Universal property of the empty set

To start us off, re­call that the empty set $$\emptyset$$ is usu­ally defined as the set which con­tains no el­e­ments. That prop­erty picks it out uniquely, be­cause we have the “Ax­iom of Ex­ten­sion­al­ity”:

Two sets are the same if and only if their el­e­ments are all the same.

If we had two sets which both con­tained no el­e­ments, then all their el­e­ments would (vac­u­ously) be the same, so by ex­ten­sion­al­ity the sets are the same.

In this ar­ti­cle, we will in­tro­duce an­other way to char­ac­ter­ise the empty set. Rather than work­ing with the el­e­ments of the set, though, we will con­sider the func­tions which have the set as their do­main. You might think at first that this is a very strange way to look at a set; and you might be right. But it turns out that if we do it this new way, char­ac­ter­is­ing the empty set by ex­am­in­ing the maps be­tween sets note”Map” is just a syn­onym for “func­tion”, in this con­text., we end up with a recipe that is much, much more widely ap­pli­ca­ble.

The empty function

When we use set the­ory to cap­ture the idea of a “func­tion”, we would usu­ally im­ple­ment $$f: A \to B$$ as a set of or­dered pairs $$(a, f(a))$$, one for each el­e­ment $$a$$ of the do­main $$A$$, with the re­quire­ment that the el­e­ments $$f(a)$$ all must lie in $$B$$. This set of or­dered pairs holds all of the in­for­ma­tion about the func­tion, ex­cept that it has omit­ted the (al­most always unim­por­tant) fact that $$B$$ is the codomain. noteWe usu­ally only care about the image of the func­tion.

This im­ple­men­ta­tion of the func­tion $$f$$ is just a set; it might look like $$\{ (0,1), (1,2), (2,3), (3,4), \dots \}$$, which would rep­re­sent the func­tion $$f: \mathbb{N} \to \mathbb{N}$$ given by $$n \mapsto n+1$$.

And we might ask, can we go the other way round? Given a set $$X$$, can we in­ter­pret it as a func­tion? Well, of course we need $$X$$ to con­tain only or­dered pairs (what could it mean for $$(0,1,2)$$ to lie in the im­ple­men­ta­tion of the func­tion $$f$$?), but also we must make sure that the func­tion is well-defined by en­sur­ing that the first co­or­di­nates of each pair are all dis­tinct. If we had the set $$\{ (0,1), (0,2) \}$$, that would in­di­cate a func­tion that was try­ing to take $$0$$ to both $$1$$ and $$2$$, and that’s am­bigu­ous as a func­tion defi­ni­tion. But those are the only re­quire­ments: for any set that obeys those two con­di­tions, we can find a func­tion which is im­ple­mented as that set.

Very well. Now look at the empty set. That only con­tains or­dered pairs: in­deed, it doesn’t con­tain any­thing at all, so ev­ery­thing in it is an or­dered pair. noteIf you’re squeamish about this, see vac­u­ous truth. And the first co­or­di­nates of each pair are dis­tinct: in­deed, there aren’t any pairs at all, so cer­tainly no two pairs have the same first co­or­di­nate.

So the empty set it­self is im­ple­ment­ing a func­tion. We call this the “empty func­tion”; it is the iden­tity func­tion on the empty set. noteDon’t get wor­ried about the empty set rep­re­sent­ing a func­tion that is the iden­tity func­tion on it­self. To worry about this is to make a type er­ror in your think­ing. The same ob­ject (the empty set) is here stand­ing for two differ­ent things, un­der two differ­ent “en­cod­ing schemes”. Un­der one en­cod­ing scheme, where we look at sets as be­ing func­tions, it’s the empty func­tion. Un­der an­other en­cod­ing scheme, where we look at sets just as be­ing sets, it’s the empty set.

Med­i­ta­tion: do­main, codomain and image

If the empty set im­ple­ments a func­tion, then that func­tion should have a do­main. What is the do­main here?

The do­main is the empty set.

