# Is $\mathbb{R}^\omega$ a completely normal space, in the box topology?

Basically, what the title says. Is $\mathbb{R}^\omega$ a completely normal space in the box topology ? ($\mathbb{R}^\omega$ is the space of sequences to $\mathbb{R}$) Thanks !

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Someone voted to close this question as an exact duplicate of How to prove that $\mathbb R^\omega$ with the box topology is completely regular. But completely regular and completely normal are two different notions, so I don't think it's a duplicate. –  Martin Sleziak Jun 29 '12 at 9:04

## 1 Answer

Quoting Munkres from his book Topology (this remark is made in the 5th exercise of section 32):

It is not known whether $\mathbb{R}^{\omega}$ is normal in the box topology. Mary-Ellen Rudin has shown that the answer is affirmative if one assumes the continuum hypothesis [RM]. In fact, she shows it satisfies a stronger condition called paracompactness.

[RM] M. E. Rudin. The box product of countably many compact metric spaces. General Topology and Its Applications , 2:293-298, 1972.

Of course this doesn't rule out the possibility that it is not completely normal.

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