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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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    $\begingroup$ If you don't mind me asking: Why do you want to do that? $\endgroup$ Commented Jan 1, 2012 at 20:38
  • $\begingroup$ @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. $\endgroup$
    – temo
    Commented Jan 2, 2012 at 16:02
  • $\begingroup$ @MichaelGreinecker But if you don't mind me asking: Why did you want to know that? $\endgroup$
    – temo
    Commented Jan 2, 2012 at 16:02
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    $\begingroup$ 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. $\endgroup$ Commented Jan 2, 2012 at 16:22
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    $\begingroup$ 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. $\endgroup$ Commented Jan 2, 2012 at 18:39

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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.

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  • $\begingroup$ @BrianMScott You're right: I did write $S$ incorrectly out. And you're also right I think, that one can't write the sets of (for example) $\mathcal{S}$ in a self-contained form. But one can write them in a form, where one first fixates the restricting indices and restriction sets and then writes the set in set-builder notation. For example the $S$ could (hopefully correctly this time) be written as: "Let $i\in I$ and $U\in \tau_i$. Then $S=\{ f:I \rightarrow \cup_{j\in I} X_j | f(i)\in U \land \forall k\in I\setminus \{ j\}: f(k)\in X_k \}$ (...) $\endgroup$
    – temo
    Commented Jan 2, 2012 at 17:39
  • $\begingroup$ (...) But even for this weaker conditions (since self-containdness isn't any more required) for writing the sets in set-builder notation, which I think should also work for the sets in $\mathcal{B}$ (by moving all the appropriate indices and open sets of the set $B$, as I have written it, outside of $B$), I can't write the open sets in set-builder notation, since it would contain infinitely many conjunction and disjunction symbols, since it would be of the form: "[specifying the fixed indices and sets] $\{f:I\rightarrow \cup_{i\in I} X_i \ | \alpha (f) \land \beta (f) \land \ldots \}$" (...) $\endgroup$
    – temo
    Commented Jan 2, 2012 at 17:46
  • $\begingroup$ (...) where $\alpha (f), \beta(f) $ look tell me that $f$ has to lie in finitely many of the sets fixed at the beginning (so there are finitely many conjunction symbols) - and since the set consists of arbitrary unions of finite intersections (every finite intersection corresponds to an $\alpha,\beta,\ldots $), I have to form arbitrary many disjunctions, to write the set in set-builder notation. (...) $\endgroup$
    – temo
    Commented Jan 2, 2012 at 17:55
  • $\begingroup$ (...) This is actually the problem I want to know the answer of. Writing the sets in a self-contained way would be a nice bonus, but that was not what I was originally after. Sorry if my malformed expressions of my sets $S,B$ misguided to think that this was the problem I was after. (Although I didn't think of the problem of self-containdness, when I was writing this question, I'm grateful to you for pointing it out, because otherwise I probably wouldn't have realised it, so I'm also glad for having written the sets in an inadequate way.) $\endgroup$
    – temo
    Commented Jan 2, 2012 at 17:57

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