Intuitively, $(\prod_{\lambda\in\Lambda}A_\lambda)\cap(\prod_{\lambda\in\Lambda}B_\lambda)=\prod_{\lambda\in\Lambda}(A_\lambda\cap B_\lambda)$. But ..

Question

I am reading "Topology 2nd Edition" by James R. Munkres.

The definition of a cartesian product of an indexed family of sets is here:^src)

Definition. Let $\{A_{\alpha}\}_{\alpha \in J}$ be an indexed family of sets; let $X = \bigcup_{\alpha \in J} A_{\alpha}$. The cartesian product of this indexed family, denoted by $$ \prod_{\alpha \in J} A_{\alpha} $$ is defined to be the set of all $J$-tuples $(x_{\alpha})_{\alpha \in J}$ of elements of $X$ such that $x_{\alpha} \in A_{\alpha}$ for each $\alpha \in J$. That is, it is the set of all functions $$ \mathbf{x} \, : \, J \to \bigcup_{\alpha \in J} A_{\alpha} $$ such that $\mathbf{x}(\alpha) \in A_{\alpha}$ for each $\alpha \in J$.

The following problem (Problem 9 on p.51) is from "Introduction to Set Theory and Topology" (in Japanese) by Kazuo Matsuzaka:

Prove that$$\left (\prod \limits _{\lambda \in \Lambda}A_\lambda \right )\cap \left (\prod \limits _{\lambda \in \Lambda}B_\lambda \right )=\prod \limits _{\lambda \in \Lambda}(A_\lambda \cap B_\lambda ).$$

Let $A_\lambda :=\mathbb{N}$.
Let $B_\lambda :=\mathbb{Z}$.
The codomain of elements of $\displaystyle \prod \limits _{\lambda \in \Lambda}A_\lambda$ is $\mathbb{N}$.
The codomain of elements of $\displaystyle \prod \limits _{\lambda \in \Lambda}B_\lambda$ is $\mathbb{Z}$.
The codomain of elements of $\displaystyle \prod \limits _{\lambda \in \Lambda}(A_\lambda \cap B_\lambda )$ is $\mathbb{N}$.
Then, $\displaystyle \left (\prod \limits _{\lambda \in \Lambda}A_\lambda \right )\cap \left (\prod \limits _{\lambda \in \Lambda}B_\lambda \right )=\varnothing$ since the codomain of elements of $\displaystyle \prod \limits _{\lambda \in \Lambda}A_\lambda$ and the codomain of elements of $\displaystyle \prod \limits _{\lambda \in \Lambda}B_\lambda$ are not the same.
Let $a$ be a function from $\Lambda$ to $\mathbb{N}$ such that $a(\lambda ):=1$ for any $\lambda \in \Lambda$.
Then, $\displaystyle a\in \prod \limits _{\lambda \in \Lambda}(A_\lambda \cap B_\lambda )$.
So, $\displaystyle \prod \limits _{\lambda \in \Lambda}(A_\lambda \cap B_\lambda )\neq \varnothing$.

But intuitively, I think $\displaystyle \left (\prod \limits _{\lambda \in \Lambda}A_\lambda \right )\cap \left (\prod \limits _{\lambda \in \Lambda}B_\lambda \right )=\prod \limits _{\lambda \in \Lambda}(A_\lambda \cap B_\lambda )$ holds.

What is the answer to this problem?

Answer 1

I think this might be down to differing definitions of tuples, if we take for instance a definition where a tuple is a set of ordered pairs, such that

$$(a, b, c, \cdots) = \{(1,a), (2,b), (3,c), \cdots\}$$

when the index set is $\mathbb N,$ then I think it's quite clear that with the given example we would have that tuples containing all natural numbers would appear in both products, matching your intuitive result.

However, using the definition quoted here where tuples are functions, because two functions can only technically be equal if their domains and codomains are the same, we don't get the same result: each of the functions in $\Pi_{\lambda \in \Lambda} \mathbb N$ map $\Lambda \to \mathbb N,$ whereas each of the functions in $\Pi_{\lambda \in \Lambda} \mathbb Z$ map $\Lambda \to \mathbb Z,$ so because $\mathbb N \neq \mathbb Z$ no pair of functions from the two products can be equal, so there is no intersection.

As a non-expert in the domain I can't say for sure whether this points to some deeper or more useful truth about the cartesian product in this setting or if it's just mathematical rules-lawyering, but that does appear to be the reason that the result they give here doesn't match the intuitive result.