[proofplan] We prove each of the four assertions separately. Part (1) rescales an $\varepsilon$-net by $\lambda$. Part (2) constructs an $\varepsilon$-net for $A + B$ from $\varepsilon/2$-nets for $A$ and $B$ via pairwise sums. Part (3) shows that the convex hull $\operatorname{conv}(A)$ is contained in the Minkowski sum of $n+1$ scaled copies of $A$ for a suitable finite-dimensional approximation, then applies parts (1) and (2) together with a direct $\varepsilon$-net construction. Part (4) follows from the [Density Characterisation of Total Boundedness](/theorems/1088). [/proofplan]

[step:Part (1): Rescale the $\varepsilon$-net by $\lambda$] Let $\lambda \in \mathbb{R}$ and let $A \subset X$ be totally bounded. If $\lambda = 0$, then $\lambda A = \{0\}$, which is finite and hence totally bounded. Assume $\lambda \neq 0$. Fix $\varepsilon > 0$. Since $A$ is totally bounded, there exists a finite set $\{a_1, \ldots, a_N\} \subset X$ with $A \subset \bigcup_{i=1}^N B(a_i, \varepsilon/|\lambda|)$. For any $\lambda a \in \lambda A$ with $a \in A$, there exists $i$ with $\|a - a_i\| < \varepsilon/|\lambda|$, so \begin{align*} \|\lambda a - \lambda a_i\| = |\lambda| \cdot \|a - a_i\| < |\lambda| \cdot \frac{\varepsilon}{|\lambda|} = \varepsilon. \end{align*} Therefore $\{\lambda a_1, \ldots, \lambda a_N\}$ is a finite $\varepsilon$-net for $\lambda A$. [/step]

[step:Part (2): Build an $\varepsilon$-net for $A + B$ from $\varepsilon/2$-nets for $A$ and $B$]Let $A, B \subset X$ be totally bounded. Fix $\varepsilon > 0$. Since $A$ is totally bounded, there exists a finite set $\{a_1, \ldots, a_N\} \subset X$ with $A \subset \bigcup_{i=1}^N B(a_i, \varepsilon/2)$. Since $B$ is totally bounded, there exists a finite set $\{b_1, \ldots, b_K\} \subset X$ with $B \subset \bigcup_{j=1}^K B(b_j, \varepsilon/2)$. Consider the finite set $F = \{a_i + b_j : 1 \le i \le N, \, 1 \le j \le K\}$, which has at most $NK$ elements. For any $a + b \in A + B$ with $a \in A$ and $b \in B$, there exist $i, j$ with $\|a - a_i\| < \varepsilon/2$ and $\|b - b_j\| < \varepsilon/2$. By the triangle inequality, \begin{align*} \|(a + b) - (a_i + b_j)\| \le \|a - a_i\| + \|b - b_j\| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $F$ is a finite $\varepsilon$-net for $A + B$.[/step]

[guided]The idea is straightforward: approximate $a$ by a net point $a_i$ and $b$ by a net point $b_j$, then $a + b$ is approximated by $a_i + b_j$. The set of all pairwise sums $a_i + b_j$ is finite (at most $N \cdot K$ elements), and the error is controlled by splitting the $\varepsilon$-budget between the two approximations. Fix $\varepsilon > 0$. Choose finite $\varepsilon/2$-nets: $\{a_1, \ldots, a_N\}$ for $A$ and $\{b_1, \ldots, b_K\}$ for $B$. Define $F = \{a_i + b_j : 1 \le i \le N, \, 1 \le j \le K\}$, so $|F| \le NK < \infty$. For $a + b \in A + B$, pick $i$ with $\|a - a_i\| < \varepsilon/2$ and $j$ with $\|b - b_j\| < \varepsilon/2$. Then \begin{align*} \|(a + b) - (a_i + b_j)\| = \|(a - a_i) + (b - b_j)\| \le \|a - a_i\| + \|b - b_j\| < \varepsilon. \end{align*} This argument uses the linear structure of $X$ in an essential way: the Minkowski sum $A + B$ is defined via vector addition, and the norm satisfies the triangle inequality. In a general metric space, there is no natural "sum" operation, so this stability result is specific to normed (or at least semi-normed) settings.[/guided]

