# $\mathbb{R}^\omega$ in the box and uniform topology is not separable

I'm trying to prove that $\mathbb{R}^\omega$ in the box and uniform topology is not separable by way of contradiction. However, I cannot really find a direction to lead to a contradiction. How can I show this?

Hint 2: For each integer-valued sequence $x \in \mathbb{Z}^\omega \subset \mathbb{R}^\omega$, find an open set $U_x$ containing $x$ and such that all $U_x$ are disjoint.
• I followed your hints please check if my understanding is correct. For the box topology case I came up with the collection of disjoint open sets $B = {A_{n}^{\omega}:A_{n}=(0,1) or (1,2), n \in \mathbb{N}}$ Then if $A$ is a dense subset, we can take an element from the intersection of each element of $B$ and $A$ then the collection of such elements would be an uncountable subset of $A$ since $B$ is uncountable. – nomadicmathematician Dec 3 '14 at 16:17
• However, I cannot find disjoint $U_x$ in the uniform topology following your second hint. Can you help me? – nomadicmathematician Dec 3 '14 at 16:21
• @user135204: For the box topology, I think you have the right idea but your notation is weird. Read literally, $B$ contains only two sets: $A_1^\omega = (0,1) \times (0,1) \times \cdots$ and $A_2^\omega = (1,2) \times (1,2) \times \cdots$. Maybe you mean $B = \{ \prod_{n} A_{a_n} : a \in \{0,1\}^\omega\}$. For the uniform topology, try taking a small ball centered at each element of $\mathbb{Z}^\omega$. – Nate Eldredge Dec 3 '14 at 16:31
• Would $B(x,1/2)$ work? For distinct each integer valued sequence $x,y$ $d(x,y) \ge 1$, where $d$ denotes the uniform metric since if two integer sequences are distinct, some coordinate has difference greater than or equal to 1. So if any integer sequence $z$ is in both $B(x,1/2)$ and $B(y,1/2)$, then by triangle inequality $d(x,y) \lt1$, which is a contradiction. Then all such balls are disjoint open sets so by the same reasoning as the previous problem, we get the result. – nomadicmathematician Dec 3 '14 at 16:33