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[Attention! This question requires some reading and it's answer probably is in form of a "soft-answer", i.e. it can't be translated into a hard mathematical proposition. (I hope I haven't scared away all readers with this.)]

Consider the following three examples:

1) [If this example seems too technical just skip it - it isn't that important for the idea I want to convey.] The set $T$ of all terms (of a functional structure) is the set $$T=\bigcap_{O\text{ closed under concatenation with function symbols}}_{O\supseteq X} O,$$where $X$ is the (countable) of all variables and "closed under concatenation with function symbol" means: If $f$ is some function symbol of arity $n$ and $x_1,\ldots,x_n\in X$, then $fx_1\ldots x_n\in T$. The above $T$ is the smallest set such that it contains the variables and is closed under concatenation with function symbols.

2) The smallest subgroup $G$ of a group $X$, containing a set $A\subseteq X$, is the set $$G:=\bigcap_{O\ \text{ is a subgroup of $X$}}_{O\supseteq A} O.$$

3) The smallest $\sigma$-algebra $\mathcal{A}$ on a set $X$ containing a set $A\subseteq X$ is the set $$\mathcal{A}:=\bigcap_{O\ \text{ is a $\sigma$-algebra on $X$}}_{O\supseteq A} O.$$

4) The set $$C:=\bigcap_{O\ \text{ is open in $X$}}_{O\supseteq A} O,$$ where $X$ is a metric space and $A\subseteq X$ is arbitrary.

Now here's the thing: The sets $T$, $G$, and $\mathcal{A}$, from examples 1),2) and 3) are also closed under the closing condition defining them, i.e. $T$ is also closed under concatenation with function symbols, $G$ is also a group and $\mathcal{A}$ is also a $\sigma$-algebra; for $G$ and $\mathcal{A}$ this is already implied by their name (the smallest subgroup, the smallest $\sigma$*-algebra*), which was the reason I also gave $T$ as an example, where it's name doesn't already imply it's closure under it's defining closure operations. But the set $C$ from example 4) need not be open, if for example $\mathbb{R}=X$ and $A=[0,1]$ (maybe there are nontrivial metric space, where it is open for nontrivial sets $A$, but I didn't want to waste time checking that). That is, $C$ isn't closed (no pun intended) under the closing condition I used to define it; or differently said: There isn't a smallest open set containing $A$.

Notice that the closure conditions in 1) and 2) have a more algebraic character, since we close under some algebraic operations, where the closure condition in 3) as a more "set-theoretical-topology"-type character, since we close under set-theoretic operations. Nonetheless in all cases the outcome is again "closed".

My question is: How do generally (abstractly) "closing conditions" $\mathscr{C}$ have to look like such that the set $$ S:=\bigcap_{O\ \text{ is closed under $\mathscr{C}$}}_{O\supseteq A} O$$ is itself closed in $X$ under $\mathscr{C}$, where $A\subseteq X$ is an arbitrary set ? Differently said: How do generally (abstractly) "closing conditions" have to look like such that there is a smallest set being closed under these conditions, containing some arbitrary fixed set.

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Yikes, when I first read this question, I thought I'd need to flag it for migration to the meta.math.se forum! –  amWhy Nov 4 '12 at 14:24
    
haha :) Maybe I should have said "criterion" instead of "condition" to make it sound more like mathematics-talk, but I think I'll leave it at "condition" for now. –  temo Nov 4 '12 at 14:33
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Let $\mathscr{C}(X)$ be the family of subsets of $X$ that are closed under $\mathscr{C}$. What you need is simply that $\mathscr{C}(X)$ be closed under arbitrary intersections; I really doubt that one can say much more than that. –  Brian M. Scott Nov 4 '12 at 16:59
    
@BrianM.Scott Well, requiring that is a bit tautological I think, since I asked how $\mathscr{C}$ must be such that it is closed under intersection: If the answer is, that $\mathscr{C}X$ is closed under arbitrary intersection, that doesn't tell me anything about $\mathscr{C}$. What I'm looking is somewhat similar to the HSP theorem (also called Birkhoff's theorem) from model theory that tells me under which circumstances an algebraic structure (like a group) is solely defined by equations its element have to obey: Said circumstances being, that the algebraic structure has to be [...] –  temo Nov 4 '12 at 18:16
    
I know that it doesn’t tell you anything about $\mathscr{C}$; my point is that I don’t think that you can say much more than that in general. –  Brian M. Scott Nov 4 '12 at 18:18

2 Answers 2

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We are given a set $A$ and a predicate $\Phi$ and take the intersection over the class of all sets $O$ with $A\subseteq O\land \Phi(O)$. One may read from the structure of that predicate $\Phi$ wether or not this is a good closure condition.

