$\Bbb{R}^\omega$ with the Uniform Metric is not 2nd Countable

Question

I'm trying to show that $\Bbb{R}^\omega$ with the uniform metric is not 2nd countable or separable (either one will work, since they are equivalent for metric spaces). My idea was to transfer the proof that $\Bbb{R}_\ell$ isn't second countable into the $\Bbb{R}^\omega$ setting. If $\mathcal{B}$ is a basis for $\Bbb{R}_\ell$, then $x \mapsto B_x \subseteq [x,x+1)$ is an injection between $\Bbb{R}$ and $\mathcal{B}$, which shows that $\mathcal{B}$ is uncountable. How do I do something similar in $\Bbb{R}^\omega$? My first guess was that $x = (x_1,x_2,...) \in (0,1) \times \{0\} \times...$ being mapped to $B(x,x_1)$ might work, but quickly discovered the contrary. I could use a hint.

Answer 1

Hint: for any subset $A \subset \omega$, consider the element $1_A$ which has 1s in the positions which are in $A$, and 0 in others. (If you think of the elements of $\mathbb{R}^\omega$ as functions from $\omega$ to $\mathbb{R}$, this is an indicator function.) There are uncountably many of these. Now find a collection of disjoint open sets, each containing one of these elements.

Answer 2

Try diagonalisation: Suppose $D$ is a countable dense set in $\mathbb{R}^\omega$. Say $D = \{d_n: n \in \mathbb{N}\}$ and each $d_n = (d_{n,1}, d_{n,2}, d_{n,3} \ldots)$. Then define $x_n = d_{n,n} + 1$ for all $n$ and this defines a point $x$ that has distance $1$ to each of the $d_n \in D$. Conclude that $D$ could not have been dense after all.