This question is likely to be trivial, but I'm confused by it. I will leave here a couple of the examples my teacher left here, so I can guide myself through the others.
Which of these subsets of $\mathbb{R}^{\omega}=\mathbb{R}\times \mathbb{R}\times\mathbb{R}\times...$ are open in product topology?
$U=\{(x_n)\in \mathbb{R}^{\omega} : x_n > 0 $ for $ n/2 \in \mathbb{N}\}$
$U_k=\{(x_n)\in\mathbb{R}^{\omega}: x_k\neq0\}$
What must happen for the subsets of the "pre-projection" to not form a subbasis?
$U_k$ is open and even belongs to the subbasis defining the product topology, since it is equal to $p_k^{-1}(\Bbb R^*)=\prod_{n\in\Bbb N}X_n$ where $X_k$ is the open subset $\Bbb R^*\subset\Bbb R$ and the other $X_n$'s are equal to $\Bbb R.$
$U$ is not open, for the same reason as a product of non-empty proper open subsets is never open in an infinite product topology: since $U\ne\varnothing,$ if it were open, it would contain some element $\bigcap_{n\le N}p_n^{-1}(X_n)=\prod_{n\in\Bbb N}X_n$ of the basis, where the $X_n$'s are non-empty open subsets of $\Bbb R$ for $n\le N,$ and $X_n=\Bbb R$ for $n>N,$ but the latter is incompatible with the definition of $U.$
"The sets $p^{-1}_\beta(A)$ with $A$ being open intervals" form a subbasis of the product topology on $\Bbb R^{\Bbb N}.$