Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

[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.

share|improve this question
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
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

up vote 0 down vote accepted

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 $O$.

share|improve this answer
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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.