Theorem 19.2 in Munkres Suppose the topology on each space $X_{\alpha}$ is given by a basis $\mathcal{B}_{\alpha}$. The collection of all sets of the form $$\prod_{\alpha \in J} B_{\alpha}$$ where $B_{\alpha} \in \mathcal{B}_{\alpha}$ for each ${\alpha}$, will serve as a basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$
I am trying to prove this theorem but I am wondering if my proof is solid. (For reminder, the definition, lemma, or any theorem used in the proof will be the ones mentioned in the Munkres.)
The way I tried to prove it is as follows:
proof. Use Lemma 13.2 in Munkres. Let $\mathcal{C}$ be the collection of all sets of the form $\prod_{\alpha \in J} B_{\alpha}$ (mentioned in the Theorem 19.2.). Then $\mathcal{C}$ is collection of open sets of the box topology since each $B_{\alpha}$ is open in $X_{\alpha}$. ($\because$ Box topology is the topology having as basis all sets of the form $\prod U_{\alpha}$, where $U_{\alpha}$ is open in $X_{\alpha}$.)
Let $U=\prod_{\alpha \in J} U_{\alpha}$ be open set in $\prod X_{\alpha}$ and $x\in U$. For $\alpha \:th$ coordinate of $x$ (denoted as $x_{\alpha}$), there is $B_{\alpha}$ s.t. $x_{\alpha} \in B_{\alpha}\subset U_{\alpha}$. ($\because \: U_{\alpha}$ is open in $X_{\alpha}$ and $B_{\alpha}$ is a basis element of $X_{\alpha}$. Thus by property of basis, the inclusion holds.) Since this can be done for all coordinates of $x$, we can say that $x$ is contained in some set $B$ which is an element of the $\prod_{\alpha \in J} B_{\alpha}$ mentioned in the Theorem 19.2 and $B$ is contained in the open set $U$ ($x\in B \subset U)$.
Hence by Lemma 13.2 in Munkres, the collection of all sets of the form $\prod_{\alpha \in J} B_{\alpha}$ is basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$. $$\tag*{$\blacksquare$}$$
The part that I am especially concerned about is the part where is start with "Let $U=\prod_{\alpha \in J} U_{\alpha}$ be open set in $\prod X_{\alpha}$" Even though I think this seems okay (since box topology is defined as the topology having as basis all sets of the form $\prod U_a$ where $U_a$ is open in $X_a$ for each $a$. I thought then arbitrary open set in the box topology will be of the form $\prod_{a \in J} U_{a}$ for open set is union of basis of the topology), I am still wondering if it was a hasty statement.
Thank you for your time.
Best Answer
Simpler: Let $O$ be open in the box topology and $x \in O$. Then by the definition of the standard base of the box topology, there are open sets $O_\alpha \subseteq X_\alpha$, $\alpha \in J$ such that $$x \in \prod_{\alpha \in J} O_\alpha \subseteq O$$
and so we have for each coordinate $\alpha$ that $x_\alpha \in O_\alpha$. We apply the fact that each $\mathcal{B}_\alpha$ is a base for $X_\alpha$ to give us $B_\alpha \in \mathcal{B}_\alpha$ such that $$x_\alpha \in B_\alpha \subseteq O_\alpha$$
for each $\alpha \in J$. Now clearly
$$x \in \prod_{\alpha \in J} B_\alpha \subseteq \prod_{\alpha \in J} O_\alpha \subseteq O$$
and the first set is in the "product base" collection defined by Munkres, and as $O$ and $x$ were arbitary, that collection has been shown to be a base for the box topology (by 13.2).