Proof of Coordinate Isometry Determined by a Finite Orthonormal Basis

For $u\in V$ and $k\in[n]$ write $\hat u_{k}=\langle e_{k},u\rangle$, so that $\Phi(u)_{k}=\operatorname{Re}\hat u_{k}$ and $\Phi(u)_{n+k}=\operatorname{Im}\hat u_{k}$. Sums of vectors are finite sums in $V$ and sums of scalars are finite sums in a field; by the definition of a tuple a $p$-tuple in a set $X$ is a map from $[p]$ to $X$, so those sums apply to tuples without change. By Modulus of a Complex Number, the modulus $|z|$ of a complex number $z$ is the nonnegative real number with $|z|^{2}=(\operatorname{Re}z)^{2}+(\operatorname{Im}z)^{2}$. By claim 3 of Canonical Form and Arithmetic of Complex Numbers a complex number is determined by its real and imaginary parts, and by claim 4 of that lemma these parts are additive, so that $\operatorname{Re}(z-w)=\operatorname{Re}z-\operatorname{Re}w$ and $\operatorname{Im}(z-w)=\operatorname{Im}z-\operatorname{Im}w$. Since $e$ is an orthonormal basis, both claims of Orthonormal Expansion and Parseval's Identity in Finite Dimensions are available for it.

Claim 1. Injectivity. Suppose $\Phi(u)=\Phi(v)$. Then $\operatorname{Re}\hat u_{k}=\operatorname{Re}\hat v_{k}$ and $\operatorname{Im}\hat u_{k}=\operatorname{Im}\hat v_{k}$ for every $k\in[n]$, hence $\hat u_{k}=\hat v_{k}$ for every $k\in[n]$. By claim 1 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions,

Surjectivity. Let $x\in\mathbb{R}^{2n}$ have coordinates $x_{i}$ for $i\in[2n]$. For $k\in[n]$ put $c_{k}=x_{k}+x_{n+k}i$, where $i$ is the imaginary unit of The Complex Numbers; by claim 3 of Canonical Form and Arithmetic of Complex Numbers and Real and Imaginary Parts of a Complex Number we have $\operatorname{Re}c_{k}=x_{k}$ and $\operatorname{Im}c_{k}=x_{n+k}$. Put $u=\sum_{k=1}^{n}c_{k}e_{k}$. By claim 1 of Elementary Properties of an Orthonormal Family, $\hat u_{j}=c_{j}$ for every $j\in[n]$, so $\Phi(u)_{k}=x_{k}$ and $\Phi(u)_{n+k}=x_{n+k}$ for every $k\in[n]$. Since every index in $[2n]$ is of one of these two forms, $\Phi(u)=x$.

Claim 2. Let $u,v\in V$ and put $w=u-v$. For $k\in[n]$, additivity and homogeneity in the second argument (conditions 2 and 3 of Complex Inner Product Space), together with $-v=(-1)v$ from Elementary Identities in a Vector Space, give $\hat w_{k}=\hat u_{k}-\hat v_{k}$, whence

Write $x=\Phi(u)$ and $y=\Phi(v)$, and let $t:[2n]\to\mathbb{R}$ be the map with $t_{i}=(x_{i}-y_{i})^{2}$. By Euclidean Distance on $\mathbb{R}^n$, $d_{E}(x,y)$ is the nonnegative real number whose square is $\sum_{i=1}^{2n}t_{i}$. Let $t'$ be the restriction of $t$ to $[n]$ and let $t'':[n]\to\mathbb{R}$ be given by $t''_{k}=t_{n+k}$. Since $2n=n+n$, Concatenation of Finite Sums gives

By claim 2 of Properties of Finite Sums (additivity of finite sums) and the identity $|z|^{2}=(\operatorname{Re}z)^{2}+(\operatorname{Im}z)^{2}$, the right-hand side equals

the last equality by claim 2 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions. Thus $d_{E}(x,y)$ and $\lVert w\rVert=d(u,v)$ are nonnegative real numbers with the same square, so they are equal by Existence and Uniqueness of the Nonnegative Square Root.

Claim 3. $S$ is nonempty. Since $e$ is in particular orthonormal, $\langle e_{1},e_{1}\rangle=1$, and $1$ is the nonnegative real number whose square is $1$, so $\lVert e_{1}\rVert=1$ by Norm Induced by a Complex Inner Product and $e_{1}\in S$.

Identification of $\Phi(S)$. For every $k\in[n]$ we have $\langle e_{k},0_{V}\rangle=0$, by homogeneity in the second argument with the scalar $0$ together with $0\,e_{k}=0_{V}$ from Elementary Identities in a Vector Space; hence $\Phi(0_{V})$ is the point $0_{E}$ of $\mathbb{R}^{2n}$ all of whose coordinates are $0$. By claim 2, $d_{E}(\Phi(u),0_{E})=d(u,0_{V})=\lVert u\rVert$ for every $u\in V$. Consequently

if $u\in S$ then $d_{E}(\Phi(u),0_{E})=\lVert u\rVert=1$; and if $d_{E}(x,0_{E})=1$ then, $\Phi$ being surjective by claim 1, $x=\Phi(u)$ for some $u\in V$, and $\lVert u\rVert=d_{E}(x,0_{E})=1$, so $u\in S$.

$\Phi(S)$ is bounded. Taking the point $0_{E}$ and the radius $1$, every $x\in\Phi(S)$ satisfies $d_{E}(0_{E},x)=1\le1$, using symmetry of the metric $d_{E}$ (Euclidean Distance is a Metric on $\mathbb{R}^n$); so $\Phi(S)$ is bounded in $(\mathbb{R}^{2n},d_{E})$.

$\Phi(S)$ is closed. Its complement relative to $\mathbb{R}^{2n}$ is

by trichotomy for the total order of the ordered field $\mathbb{R}$ (Ordered Field). This complement is open in $(\mathbb{R}^{2n},d_{E})$. Indeed, if $d_{E}(x,0_{E})<1$, put $r=1-d_{E}(x,0_{E})>0$; every $y$ with $d_{E}(x,y)<r$ satisfies $d_{E}(y,0_{E})\le d_{E}(y,x)+d_{E}(x,0_{E})<r+d_{E}(x,0_{E})=1$ by the triangle inequality. If instead $1<d_{E}(x,0_{E})$, put $r=d_{E}(x,0_{E})-1>0$; every $y$ with $d_{E}(x,y)<r$ satisfies $d_{E}(x,0_{E})\le d_{E}(x,y)+d_{E}(y,0_{E})$, hence $1=d_{E}(x,0_{E})-r<d_{E}(x,0_{E})-d_{E}(x,y)\le d_{E}(y,0_{E})$. In both cases an open ball about $x$ is contained in the complement, so the complement belongs to the topology determined by $d_{E}$ and $\Phi(S)$ is closed in $\mathbb{R}^{2n}$.

Conclusion. By Heine-Borel Theorem in $\mathbb{R}^n$, $\Phi(S)$ is compact in $\mathbb{R}^{2n}$. By claims 1 and 2, $\Phi$ is a distance-preserving bijection from the metric space $(V,d)$ onto the metric space $(\mathbb{R}^{2n},d_{E})$, so claim 3 of A Distance-Preserving Bijection is a Homeomorphism shows that $S$ is compact in $V$.