Let $X$ be a compact infinte metric space and $f:X\to X$ an expansive homeomorphism. (Mapping $f:X\to X$ is expansive if $(\exists \epsilon >0)(\forall x\neq y)(\exists n\in \mathbb{Z}) d(f^n(x),f^n(y)) > \epsilon$ or, equivalent: $(\exists \epsilon >0)((\forall n\in \mathbb{Z}) d(f^n(x),f^n(y)) \leq \epsilon \implies x=y)$.) Prove that there are 2 different points $x,y\in X$ such that $lim_{n\to \infty} d(f^n(x),f^n(y))=0$.
My idea was: there must exist a point that isn't a fixed point (because otherwise $f$ would be identity and that isn't expansive). We observe that point $x$ and point $y=f(x)$.
$lim_{n\to \infty} d(f^n(x),f^n(y))=\\ lim_{n\to \infty} d(f^n(x),f^{n+1}(x))=\\ d(lim_{n\to \infty}f^n(x),lim_{n\to \infty} f^{n+1}(x))=\\ d(lim_{n\to \infty}f^n(x),lim_{n\to \infty} f^n(x))=0$
But I'm not sure that this calculation is correct.
You know $f$ is not positive expansive, which implies there exist $x\neq y$ and $0<\delta<\epsilon$ such that $d(f^n(x),f^n(y))\leq \delta$ for all $n\geq 0$ (you can take $\delta=\epsilon/2$). If the quantity$$L:=\limsup_{n\to\infty} d(f^n(x),f^n(y))$$ is zero, we're done. Otherwise, pick a sequence $n_k$ such that $\lim d(f^{n_k}(x),f^{n_k}(y))=L$ and such that $f^{n_k}(x)\to x^*$ and $f^{n_k}(y)\to y^*$ for some $x^*,y^*.$ We necessarily have $d(x^*,y^*)=L,$ so $x^*\neq y^*.$ For any $r\in\mathbb Z,$ $$d(f^r(x^*),f^r(y^*))=\lim_{k\to\infty} d(f^{r+n_k}(x),f^{r+n_k}(y))\leq\delta,$$ which contradicts expansivity of $f.$