Why does intersection always preserve "closed" structures?

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 $\mathcal{F} = \{A | \operatorname{cl}(A) = A\}$.

Short answer: a closure operator is probably the notion you are looking at.

Definitions. Let $E$ be a set. A map $X \to \overline{X}$ from ${\cal P}(E)$ to itself is a closure operator if it is extensive, idempotent and isotone, that is, if the following properties hold for all $X, Y\subseteq E$:

1. $X\subseteq\overline{X}$ (extensive)
2. $\overline{\overline{X}} = \overline{X}$ (idempotent)
3. $X\subseteq Y$ implies $\overline{X}\subseteq\overline{Y}$ (isotone)

A set $F\subseteq E$ is closed if $\overline{F} = F$. If $F$ is closed, and if $X\subseteq F$, then $\overline{X}\subseteq \overline{F} = F$. It follows that $\overline{X}$ is the least closed set containing $X$. This justifies the terminology closure. Actually, closure operators can be characterised by their closed sets.

Theorem. A set of closed subsets for some closure operator on $E$ is closed under (possibly infinite) intersection. Moreover, any set of subsets of $E$ closed under (possibly infinite) intersection is the set of closed sets for some closure operator.

Proof. Let $X\to \overline{X}$ be a closure operator and let $(F_i)_{i\in I}$ be a family of closed subsets of $E$. Since a closure is isotone, $\overline{\bigcap_{i\in I}F_i} \subseteq \overline{F_i} = F_i$. It follows that $\overline{\bigcap_{i\in I}F_i} \subseteq \bigcap_{i\in I}F_i$ and thus $\bigcap_{i\in I}F_i$ is closed.

Given a set $\cal F$ of subsets of $E$ closed under intersection, denote by $\overline{X}$ the intersection of all elements of $\cal F$ containing $X$. Then the map $X\to \overline{X}$ is a closure operator for which $\cal F$ is the set of closed sets.

Maybe a "naive" reason may be due to the interpretation of intersection as "and". If $x,y\in A\cap B$, then $x, y$ are in both $A$ and $B$.

By virtue of the fact that $x,y\in A$ alone, it is guaranteed (by the relevant closure property) that $x\cdot y\in A$, where $\cdot$ is the binary operation. Similarly, $x\cdot y\in B$. Hence, $x\cdot y\in A\cap B$.

In contrast, for the case of union, $x,y\in A\cup B$, it may be the case where $x\in A$ while $y\in B$. Hence, it is not guaranteed (a priori) that $x$ and $y$ interact compatibly with each other, since they are from different sets to begin with.

A similar phenomenon (with similar reasoning) is why "restrictions" of functions/morphisms behave so well:

• restriction of a homomorphism to a subgroup is a homomorphism

• restriction of homeomorphism is homeomorphism

A more sophisticated answer I suspect may come from category theory, which is the field to look for when uniting these phenomena that transcends across different areas of math.