In­deed, there are no el­e­ments in the do­main, be­cause an el­e­ment of the do­main is just a thing in the first co­or­di­nate of one of the or­dered pairs, but there aren’t any or­dered pairs so there can’t be any el­e­ments of the do­main. <div><div>

What is the image?

The image is also the empty set.

In­deed, there are no el­e­ments in the image, be­cause an el­e­ment of the image is just a thing in the sec­ond co­or­di­nate of one of the or­dered pairs, but there aren’t any or­dered pairs so there can’t be any el­e­ments of the image. <div><div>

What is the codomain?

Aha! Trick ques­tion. We said ear­lier that in look­ing at func­tions be­ing im­ple­mented as sets, we threw away in­for­ma­tion about the codomain. We can’t ac­tu­ally tell what the codomain of the func­tion im­ple­mented by $$\emptyset$$ is.

Im­por­tant fact

From the do­main/​image/​codomain med­i­ta­tion above, we can now note that for ev­ery set $$A$$, there is a func­tion from $$\emptyset$$ to $$A$$: namely, the empty func­tion. Put an­other way, we can always in­ter­pret the empty set as be­ing a func­tion from $$\emptyset$$ to $$A$$, what­ever $$A$$ is.

Let $$A = \{ 1 \}$$, so we’re show­ing that there is a func­tion from $$\emptyset$$ to $$\{1\}$$: namely, the empty func­tion, which is im­ple­mented as a set by $$\emptyset$$. This is a valid im­ple­men­ta­tion of a func­tion from $$\emptyset$$ to $$\{1\}$$, be­cause $$\emptyset$$ is a set which satis­fies the three prop­er­ties that are re­quired for us to be able to in­ter­pret it as a func­tion from the do­main $$\emptyset$$ to the codomain $$\{1\}$$:

• Every­thing in $$\emptyset$$ is an or­dered pair (vac­u­ously).

• Every or­dered pair in $$\emptyset$$ con­sists of one el­e­ment from the do­main $$\emptyset$$, fol­lowed by one el­e­ment from the codomain $$\{1\}$$ (again vac­u­ously).

• Every el­e­ment of the do­main $$\emptyset$$ ap­pears ex­actly once in the first slot of an or­dered pair in the func­tion-im­ple­men­ta­tion $$\emptyset$$.

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More­over, for ev­ery set $$A$$, there is only one func­tion from $$\emptyset$$ to $$A$$. In­deed, if we had a differ­ent func­tion $$f: \emptyset \to A$$, then there would have to be some el­e­ment of the do­main $$\emptyset$$ on which $$f$$ differed from the empty func­tion. But there aren’t any el­e­ments of the do­main at all, so there can’t be one on which $$f$$ and the empty func­tion differ. Hence $$f$$ is ac­tu­ally the same as the empty func­tion.

The uni­ver­sal prop­erty of the empty set

The uni­ver­sal prop­erty of the empty set is as fol­lows:

The empty set is the unique set $$X$$ such that for ev­ery set $$A$$, there is a unique func­tion from $$X$$ to $$A$$. To bring this prop­erty in line with our usual defi­ni­tion, we de­note that unique set $$X$$ by the sym­bol $$\emptyset$$.

We’ve just proved that our stan­dard defi­ni­tion of $$\emptyset$$ does satisfy that uni­ver­sal prop­erty; that was the Im­por­tant Fact just above.

Aside: why “uni­ver­sal”?

The prop­erty is a “uni­ver­sal prop­erty” be­cause it’s not “lo­cal” but “uni­ver­sal”. Rather than con­sid­er­ing the in­di­vi­d­ual things we can say about the ob­ject $$\emptyset$$, the prop­erty talks about it in terms of ev­ery other set. That is, the prop­erty defines $$\emptyset$$ by refer­ence to the “uni­verse” of sets.

In gen­eral, a “uni­ver­sal prop­erty” is one which defines an ob­ject by spec­i­fy­ing some way that the ob­ject in­ter­acts with ev­ery­thing else, rather than by look­ing into the ob­ject for some spe­cial iden­ti­fy­ing fea­ture.

