Is the supremum of expandable constant is expandable constant?

Question

A homeomorphism $T:X \rightarrow X $on compact metric space$(X,d)$ is called expandable if there is a constant $e>0$ that satisfies$\forall x\neq y\in X,\exists n\in\mathbb{Z},s.t. d(T^nx,T^ny)>e$,And $e$ is called expandable constant.
We call the supremum of all the expandable constants of $T$ as $r(T)$, is $r(T)$ an expandable constant?
I tried to think about $T: S^1\rightarrow S^1, Tx=\pi x$ My intuition is that all of its trajectories are dense then $\forall x\in (0,1/2)$is expandable constant, but I can't prove it

Answer 1

Here's a counterexample.

Let $X=\{0,1\}^{\mathbb{Z}}$ equipped with the metric $d(x,y) = \sum_{n\in \mathbb{Z}} \frac{1}{2^{|n|}} |x_n-y_n|$. Let $T((x_n)_{n\in\mathbb{N}}) = (x_{n+1})_{n\in\mathbb{N}}$ be the shift by $1$.

First observe that if $x = \delta_0$ (i.e. the sequence that is always zero and $1$ only in the $0$'th coordinate) and $y=0$, then $d(x,y) = 1$ and for every $n$, $d(T^nx,T^ny) = \frac{1}{2^{|n|}}<1$. In particular, no expandable constant can be greater than or equal to $1$.

Now let $x,y$ be arbitrary so that $x\not = y$, then there exists some coordinate $n$ so that $x_n\not = y_n$. We then see that $T^{-n} x$ and $T^{-n} y$ disagree in the first coordinate and so $d(T^{-n}x, T^{-n} y) \geq 1$.

We deduce that any number less than $1$ is an expandable constant.

This completes the proof.