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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!

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    If you don't mind me asking: Why do you want to do that?2012-01-01
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    @MichaelGreinecker I don't. I was trying to figure out how the product topology on some family of sets endowed with the discrete topology looks like - meaning I was trying to see how a generic open set would look like. And trying to see how something looks like (in this case) means (among other things) describing in set-builder notation; and in the process of doing that it hit me, that I can't actually describe the set in set-builder notation. And this I found to be rather strange (since I thought that any well-formed set can be described like that), so I posted this problem here.2012-01-02
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    @MichaelGreinecker But if you don't mind me asking: Why did you want to know that?2012-01-02
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    To me, this is a somewhat unusual way to describe a set. I know how a circle looks like, but I don't think I understand it better by viewing it as the complement of a union of open rectangles in $\mathbb{R}^2$. In the case of the discrete topology, a subset of $\prod_i X_i$ looks essentially like $S\times\prod_{j\in J}X_j$ where $J\subseteq I$ has a finite complement and $S$ is an arbitrary subset of $\prod_{i\in I\backslash J}$. Since you only have to construct points in the first step, set builder notation actually works quite well for products of discrete spaces.2012-01-02
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    @MichaelGreinecker Yes, it is unusual - but even if it unusual to write a circle like that, one should be able to do it! That was the motivation behind all of this. Btw, did you mean, in the case of the discrete topology, that an **open set** of $\prod_i X_i$ had to essentially look like $S \times \prod_j X_j $ ? I didn't really understood what you meant by that. (...)2012-01-02
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    (...) An open set of $\prod_i X_i$, as I know it, consists of all functions for which (possibly arbitrary many) sets of subset of our spaces are fixed such that for selections of finitely many points of the domain of your function we have a set of such subsets, such that the values at those points that are those subset (which corresponding in the index to that point); (...)2012-01-02
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    (...) or shortly said: all those function sets that consist of arbitrary unions of function sets that at finitely many point have values in some prescribed subset (since we have the discrete topology) of the space corresponding to that point. (...)2012-01-02
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    (...) Am I maybe missing the point in "...looks **essentially** like..." (vs. **is**) ?2012-01-02
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    Yes, I meant an open set. Silly me. The "looks essentially like" refers to the fact, that $S\times\prod_{i\in J}X_i$ is in principle a set of ordered pairs and not a set of functions with domain $I$. This is one of the cases where rigor can lead to rigor mortis.2012-01-02

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