Let $\vert \vert -\vert \vert$ be the standard Euclidean norm on $\mathbb R^n$ given by $\vert \vert x \vert \vert = \sqrt{x_1^2 + \cdots x_n^2}$ where $x\in \mathbb{R}^n$. Let $m$ and $n$ be positive integers. If $(\mathbb R^m, \vert \vert - \vert \vert)$ is isometric to $(\mathbb R^n, \vert \vert - \vert \vert )$, then show that $m =n$.
I am struggling with the proof. I have dealt with linear isometries in Linear Algebra. In that case, assuming a linear mapping $\phi : \mathbb{R}^m \to \mathbb R^n$ would require $\phi$ to be bijective. Since the mapping would be both injective and surjective, it would imply $m = n$. However, $\phi$ is not necessarily linear.
Using properties of isometry, I know $\phi$ is bijective continuous, preserves distance and has a continuous inverse. Here's what I tried. If $\phi(x)=y$, then $\vert \vert \phi(x) \vert \vert = \vert \vert y\vert \vert = \vert \vert x \vert \vert$. Then using standard notation,
\begin{align} &\sum\limits_{i = 1}^{n} y_i^2 = \sum\limits_{i = 1}^{m} x_i^2\\ &\vert \vert \phi(x + h) \vert \vert = \vert \vert y + h'\vert \vert = \vert \vert x + h \vert \vert \\ \implies & \lim\limits_{h' \to 0} \sum\limits_{i = 1}^{n}(y_i + h'_i)^2 = \lim\limits_{h\to 0} \sum\limits_{i = 1}^{m} (x_i + h_i)^2 \end{align}
I can make $h', h$ arbitrarily small so continuity of $\phi$ holds irrespective of dimensions $m, n$. What can I do at this point? I don't know if I can use the condition of continuous inverse somehow. However, I feel since the inverse argument would be completely symmetrical, this proof should use contradiction.
