Determine whether $\mathbb{R}^\omega$ is separable or complete 
Let $d_\omega$ be the metric on $\mathbb{R}^\omega := \{(a_k)_{k\geq 1} : a_k\in \mathbb{R}\}$ given by $d_\omega(a,b) = \sum_{k=1}^\infty \dfrac{|a_k-b_k|}{2^k(1+|a_k - b_k|)}.$


Determine whether $(\mathbb{R}^\omega, d_\omega)$ is separable.


Determine whether $(\mathbb{R}^\omega, d_\omega)$ is complete.

I know that for a sequence $(a_n)$ in $\mathbb{R}^\omega, d_\omega(a_n,b) \to 0$ if and only if $\lim\limits_{n\to\infty} |a_{n,k} - b_k|$ as $n\to\infty$ for all $k\in\mathbb{Z}^+$, but I'm not sure if this is useful for showing that the metric space is separable. I also know that the rational numbers are countable, countable unions of countable sets are countable, and that the rational numbers are dense in $\mathbb{R}.$ These facts are used to show $(\ell_1, d_1)$ and $(\ell_2, d_2)$ are separable, but I'm not sure if they're useful here.
As for the second part, to show the metric space is complete, it suffices to show that every Cauchy sequence in the metric space converges in the metric space. Suppose $(a_n)$ is Cauchy in $\mathbb{R}^\omega$. Let $\epsilon > 0.$ Choose $M\in\mathbb{Z}^+$ so that $k,l\geq M\Rightarrow d_\omega( a_k, a_l) < \epsilon.$ I was wondering if I might be able to mimic the proof that $(\ell_\infty, d_\infty)$ is complete using the fact that $\mathbb{R}$ is complete?
 A: Separability
Let $c_{0,0}^{\mathbb{Q}}$ be the sequences with finite support with rational coefficients. This set is countable. For density, let $x = (x_n)_{n=1}^\infty \in \mathbb{R}^\omega$ and $\epsilon > 0$ be given. Since $$\frac{|x-y|}{1 + |x-y|} \leq 1$$
the comparison test for series tells us that the metric is well defined. Thus, we may get an $N$ so large that $x^N = (x_1, \ldots , x_N, 0, 0, \ldots )$ is within $\frac{\epsilon}{2}$ of $x$. Now, get rationals $q_1, \ldots , q_N$ with $|x_i - q_i| < \frac{\epsilon}{N}$. I claim $q = (q_1, \ldots , q_N, 0, 0, \ldots )\in c_{0,0}^{\mathbb{Q}}$ satisfies $d(x,q) < \epsilon$. Indeed:
$$d(x,q) \leq d(x,x^N) + d(x^N, q) < \frac{\epsilon}{2} + \sum_{j=1}^N \frac{|x_i-q_i|}{2^j(1+|x_i-q_i|)} \leq \frac{\epsilon}{2} + \sum_{j=1}^N |x_i-q_i| < \epsilon$$
As desired.
Completeness
Let $(x^n)_{n=1}^\infty \subset \mathbb{R}^\omega$ be Cauchy. Then, for each fixed $j$: $$\frac{|x_j^n - x_j^m|}{2^j(1 + |x_j^n-x_j^m|)} < d(x^n,x^m)$$
so that $(x_j^n)_{n=1}^\infty $ is Cauchy in $\left( \mathbb{R}, \frac{|\cdot |}{1+|\cdot|} \right)$, which is a complete metric space. Hence, $(x_j^n)_{n=1}^\infty$ converges, say to $x_j$. Let $x = (x_1,x_2,x_3, \ldots) \in \mathbb{R}^\omega$. I claim $x^n \to x$. Given $\epsilon > 0$, pick $R$ so large that $$\sum_{j=R+1}^\infty \frac{|x_j^n - x_j|}{2^j(1 + |x_j^n - x_j|)} < \frac{\epsilon}{2}$$ Now, pick $k$ such that, for $1 \leq j \leq R$ $$\frac{|x_j^k - x_j|}{1 + |x_j^k-x_j|} < \frac{\epsilon 2^{j-1}}{R}$$ which we can do by coordinate-wise convergence. Finally, $$d(x^k, x) = \sum_{j=1}^R \frac{|x_j^n - x_j|}{2^j(1 + |x_j^n - x_j|)} + \sum_{j=R+1}^\infty \frac{|x_j^n - x_j|}{2^j(1 + |x_j^n - x_j|)} < \sum_{j=1}^R \frac{\epsilon}{2R} + \frac{\epsilon}{2} = \epsilon$$
Which shows that the given space is complete.
