Definition Of $J$-tuple, $J$ is an Index Set 
Let J be an index set. Given a set $X$, we define a J-tuple of elements of $X$ to be a function $x:J\to X$.

Question: When we define the notion of  J-tuple in terms of map, don’t we loose the notion of “order”(order list) on J-tuples of elements of $X$? From https://en.wikipedia.org/wiki/Tuple#Tuples_as_functions definition of J-tuple, when J is finite, map $F$ should be surjective. That is not the case, when J is an arbitrary index set. This is the reason, why I don’t like to define sequence in terms of map, even though it is formal way to define sequence. And saying a map is an element of $\Pi_{i \in I}X_{i}$ set, sound really absurd/unnatural to me.
Note I have mixed some notation from Wikipedia definition and Munkres’ definition.
 A: If $J$ has an order (say if $J=\{1,2,3,\dots, n\}$, or $J=\Bbb N$) then we can sort of say the $J$-tuple $f: J \to X$ has an order too. The notation $(x_1,x_2)$ in a $2$-tuple in $X^2$ is just a shorthand for the function $\{1,2\} \to X$ defined by $f(1)=x_1$ and $f(2)=x_2$. The subscript notation or tuple notation is just another way of writing the function compactly.
The same way a sequence $(x_n)_n$ is just a function $f: \Bbb N \to X$ as well. The order (as used in convergence) purely comes from the index set/domain.
The function can be totally arbitrary (even constant or almost constant, or $f(x)=\sin(x)$ for an $\Bbb R$-tuple with values in $[0,1]$ e.g.
Products in set theory (hence in topology, analysis and algebra too) are just sets of functions. It is what it is.
Added after comments
We want to show $\prod_{\alpha \in J} U_\alpha \cap  \prod_{\alpha \in J} V_\alpha = \prod_{\alpha \in J} (V_\alpha \cap V_\alpha)$.
So let $f$ be in the LHS. So it is a function (i.e. a set of pairs) and as $f \in \prod_{\alpha \in J} U_\alpha$ for each fixed $\alpha$ there is a unique pair $(\alpha, x) \in f$ with $x \in U_\alpha$. Likewise there is a unique pair $(\alpha, x') \in f$ with $x' \in V_\alpha$. By unicity $x=x' \in U_\alpha \cap V_\alpha$. So $f$ has a unique pair with first component $\alpha$ and second component in $U_\alpha \cap V_\alpha$, and as this holds for all $\alpha \in J$, $f \in \prod_{\alpha \in J} (U_\alpha \cap V_\alpha)$. The reverse inclusion is similar.
The codomain argument is nonsense: $f \in \prod_{\alpha \in J} X_\alpha$ iff $f$ is a set of pairs such that
$$\forall \alpha \in J: \exists! (a,b) \in f: \exists x \in X_\alpha: (a,b)=(\alpha, x)$$
(this is set theory, we don't have to specify a codomain in advance like in category theory or topology (when considering a map between spaces) e.g.). A possible codomain can be computed from the set of pairs using standard axioms.
