Separability of the space of sequences of real numbers $R^\omega$ with the uniform topology

Question

Consider the space of sequences of real numbers $\mathbb{R}^\omega$. I am trying to figure out whether this space is separable in the uniform topology, which is the topology generated by the metric $\rho (x,y) = \sup d(x_i,y_i)$, where $d(x_i,y_i) = \min \{|x_i-y_i|,1\}$.

For this I need to find a countable dense subset, but I cannot find any (I have been trying various versions of sets of points with rational coordinates)... So does it mean that it is non-separable?

Answer 1

For a subset $A$ of $\mathbb{N}$ define the point $x_A$ where $x_A(n) = 0$ for $n \notin A$, $x_A(n) =1$ for $n \in A$. Then what is $d(x_A, x_B)$ for $A \neq B$?

Answer 2

In fact, this space is not separable. Suppose $\mathbb{R}^\omega$ is separable, then there is a countable subset of $\mathbb{R}^\omega$, let's say $A=\{x^n\}_{n\in\mathbb{N}}$, which is dense in the uniform topology. Now let $x^n_k$ represent the $k$-th element of the sequence $x^n$. Let $1>\varepsilon>0$ and define $y\in\mathbb{R}^\omega$ in the following way:

Pick $y_k$ such that $1>|y_k-x^k_k|\geq\varepsilon$.

We have, then:

$$d(y,x^n)=\sup_{k\in\mathbb{N}}d(y_k,x^n_k)\geq d(y_k,x^k_k)\geq\varepsilon$$

Then we have that $x^n\notin B(y,\varepsilon)$ for every $n\in\mathbb{N}$, which is a contradiction.