# How to prove that $\mathbb R^\omega$ with the box topology is completely regular

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.

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For problems like these there's no reason not to put the entire question in the title. –  Qiaochu Yuan May 2 '11 at 0:28
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. –  Brian M. Scott Oct 20 '11 at 7:18
First, it suffices to only consider the case where $a = (0, 0, \dots)$ and the open neighborhood $(-1,1)^\mathbb{N}$ of $a$ is disjoint from $F$ (why?).
Hint: Now, having reduced the general case to this one, note that the uniform topology on $\mathbb R^\mathbb{N}$ is coarser than the box topology. Hence any function continuous on $\mathbb R^\mathbb{N}$ in the uniform topology is also continuous with respect to the box topology.