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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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    If your theory has any existential axioms then it becomes much harder for the intersection of submodels to still be a submodel. If your axioms are all of the form "for all [...] there exists _unique_ [...] such that [...]" (more precisely, if you have a "cartesian" theory in the sense of Johnstone, [_Sketches of an elephant_, Part D]) then the intersection of submodels should still be a submodel.2012-11-18

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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    @temo: I'm not completely sure, but $\exists$ and $O\in$ can be severe obstacles, though I'm not sure how to formalize that. My gut feeling about the mishap with "open" is: If $T$ is *any* subset of $2^X$, we can consider $x\in O\rightarrow \exists U\in T\colon x\in U\subseteq O$, which happens to coincide with "open" if $T$ is a topology (or a basis of one)2012-11-20
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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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    ...All that ordinal stuff is just a fancy way of saying: Keep on adding stuff until you are no longer adding anything new.2012-11-20