Why the bilateral shift map is not an expanding map?

Question

A continuous transformation $f: M → M$ in a compact metric space $M$ is called expanding if there are constants $σ> 1$ and $ρ> 0$ such that for all $p ∈ M$, the image of the ball $B (p, ρ)$ contains a neighborhood of the closure of $B (f (p), ρ)$ and

$d (f (x), f (y)) ≥ \sigma d (x, y) \,\,\,for\,\,\,\ all \,\,\, x, y ∈ B (p, ρ).$

If $f: Σ_n → Σ_n$ is the unilateral shift: $f((x_n)_n) = (x_{n+1})_n$ for all $n \in \mathbb{N}$, it is true that $f$ is a expanding map.

Now, if $f: Σ_n → Σ_n$ is the bilateral shift, it is not true that $f$ is a expanding map. Does anyone know a counterexample?

Answer 1

Simply, in general there exists no $\sigma$ as in your definition. For example:

Take any two sequences $x$ and $y$ differing only by one element, say $x_k\ne y_k$. Then $$d(f^n (x), f^n (y))\to0\quad \text{when} \ n\to+\infty.$$

Indeed, computing explicitly with any distance giving the Tychonoff topology you get this property. I suggest that you use the simple distance $\beta^{-m}$ for a fixed $\beta>1$, where $m$ is the smallest integer such that $x_m\ne y_m$ or $x_{-m}\ne y_{-m}$. Then $$d(f^n (x), f^n (y))=\beta^{-n+l(k)}$$ for any sufficiently large $n$, with $l(k)$ depending only on $k$ (take $k\ge0$ and $k\le0$ to see the difference).