Explicit description in set-builder notation of an arbitrary open set of the product topology Short version: Is it possible to explicitly describe the open sets of the product topology (of arbitrary topological spaces) via set-builder notation? (Or differently formulated: What do to if set set I want to describe contains arbitrary many disjunction symbols ?)
Long  version: Given a set $I$ and for every $i\in I$ a topological space $(X_i,\tau _i)$, then one can endow $\prod X_i $ with a topology by specifying  a set $$\mathcal{S}=\{\prod Y_i \ |\  \exists j\in I \ \ \forall i \in I\setminus \{j\}: Y_i=X_i \textrm{ and } Y_j\in \tau _j\}.$$
Then there is only one topology, the product topology, for which this set is a subbase. To obtain the open sets in this topology, one has to go through the following process: One first has to intersect the elements from $S$ finitely many times - then one has obtained a base, $\mathcal{B}$, for the topology - and then one has to form arbitrary unions of the set in the base.
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Let's warm up - I can describe the an element to be in $\mathcal{S}$ either by its construction, meaning it satisfies:
$S\in \mathcal{S} \Leftrightarrow \exists j\in I \ \ \exists O\in \tau_j \ \ \forall i \in I\setminus \{j\}: Y_i=X_i \textrm{ and } Y_j=O,$
or if it is of the form
$S=\{f:I\rightarrow \cup_{i\in I} X_i \ | \ \exists i\in I \ \ \exists O\in \tau_i: f(i)\in O  \}.$
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The same (only a bit more tedious to write out) goes for $\mathcal{B}$:
$B\in \mathcal{B}  \Leftrightarrow  \exists J, \ J \textrm{ finite set, such that } B=\cap_{j\in J} S_j $ with $S_j\in S,$
or in set-builder notation:
$B = \{  f:I\rightarrow \cup_{i\in I} X_i \ | \ \exists n\in \mathbb{N} \ \ \exists i_1,\ldots,i_n \in I \ \ \exists O_{i_1}\in T_{i_1},\ldots,O_{i_n}\in T_{i_n} $ such that $f(i_1)\in O_{i_1} \land ,\ldots, \land f(i_n)\in O_{i_n} \}. $
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But if I want to describe an open set like above, I can only describe it via its contruction: 
$O\in \mathcal{O}  \Leftrightarrow  \exists F, F \textrm{ arbitrary set},\  \ \exists J, \ J \textrm{ finite set, such that } O=\cup_{f \in F} \, \cap_{j\in J} S_{fj} $ with $S_{fj}\in S$.
If I wanted to describe it in set-builder notation I would have to have use infinitely many disjunction symbols. 
To exemplify this, see the following: Since an union of, lets say two base sets, $B_1\cup B_2$, can be written in set-builder notation as - this gets nasty - 
$B_1\cup B_2= \{  f:I\rightarrow \cup_{i\in I} X_i \ | \ \  \exists n_1,n_2 \in \mathbb{N} \ \ \exists i_{1,n_1},\ldots ,i_{n_1,n_1}\in I, \ \  [\ldots ]$ 
$ [\ldots ]  i_{1,n_2},\ldots ,i_{2,n_2} \in I \ \ \ \exists O_{1,n_1} \in \tau_{i_{1,n_1}} ,\ldots ,O_{n_1,n_1} \in \tau_{i_{n_1,n_1}}, \ [\ldots ]  $
$[\ldots ]  O_{1,n_2} \in \tau_{i_{1,n_2}}  ,\ldots ,O_{n_2,n_2} \in \tau_{i_{n_2,n_2}}  \textrm{ such that }  [\ldots ]$
$  [\ldots ]  ( f(i_{1,n_1})\in O_{1,n_1} \land ,\ldots , \land f(i_{n_1,n_1})\in O_{n_1,n_1}) \ \ \lor ( f(i_{1,n_2})\in O_{1,n_2} \land ,\ldots , \land f(i_{n_2,n_2})\in O_{n_2,n_2})$}
I would get for arbitrary unions  arbitrary many disjunctions - but I can't write that out!
 A: Your set-builder expression for a typical subbase element is inadequate: it doesn’t even ensure that elements of $S$ belong to the product of the $X_i$, since it doesn’t require $f(j)\in X_j$ for all $j\in I$. You could remedy that shortcoming with 
$$S=\left\{f\in ^I(\cup_{i\in I}X_i):\exists i\in I\,\exists U\in\tau_i\Big(f(i)\in U\land\forall j\in I\big(f(j)\in X_j\big)\Big)\right\}\;,$$
but it still wouldn’t do what you want: every member of $\prod\limits_{i\in I}X_i$ satisfies the set-building condition, so this is just a fancy description of the product.
The problem is that the definition allows a different choice of the restricting index $i$ and the restriction $U$ for each $f\in S$, and to get a member of the subbase you have to fix a single restricting index $i$ and restriction $U$ that applies to all of them. I see no way to write a self-contained expression
$$S=\left\{f\in ^I(\cup_{i\in I}X_i):\varphi(f)\right\}$$
that actually describes all subbasic sets and nothing else; external parameters $i$ and $U$ specifying the restriction on the members of $S$ seem to be necessary.
Your set-builder version of $B$ has similar problems. Unless you can find a way to overcome them $-$ and as I said, I don’t at the moment see one $-$ there’s not much point worrying about the next stage.