Does the uni­ver­sal prop­erty uniquely pick out $$\emptyset$$?

I sneak­ily slipped the words “is the unique set” into the prop­erty, with­out prov­ing that they were jus­tified. What use would it be if our uni­ver­sal prop­erty didn’t ac­tu­ally char­ac­ter­ise the good old $$\emptyset$$ we know and love? noteWell, it might still be of some use. But it would mean the uni­ver­sal prop­erty might not work as a defi­ni­tion of $$\emptyset$$. Let’s see now that at least the prop­erty isn’t to­tally stupid: there is a set which doesn’t have the uni­ver­sal prop­erty.

$$\{1\}$$ doesn’t satisfy the uni­ver­sal property

We need to show that the fol­low­ing is not true:

For ev­ery set $$X$$, there is a unique func­tion from $$\{1\}$$ to $$X$$.

Let $$X$$ be any set at all with more than one el­e­ment. For con­crete­ness, let $$X$$ be $$\{ a, b \}$$.

Now, there are two func­tions from $$\{1\}$$ to $$\{a,b\}$$: namely, $$f: 1 \mapsto a$$, and $$g: 1 \mapsto b$$.

So $$\{1\}$$ fails to satisfy the uni­ver­sal prop­erty of $$\emptyset$$, and in­deed it fails mas­sively: for ev­ery set $$X$$ which has more than one el­e­ment, there is more than one func­tion $$\{1\} \to X$$. (Though all we needed was for this to hold for some $$X$$.) <div><div>

Only the empty set satis­fies the uni­ver­sal property

It’s ac­tu­ally the case that the empty set is the only set which satis­fies the uni­ver­sal prop­erty. There are three proofs, none of them very com­pli­cated and all of them ped­a­gog­i­cally use­ful in differ­ent ways. Here, we will du­pli­cate one of the most “cat­e­gory-the­ory-like” proofs, be­cause it’s re­ally rather a new way of think­ing to a stu­dent who has not met cat­e­gory the­ory be­fore, and the style turns up all over cat­e­gory the­ory. To dis­t­in­guish it from the other two proofs (which, re­mem­ber, are de­tailed here), we’ll call it the “maps” proof.

The “maps” way

We’ll ap­proach this in a slightly sneaky way: we will show that if two sets have the uni­ver­sal prop­erty, then there is a bi­jec­tion be­tween them. noteThe most use­ful way to think of “bi­jec­tion” in this con­text is “func­tion with an in­verse”. Once we have this fact, we’re in­stantly done: the only set which bi­jects with $$\emptyset$$ is $$\emptyset$$ it­self.

Sup­pose we have two sets, $$\emptyset$$ and $$X$$, both of which have the uni­ver­sal prop­erty of the empty set. Then, in par­tic­u­lar (us­ing the UP of $$\emptyset$$) there is a unique map $$f: \emptyset \to X$$, and (us­ing the UP of $$X$$) there is a unique map $$g: X \to \emptyset$$. Also there is a unique map $$\mathrm{id}: \emptyset \to \emptyset$$. noteWe use “id” for “iden­tity”, be­cause as well as be­ing the empty func­tion, it hap­pens to be the iden­tity on $$\emptyset$$.

The maps $$f$$ and $$g$$ are in­verse to each other. In­deed, if we do $$f$$ and then $$g$$, we ob­tain a map from $$\emptyset$$ (be­ing the do­main of $$f$$) to $$\emptyset$$ (be­ing the image of $$g$$); but we know there’s a unique map $$\emptyset \to \emptyset$$, so we must have the com­po­si­tion $$g \circ f$$ be­ing equal to $$\mathrm{id}$$.

We’ve checked half of “$f$ and $$g$$ are in­verse”; we still need to check that $$f \circ g$$ is equal to the iden­tity on $$X$$. This fol­lows by iden­ti­cal rea­son­ing: there is a unique map $$\mathrm{id}_X : X \to X$$ by the fact that $$X$$ satis­fies the uni­ver­sal prop­erty noteAnd we know that this map is the iden­tity, be­cause there’s always an iden­tity func­tion from any set $$Y$$ to it­self., but $$f \circ g$$ is a map from $$X$$ to $$X$$, so it must be $$\mathrm{id}_X$$.