For example:

  • transitive closure of relations: $\Phi(O)\equiv \forall x,y\in O\colon\phi(x,y)\to f(x,y)\in O$ where $\phi(\langle x_1,x_2\rangle,\langle y_1,y_2\rangle)\equiv x_2=y_1$ and $f(\langle x_1,x_2\rangle,\langle y_1,y_2\rangle)=\langle x_1,y_2\rangle$.
  • generated subgroup: $\Phi(O)\equiv \forall x,y\in O\colon\phi(x,y)\to f(x,y)\in O$ with $\phi(x,y)\equiv x\in X\land y\in X$ and $f(x,y)=xy^{-1}$.
  • topological closure: $\Phi(O)\equiv \forall S\subset O, x\in O\colon\phi(x,S)\to f(x,S)\in O$ with $\phi(x,S)\equiv S\subseteq X\land x\text{ is a limit point of }S$ and $f(x,S)=x$.

and so on. In general there may occur elements of $O$, subsets of $O$ and many other higher structures (elations, functions, ...). In all these cases we obtain closure of the condition under arbitrary intersection: If for all $O_i$ we have that e.g. $\phi(x,S)$ implies $f(x,S)\in O$ then for $O=\bigcap O_i$ we have that $\phi(x,S)$ for $x\in O, S\subset O$ implies the same for all $O_i$, hence $f(x,S)$ in all $O_i$, hence in $O$.

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But what fails (in the structure of the predicate) in the case of arbitrary intersection of open sets (as in example 4) of my question) ? –  temo Nov 19 '12 at 12:25
    
There is an existential quantifier in the ways –  Hagen von Eitzen Nov 19 '12 at 12:28
    
Could you please be a little more explicit ? I can't see where this quantifier has to be, since one defines a set $O\subseteq X$ to be open, if it is in some a priori defined topology on $X$; and in the definition of topology we don't use an existential quantifier, as far as I can see... –  temo Nov 19 '12 at 12:37
    
I think the main difference is: If you simply take the topology $T$ on $X$ as given data, then $\Phi(O)$ has simply the form $\Phi(O)\equiv O\in T$, whereas all my examples use only $\in O$, not $O\in$. –  Hagen von Eitzen Nov 19 '12 at 16:55
    
But where is then the existential quantifier ? (And could you please also answer my comment below the answer of the other question I had a bounty on: math.stackexchange.com/questions/222855/…) –  temo Nov 20 '12 at 11:02

Suppose that $X$ is a set, $\mathcal{A}$ is a distinguished family of subsets of $A$, and $f \colon \mathcal{A} \rightarrow \mathbf{Pow}\,X$. Here $\mathbf{Pow}\,X$ is the power set of $X$. Now define a class of functions indexed by the ordinals. In all cases the domain of the function is $\mathbf{Pow}\,X$. \begin{align} g_{0} : A &\mapsto A \\ g_{\alpha + 1} \colon A &\mapsto g_{\alpha}(A) \cup (\cup \{ f (B) \colon B \subseteq g_{\alpha}(A) \} )\\ g_{\alpha} \colon A &\mapsto \cup \{ g_{\beta}(A) \colon \beta < \alpha \} \text{ if $\alpha$ is a limit ordinal.} \end{align} There is an ordinal $\alpha^*$ satisfying for all $A \subseteq X$ and all $\alpha \geq \alpha^*$ we have $g_{\alpha}(A) = g_{\alpha^*}(A)$. The function $g_{\alpha^*}$ is a closure operator. For all $A,B \subseteq X$ we have $$A \subseteq g_{\alpha^*}(A) = g_{\alpha^*}(g_{\alpha^*}(A))$$ and, if $A \subseteq B$ we have $$g_{\alpha^*}(A) \subseteq g_{\alpha^*}(B).$$

You may wish to check out my answer to this question and this question.

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I'm afraid that was a little bit too high for me, since I know almost nothing about ordinals. I also couldn' figure out, how the $g_{\alpha^*}$ is related to my closure condition. Is it, that my closure conditions $\mathscr{C}$ are "good" if there existsan ordinal $\alpha^*$ such that $g_{\alpha^*}$ becomes a closure operator ? If that is so, what does that tell me about $\mathscr{C}$ ? –  temo Nov 19 '12 at 12:34
    
Suppose you have a set $A$ and you wish to find the closure of $A$. In general the closure of $A$ is bigger than $A$. To be concrete suppose that $X$ is a group and the closure of a set $A \subseteq X$ is the smallest subgroup that contains $A$. The smallest subgroup is $\{ e \} $, the trivial subgroup so we might want $f(\varnothing) = \{ e \} $. Since subgroups are closed under products of elements and taking inverses we might want to say $f(A) = \{ ab \colon a, b \in A \} \cup \{ a^{-1} \colon a \in A \} $... –  Jay Nov 20 '12 at 12:56
    
...All that ordinal stuff is just a fancy way of saying: Keep on adding stuff until you are no longer adding anything new. –  Jay Nov 20 '12 at 13:02

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