[proofplan] We first verify that the maximum formula defines a metric on the product: each metric axiom follows from the corresponding coordinatewise metric axiom and the fact that the index set $\{1,\dots,n\}$ is finite and nonempty. To prove completeness, we take an arbitrary $d$-[Cauchy sequence](/page/Cauchy%20Sequence) in $X$ and show that each coordinate sequence is Cauchy in its complete coordinate space. The coordinate limits assemble into a point of the product, and finiteness of the number of coordinates lets us pass from coordinatewise convergence to convergence in the maximum metric. [/proofplan]

[step:Verify that the maximum formula defines a metric on the product] Let $I := \{1,\dots,n\}$. Since $n \geq 1$, the set $I$ is finite and nonempty, so the maximum in the definition of $d$ is taken over a finite nonempty set of [real numbers](/page/Real%20Numbers). For all $x=(x_1,\dots,x_n) \in X$ and $y=(y_1,\dots,y_n) \in X$, each number $d_i(x_i,y_i)$ is nonnegative because $d_i$ is a metric. Hence \begin{align*} d(x,y)=\max_{i \in I} d_i(x_i,y_i) \geq 0. \end{align*} If $x=y$, then $x_i=y_i$ for every $i \in I$, so $d_i(x_i,y_i)=0$ for every $i \in I$, and therefore $d(x,y)=0$. Conversely, suppose $d(x,y)=0$. For every $i \in I$, \begin{align*} 0 \leq d_i(x_i,y_i) \leq \max_{j \in I} d_j(x_j,y_j)=0. \end{align*} Thus $d_i(x_i,y_i)=0$ for every $i \in I$. Since each $d_i$ is a metric, this implies $x_i=y_i$ for every $i \in I$, and hence $x=y$. For symmetry, let $x=(x_1,\dots,x_n) \in X$ and $y=(y_1,\dots,y_n) \in X$. Since each $d_i$ is symmetric, \begin{align*} d(x,y)=\max_{i \in I} d_i(x_i,y_i)=\max_{i \in I} d_i(y_i,x_i)=d(y,x). \end{align*} For the triangle inequality, let $x=(x_1,\dots,x_n)$, $y=(y_1,\dots,y_n)$, and $z=(z_1,\dots,z_n)$ be elements of $X$. For each $i \in I$, the triangle inequality for $d_i$ gives \begin{align*} d_i(x_i,z_i) \leq d_i(x_i,y_i)+d_i(y_i,z_i). \end{align*} By the definition of $d$, we also have \begin{align*} d_i(x_i,y_i) \leq d(x,y) \end{align*} and \begin{align*} d_i(y_i,z_i) \leq d(y,z). \end{align*} Therefore, for every $i \in I$, \begin{align*} d_i(x_i,z_i) \leq d(x,y)+d(y,z). \end{align*} Taking the maximum over $i \in I$ gives \begin{align*} d(x,z)=\max_{i \in I} d_i(x_i,z_i) \leq d(x,y)+d(y,z). \end{align*} Thus $d$ is a metric on $X$. [/step]

[step:Pass from a Cauchy sequence in the product to Cauchy sequences in every coordinate]Let $(x_k)_{k \in \mathbb{N}}$ be a Cauchy sequence in the [metric space](/page/Metric%20Space) $(X,d)$. For each $k \in \mathbb{N}$, write \begin{align*} x_k=(x_{k,1},\dots,x_{k,n}), \end{align*} where $x_{k,i} \in X_i$ for each $i \in I$. Fix $i \in I$. We claim that $(x_{k,i})_{k \in \mathbb{N}}$ is Cauchy in $(X_i,d_i)$. Let $\varepsilon>0$. Since $(x_k)_{k \in \mathbb{N}}$ is Cauchy in $(X,d)$, there exists $N_i \in \mathbb{N}$ such that for all $k,m \in \mathbb{N}$ with $k \geq N_i$ and $m \geq N_i$, \begin{align*} d(x_k,x_m)<\varepsilon. \end{align*} For such $k$ and $m$, the coordinate distance is bounded by the maximum: \begin{align*} d_i(x_{k,i},x_{m,i}) \leq \max_{j \in I} d_j(x_{k,j},x_{m,j})=d(x_k,x_m)<\varepsilon. \end{align*} Thus $(x_{k,i})_{k \in \mathbb{N}}$ is Cauchy in $(X_i,d_i)$.[/step]

[guided]The product metric controls every coordinate because it is defined as the largest coordinate distance. We make that control precise. Let $i \in I$ be fixed. To prove that $(x_{k,i})_{k \in \mathbb{N}}$ is Cauchy in $(X_i,d_i)$, take an arbitrary $\varepsilon>0$. Since $(x_k)_{k \in \mathbb{N}}$ is Cauchy in $(X,d)$, there exists $N_i \in \mathbb{N}$ such that whenever $k,m \geq N_i$, \begin{align*} d(x_k,x_m)<\varepsilon. \end{align*} Writing $x_k=(x_{k,1},\dots,x_{k,n})$ and $x_m=(x_{m,1},\dots,x_{m,n})$, the definition of $d$ gives \begin{align*} d(x_k,x_m)=\max_{j \in I} d_j(x_{k,j},x_{m,j}). \end{align*} A maximum is at least each one of the numbers over which it is taken. Therefore the fixed coordinate distance satisfies \begin{align*} d_i(x_{k,i},x_{m,i}) \leq \max_{j \in I} d_j(x_{k,j},x_{m,j})=d(x_k,x_m)<\varepsilon. \end{align*} This is exactly the Cauchy condition for the sequence $(x_{k,i})_{k \in \mathbb{N}}$ in the metric space $(X_i,d_i)$. Since $i$ was arbitrary, every coordinate sequence is Cauchy in its own coordinate metric.[/guided]

[step:Use coordinate completeness to construct the candidate limit in the product] For each $i \in I$, the sequence $(x_{k,i})_{k \in \mathbb{N}}$ is Cauchy in $(X_i,d_i)$ by the preceding step. Since $(X_i,d_i)$ is complete, there exists an element $x_i \in X_i$ such that \begin{align*} \lim_{k \to \infty} d_i(x_{k,i},x_i)=0. \end{align*} Define \begin{align*} x:=(x_1,\dots,x_n) \in X. \end{align*} This point is the only possible product limit of the sequence $(x_k)_{k \in \mathbb{N}}$ under coordinatewise convergence. [/step]

[step:Use finiteness of the product to prove convergence in the maximum metric] Let $\varepsilon>0$. For each $i \in I$, the convergence $d_i(x_{k,i},x_i) \to 0$ gives an integer $M_i \in \mathbb{N}$ such that for all $k \in \mathbb{N}$ with $k \geq M_i$, \begin{align*} d_i(x_{k,i},x_i)<\varepsilon. \end{align*} Because $I$ is finite and nonempty, the integer \begin{align*} M:=\max_{i \in I} M_i \end{align*} is well-defined. If $k \geq M$, then $k \geq M_i$ for every $i \in I$, and hence \begin{align*} d_i(x_{k,i},x_i)<\varepsilon \end{align*} for every $i \in I$. Taking the maximum over $i \in I$ yields \begin{align*} d(x_k,x)=\max_{i \in I} d_i(x_{k,i},x_i)<\varepsilon. \end{align*} Therefore $x_k \to x$ in the metric space $(X,d)$. Since every Cauchy sequence in $(X,d)$ converges to an element of $X$, the metric space $(X,d)$ is complete. [/step]