Showing a homeomorphism has dense orbit for every x in a compact metric space X, if $d(x,y)=d(T(x),T(y))$

Question

I am self studying entropy/topological dynamics, and came across this problem.

Note: Two definitions:

1)We say T is topologically transitive if there exists an $x\in X $ s.t.$ Orb(x):= \{T^k(x): k\in \mathbb{Z}\}$ is dense in X, i.e. if its closure is $X$.

2)We say T is minimal if every x in X has dense orbit.

The problem: Let X be a compact metric space with metric d and let $T : X → X$ be a topologically transitive homeomorphism. Show that if T is an isometry (i.e. $d(T(x), T(y)) = d(x, y)$, for all $x, y ∈ X$) then T is minimal.

I have tried the following. Choose an arbitrary y in X, let $\epsilon>0$ arbitrary and then show that for arbitrary z in X, there exists a $k\in \mathbb{Z}$ s.t. $d(T^k(y),z)<\epsilon$ by using triangle inequality and the isometry but I just can't get anywhere. Could anyone help? Thanks in advance!

Answer 1

Let $x\in X$ have dense orbit, $y,z\in X$, and $\varepsilon>0$. Consider the sequence $(T^k(y))_{k=1}^\infty$. Due to sequential compactness, we can find natural numbers $n_1<n_2$ such that $d(T^{n_1}(y),T^{n_2}(y))<\varepsilon/3$. Using the fact that $T$ is an isometry, we deduce $d(T^k(y),y)<\varepsilon/3$ for some natural number $k$. Since $x$ has dense orbit, there exists integers $m$ and $n$ satisfying $d(T^m(x),y)<\varepsilon/3$ and $d(T^n(x),z)<\varepsilon/3$. Then $$ d(T^{k+n-m}(y),z)\leq d(T^{k+n-m}(y),T^{n-m}(y))+d(T^{n-m}(y), T^{m+n-m}(x)) +d(T^n(x),z) \\ = d(T^k(y),y)+d(y, T^m(x)) +d(T^n(x),z) < \varepsilon. $$ Therefore $T$ is minimal.

Answer 2

Assume $T$ is a transitive isometry.

The goal is to show that for all $x \in X$, the orbit of $x$ is dense in $X$.

Fix $x,y \in X$.

We want to show $y$ is in the closure of the orbit of $x$.

Equivalently, we want to show that for all $\epsilon>0$, there exists $p \in \mathbb{Z}$ such that $d(T^p(x),y) < \epsilon$.

Fix $\epsilon>0$.

Since $T$ is transitive, there exists $w \in X$ such that the orbit of $w$ is dense in $X$.

Hence there exist $m,n \in \mathbb{Z}$ such that $$d(T^m(w),x) < \frac{\epsilon}{2}$$ $$d(T^n(w),y) < \frac{\epsilon}{2}$$ Let $p=n-m$. \begin{align*} \text{Then}\;\;d(T^p(x),y) &= d(T^{n-m}(x),y)\\[4pt] &=d(T^{-m}(x),T^{-n}(y))&&\text{[since $T$ is an isometry]}\\[4pt] &\le d(T^{-m}(x),w)+d(w,T^{-n}(y))\\[4pt] &=d(w,T^{-m}(x))+d(w,T^{-n}(y))\\[4pt] &=d(T^m(w),x)+d(T^n(w),y)&&\text{[since $T$ is an isometry]}\\[4pt] &<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\[4pt] &=\epsilon\\[4pt] \end{align*} Hence $y$ is in the closure of the orbit of $x$, as required, which thus completes the proof.

Note: I never used compactness.