Shift invariant metric on countable product of $\mathbb{R}$.

Question

Suppose we examine $\mathbb{R}$ with the usual topology and define $(X,T)$ where $$ X = \prod_{t\in\mathbb{Z}}\mathbb{R} $$ and $T$ is the product topology. I know from this and this question that this topology is completely metrizable via the metric $$ \delta(x,y) = \sum_{n=1}^{\infty}2^{-n}\frac{|x_n-y_n|}{1+ |x_n-y_n|}, $$ where we used some bijective mapping $\mathbb{Z}\rightarrow\mathbb{N}$ to reorder.

Question: Define the shift to the left $\tau$ on $X$ by $$ \tau x = \tau(\ldots,x_{-1},x_0,x_1,\ldots) = (\ldots,x_{0},x_1,x_2,\ldots) $$ Is $(X,T)$ completely metrizable by a metric $d$ that satisfies $$ d(\tau x,\tau y) = d(x,y). $$

Answer 1

Probably not. Under $\delta$, orbits will behave very differently topologically than they would under any topology induced by a metric satisfying $d$. For instance, $\delta$ has 'asymptotic' orbits - pairs of elements $x \neq y$ such that $\displaystyle\lim_{n \to \infty}\delta(\tau^nx,\tau^ny) = 0$ - and the existence of such orbits is (I believe) a topological property of the space together with the shift $\tau$, rather than a metric property.

On the other hand, $d$ has no such orbits.

The problem is that $\delta$ is really a metric that gives weight to the 'origin' of a sequence. So take for example the element $x = (\ldots ,0,0,1,0,0,\ldots)$. Under $\delta$ the closure of the orbit $\{\tau^n(x) \mid n\in\mathbb{Z}\}$ also contains the element $ y = (\ldots ,0,0,0,\ldots)$. On the other hand, because $d$ is a 'non-local' metric, there exists an $\epsilon >0$ such that for all $n$, $d(\tau^n(x),y)\geq \epsilon$, and so $y \notin \overline{\{\tau^n(x) \mid n\in\mathbb{Z}\}}$ with respect to the $d$ metric.

Obviously, my claim that the existence of asymptotic pairs being a topological property still needs to be checked, but my example above in terms of orbit closures should be enough I think.