# Why does intersection always preserve “closed” structures?

In many areas of math one can talk about types of closure: Subsets of sets with binary operations can be closed under that binary operation, subsets of topological spaces can be closed, sets of ordinals can be closed.

There seems to be a common thread between many of these: the intersection of these structures is always a structure of the same kind. For example, if $$A,B\subseteq(S,*)$$ are closed under the binary operation $$*$$, then so is their intersection. Often we can even say more: the arbitrary intersection of subgroups is a subgroup, etc. The intersection of $$<\kappa$$ club subsets of $$\kappa$$ is club, although the “ub” is unimportant – the intersection can probably be arbitrary if we only require closure. The intersection of $$\sigma$$-algebras is a $$\sigma$$-algebra, although I think this just a consequence of the binary operation example. Filters have the finite intersection property. The intersection of closed sets in a topological space is closed (one might simply see this as a consequence of De Morgan, but I think it is similar to the other examples when viewing closed sets as those which contain all their limit points as opposed to complements of open sets).

Many examples of these kinds are very, very easy to prove, often following straight from the definitions. So much so that I might hesitate to make any comment on them in the first place, were it not for my inability to formally identify what exactly it is in all these structures that forces them to have this intersection-closure property. And maybe it’s nothing at all, and I’m just cherry picking (after all, several structures aren’t closed under intersection, like open sets, cardinality, etc).

So my question: Is there a generalized “closedness” property which encompasses these examples as well as several others? Maybe the property is more general than intersection of sets? I gave several set-theoretic examples but that is only due to my mathematical exposure, and I’m not just asking about those in set theory. Maybe there are even equivalent notions of “intersection” and “closure” outside of a set-theoretic context.

Edit: As user yoyostein mentioned, maybe there is a categorical perspective on this. At the risk of exposing my severe lack of expertise: my thoughts are to define a “categorical inclusion morphism” generalizing the inclusion morphism from a subset to a set. Then fixing $$A,B$$ we take the category whose objects $$(f_{1},g_{1},X)$$ consist of these inclusion maps $$f_{1}:X\rightarrow A$$, $$f_{2}:X\rightarrow B$$ and whose morphisms are the usual commutative diagrams. Then $$A\cap B$$ would be final in this category, and so these “closed” structures would really be those for which this intersection construction exists in their respective categories. Any chance this is going anywhere?

In the answer here user Stahl gives a categorical explanation for why this is the case for many algebraic structures. Unfortunately, I’m not familiar enough with category theory to tell if what Stahl has written generalizes to “less algebraically-motivated” structures like topological spaces or club sets (actually, I think those are topological), but I’d guess in many cases the properties of the categories he’s mentioning hold elsewhere like in $$\mathsf{Top}$$.

Many of these examples can be generalized by the notion of closure. Say in your universe $$U$$ you have a mapping $$\operatorname{cl}: \mathcal{P}(U) \rightarrow \mathcal{P}(U)$$ with the properties that

i) $$A \subseteq \operatorname{cl}(A)$$ for all $$A$$

ii) if $$A \subseteq B$$ then $$\operatorname{cl}(A) \subseteq \operatorname{cl}(B)$$ (monotonicity.)

Then defined the “closed” sets $$S$$ to be those for which $$\operatorname{cl}(S) = S$$. Usually $$\operatorname{cl}(S)$$ is thought of as the object ‘generated’ by $$S$$. For example, other than the usual closure from topology, $$cl$$ could be the span of vectors, or the subgroup/subring/submodule/$$\sigma$$-subalgebra etc. generated by $$S$$; or the connected components $$S$$ belongs to, or the convex hull of $$S$$. We want to be able to combine elements of $$S$$ in various ways, and by taking $$\operatorname{cl}(S)$$ we add in all the extra elements of $$U$$ to do whatever it is we need, but no more.

I claim that if $$A,B$$ are closed then $$A \cap B$$ is closed. Let $$A,B$$ be closed; then

$$\operatorname{cl}(A \cap B) \subseteq \operatorname{cl}(A)$$ $$\operatorname{cl}(A \cap B) \subseteq \operatorname{cl}(B)$$ by (ii), implying $$\operatorname{cl}(A \cap B) \subseteq \operatorname{cl}(A) \cap \operatorname{cl}(B)$$; by definition, $$\operatorname{cl}(A) = A$$ and $$\operatorname{cl}(B) = B$$, so $$\operatorname{cl}(A \cap B) \subseteq A \cap B$$. Furthermore,

$$\operatorname{cl}(A \cap B) \supseteq A \cap B$$ by (i); so $$\operatorname{cl}(A \cap B) = A \cap B.$$ So $$A \cap B$$ is closed. And the same proof works for showing intersections of arbitrary families of closed sets are closed.

Conversely, if we have a family of ‘closed’ objects $$\mathcal{F} \subseteq \mathcal{P}(U)$$ that is closed under intersection, then we can define $$\operatorname{cl}(A) = \bigcap \{S \in \mathcal{F} | S \supseteq A\}$$. In this case, $$cl$$ clearly obeys (i) and (ii), and $$\mathcal{F} = \{A | \operatorname{cl}(A) = A\}$$.