6
$\begingroup$

How do I prove that $\mathbb R^\omega$ with the box-topology (i.e., the basis are of the form $\prod_n G_n$, where $G_n$ are open in $\mathbb R$) is Completely Regular (i.e. Given a point $a$ and a closed set $F$; one can find a continuous function $f:\mathbb R^\omega \to [0,1]$ such that $f(a)=0$ and $f(F)=1$). Thank you.

Note: It is not known whether $\mathbb R^\omega$ with the box-topology is Normal.

  • 1
    For problems like these there's no reason not to put the entire question in the title.2011-05-02
  • 2
    It’s been known since 1972 that CH implies that $\square^\omega\mathbb{R}$ is not just normal, but paracompact: M. E. Rudin, *The box product of countably many compact metric spaces*, General Topology and Appl. 2 (1972), 293-298. MR 48:2969.2011-10-20

2 Answers 2