# 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 $$\mathsf{Top}$$.

• These are the kinds of questions I get excited about. +1! – Don Thousand Mar 30 at 21:15
• But, the intersection if infinitely many closed set is not necessarily open. I always thought that this was an interesting one, and part of why infinity is so messy. – Doug M Mar 30 at 21:19
• @DougM I assume you meant not necessarily closed? But in that case, I thought it was by De Morgan. – P-addict Mar 30 at 21:27
• Sorry, I meant to say that the intersection of infinitely many open sets are not necessarily open. – Doug M Mar 30 at 21:32
• @DougM : Or maybe you meant that that the intersection of infinitely many open sets IS not necessarily open? – Michael Hardy Mar 30 at 22:08

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

• Thank you, this (or something like it) is definitely what I was looking for, at least as far as the set-theoretic examples I'm familiar with go. It's surprisingly simple, can't believe I didn't think of this. – P-addict Mar 31 at 5:37
• Sure, it was a nice question that I had to think about a bit! – Jair Taylor Mar 31 at 6:31
• Also $\operatorname {cl}(\operatorname {cl}(A))=\operatorname {cl}(A)$. – lhf Apr 18 at 11:05
• @lhf I think that requirement is needed to guarantee that $\operatorname{cl}(A) = \bigcap \{S \in \mathcal{F} | S \supseteq A\}$ so that there is a bijection between closure operators and intersection-closed families. But if you only want to prove $\mathcal{F}$ is intersection-closed iff there is a (not necessarily unique) closure operator, that axiom isn't necessary. – Jair Taylor Apr 19 at 18:31

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.

• Beat you by a few seconds, I think. – Jair Taylor Mar 31 at 5:22
• @jair-taylor This happens sometimes... :=) – J.-E. Pin Mar 31 at 5:25

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.

This works for objects and properties that are of the form "if something is in the set then something else is in the set", i.e.,

If $$A\subseteq S$$ then $$a\in S$$

where (possibly many) pairs $$(A,a)$$ are given. If this statement holds for each $$S_i, i\in I$$, then it also holds for $$S:=\bigcap_{i\in I}S_i$$. Namely, if $$A\subseteq S$$, then $$S\subseteq S_i$$ for all $$i$$, then $$a\in S_i$$ for all $$i$$, then $$a\in S$$.

For example, given a group $$G$$, the concept of subgroup $$H$$ can be defined by

• If $$\{a,b\}\subseteq H$$ then $$ab\in H$$
• If $$\{a\}\subseteq H$$ then $$a^{-1}\in H$$
• If $$\emptyset\subseteq H$$, then $$e\in H$$

where $$a,b$$ run over all of $$G$$. These are all of the form above. Therefore, the intersection of subgroups is a subgroup.

For an ideal of a ring $$R$$, we can use (the above for subgroup of additive group together with)

• If $$\{a\}\subseteq I$$, then $$ca\in I$$

where $$a,c$$ run over $$R$$. It follows that the intersection of ideal is an ideal.

For closed sets of a topological space $$X$$, we can use

• If $$A\subseteq S$$, then $$a\in S$$

where $$A$$ runs over all subsets of $$X$$ having $$a$$ as a limit point.

(You can also spell out the corresponding conditions for $$\sigma$$-algebras, for ordinals, for much more).

• This is interesting notation. Writing "if $\{a,b\} \subseteq H$," I mean. Out of curiosity, what inspired you to change from "if $a,b \in H$" to this new notation? – goblin Apr 18 at 12:55

The top-voted answers are OK, but also very incomplete and a bit circular.

I mean sure, there's a standard theory about how closure operators (which are monotone functions satisfying $$A \leq \mathrm{cl}\, A, \: \mathrm{cl} \,\mathrm{cl}\, A \leq \mathrm{cl} A$$) and closure systems (which are collections of sets closed under intersection, aka Moore families) are the same thing. And, yes it's good to know about this bijection.

At the same time, merely describing this bijection doesn't really explain why we ended up calling these things closure operators and/or closure systems in the first place. Here's a hypothetical Q&A to illustrate my point:

Q: So why are they called closure systems?

A: Because they induce closure operators.

Q: But why are they called closure operators?

A: Well, the closed sets of a closure operator always form a closure system.

Q: But why are they called closed sets as opposed to, say, flabby sets? And what does any of this have to do with the intuitive notion of a set being "closed" or otherwise "inescapable" with respect to some functions or operations?

