5

Suppose we specify subsets of a reference set by pairs, where the first co-ordinate specifies a member of the universe of discourse, and the second co-ordinate specifies the value that the characteristic function yields for that member of the reference set.

E. G., if we have $\{a, b\}$ as the universe of discourse, then we have subsets $\{(a, 0), (b, 0)\}, \{(a, 0), (b, 1)\}, \{(a, 1), (b, 0)\}, \{(a, 1), (b, 1)\}$, where $(x, 0)$ indicates that $x$ does not belong to the set, while $(x, 1)$ indicates that $x$ belongs to the set.
Now, consider the empty set compared across different universes of discourses, for example $\{a, b\}$, and $\{a\}$.
For $\{a, b\}$ we have $\{(a, 0), (b, 0)\}$ as the empty set under this specification, and for $\{a\}$ we have $\{(a, 0)\}$ as the empty set under this specification.

So, it would seem that we have a relative notion of an empty set in this context, but oftentimes books talk of "the empty set" which suggests it as absolute.

So, does the concept of the empty set come as absolute, or relative?

Doug Spoonwood
  • 11,211

3 Answers3

7

If you require extensionality from $\in$, i.e. two sets are equivalent if and only if they have the same memebers, and a model is transitive (a member of a set in the model is also in the model) - then the empty set is absolute.

This is quite simple to prove, since all the elements of all the sets are also sets, and the empty set is such that no one is a member of it.

On the other hand, if you consider $V$ a model of $ZF$, pick $x\in V$, and declare $\in^*$ to be the relation defined only on sets generated from iterations of $\mathcal P(x)$ (i.e. repeating the power set operation), no one will be in $x$, you will have a model of $ZF$ and the empty set of this model will be $x$.

However, as the intuition guiding us with topological spaces is mostly $\mathbb R^n$ the same happens with set theory, and the intuition that guides us is that of well founded and transitive models of $ZF$, in which the empty set is somewhat absolute (with respect to inner models and forcing extensions).

So essentially this all boils to your theory of sets, and how you define $\in$, and which models you are taking.

Addendum: An important point is internal and external view of the empty set. If a model is extensional (i.e. $\in$ satisfies extensionality) then the empty set is unique for the model, and all the models perceive their empty sets the same way -- the only set that has no elements (and this is unique by extensionality).

However, if we consider a model from an external point of view we have that it could have an empty set different than the model we work in, as the example I gave above.

Asaf Karagila
  • 393,674
6

There are various ways to answer this question depending on what kind of perspectives you want to take. First, I wouldn't call what you're describing an empty set, exactly: since you're specifying the superset, you're really describing the unique inclusion map $\emptyset \to A$, which has $A$ as part of its data.

Taking a foundational point of view, it is true that in ZF there is literally a unique empty set $\{ \}$ (it exists by the empty set axiom and is unique by the axiom of extension). The empty subset of any set is precisely this empty set, although in ZF subsets do not come with an identification of their parent set, and if you provide such an identification you are specifying a function of some kind like I said above.

Taking a more categorical view, "the" empty set is "the" initial object in the category of sets. Initial objects are not literally unique in general, but an initial object is more than an object: it comes with distinguished maps $\emptyset \to A$ for every set $A$ satisfying the appropriate universal property, and any two initial objects (together with those maps) are isomorphic via a unique isomorphism. This is one way to rigorously justify the use of "the" empty set even if you aren't taking a foundational point of view.

From the categorical point of view, one can think of subsets of a set $A$ as monomorphisms $S \to A$, and then associated to any set $A$ is the unique morphism $\emptyset \to A$, which one can think of as the initial object in the category of monomorphisms into $A$ (or more generally the category of morphisms into $A$). So in some sense it is the "relative empty set" over $A$.

