Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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

1 Answer 1

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.

What would be a canonical choice for your function?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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