This is from an example in one of my topology lectures. I am not sure if I am misunderstanding the statement.
Suppose $X = \prod_{i \in I} X_i$, where for each $X_i$ we have some topological space. Suppose also that $\forall i \in I (A_i \subset X_i)$. We endow each $A_i$ with the subspace topology with respect to $X_i$. Finally, we define $A := \prod_{i \in I} A_i)$. Is it true that the product topology of $A$ coincides with the subspace topology $A \subset X$, regardless of whether $X$ is endowed with the box or product topology? In our lecture, I think we said it's true, but I can't prove it.
My attempt:
First I note for the product topology on $A$, it is generated by the basis $$\beta_A = \big\{\prod_{i=1}^n Z_i \times \prod_{i >n} A_i: \forall i \leq n(Z_i \text{ is open in } A_i)\big\}$$
Consider the case when $X$ is endowed with the product topology. I note that for any open $U$ in $X$, we have as an element of the subspace topology $A \subset X$:
$$\begin{align} U \cap A & = \big(\prod_{i=1}^n U_i \times \prod_{i>n} X_i\big) \bigcap \prod_{i \in I} A_i \\&= \big(\prod_{i=1}^n U_i \bigcap A_i\big) \times \big(\prod_{i>n} A_i \bigcap X_i\big) \\ &= \big(\prod_{i=1}^n U_i \bigcap A_i\big) \times \prod_{i>n} A_i\end{align}$$
Where in the first line, $U_i$ is an open set in $X_i$.
In this case, I can see that the 'set-structure' of $\beta_A$ and an element of the subspace topology are the same. That is, since each $A_i$ is a subspace of $X_i$, the open set $Z_i$ must be of the form $U_i \cap A_i$. At least at as a cursory argument, this makes some sense to me that the topologies will coincide).
I run into a problem when $X$ is endowed with the box topology. I note that in the box topology, basis elements can be of the form $\prod_{i \in I} U_i$ in which $U_i$ is open in every corresponding $X_i$, and there can be infinitely many $U_i \neq X_i$. So with similar reasoning to above, elements of the subspace topology $A \subset X$ look like:
$$\begin{align} U \cap A &= \prod_{i \in I} U_i \bigcap \prod_{i \in I} A_i \\ &= \prod_{i \in I} \big(U_i \bigcap A_i \big) \\ &= \prod_{i\in I} H_i \end{align}$$
The last equality occurring in the case that the family $U_i \subset A_i$ for all $U_i$ open in $X_i$. This shows that in general, the subspace topology of $A \subset X$ may contain elements that are products of infinitely many open sets not equal to $A_i$, which obviously cannot occur in the product topology.
Question: Is the bolded statement in the first paragraph true? Does the bolded statement in the previous paragraph disprove it?
[Note: I know my proof for $X$ having the product topology is at best incomplete, since I am comparing basis elements to topology elements, rather than comparing the actual topologies. I have been stuck on how to proceed unfortunately. Nonetheless, I think my reasoning might work as a disproof in the case that $X$ has the box topology, but I am not sure.]
