What is the topological dimension of $\mathbb{R}^\mathbb{Z}$? it is a separable topological space?

Question

Initially, consider $\mathbb{R}$ endowed with a discrete topology, and discrete metric $\delta(x,y)=0 $ if $x=y$ and $\delta(x,y)=1 $ if $x\neq y$.

Next, we endowing $~~\mathbb{R}^\mathbb{Z}=\prod_{-\infty}^{+\infty} \mathbb{R}~~$ with the metric $$d(x,y)=\sum_{i=-\infty}^{+\infty}\frac{\delta(x_i,y_i)}{2^{|i|}},$$ where $x=(x_i),~ y=(y_i)\in \mathbb{R}^\mathbb{Z}$.

What is the topological dimension of $\mathbb{R}^\mathbb{Z}$?

Is $\mathbb{R}^\mathbb{Z}$ a discrete topological space?

Is $\mathbb{R}^\mathbb{Z}$ a separable topological space?

Please help me! Thanks.

Answer 1

It is metric, but not separable (even one factor is not) and certainly not discrete (just like the Cantor set $\{0,1\}^\mathbb{Z}$ is not: basic open sets are never singletons). It is zero-dimensional as a countable product of zero-dimensional spaces.