Uncountable product of second countable spaces

Question

Let $(X_i, \mathcal{T}_i)_{i \in I}$ be a family non indiscrete topological spaces and equip the product with the product topology $\mathcal{T}$. If $\prod X_i$ is second countable, prove that $|I| \leq |\mathbb{N}|$

The proof my books provides goes as follows:

By non indiscreteness of the spaces, we can find $\emptyset \neq B_k \neq X_k$ for every $k \in I$ where $B_k \in \mathcal{T}_k$.

Fix a basis $\mathcal{A}$ for the product topology, and choose $A_k \neq \emptyset$ in $\mathcal{A}$ such that $A_k \subseteq \operatorname{pr}_k^{-1}(B_k)$.

Define $f: I \to \mathcal{A}: k \mapsto A_k$

It then says that if $f^{-1}(\{A\})$ is infinite for $A \in \mathcal{A}$, then $$\operatorname{int}\left(\bigcap_{j \in f^{-1}(\{A\})} \operatorname{pr_j^{-1}(B_j)}\right) = \emptyset$$

and then the proof goes on. I can't understand why this set has empty interior. Can anyone help?

Answer 1

Note that every basic open set is a finite intersection of preimages of open sets in each $X_\alpha$ by the canonical projections. If the interior is nonempty, then there is an interior points, which has a basic open set contained in given set $$\operatorname{int}\left(\bigcap_{j \in f^{-1}(\{A\})} \operatorname{pr}_j^{-1}(B_j)\right),$$ which is impossible, since given set is an intersection of infinitely many such preimages.

Answer 2

I would set it up slightly differently. Denoting the projections by $\pi_i, i \in I$, for each $i$ we pick $O_i$ open with $\emptyset \neq O_i \neq X_i$, which we can do because all $X_i$ are not insdicrete. Also, pick $p_i \in O_i$ and $q_i \in X_i\setminus O_i$. Let $q(i)$ be the point in the product that is $q_j$ for all $j \neq i$, and with $q_i=p_i$.

If $\mathcal{B}$ is a base for $\prod_{i \in I} X_i$, then for each $i$ we can pick $B_i \in \mathcal{B}$ such that $q(i) \in B_i \subseteq \pi_i^{-1}[O_i]$, which can be done by the definition of a base and the fact that $q(i) \in \pi_i^{-1}[O_i]$.

If now $i_1 \neq i_2$ we have that $q(i_1) \in B_{i_1}$ by construction, while $q(i_2) \notin B_{i_1}$ (otherwise $\pi_{i_1}(q(i_2)) = q_{i_1} \in O_{i_1}$, which is not the case.), and also $q(i_2) \in B_{i_2}$ ,$q(i_1) \notin B_{i_2}$, so $B_{i_1} \neq B_{i_2}$ and the mapping $i \to \mathcal{B}$ defined by $i \to B_i$ is injective. Hence $|I| \le |\mathcal{B}|$ and as we can start with a countable base, $I$ is at most countable.

Your final question follows from the fact that if $A = \prod_{i \in I} A_i$ and the interior of $A$ is non-empty, then $\{i : A_i \neq X_i\}$ is finite, as it must contain a standard basic open set.