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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?

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    $\begingroup$ Trivial, Thank you very much for your edit. $\endgroup$ Commented Jun 27, 2022 at 23:31
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    $\begingroup$ Some authors includes codomain as part of the component of a function, so that a function from $A$ to $B$ is a triple $(A, B, f)$, where $f\subseteq A\times B$ is a function relation. However, other authors regard functions as the same thing as function relations. (See this discussion, for instance.) If we adopt the second convention, then Munkres's definition allows to prove the desired equality. $\endgroup$ Commented Jun 27, 2022 at 23:47
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    $\begingroup$ In short, all this confusion has stemmed from the fact that Munkres did not make a strict distinction between function (domain, codomain, function relation combined) and function relation. Unfortunately, confusing these two notions, either mistakenly or intentionally, is a rather common convention in mathematics. $\endgroup$ Commented Jun 27, 2022 at 23:59
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    $\begingroup$ @tchappy ha I'm glad you like it! Thank you for making me know it, I hope you have a joyful math journey! $\endgroup$ Commented Jun 28, 2022 at 0:31
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    $\begingroup$ @SangchulLee Thank you very much for your answer and your edit. $\endgroup$ Commented Jun 28, 2022 at 0:44

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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.

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