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!