MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

You might think that the title is an overstatement of a well-known fact but it is the best title I can come up with for the wonders the intersection operator does in some fields of math.

Recently,(on summer vacation) I was studying one subject after another and after changing about three subjects, I began to notice that in all these the set-theoretic intersection operator always carried over some property of the parent sets to the one obtained after the intersection.

To summarize briefly:

Let $A$ and $B$ be two sets, say with property P. Then, $A \cap B$ has property P.

Evidences, some trivial:


  • The intersection of two open sets is open.

  • The intersection of two closed sets is closed.

  • The nonempty intersection of two subspaces of a metric space is a metric space.

......and so forth.


  • The intersection of two subspaces of a vector space is a vector.

  • The intersection of two subgroups of a group is a group(w.r.t the same binary operation and clearly the intersection is between the underlying sets).

  • The intersection of two sub-fields of a field is a field.

......the list continues.

The third subject was Graph Theory, but I haven't yet come across the notion of intersection.

Now I would like to ask whether this trend always holds or whether there is some underlying principle each discipline abides by when using the notion of intersection. Is there any property deviating from this trend? What are the reasons for the ubiquity of the quoted property?

share|cite|improve this question
Well, the intersection of two subsets of a group which are not subgroups does not necessarily have that same property... – Mariano Suárez-Alvarez Jul 26 '10 at 7:59
@Mariano,in your case P $=$ being a subset. So, taking two subsets of a group, their intersection is also a subset. – Unknown Jul 26 '10 at 8:16
No: in my case P = being a subset which is not a subgroup. – Mariano Suárez-Alvarez Jul 26 '10 at 8:17
But I will be more pleased if I see counterexamples that are not 2nd derivatives. That is when P is an atomic property and not a combination of two. – Unknown Jul 26 '10 at 8:41
The property of being a non-atomic property is well-defined. There is then a subjective question of which decompositions of a non-atomic property are nontrivial, and a related question of which properties (when presented without a given decomposition) also have a decomposition rendering them non-atomic. The desire for counterexamples that are minimal as possible is a reasonable one. – T.. Jul 27 '10 at 18:42
up vote 15 down vote accepted

When property P is universal ($\forall ...$) it is likely to correspond to closed sets, and thus be preserved under intersection. Examples: axioms of a group, ring, field, directed graph; having symmetry under a given group.

However, if P is existential ($\exists ...$) it corresponds to open sets and is more likely to be preserved by unions (or products), not intersections. Examples: being algebraically closed, having at least 53 elements. (Well, algebraic closure is $\forall \exists$ so of course it is even more complicated. But falling out of the pure $\forall$ class it fails the intersection property.)

The first situation is possibly more common because we want structures to satisfy some, well, structural properties. Properties expressed by equations usually correspond to closed sets.

To some extent this is formalized in Birkhoff's theorem On equational presentations. Any book on Universal Algebra will discuss it.

Also, the sample of concepts is biased, because definitions that become standard are often selected for their useful formal properties. Concepts not having stability under intersection (or union, or inheritance by sub- or super-structures) are less likely to be used.

share|cite|improve this answer

There is a general sense in which any property that is closed under arbirtrary intersection is exactly a closure property.

To explain what I mean, suppose that $X$ has property $P$ and that the collection of subsets of $X$ with property $P$ is closed under arbitrary intersection. Then I claim that there is a function $cl$, a closure operator, defined on subsets of $X$ such that $A\subset cl(A)=cl(cl(A))$ and $A\subset B\to cl(A)\subset cl(B)$, for which the sets with property $P$ are exactly the sets $A$ that are closed with respect to $cl$, meaning that $A=cl(A)$.

To see this, simply let $cl(A)$ be the intersection of all $B$ with property $P$ such that $A\subset B$.

Apart from this, there many examples of natural properties that are not closed under intersection.