Qiaochu Yuan
  • 419,620
  • 4
    In $ZF$ the empty set is unique for transitive (and well founded) models, if you do not require this then you can have different models with different empty sets. Just as well, if you change the way $\in$ works, you will get a different $\emptyset$ too. – Asaf Karagila Jun 13 '11 at 21:24
  • @Asaf: could you post an answer, elaborating on your comment? That would be great! – amWhy Jun 13 '11 at 21:34
  • 2
    @amWhy: I did... – Asaf Karagila Jun 13 '11 at 21:34
  • Asaf, there is no need for this equivocation. Set theorists prove, in their background theory ZF, that the empty set is unique. Full stop. – JDH Jun 13 '11 at 21:43
  • @JDH: I thought I understand what Asaf was saying, but it seems I didn't. Suppose we have a non-transitive model containing some set all of whose elements are not in the model. Then isn't that set empty and also not identical to the "standard" empty set? – Qiaochu Yuan Jun 13 '11 at 21:47
  • @Joel: I recall Jech's description about constructing a model of $ZFA$ from a model of $ZFC$ in which the empty set is different. If you take the kernel of the $ZFA$ model, you have a new model of $ZF$ that externally has a whole other different empty set. Of course that internally every model thinks that the empty set is the same, but the whole point of the question was external definition (at least this is how I read it). I will add this to my answer, as I think it is very relevant. – Asaf Karagila Jun 13 '11 at 21:49
  • I think the confusion here might be about something like the difference between $\mathrm{ZFC}\vdash \exists ! x \forall y(\neg y \in x)$ and "for any two models $\mathcal{M}$ and $\mathcal{N}$ of $\mathrm{ZFC}$, the interpretations $\emptyset_\mathcal{M}$ and $\emptyset_\mathcal{N}$ are isomorphic (in some way)". – matt Jun 13 '11 at 21:58
  • 1
    @Qiaochu: As I wrote in the addendum to my answer, the empty set is the same internally for every extensional model, however externally it can be different. – Asaf Karagila Jun 13 '11 at 22:10
  • 3
    To my way of thinking, the question is whether there is a unique empty set. We can prove from the fundamental axioms that indeed, the empty set is unique. I take this as the answer to the question, regardless of whatever models of some theory that might exist. That is, when considering set-theoretic questions, we prefer to answer them inside the theory, rather than in the meta-theory, and it would seem to be an unnecessary intrusion of the meta-theory here to take Asaf's line. (Of course, I totally agree that different models of ZF can have different objects for $0$, but find this irrelevant.) – JDH Jun 13 '11 at 22:38
  • 2
    @JDH: As I said, I completely agree that internally this is a unique set. However the notion of an empty set can be defined outside of $ZF$, and the $\in$ relation need not be extensional, and models need not be transitive. Indeed in the framework of $ZF$ the empty set is very unique (very in the sense that inner models share the same empty set), however in a general settings where the $\in$ is not assuming to be extensional, and maybe not even well founded, I fail to see why there must always be only one set with no elements. – Asaf Karagila Jun 13 '11 at 22:54
  • @JDH: ah, I see. I was confusing the internal and external descriptions. Of course internally any two empty sets are identical by extension. – Qiaochu Yuan Jun 13 '11 at 23:04
  • But Asaf, would you also say that we can't really say that there is a unique real number $\pi$, even after having fixed the set-theoretic meaning of real numbers as, say, Dedekind cuts in the rationals (and having fixed the meaning of $\mathbb{Q}$, etc.)? After all, different models of set theory could have different objects representing $\pi$... – JDH Jun 13 '11 at 23:32
  • @JDH If you can interpret numbers by equalities, saying that there exists a "unique real number pi" certainly sounds strange (at least to me), since many different infinite series equal pi (in the sense of which an infinite series equals anything, i. e. convergence of partial sums). If we interpret "pi" by these series, and consider these series as series, they certainly seem very different. So, what problem exists with considered pi as many different numbers in that sense? – Doug Spoonwood Jun 14 '11 at 01:58
  • 3
    Doug, my point was that once we have fixed a particular completion of $\mathbb{Q}$ as the real numbers, then we can easily prove that the number $\pi$ is unique. You wouldn't say that there are at least two real numbers that are the ratio of the unit circle's circumference with its diameter, would you? – JDH Jun 14 '11 at 02:47
  • @JDH: Of course that after choosing a model of $ZF$ the empty set is unique. I agreed to that earlier. However just as forcing can be described "similar to building an algebraic closure" and yet it is different in forcing because different filters yield different models. The same applies here, most branches of mathematics do not care for internal and external definitions, they don't mind about the logic behind it. They just assume several assumptions and draw conclusions. In set theory, we do care about definability, internal and external definitions, and so on. (cont...) – Asaf Karagila Jun 14 '11 at 04:42
  • @JDH: (cont) But what do I know? You're the set theorist, I'm just a grad student :-) – Asaf Karagila Jun 14 '11 at 04:43
  • @JDH I wouldn't even talk about it as a unique real number when talking about the ratio of the unit circle's circumference with its diameter. I'd take a clue from Archimedes and talk about it lying between two rational numbers, or belonging to some continuous interval like [3.1415, 3.1416]. Rational approximations to pi almost surely matter far more than its exact value (wouldn't you agree to this?), and those rational approximations are NOT unique and by no means always equivalent. I would say that two rational approximations to pi often works out better than a less concrete real number. – Doug Spoonwood Jun 14 '11 at 05:40
  • 1
    @Doug: You are missing the point in his comment. Every model of the reals thinks of $\pi$ as the same number, as do models of $ZF$ with the empty set. I said on the other hand that the models might not know it is indifferent set, or worse - they are not models of $ZF$. I think that this is an important point but Joel seems to disagree with me (I guess for didactic reasons). – Asaf Karagila Jun 14 '11 at 07:16
  • 1
    It seems the two different attitudes here, Asaf, result from two ways of taking set theory. If we understand set theory as the background theory of mathematics, then we answer the question by proving in that background theory that the empty set is unique. If instead were are doing set theory like a group theorist does group theory, by looking at all the models of set theory, then we are led to emphasize that different models of set theory can have different objects representing the empty set. But what is the background theory of that enterprise, and does it prove that the emptyset is unique? – JDH Jun 14 '11 at 10:42
3

It is absolute. What you're seeing is an image of the empty set under a map and this can be different for different maps.

lhf
  • 216,483