Let $f:X\rightarrow X$ be a homeomorphism. $f$ is called an $c$-expansive homeomorphism, whenever for every $x\neq y$, there is an integer $n$ with $d(f^{n}(x), f^{n}(y)) >c$.

Question. Is there an expansive homeomorphism with expansive constant $c>0$ on compact metric space $X$ such that for $x\neq y$ there are only finite integer $n$, with $d(f^{n}(x), f^{n}(y)) >c$?

• Is the integer $n$ in the definition required to be positive? – Eric Wofsey Apr 24 '16 at 4:45

Here is an example. Let $X=\{0\}\cup\{1/n:n\in\mathbb{Z}_+\}$, and let $g:\mathbb{Z}\to\{1/n:n\in\mathbb{Z}_+\}$ be a bijection such that $g(0)=1$. Define $f:X\to X$ by $f(0)=0$ and $f(g(n))=g(n-1)$ for each $n\in\mathbb{Z}$. Then $f$ is $1/3$-expansive: for any $x\neq y$, you may assume WLOG that $x\neq 0$ so $x=g(n)$ for some $n\in\mathbb{Z}$, and then $f^{n}(x)=1$ and $f^{n}(y)\leq 1/2$ so $d(f^{n}(x),f^{n}(y))> 1/3$. But for any $x\neq y$, there are only finitely many values of $n$ such that this happens, since $d(a,b)>1/3$ implies at least one of $a$ and $b$ must be either $1$ or $1/2$, and there is at most one value of $n$ such that $f^n(x)=1$ and at most one value of $n$ such that $f^n(x)=1/2$ and similarly for $y$.
On the other hand, if in the definition of $c$-expansive you require $n$ to be positive, there are no examples. Let $x\neq y$ and let $n>0$ be such that $d(f^n(x),f^n(y))>c$. Setting $x'=f^n(x)$ and $y'=f^n(y)$, there then exists some $n'>0$ such that $d(f^{n'}(x'),f^{n'}(y'))>c$. But $f^{n'}(x')=f^{n+n'}(x)$ and $f^{n'}(y')=f^{n+n'}(y)$. This shows that for any $n$ that works for $x$ and $y$, there is a larger integer $n+n'$ which also works. Thus there must be infinitely many $n$ such that $d(f^n(x),f^n(y))>c$.