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I have seen in Munkres Topology that $\mathbb R^I$ is regular but not normal if $I$ is uncountable. Whereas, if $I$ is countable, then $\mathbb R^I$ is normal.

I want to prove that $(0,1)^{\mathbb R}$ is regular but not normal.

We can replace $(0,1)$ by $\mathbb R$ since they are homeomorphic. And here $I = \mathbb R$, thus $I$ is uncountable. And hence $(0,1)^{\mathbb R}$ is not normal. But how can I prove this as well as the fact that $(0,1)^{\mathbb R}$ is regular? Thank you.

Fail purpose 4fun
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    https://math.stackexchange.com/questions/191646/mathbbr-mathbbr-is-not-normal – Alessandro Codenotti Dec 18 '20 at 13:28
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    For the regularity, show that an arbitrary product $X=\prod_{i\in I}X_i$ of regular spaces is regular. It suffices to show that if $x\in X$, and $B_0$ is a basic open nbhd of $x$, then there is a basic open nbhd $B_1$ of $x$ such that $\operatorname{cl}B_1\subseteq B_0$. If $B_0=\prod_{i\in I}U_i$, where $I_0={i\in I:U_i\ne X_i}$ is finite, you use regularity of the factors $X_i$ for $i\in I_0$ to do this. – Brian M. Scott Dec 18 '20 at 18:26

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