  • The intersection of two groups is not necessarily a group. (e.g. when they are not subgroups of a larger group.)
  • Same for almost any other type of algebraic structure.
  • The intersection of two nonempty sets may not be nonempty.
  • The intersection of two ultrafilters on a set is not necessarily an ultrafilter.
  • The intersection of two maximal ideals in a ring is not necessarily a maximal ideal.
  • The intersection of two unbounded subsets of the plane may not be unbounded.
  • etc.
share|cite|improve this answer
In Introduction to Topology by Bert Mendelson, around pp50's, it actually asks to prove the above notion of closure for closed sets. – Unknown Jul 26 '10 at 11:39
Here is an example of a $\forall$ property that is not closed under intersection: the intersection of two linear orders on a set is not necessarily a linear order. Rather, it is the partial order that the two linear orders have in common. But linearity is $\forall$-expressible, as $\forall p,q\, (p\leq q\vee q\leq p)$. – Joel David Hamkins Jul 26 '10 at 15:07
Joel, the $\forall$ property of being a linear order is closed in the usual sense, since it is expressed by equations on the $Z/2$-valued function on pairs that records the relation, such as $f(p,q) + f(q,p) = f(p,p)=1$. It satisfies the same intersection property as closed sets in topology: if $A$ and $B$ are substructures of $X$ having property $P$ (as defined in $X$ or inherited from it, e.g., both are closed sets in a given topology on $X$, or both are subsets of a linearly ordered set $X$) then so is $A \cap B$. Closed sets in different topologies need not have closed intersection. – T.. Jul 27 '10 at 3:02
I agree with that, although I would describe it differently. First order universal formulas used to define sets of points are of course closed under arbitrary intersection (or even just subsets). I was in contrast using a universal formula as a second order definition, to define a property of the relation appearing in it (linearity), and these are not necessarily closed under intersection. – Joel David Hamkins Jul 27 '10 at 3:27
Also, in all six of the examples of intersection-closed properties given in the question, the intersection was of substructures of a given structure. Under this interpretation it would be very interesting to see an $\forall$ property not closed under intersection. Being a field is an interesting example, since existence of inverses is a priori a $\forall \exists$ statement. However, direct sums of fields do have an equational presentation (by Birkhoff's theorem), and any subset of a field that is of this direct-sum form is a field, so the inverses axiom doesn't stop intersection-closure. – T.. Jul 27 '10 at 3:33

So it's pretty clear, I think, that this trend doesn't always hold. For example, the intersection of two simply connected subspaces of a space does not have to be simply connected. (For example, the intersection of hemispheres on the sphere is a circle).

I think that the reason it holds in all the cases you mention except the case of open sets, is that the property is somehow a "closure" property. Closed sets are closed under limits (or nets, more correctly), subobjects of algebraic objects are closed under some operations, etc. The pervasiveness you see is due to the fact that "closure" type properties behave well under intersection. The open set example, is, I think, a red herring. There, notice that only finite intersections maintain the property in question, and that this behavior with respect to intersections is part of the definition... Probably to avoid having too few open sets.

Anyway, I'm not sure what else there is to say about this question- it's rather vague, so I don't even know if this is the kind of answer you wanted.

share|cite|improve this answer
Thanks, Dylan. But can you come up with two set having just one "big" property and not following the trend? In your case, though acceptable an answer, I get the feeling that simple-connectedness and being a subspace are two properties. See my comment above. – Unknown Jul 26 '10 at 8:45
Well. I think the fact that simple connectedness and being a subspace are two properties is just a notational issue. Moreover, if you consider an ambient space, what should be simply connected if not a subspace? Being a subfield of a field also means to be closed under various operations, and hence could be regarded as not just one property. – Stefan Geschke Jul 26 '10 at 9:19

Mulling this over, I thought: "Compared to what?". So another question is "Why are more properties closed under intersection than under union"? While that may be arguable, there's something to be said for looking at it (e.g. a union of two non-trivial subgroups of the same subgroup is not a subgroup).

My intuition is that it has to do with the fact that, constructively, $(P \rightarrow X \wedge Y) \Leftrightarrow (P \rightarrow X) \wedge (P \rightarrow Y)$ but $(P \rightarrow X \vee Y) \nLeftrightarrow (P \rightarrow X) \vee (P \rightarrow Y)$. Here $P$ restricts the properties $X$ and $Y$ to an interesting 'sub-universe'.

I'm sure someone with more knowledge of lattice or even category theory can restate this more succinctly.

share|cite|improve this answer
The second paragraph of yatima2975's answer implicitly involves quantification. In (classical) propositional logic, $P\to(X\vee Y)$ is equivalent to $(P\to X)\vee(P\to Y)$; to see this, just check the truth tables. What yatima2975's intuition really seems to depend on is that universal quantification distributes over "and" but not over "or". (Since existential quantification distributes over "or" but not over "and", I don't see much clarification here.) – Andreas Blass Jul 26 '10 at 15:05
Andreas, your parenthetical addition of 'classical' is indeed right, and relevant! I've been doing functional progamming and constructive logic for far too long. I have edited my post accordingly; thank you for catching my mistake. – yatima2975 Jul 26 '10 at 15:50

You might want to check the definition of a filter:

I think the definition of a filter exemplifies the "intersection property" you talk about.

share|cite|improve this answer

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.