Proof of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space

Throughout, $\pi$ denotes the map described in claim 1, and coordinates of points of $\mathbb{R}^{m+n}$ are compared using the index decomposition recorded in the statement: every $k\in[m+n]$ lies in $[m]$ or equals $m+j$ for exactly one $j\in[n]$, and not both.

Claim 1. Let $w\in\mathbb{R}^{m+n}$ and let $\pi(w)=(\xi,\eta)$ with $\xi_k=w_k$ for $k\in[m]$ and $\eta_j=w_{m+j}$ for $j\in[n]$. Then $\iota(\pi(w))$ and $w$ have the same coordinate at every index of $[m+n]$, by the two defining clauses of $\iota$, so $\iota(\pi(w))=w$. Conversely, for $(\xi,\eta)\in\mathbb{R}^m\times\mathbb{R}^n$ the first entry of $\pi(\iota(\xi,\eta))$ has $k$th coordinate $\iota(\xi,\eta)_k=\xi_k$ for $k\in[m]$, and its second entry has $j$th coordinate $\iota(\xi,\eta)_{m+j}=\eta_j$ for $j\in[n]$; hence $\pi(\iota(\xi,\eta))=(\xi,\eta)$. So $\iota$ and $\pi$ are mutually inverse, and $\iota$ is a bijection.

Claim 2. Two points of $\mathbb{R}^{m+n}$ are equal exactly when they agree at every index of $[m+n]$, so it suffices to compare coordinates at $k\in[m]$ and at $m+j$ with $j\in[n]$.

By Sum of Points of $\mathbb{R}^n$ the $k$th coordinate of $\iota(\xi,\eta)+\iota(\xi',\eta')$ is $\iota(\xi,\eta)_k+\iota(\xi',\eta')_k$. For $k\in[m]$ this is $\xi_k+\xi'_k$, which is the $k$th coordinate of $\xi+\xi'$ and hence of $\iota(\xi+\xi',\eta+\eta')$; for the index $m+j$ with $j\in[n]$ it is $\eta_j+\eta'_j$, which is the $j$th coordinate of $\eta+\eta'$ and hence the $(m+j)$th coordinate of $\iota(\xi+\xi',\eta+\eta')$. This proves the first identity. The second is identical, using Scalar Multiple of a Point of $\mathbb{R}^n$ and that the $k$th coordinate of $\lambda z$ is $\lambda z_k$; the third is identical, using the coordinatewise description of the difference in Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$.

Claim 3. Let $c:[m+n]\to\mathbb{R}$ be the family $c_k=\iota(\xi,\eta)_k\,\iota(\xi',\eta')_k$, so that by Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$

By Splitting a Finite Sum at an Index this equals $\sum_{k=1}^{m}c'_k+\sum_{j=1}^{n}c''_j$, where $c'$ is the restriction of $c$ to $[m]$ and $c''_j=c_{m+j}$. For $k\in[m]$ we have $c'_k=\xi_k\xi'_k$, so $\sum_{k=1}^{m}c'_k=\xi\cdot\xi'$; and for $j\in[n]$ we have $c''_j=\eta_j\eta'_j$, so $\sum_{j=1}^{n}c''_j=\eta\cdot\eta'$. This proves the first identity of claim 3.

Taking $\xi'=\xi$ and $\eta'=\eta$ and using claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, which gives $\lVert z\rVert^{2}=z\cdot z$ in each Euclidean space, yields $\lVert\iota(\xi,\eta)\rVert^{2}=\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}$.

Claim 4. By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, $d_E(u,u')=\lVert u-u'\rVert$ in each Euclidean space. By claim 2 of the present lemma, $\iota(z)-\iota(z')=\iota(\xi-\xi',\eta-\eta')$, so by claim 3,

which is the first assertion.

Write $\alpha=d_E(\xi,\xi')$, $\beta=d_E(\eta,\eta')$, $\gamma=d_E(\iota(z),\iota(z'))$ and $\mu=d_{\times}(z,z')$, all nonnegative because a metric takes nonnegative values. By Product Metric on the Cartesian Product of Two Metric Spaces, $\mu$ is the maximum of $\alpha$ and $\beta$, so $\alpha\le\mu$ and $\beta\le\mu$ by claim 1 of Elementary Properties of the Maximum of Two Elements, and $\mu$ equals $\alpha$ or $\beta$ by claim 2 of that lemma.

Since $\mu$ is one of $\alpha,\beta$, we get $\mu^{2}\le\alpha^{2}+\beta^{2}=\gamma^{2}$, using that the other square is nonnegative by Nonnegativity of Squares in an Ordered Field and claim 3 of Elementary Arithmetic in an Ordered Field. As $\mu$ and $\gamma$ are nonnegative, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\mu\le\gamma$.

For the upper bound, $\alpha\le\mu$ and $\beta\le\mu$ give $\alpha^{2}\le\mu^{2}$ and $\beta^{2}\le\mu^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, so

the middle inequality holding because $\mu\,\mu$ is nonnegative, and the last equality being field arithmetic. Since $\gamma$ and $\mu+\mu$ are nonnegative, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\gamma\le\mu+\mu$.