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 1: Try finding uncountably many disjoint open sets.

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.

  • $\begingroup$ 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. $\endgroup$ – nomadicmathematician Dec 3 '14 at 16:17
  • $\begingroup$ However, I cannot find disjoint $U_x$ in the uniform topology following your second hint. Can you help me? $\endgroup$ – nomadicmathematician Dec 3 '14 at 16:21
  • $\begingroup$ @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 $\mathbb{Z}^\omega$. $\endgroup$ – Nate Eldredge Dec 3 '14 at 16:31
  • $\begingroup$ 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. $\endgroup$ – nomadicmathematician Dec 3 '14 at 16:33
  • $\begingroup$ @user135204: That's it! $\endgroup$ – Nate Eldredge Dec 3 '14 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.