[step:Part (3): Approximate convex combinations using a finite $\varepsilon/2$-net]Let $A \subset X$ be totally bounded. Fix $\varepsilon > 0$. By total boundedness, there exists a finite set $F = \{a_1, \ldots, a_N\} \subset X$ with $A \subset \bigcup_{i=1}^N B(a_i, \varepsilon/2)$. Every element of $\operatorname{conv}(A)$ has the form $\sum_{k=1}^m \lambda_k x_k$ for some $m \in \mathbb{N}$, $x_1, \ldots, x_m \in A$, and $\lambda_1, \ldots, \lambda_m \ge 0$ with $\sum_{k=1}^m \lambda_k = 1$. For each $x_k$, choose $a_{i_k} \in F$ with $\|x_k - a_{i_k}\| < \varepsilon/2$. Then \begin{align*} \left\|\sum_{k=1}^m \lambda_k x_k - \sum_{k=1}^m \lambda_k a_{i_k}\right\| = \left\|\sum_{k=1}^m \lambda_k(x_k - a_{i_k})\right\| \le \sum_{k=1}^m \lambda_k \|x_k - a_{i_k}\| < \sum_{k=1}^m \lambda_k \cdot \frac{\varepsilon}{2} = \frac{\varepsilon}{2}. \end{align*} The point $\sum_{k=1}^m \lambda_k a_{i_k}$ is a convex combination of elements of $F$, so it lies in $\operatorname{conv}(F)$. Since $F$ is finite, $\operatorname{conv}(F)$ is a convex hull of finitely many points in $\mathbb{R}^N$-many dimensions, and is contained in the affine hull of $F$, which has dimension at most $N - 1$. We now cover $\operatorname{conv}(F)$. The set $\operatorname{conv}(F)$ is a compact subset of the finite-dimensional affine subspace $\operatorname{aff}(F)$ (as a continuous image of the compact simplex $\Delta_{N-1} = \{(\mu_1, \ldots, \mu_N) \in \mathbb{R}^N : \mu_i \ge 0, \, \sum \mu_i = 1\}$ under the map $\mu \mapsto \sum_{i=1}^N \mu_i a_i$). In particular, $\operatorname{conv}(F)$ is totally bounded: there exists a finite set $G \subset X$ with $\operatorname{conv}(F) \subset \bigcup_{g \in G} B(g, \varepsilon/2)$. For any $v = \sum_{k=1}^m \lambda_k x_k \in \operatorname{conv}(A)$, the point $w = \sum_{k=1}^m \lambda_k a_{i_k} \in \operatorname{conv}(F)$ satisfies $\|v - w\| < \varepsilon/2$, and there exists $g \in G$ with $\|w - g\| < \varepsilon/2$. By the triangle inequality, \begin{align*} \|v - g\| \le \|v - w\| + \|w - g\| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $G$ is a finite $\varepsilon$-net for $\operatorname{conv}(A)$.[/step]

[guided]The challenge in proving total boundedness of $\operatorname{conv}(A)$ is that convex combinations involve arbitrary weights and arbitrary numbers of points: even though $A$ can be covered by $N$ balls, a convex combination may involve thousands of points from $A$. The key insight is that the convexity of the weights ($\lambda_k \ge 0$, $\sum \lambda_k = 1$) means that replacing each $x_k$ by a nearby net point introduces a controlled error. Fix $\varepsilon > 0$ and choose a finite $\varepsilon/2$-net $F = \{a_1, \ldots, a_N\}$ for $A$. **Approximation step.** For $v = \sum_{k=1}^m \lambda_k x_k \in \operatorname{conv}(A)$, replace each $x_k$ by the nearest net point $a_{i_k}$. The error is \begin{align*} \left\|v - \sum_{k=1}^m \lambda_k a_{i_k}\right\| \le \sum_{k=1}^m \lambda_k \|x_k - a_{i_k}\| < \frac{\varepsilon}{2}, \end{align*} where the first inequality uses the triangle inequality and the convexity condition $\lambda_k \ge 0$, and the second uses $\|x_k - a_{i_k}\| < \varepsilon/2$ together with $\sum \lambda_k = 1$. This shows $\operatorname{conv}(A) \subset \operatorname{conv}(F) + B(0, \varepsilon/2)$, where the inclusion is understood in terms of Minkowski sums. **Covering the finite convex hull.** The set $\operatorname{conv}(F)$ is the image of the standard simplex $\Delta_{N-1} \subset \mathbb{R}^N$ under the continuous linear map $\mu \mapsto \sum_{i=1}^N \mu_i a_i$. Since $\Delta_{N-1}$ is compact (closed and bounded in $\mathbb{R}^N$), its image $\operatorname{conv}(F)$ is compact and hence totally bounded. Choose a finite $\varepsilon/2$-net $G$ for $\operatorname{conv}(F)$. **Combining.** For $v \in \operatorname{conv}(A)$, the above approximation gives $w \in \operatorname{conv}(F)$ with $\|v - w\| < \varepsilon/2$, and the covering of $\operatorname{conv}(F)$ gives $g \in G$ with $\|w - g\| < \varepsilon/2$. The triangle inequality yields $\|v - g\| < \varepsilon$, so $G$ is a finite $\varepsilon$-net for $\operatorname{conv}(A)$.[/guided]

[step:Part (4): Total boundedness of $\overline{A}$ from the Density Characterisation] Let $A \subset X$ be totally bounded. By the [Density Characterisation of Total Boundedness](/theorems/1088) (equivalence of conditions (1) and (3)), the total boundedness of $A$ implies the total boundedness of $\overline{A}$. Alternatively, a direct argument: fix $\varepsilon > 0$. Since $A$ is totally bounded, there exists a finite set $F = \{a_1, \ldots, a_N\} \subset X$ with $A \subset \bigcup_{i=1}^N B(a_i, \varepsilon/2)$. For any $y \in \overline{A}$, there exists $a \in A$ with $\|y - a\| < \varepsilon/2$ (by the definition of closure), and there exists $a_i \in F$ with $\|a - a_i\| < \varepsilon/2$. The triangle inequality gives $\|y - a_i\| < \varepsilon$, so $F$ is a finite $\varepsilon$-net for $\overline{A}$. [/step]