So $$f$$ and $$g$$ are bi­jec­tions from $$\emptyset \to X$$ and $$X \to \emptyset$$ re­spec­tively.

Recap

To sum­marise the dis­cus­sion above, we have shown that the uni­ver­sal prop­erty of the empty set uniquely char­ac­ter­ises the empty set. We could ac­tu­ally use it as a defi­ni­tion of the empty set:

We define the empty set to be the unique set $$X$$ such that for ev­ery set $$A$$, there is a unique func­tion from $$X$$ to $$A$$.

This is a bit of a strange defi­ni­tion when we’re talk­ing about sets, but it opens the door to many differ­ent and use­ful con­struc­tions. You might like to think about the fol­low­ing similar uni­ver­sal prop­er­ties, and what they define.

• The set $$X$$ such that for ev­ery set $$A$$, there is a unique func­tion from $$A$$ to $$X$$. (That is, the “re­verse” of the empty set’s UP.)

• The group $$G$$ such that for ev­ery group $$H$$, there is a unique ho­mo­mor­phism from $$H$$ to $$G$$.

• The ring $$R$$ such that for ev­ery ring $$S$$, there is a unique ho­mo­mor­phism from $$R$$ to $$S$$. (This one is more in­ter­est­ing!)

Aside: defi­ni­tion “up to iso­mor­phism”

With uni­ver­sal prop­er­ties, it is usu­ally the case that we ob­tain a char­ac­ter­i­sa­tion of ob­jects only “up to iso­mor­phism”: the uni­ver­sal prop­erty doesn’t care about bi­jec­tions. If the ob­ject $$X$$ satis­fies a uni­ver­sal prop­erty, and the ob­ject $$Y$$ is iso­mor­phic to $$X$$, then $$Y$$ will prob­a­bly satisfy the uni­ver­sal prop­erty as well. (Nor­mally this doesn’t re­ally mat­ter, since ob­jects which are iso­mor­phic are “ba­si­cally the same” in most of the ways we care about.)

In this case, though, we do end up with a uniquely-defined ob­ject, be­cause it so hap­pens that the only ob­ject iso­mor­phic noteIn this case, when talk­ing about sets, “iso­mor­phic to” means “bi­ject­ing with”. to the empty set is the empty set it­self.

An ex­am­ple where we don’t end up with a uniquely-defined ob­ject is the “re­verse” of the empty set’s uni­ver­sal prop­erty: “the set $$X$$ such that for ev­ery set $$A$$, there is a unique func­tion from $$A$$ to $$X$$”. (You may have thought about this prop­erty in the pre­vi­ous sec­tion.) In this case, the sets which satisfy that “re­verse” uni­ver­sal prop­erty are ex­actly the sets with one el­e­ment.

You should try to go through the “maps” way of show­ing that the uni­ver­sal prop­erty of the empty set picks out a unique set, but in­stead of us­ing the UP of the empty set, try us­ing the “re­verse” prop­erty. The same struc­ture of proof will show that if two sets satisfy the “re­verse” uni­ver­sal prop­erty, then there is a bi­jec­tion be­tween them. So we do still re­tain that if two sets satisfy the “re­verse” UP, then they are iso­mor­phic; and in­deed if we have two sets with one el­e­ment, there re­ally is a bi­jec­tion be­tween them (given by match­ing up the sin­gle el­e­ment of each set).

This is a gen­eral phe­nomenon: most uni­ver­sal prop­er­ties don’t define an ob­ject uniquely, but they do give us the fact that if any two ob­jects satisfy the uni­ver­sal prop­erty, then the two ob­jects are iso­mor­phic.

Children:

Parents:

• Universal property

A uni­ver­sal prop­erty is a way of defin­ing an ob­ject based purely on how it in­ter­acts with other ob­jects, rather than by any in­ter­nal prop­erty of the ob­ject it­self.