For $a,b \in \mathbb{R}^n$, prove that there is an isometry of $\mathbb{R}^n$ onto itself which maps $a$ to $b$.
I am working with the following definition of an isometry:
Metric spaces $(X,d)$ and $(Y,d')$ are metrically equivalent, or isometric, if there is a one-to-one function $f:X \to Y$ from $X$ onto $Y$ such that for all $a,b \in X, \quad d(a,b) = d'(f(a),f(b))$.
Based on what I have read, maybe a particularly simple choice for a function to satisfy the above requirements will be the identity function $i: (X,d) \to (Y,d')$ since it is continuous. However, this seems like too simple of a solution since there is not much proof involved. Is this even the right way of approaching this problem? Any advice and suggestions will be greatly appreciated. Thanks in advance.
Hint: Let $a,b\in\Bbb R$ and let $f:\Bbb R\to\Bbb R$ be the map $f(x)=x+(b-a)$. Then $f(a)=b$. Is $f$ an isometry? Can this be generalized?
Let's develop this hint a bit more, first by showing that $f$ is indeed an isometry.
Claim 1. $f$ is invertible
Proof. Let $g:\Bbb R\to\Bbb R$ be the map $g(x)=x+(a-b)$. Then \begin{align*} f(g(x)) &= g(x)+(b-a) & g(f(x)) &= f(x)+(a-b) \\ &= x+(a-b)+(b-a) & &=x+(b-a)+(a-b) \\ &= x & &= x \end{align*} This proves that $f$ is invertible with $f^{-1}=g$. $\Box$
Claim 2. $\lVert f(x)- f(y)\rVert=\lVert x-y\rVert$
Proof. Note that \begin{align*} \lVert f(x)-f(y)\rVert &= \big\lVert x+(b-a)-\big(y+(b-a)\big)\big\rVert \\ &= \lVert x+(b-a)-y-(b-a)\rVert \\ &= \lVert x-y\rVert \end{align*} as advertised. $\Box$
Do you see how Claim 1 and Claim 2 combined prove that $f$ is an isometry?