I would like to prove the following theorem from Mendelson's Introduction to Topology:
For each $a,b\in\Bbb R^n$, prove that there is a topological equivalence between $(\Bbb R^{n},d)$ and itself defined by inverse functions $f\colon\Bbb R^n\longrightarrow\Bbb R^n$ and $g\colon\Bbb R^n\longrightarrow\Bbb R^n$ such that $f(a)=b$. [Hint: If $a=(a_{1}, a_{2}, ..., a_{n}), b = (b_{1}, b_{2}, ..., b_{n})$, define $f$ by setting $f(x_{1}, x_{2}, ..., x_{n}) = (x_{1}+b_{1}-a_{1}, ..., x_{n}+b_{n}-a_{n})$.]
I'm not totally certain what is meant by $(\Bbb R^n,d)$ since the text has somewhat ambiguously used this notation to sometimes specifically mean that $d(a,b) = \max{|b_{i}-a_{i}|}_{i\leq n}$, and then sometimes used this to mean any unspecified distance metric. I've been trying to prove it for an unspecified distance metric, which might be the problem here.
What I've done so far: Trivially, this definition of $f$ makes $g(x_{1}, ..., x_{n})=(x_{1}-b_{1}+a_{1}, ..., x_{n}-b_{n}+a_{n})$ an inverse. Let $\delta_{1}=d(a,b), \quad \delta_{2}=d(f(a),f(b))=\left( (b_{1}, ..., b_{n}), (2b_{1}-a_{1}, ..., 2b_{n}-a_{n})\right)$. To prove topological equivalence from the definition would be to prove that $f$ and $g$ are continuous.
So let $\varepsilon>0,\quad x\in\Bbb R^n$ be given. We want to find a neighborhood of $x$ such that $y\in B(x,\delta)$ implies $d(f(x),f(y))<\varepsilon$. At this point I've just been trying to wrangle $d(f(x),f(y))$ by Algebra to get into a format where I could constrain $x,y$ by some amount and get the desired result. My Algebra so far:
For easy notation, let $(x_{1}+b_{1}-a_{1}, ..., x_{n}+b_{n}-a_{n})$ be written $x+b-a$, for instance. Then
$d(f(x),f(y)) = d(x+b-a, y+b-a) \leq d(x+b-a, y) + d(y, y+b-a) \leq d(y, x) + d(x, x+b-a) + d(y,y+b-a)$.
At this point I don't see where to go from here.
I assume that $d$ is the metric induced by a norm, as usual in $\mathbb{R}^n$. Then $d(f(x),f(y))=d(x+b-a,y+b-a)=d(x,y)$ since norms are invariant under translation.