2
$\begingroup$

One of the axioms of ZF set theory is the axiom of union: $$(\forall x)(\exists y)(\forall z)(z\in y \iff (\exists t)(z\in t\ \&\ t\in x)).$$ The axiom of union (together with the axiom of extensionality) guarantees that the operation $$x\mapsto\bigcup x$$ can be defined for all sets and indeed captures our intuition of the union of all elements of $x$.

On the other hand intersection, understood as a common part of all members of a set, cannot be defined since $\bigcap\emptyset$ poses a problem. However, what is stopping us from using the axiom scheme of separation and defining $$\bigcap x :=\{y\in\bigcup x\ |\ (\forall z)(z\in x\Rightarrow y\in z)\},$$ which is a correct instance of the axiom scheme of separation and captures the notion of the intersection of all elements for nonempty sets $x$, while for an empty set we have clearly $\bigcap\emptyset = \emptyset$ (since $\bigcup\emptyset = \emptyset$)?

Addressing comments and to be more precise: how to define an intersection of all members of a set in ZF (please provide a formula in a language of ZF) such that the induced operation of intersection is undefined for the empty set?

I ask because I saw that people have this definition in mind (instead of "mine" presented above), but rarely write it down and just go on just claiming that $$\bigcap x$$ is possible to define for nonempty sets $x$, and undefined for $\emptyset$ (i.e. see chapter 5 in "Notes on logic and set theory" by P. T. Johnstone).

$\endgroup$
  • $\begingroup$ What do you mean by claiming that intersection can't be defined? You just defined it quite well. $\endgroup$ – DRF Nov 29 '18 at 13:42
  • $\begingroup$ @drhab That is an unfortunate side effect of wanting a set sized intersection. To be honest though I'm not exactly sure what the OP is exactly asking. It's quite consistent to either define intersection as he did or leave it undefined for the empty family (empty set). The only thing you can't reasonably do is to define the intersection of the empty family to be the universe as then it's not a set anymore. $\endgroup$ – DRF Nov 29 '18 at 13:50
0
$\begingroup$

Let's pretend that you are a new student to set theory. Let's assume that you understood the primitive notions of membership and equality, what formulas are (in the language of Set Theory), and the following nonlogical axioms: extensionality, existence of empty set, separation, pairing, union.

Now let's develop the notion of intersection.

Remark. For each nonempty collection $\mathcal{C}$ of sets, there exists a unique set, denoted $$\bigcap\mathcal{C}\qquad\text{or}\qquad\{x:(\forall A\in \mathcal{C})[x\in A]\},$$ whose elements are exactly those elements $x$ such that $x\in A$ for every $A\in\mathcal{C}$.

Indeed, consider $\{x\in A_0:(\forall A\in \mathcal{C})[x\in A]\}$ where $A_0$ is any fixed element of $\mathcal{C}$.

Definition. Let $\mathcal{C}$ be a nonempty collection of sets. The intersection of $\mathcal{C}$ is $\bigcap\mathcal{C}$.

Definition. Let $X$ and $Y$ be sets. The intersection of $X$ and $Y$, denoted $X\cap Y$, is $\bigcap\{X,Y\}$.

End of development.

Notice that $\bigcap\mathcal{C}$ is only defined once it is established that it is nonempty.

Side Note: It is unreasonable to define an infimum operator for an arbitrary set $X$. The set $X$ does not even have an ordering! It is pointless to try. Only once $X$ has an established partial ordering should one continue to define the infimum operator. But even then the infimum may not exist in all cases. That is why the typical work-around is to instead focus one's attention to a complete lattice, or even better, consider the completion of $(X,\preceq)$ (if it is unique up to isomorphism), or resign oneself that some collections have no infimum.

I use infimum here because that is essentially what $\bigcap$ is---an infimum of a collection of sets under the established ordering $\subseteq$. However, as we are quite aware, there is no infimum for $\varnothing$.

$\endgroup$
1
$\begingroup$

I'm not exactly sure if I'm understanding the question correctly but I will try and give an explanation based on my understanding.

We certainly can define $\bigcap$ as you have nicely shown. I don't think you even need the axiom of union since the intersection is in every set in $x$. Now the reason it can be defined this way means we don't need to have it as an axiom (just pointing that out in case there was confusion).

The reason we don't (usually) define the intersection in the way we do for union (i.e. as an intersection of a single set, understood as a family of sets, as it is done for union) is IMO that it's rarely used and also precisely because it's not an axiom.

Writing $\bigcap_{x\in X}x$ is easy enough and we could do it for union aswell. But the notation used for the axiom is a little neater.

But we fairly rarely take intersection of whole families, or at least I can't really think of many situations we do so.

Another reason why we do have the $\bigcup x$ notation for union is that there is parts of set theory where it does help. That is when we create transitive closures of different kinds. Essentially sometimes we are trying to make sure we have a nice big enough set where everything we need to do can be done. That is that the set we work in contains everything we could possibly need to talk about has nice properties and is still a Set as opposed to a proper class.

$\endgroup$
1
$\begingroup$

I reject this as counterintuitive.

The smaller the set $a$ the larger the set $\cap a$ because the number of demands on $x$ to be element of $\cap a$ decreases if $a$ gets smaller.

But then suddenly if $a$ is the smallest set (empty) and all demands are vanished then no $x$ can pass anymore?... Weird situation!

What also goes lost (and is in my eyes very valuable) is the convention that: $$\cap\varnothing:=X$$ when we are working in some universal set $X$.

Next to that I do not see (yet) any advantages.


PS. I use $\cap$ instead of $\bigcap$ here, seeing $\cap$ as an operator on non-empty sets and seeing $\bigcap_{x\in a}x$ as another notation of $\cap a$.

That is also a matter of taste of course.

$\endgroup$
  • $\begingroup$ I wonder when you say you find it counterintuitive do you in general expect to only perform intersections of families of sets when you have a set sized universe? What I mean is that as far I can see if I wrote $\cap x$ for a non-empty set $x$ I think everyone would consistently understand it as the intersection of the family of sets $x$. To be fair we rarely use the convention of $\cup x$ as the union of elements of $x$ outside set theory and even there usually in cases where we are building something (e.g. V, L, H) usually models. $\endgroup$ – DRF Nov 29 '18 at 13:55
  • $\begingroup$ @DRF I changed "universe" into "universal set" to avoid misunderstanding. $\cap\varnothing$ is also used in e.g. topology: if $\mathcal D$ is a collection of subsets of $X$ then the finite intersections (among them $\cap\mathcal\varnothing=X$) is known to be a basis. $\endgroup$ – drhab Nov 29 '18 at 14:04
  • $\begingroup$ I realize that, and there certainly are reasonable uses for defining the intersection of the empty set to be "some" universal set that is useful at the time. Of course the fact that the intersection of the empty set then depends on "where we are" means we can't have a single definition of intersection. This is an interesting example of the fact that intersection and union are actually not quite as complementary as one might want since the whole universe is not a "nice enough" boolean algebra. $\endgroup$ – DRF Nov 29 '18 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.