \ldots, q_L) \in \{1, \ldots, P+1\}^L$ that is realised by some $f \in \mathcal{F}$, choose one representative $f_{(q_1,\ldots,q_L)} \in \mathcal{F}$. The collection of all representatives forms a finite set $\mathcal{G} \subset \mathcal{F}$ with $|\mathcal{G}| \le (P+1)^L$. We verify that $\mathcal{G}$ is an $\varepsilon$-net for $\mathcal{F}$. Let $f \in \mathcal{F}$ with approximate profile $(q_1, \ldots, q_L)$, and let $g = f_{(q_1,\ldots,q_L)} \in \mathcal{G}$ be the representative sharing the same profile. For each $j \in \{1, \ldots, L\}$, \begin{align*} |f(t_j) - g(t_j)| \le |f(t_j) - v_{q_j}| + |v_{q_j} - g(t_j)| \le \frac{\varepsilon}{6} + \frac{\varepsilon}{6} = \frac{\varepsilon}{3}. \end{align*} For any $x \in K$, choose $j$ with $d_K(x, t_j) < \delta$. Equicontinuity gives $|f(x) - f(t_j)| < \varepsilon/3$ and $|g(x) - g(t_j)| < \varepsilon/3$. Therefore \begin{align*} |f(x) - g(x)| &\le |f(x) - f(t_j)| + |f(t_j) - g(t_j)| + |g(t_j) - g(x)| \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \end{align*} Since this holds for all $x \in K$, we have $\|f - g\|_\infty < \varepsilon$. Therefore $\mathcal{G}$ is a finite $\varepsilon$-net for $\mathcal{F}$, and $\mathcal{F}$ is totally bounded. [guided] Let us carefully track the constants. The goal is to show that for any $\varepsilon > 0$, there exists a finite $\varepsilon$-net for $\mathcal{F}$ in the sup-norm. The $\varepsilon$-budget must be split three ways: two parts for equicontinuity (oscillation of $f$ and $g$ from $x$ to the nearest net point $t_j$) and one part for the profile discretisation error (the gap between $f(t_j)$ and $g(t_j)$). **Setup.** Given $\varepsilon > 0$, apply equicontinuity with parameter $\varepsilon/3$: there exists $\delta > 0$ such that $d_K(x, y) < \delta$ implies $|f(x) - f(y)| < \varepsilon/3$ for all $f \in \mathcal{F}$. Choose a finite $\delta$-net $\{t_1, \ldots, t_L\} \subset K$. Discretise $[-M, M]$ with mesh $\varepsilon/3$ (so the nearest grid point is within $\varepsilon/6$), using $P + 1$ grid points. **Profile assignment.** For each $f \in \mathcal{F}$ and each $j$, choose $q_j$ with $|f(t_j) - v_{q_j}| \le \varepsilon/6$. If $g$ shares the same profile, then $|f(t_j) - g(t_j)| \le |f(t_j) - v_{q_j}| + |v_{q_j} - g(t_j)| \le \varepsilon/6 + \varepsilon/6 = \varepsilon/3$. **Pointwise bound.** For $x \in K$, choose $t_j$ with $d_K(x, t_j) < \delta$. Then \begin{align*} |f(x) - g(x)| &\le |f(x) - f(t_j)| + |f(t_j) - g(t_j)| + |g(t_j) - g(x)| \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \end{align*} Since this holds for all $x \in K$, we have $\|f - g\|_\infty < \varepsilon$. The number of distinct profiles is at most $(P+1)^L$, so we obtain a finite $\varepsilon$-net of size at most $(P+1)^L = \lceil 6M/\varepsilon + 1 \rceil^L$. Since $\varepsilon > 0$ was arbitrary, $\mathcal{F}$ is totally bounded. The proof reveals the quantitative content: the covering number satisfies $\mathcal{N}(\mathcal{F}, \varepsilon) \le \lceil 6M/\varepsilon + 1 \rceil^L$, where $L$ is the size of a $\delta(\varepsilon)$-net for $K$ and $\delta(\varepsilon)$ is the modulus of equicontinuity. When $\mathcal{F}$ consists of $\Lambda$-Lipschitz functions, $\delta = \varepsilon/(3\Lambda)$ and $L \le \lceil 3\Lambda \operatorname{diam}(K)/\varepsilon \rceil^{\dim(K)}$, recovering the polynomial growth of covering numbers in the Lipschitz case. [/guided] [/step]