A: I have no idea.

To complete the discussion, what we need is theorem to explain why we keep getting Moore families whenever we're interested in subsets that are closed under certain operations. This will only make sense if you know some category theory, so make sure you look into that.

Whenever $$X$$ is a set and $$A$$ is a subset, write $$\eta_{X,A} : A \rightarrow X$$ for the inclusion function defined by $$a \in A \mapsto a \in X$$. With that notation in place, here's the theorem you're look for:

Moore family master theorem.

Let $$X$$ denote a set (think of $$X$$ as equipped with some operations.)

Let $$I$$ denote a set (think of $$I$$ as an index set.)

For each $$i \in I$$, let $$F_i$$ denote an endofunctor on $$\mathbf{Set}$$ and let $$f_i$$ denote a function $$f_i : F_i(X) \rightarrow \mathcal{P}(X)$$.

Call $$A \subseteq X$$ closed if and only if, for all $$i \in I$$, the function $$f^A_i := f_i \circ F_i(\eta_{X,A}) : F_i(A) \rightarrow \mathcal{P}(X)$$ satisfies $$\forall t \in F_i(A) : f^A_i(t) \subseteq A.$$

Fact: The collection of closed subsets always forms a Moore family.

Example 1. To show that subgroups of a group $$G$$ form a Moore family, let $$I = \{\mathrm{law},\mathrm{identity},\mathrm{inverse}\}.$$ Let the $$F_i$$ denote the following endofunctors on $$\mathbf{Set}$$ respectively: $$F_\mathrm{law} = \Box^2, F_\mathrm{identity} = \Box^0, F_\mathrm{inverse} = \Box^1.$$ Let the $$f_i$$ denote the following functions respectively $$f_{\mathrm{law}} = (x,y \in G \mapsto \{xy\})$$ $$f_{\mathrm{identity}} = (\{1_G\})$$ $$f_{\mathrm{inverse}} = (x \in G \mapsto \{x^{-1}\})$$

It can be seen that a subset of $$G$$ is closed with respect to this data if and only if it's a subgroup in the usual sense of the word. Hence by the Moore family master theorem, the collection of subgroups of $$G$$ necessarily forms a Moore family.

Example 2. To show that the closed subsets of a convergence space $$X$$ form a Moore family, let $$I = \{\mathrm{lim}\}$$. Let $$F_\mathrm{lim}$$ denote the filter endofunctor $$\Phi$$. Let $$f_\mathrm{lim} : \Phi X \rightarrow \mathcal{P}(X)$$ denote the function that returns the set of all limit points of a filter. Then the closed sets with respect to this data are precisely the closed sets of the convergence space in the usual sense of the word, and we conclude these form a Moore family by the master theorem.

Example 3. I claim that the uppersets of a poset $$P$$ form a Moore family. Let $$I = \{\mathrm{up}\}$$ and let $$F_\mathrm{up} = \mathrm{id}_\mathbf{Set}$$. Let $$f_{\mathrm{up}} : P \rightarrow \mathcal{P}(P)$$ by defined by $$p \mapsto \{x \in P : x \geq p\}$$. By the Moore family master theorem, the desired result follows.

Proof of the master theorem. Let $$J$$ denote a set and suppose $$A_j$$ is a family of closed subsets of $$X$$. We need to show that $$C := \bigcap_{j \in J} A_j$$ is closed. Consider $$i \in I$$. Our goal is to prove that $$\forall t \in F_i(C) : f^C_i(t) \subseteq C.$$ Consider $$t \in F_i(C)$$. We need to show that $$f^C_i(t) \subseteq C.$$ That is, we're trying to show that $$f^C_i(t) \subseteq \bigcap_{j \in J} A_j.$$ Becayse of what intersection means, it's enough to show that $$\forall j \in J : f^C_i(t) \subseteq A_j.$$ So consider $$j \in J$$. It's enough to prove $$f^C_i(t) \subseteq A_j.$$ Since $$A_j$$ is closed, we know that $$f^{A_j}_i(F_i(\eta_{A_j,C})(t)) \subseteq A_j.$$ Thus, it's enough to show that $$f^C_i(t) \subseteq f^{A_j}_i(F_i(\eta_{A_j,C})(t)).$$ But if you unpack the definitions, you'll see that this reduces to showing $$F_i(\eta_{X,C}) = F_i(\eta_{X,A_j} \circ \eta_{A_j,C}),$$ which is trivial. QED.

