Expansive homeomorphism on $\mathbb{R}^{2}$ Homeomorphism $f:(X, d)\to (X, d)$ is called $c$-expansive whenever for  $x\neq y$ in $X$,  there is $n\in \mathbb{Z}$ such that $d(f^{n}(x), f^{n}(y))>c$.
Let $A=\left(
  \begin{array}{cc}
    2 & 0 \\
    0 & \frac{1}{2} \\
  \end{array}
\right)
$ and $f:\mathbb{R}^{2}\to \mathbb{R}^{2}$ be defined by $f(x)=Ax$. Consider the stereographic projection $P:S^{2}-\{(0, 0, 1)\}\to \mathbb{R}^{2}$ defined by $P(x, y, z)=\frac{(x,y)}{1-z}$.
Is it true that $f$ is expansive when $\mathbb{R}^{2}$ has the
metric induced by $P$? 
I know that $f$ is not expansive, but its proof is not clear for me.
 A: Let $(S^2,d)$ be a metric space with $d$ being the Euclidean metric inherited from $\mathbb{R}^3$. Using the stereographic projection $P: S^2-\{(0,0,1)\}\to \mathbb{R}^2$ we define the metric on $\mathbb{R}^2$ to be given by $d'(P(x_1,y_1,z_1), P(x_2, y_2, z_2))=d((x_1,y_1,z_1),(x_2, y_2, z_2))$. Note that $P^{-1}: \mathbb{R}^2\to S^2-\{(0,0,1)\}$ is given by
$$
(X,Y)\mapsto \left(\frac{2X}{X^2+Y^2+1}, \frac{2Y}{X^2+Y^2+1}, \frac{X^2+Y^2-1}{X^2+Y^2+1}\right)
$$
Now consider a linear function $f:\mathbb{R}^2\to \mathbb{R}^2$ determined by a matrix $M=\mathrm{diag}(e^\lambda, e^{\lambda'})$  for some $\lambda,\lambda'\neq 0$. Consider the points
$A_0=(e^\mu, 0), B_0=(e^{\nu}, 0)$, also for simplicity let $A_n=f^n(A_0)$ and $B_n=f^n(B_0)$.
Then (by a bit of calculation)
$$
P^{-1}(A_n)=\left(\frac{1}{\cosh (n\lambda+\mu)}, 0, \tanh (n\lambda+\mu)\right), \quad
P^{-1}(B_n)=\left(\frac{1}{\cosh (n\lambda+\nu)},0, \tanh (n\lambda+\nu)\right)
$$
Then with a bit of calculation
$$d'(A_n, B_n)=\sqrt{\frac{2[\cosh(\mu-\nu)-1]}{\cosh(n\lambda+\mu)\cosh(n\lambda+\nu)}}\leq \sqrt{2[\cosh(\mu-\nu)-1]}$$
So $f$ cannot possibly be $\epsilon$-expansive for any $\epsilon>0$. Because by chooisng $\nu=0$ and $\mu$ such that $\cosh \mu < 1+\epsilon^2/2$ the points $A_0=(e^\mu,0)$ and $B_0=(1,0)$ are such that for all $n\in \mathbb{Z}$ one has $d'(A_n, B_n)< \epsilon$. If you take a closer look at the above argument, if $M$ is any invertible diagonal matrix then $f$ is not $\epsilon$-expansive for any $\epsilon>0$. In fact more generally if (the invertible matrix) $M$ has any eigenvectors (over $\mathbb{R}$ that is) then $f$ is not expansive [If $v$ is that eigenvector take $A_0=e^\mu v$ and $B_0=v$.]
