[proofplan] We prove [total boundedness](/page/Total%20Boundedness) directly from the definition. Given $\varepsilon > 0$, boundedness places $A$ inside a large cube $[-R,R]^n$. We divide that cube into finitely many congruent smaller cubes whose Euclidean radius from center to corner is less than $\varepsilon$, and the finitely many centers of these cubes give an $\varepsilon$-cover of $A$. [/proofplan] [step:Place the bounded set inside a Euclidean cube] If $A = \varnothing$, then $A$ is covered by the empty finite family of balls for every $\varepsilon > 0$, so $A$ is [totally bounded](/page/Totally%20Bounded). Assume henceforth that $A \neq \varnothing$. Because $A$ is bounded in the [Euclidean metric](/page/Euclidean%20Metric), there exist $a_0 \in A$ and $S > 0$ such that $|x-a_0| \leq S$ for every $x \in A$. Define $R := |a_0| + S + 1$. Then $R > 0$, and for every $x \in A$ the triangle inequality gives $|x| \leq |x-a_0| + |a_0| \leq S + |a_0| < R$. Hence each coordinate of $x$ satisfies $|x_i| \leq |x| < R$, so $A \subset [-R,R]^n$. [/step] [step:Choose a finite grid whose cells have radius less than $\varepsilon$] Fix $\varepsilon > 0$. By the [Archimedean property](/theorems/737) of $\mathbb{N}$, choose $m \in \mathbb{N}$ such that \begin{align*} \frac{\sqrt{n}R}{m} < \varepsilon. \end{align*} Let $J$ denote the finite index set \begin{align*} J := \{0,1,\dots,m-1\}^n. \end{align*} For each $j = (j_1,\dots,j_n) \in J$, define the grid center $c_j \in \mathbb{R}^n$ by declaring its $i$th coordinate to be \begin{align*} (c_j)_i := -R + \frac{2R}{m}\left(j_i + \frac{1}{2}\right) \end{align*} for each $i \in \{1,\dots,n\}$. The set of grid centers \begin{align*} F := \{c_j : j \in J\} \end{align*} is finite because $J$ is finite. [guided] Now we build the finite set that will serve as the centers of the covering balls. The reason to use a cube rather than a ball is that a cube has a simple coordinate grid. Fix $\varepsilon > 0$. We already know that $A \subset [-R,R]^n$ for some $R > 0$. We choose the number of subdivisions $m \in \mathbb{N}$ large enough that the distance from the center of each small cube to any of its corners is less than $\varepsilon$. Since that distance will be $\sqrt{n}R/m$, the Archimedean property lets us choose $m$ such that \begin{align*} \frac{\sqrt{n}R}{m} < \varepsilon. \end{align*} Define the finite index set \begin{align*} J := \{0,1,\dots,m-1\}^n. \end{align*} Each element $j = (j_1,\dots,j_n) \in J$ labels one small cube in the coordinate subdivision of $[-R,R]^n$. For each such $j$, define a point $c_j \in \mathbb{R}^n$ by \begin{align*} (c_j)_i := -R + \frac{2R}{m}\left(j_i + \frac{1}{2}\right) \end{align*} for every coordinate index $i \in \{1,\dots,n\}$. This point is the center of the corresponding small cube. Finally define \begin{align*} F := \{c_j : j \in J\}. \end{align*} Because $J$ has exactly $m^n$ elements, $F$ is finite. This finiteness is the key point: total boundedness requires a finite family of balls, and the grid gives finitely many candidate centers. [/guided] [/step] [step:Show every point of $A$ lies within $\varepsilon$ of a grid center] Let $x \in A$. Since $A \subset [-R,R]^n$, we have $x_i \in [-R,R]$ for each $i \in \{1,\dots,n\}$. For each coordinate index $i$, choose $j_i \in \{0,1,\dots,m-1\}$ such that $x_i$ lies in the $j_i$th subinterval of the subdivision of $[-R,R]$ into $m$ intervals of length $2R/m$, choosing $j_i=m-1$ if $x_i=R$. Let $j := (j_1,\dots,j_n) \in J$. By construction, for each $i \in \{1,\dots,n\}$, $|x_i - (c_j)_i| \leq \frac{R}{m}$. Therefore, using the Euclidean norm formula, $|x-c_j|^2 = \sum_{i=1}^n |x_i-(c_j)_i|^2 \leq \sum_{i=1}^n \left(\frac{R}{m}\right)^2 = \frac{nR^2}{m^2}$. Taking square roots gives $|x-c_j| \leq \frac{\sqrt{n}R}{m} < \varepsilon$. Thus every $x \in A$ belongs to some Euclidean open ball $B(c_j,\varepsilon)$ with $c_j \in F$. [/step] [step:Conclude total boundedness from the finite ball cover] For the arbitrary $\varepsilon > 0$ fixed above, define the Euclidean open ball centered at $c \in \mathbb{R}^n$ with radius $\varepsilon$ by $B(c,\varepsilon) := \{y \in \mathbb{R}^n : |y-c| < \varepsilon\}$. We have constructed a finite set $F \subset \mathbb{R}^n$ such that $A \subset \bigcup_{c \in F} B(c,\varepsilon)$. This is exactly the definition of total boundedness for $A$ as a subset of the Euclidean [metric space](/page/Metric%20Space) $\mathbb{R}^n$. Hence $A$ is totally bounded. [/step]