• This is an interesting perspective (I'm still working through some of the categorical details, as I'm a bit new to the field). By the way, in Example 3, do you mean to write "Let $f_{\mathrm{up}}:P\rightarrow\mathcal{P}(P)$..."? – P-addict Apr 1 at 4:02
• @P-addict, good catch! – goblin Apr 1 at 4:59
• "what does any of this have to do with the intuitive notion of a set being "closed" or otherwise "inescapable" with respect to some functions or operations?" - Hmm, this seems fairly intuitive to me. Since $\operatorname{cl}(A) \subseteq A$, you can think of $\operatorname{cl}$ as trying to add some stuff to $A$. And if you can't add anything, then you could say that you can't "escape" from $A$ using that operator, so it seems reasonable to call it closed. It seems pretty similar to the requirements $f^A_i(t) \subseteq A.$ – Jair Taylor Apr 2 at 21:30
• @JairTaylor, your thesis looks interesting. You may find this question of mine to your liking. I'd also be interested in getting feedback on this question. – goblin Apr 18 at 13:27
• @goblin Thanks! I asked a question about it here and got a nice answer (although I never found the combinatorial interpretation I wanted.) That's a neat formula for Eulerian polynomials that I didn't know. And I remember seeing the second Q of yours and thinking it was interesting - but $x \oplus y$ is not an FGL since it can't be expanded into a power series in $x,y$ so I don't have much to say about it. – Jair Taylor Apr 20 at 6:16

If your structure is defined by universal statements (statements of the form "for all ..."), then it will be closed under intersection. For example, a subgroup is a subset $$H \subseteq G$$ satisfying $$\forall a,b \in H\colon ab^{-1} \in H,$$ and a closed set in a metric space satisfies "for all convergent sequences of points in the set, the limit is also in the set".

In contrast, open sets in a metric space satisfies "for each point in the set there exists a neighborhood contained in the set", which is of type $$\forall\exists$$ rather than $$\forall$$.

• +1 because the broader idea here is good. The specific idea that a subgroup is a subset satisfying $\forall a,b \in H : ab \in H$ isn't quite right. For example, consider $H = \{x^n : n \in \mathbb{N}\}$ and $G = \{x^n : n \in \mathbb{Z}\}$. On the other hand, the condition $ab^{-1} \in H$ does the trick. I guess what you were thinking is to use that a function $\phi$ between groups that satisfies $\phi(ab) = \phi(a)\phi(b)$ is automatically a group homomorphism. Then the idea is to apply this to inclusions functions. The problem is, if $\varphi : M \rightarrow G$ is a function satisfying... – goblin Apr 18 at 9:49
• $\phi(ab) = \phi(a)\phi(b)$, we can't use that $G$ is a group to infer that $M$ is a group (e.g. imagine $M$ is a magma, or even a monoid; in either case). – goblin Apr 18 at 9:50
• Thanks, corrected. – Yuval Filmus Apr 18 at 9:51
• No worries. $\;\;$ – goblin Apr 18 at 9:52
• I think there's still a minor issue here (though it might just be a linguistic thing). Let $M$ denote a magma. If I'm interpreting your claim correctly, you're saying that the collection of all "sub-semigroups" of $M$ (defined as submagmas that happen to satisfy the associativity identity) should be closed under intersection. This is kind of true, but note that the empty intersection should really be taken as $M$, since we're working in $\mathcal{P}(M)$. And of course, $M$ might not be a semigroup. – goblin Apr 18 at 12:50

If closedness is thought of as containing its own [boundary/range/span/closure/set of inverse elements/whatever]: each participant of an intersection likely contains the [whatever] of the full intersection, and thus the intersection also includes its own [whatever].

As my mathematical education is rather limited I've only seen this phenomenon in measure theory (& probability theory): if $$a,b \in \sigma(F) = \cap_{\textrm{C: sigma-algebra on F}}C$$ i.e. if $$a,b$$ are in the smallest sigma algebra of $$F$$, then $$a,b$$ belong to all sigma algebras $$C$$ on $$F$$, which I think gives a nice interpretation of $$\sigma(F)$$ as the set of all subsets of $$F$$ which are always measurable (in any "configuration" of the world). For probability, you can use the equivalent English statement that $$\sigma(\Omega)$$ is the set of all events for which you can always say if they happened or not.