[proofplan] We prove the identity by expanding both sides from the definitions. The Euclidean [product metric](/page/Product%20Metric) squares to the sum of the squared Euclidean distances in the two factors. Under the coordinate-concatenation map, that same sum is exactly the squared Euclidean distance in $\mathbb{R}^{m+n}$. Since both distances are nonnegative, equality of their squares implies equality of the distances. [/proofplan] [step:Expand the product distance in coordinates] Let $(x,y),(x',y') \in \mathbb{R}^m \times \mathbb{R}^n$ be arbitrary. Write \begin{align*} x=(x_1,\dots,x_m), \quad x'=(x_1',\dots,x_m'), \quad y=(y_1,\dots,y_n), \quad y'=(y_1',\dots,y_n'). \end{align*} By the definition of the [Euclidean metric](/page/Euclidean%20Metric) on $\mathbb{R}^m$ and $\mathbb{R}^n$, \begin{align*} d_m(x,x')^2=\sum_{i=1}^{m}(x_i-x_i')^2. \end{align*} Also, \begin{align*} d_n(y,y')^2=\sum_{j=1}^{n}(y_j-y_j')^2. \end{align*} Using the definition of the Euclidean product metric $d_2$, we obtain \begin{align*} d_2((x,y),(x',y'))^2=\sum_{i=1}^{m}(x_i-x_i')^2+\sum_{j=1}^{n}(y_j-y_j')^2. \end{align*} [guided] We start with two arbitrary points of the product space because the theorem asserts an equality of distances for every pair of points. Let \begin{align*} x=(x_1,\dots,x_m), \quad x'=(x_1',\dots,x_m'), \quad y=(y_1,\dots,y_n), \quad y'=(y_1',\dots,y_n'). \end{align*} The Euclidean metric on $\mathbb{R}^m$ is defined by taking the square root of the sum of squared coordinate differences, so its square is \begin{align*} d_m(x,x')^2=\sum_{i=1}^{m}(x_i-x_i')^2. \end{align*} Likewise, the Euclidean metric on $\mathbb{R}^n$ gives \begin{align*} d_n(y,y')^2=\sum_{j=1}^{n}(y_j-y_j')^2. \end{align*} The metric $d_2$ is the $\ell^2$ product metric, so it combines the two factor distances by the same square-sum rule: \begin{align*} d_2((x,y),(x',y'))=\left(d_m(x,x')^2+d_n(y,y')^2\right)^{1/2}. \end{align*} Squaring this nonnegative quantity and substituting the two Euclidean distance formulas gives \begin{align*} d_2((x,y),(x',y'))^2=\sum_{i=1}^{m}(x_i-x_i')^2+\sum_{j=1}^{n}(y_j-y_j')^2. \end{align*} This is the expression we will compare with the Euclidean distance after concatenating coordinates. [/guided] [/step] [step:Identify the same coordinate sum in $\mathbb{R}^{m+n}$] Define \begin{align*} z:=C(x,y)=(x_1,\dots,x_m,y_1,\dots,y_n) \end{align*} and \begin{align*} z':=C(x',y')=(x_1',\dots,x_m',y_1',\dots,y_n'). \end{align*} By the definition of the Euclidean metric on $\mathbb{R}^{m+n}$, \begin{align*} d_{m+n}(z,z')^2=\sum_{i=1}^{m}(x_i-x_i')^2+\sum_{j=1}^{n}(y_j-y_j')^2. \end{align*} Comparing this identity with the formula obtained for $d_2((x,y),(x',y'))^2$, we have \begin{align*} d_2((x,y),(x',y'))^2=d_{m+n}(C(x,y),C(x',y'))^2. \end{align*} [/step] [step:Take nonnegative square roots to conclude the metric identity] Both $d_2((x,y),(x',y'))$ and $d_{m+n}(C(x,y),C(x',y'))$ are nonnegative [real numbers](/page/Real%20Numbers) because they are distances. Therefore equality of their squares implies equality of the numbers themselves: \begin{align*} d_2((x,y),(x',y'))=d_{m+n}(C(x,y),C(x',y')). \end{align*} Since $(x,y)$ and $(x',y')$ were arbitrary points of $\mathbb{R}^m \times \mathbb{R}^n$, the Euclidean product metric agrees with the usual Euclidean metric on $\mathbb{R}^{m+n}$ under coordinate concatenation. [/step]