$\newcommand{\Int}{\operatorname{Int}}$I need help to show that in the product space $R^\omega$, $\Int((0,1)^\omega)=\emptyset$ Hence $$\Int\left(\prod A_ {\alpha}\right)\neq \prod \Int\left( A_ {\alpha}\right)$$ does not hold true, where $\alpha \in A$.
Thank you.
If $U$ is any non-empty open subset of $\mathbb{R}^\omega$, it contains some open subset from the standard base of that product, namely a set of the form $\prod_{\alpha} U_\alpha$, where $U_\alpha = \mathbb{R}$ for all but finitely many $\alpha$, and equals some non-empty open subset of $\mathbb{R}$ otherwise. Such a set can never be a subset of $(0,1)^\omega$.
$\Bbb R^\omega$ has a base of open sets of the form $B=\prod_{n\in\omega}U_n$, where each $U_n$ is open in $\Bbb R$, and $$\operatorname{supp}(B)=\{n\in\omega:U_n\ne\Bbb R\}$$ is finite.$\newcommand{\Int}{\operatorname{Int}}$ Suppose that $\Int\left((0,1)^\omega\right)\ne\varnothing$; then there is a non-empty basic open set $B\subseteq\Int\left((0,1)^\omega\right)$. Let $x\in B$ be arbitrary, fix $m\in\omega\setminus\operatorname{supp}(B)$, and define $y\in\Bbb R^\omega$ by
$$y_k=\begin{cases} x_k,&\text{if }k\ne m\\ 2,&\text{if }k=m\;. \end{cases}$$
Do you see the contradiction here?
