Isometry of two metric spaces How to prove that $(\Bbb R^2, d_1)$ and $(\Bbb R^2, d_\infty)$ are isometric?
My approach
Let $f:(\Bbb R^2, d_1)\to(\Bbb R^2, d_\infty)$ be a function defined by $f(x, y) = (x+y,x-y)$. I can easily prove that $f$ is bijective, but I can not prove the isometry of $f$.
 A: Since both metrics are induced by norms and your function is linear, it is enough to check that $f$ preserves norms, i.e. we must check that
$$\max \{|x+y|,|x-y|\}=|x|+|y|$$
for all $x,y\in \Bbb R$.
We can assume WLOG that $x$ and $y$ and non-negative, since changing their sign would not change the absolute values in the RHS and would only swap the two terms in the LHS.
Then the RHS would be equal to $\max\{x+y,|x-y|\}$; but since
$$|x-y|\leq |x|+|y|=x+y,$$
both sides must then be equal to $x+y$.
A: The map $f$ maps the square $Q_1$ with vertices $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$ onto the square $Q_\infty\subset({\mathbb R}^2,d_\infty)$ with vertices $(1,1)$, $(1,-1)$, $(-1,-1)$, $(-1,1)$.
Since $Q_1$ is the closed unit ball in $({\mathbb R}^2,d_1)$, and $Q_\infty$ is the closed unit ball in $({\mathbb R}^2,d_\infty)$, it follows from general principles about norms and linear maps that $f$ is an isometry.
A: Here it is important to keep indices and dimensions straight. Remember that both the $d$'s take in two pairs of numbers as arguments, while $f$ only takes in one pair. Both a basic pair (vector / point) and a pair of pairs is tempting to write as $(x, y)$, so this is a breeding ground for confusion. I will try to keep it straight in my answer, using indexed variables for numbers and unindexed variables for points.
Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$. Then $d_1(x, y) = |x_1-y_1| + |x_2-y_2|$, while
$$d_\infty(f(x), f(y)) = \max(|f(x)_1 - f(y)_1|, |f(x)_2 - f(y)_2|)\\
= \max(|(x_1 + x_2) - (y_1+y_2)|, |(x_1-x_2) -(y_1-y_2)|)\\
= \max(|(x_1 -y_1) + (x_2-y_2)|, |(x_1-y_1) - (x_2-y_2)|)$$
which is now of the form $\max(|a+b|, |a-b|)$, which is equal to $|a|+|b|$ (you can, for instance, argue by cases: do $a$ and $b$ have the same sign or not?), and we're done.
A: Let $(x,y), (u,v) \in \mathbb{R}^2$.
$$ D =d_\infty(f(x,y),f(u,v)) $$
$$=d_\infty((x+y,x-y),(u+v,u-v))$$
$$ = \|x+y-u-v, x-y-u+v||_{\infty}$$
As, $$\|(A,B)\|_{\infty}= \max(A,B) =\frac{|A+B|+|A-B|}{2} $$
we can write
$$ D = \frac{1}{2}(|(x+y-u-v)+(x-y-u+v)|+|(x+y-u-v)-(x-y-u+v)|)$$
$$= \frac{1}{2}(|2x-2u|+|2y-2v|)$$
$$ = |x-u|+|y-v|$$ 
$$ =d_1((x,y),(u,v))                $$
