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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?

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Your acceptance rate is only 33% at the moment of writing this comment, this is somewhat frowned upon in this site. If your previous questions have satisfactory answers please accept them, and if not please revise the question to add what is it that you are looking for in an answer. –  Asaf Karagila Jun 13 '11 at 22:20
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@ASAF What gets frowned upon here? I thought I had checked both your response and Qiaochu's response, since I find both of them acceptable. Edit: Looking over things more, it seems I can't keep a green check mark on both of them. If I can't accept more than one answer here, I'd rather check neither since I don't necessarily prefer one over the other. –  Doug Spoonwood Jun 14 '11 at 14:19

3 Answers 3

up vote 7 down vote accepted

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.

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Is today the official empty set day or what? –  t.b. Jun 13 '11 at 23:25
    
@Theo: it seems so, yes. I have been seriously considering adding a new tag for these kinds of questions: "nullology" or somesuch. –  Pete L. Clark Jun 13 '11 at 23:55
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@Pete: how about [nonsuch], or [empty]? (-: –  Arturo Magidin Jun 14 '11 at 3:51
    
@Theo: You should file an official holiday with the UN :-) –  Asaf Karagila Jun 14 '11 at 4:44
    
I actually read this math.stackexchange.com/questions/45120/… thread and remembered this question which I more-or-less thought up years ago. And there's no worries when thinking about empty set. After all... there's nothing to it! –  Doug Spoonwood Jun 14 '11 at 5:24

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 $A$.

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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
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@amWhy: I did... –  Asaf Karagila Jun 13 '11 at 21:34
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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
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@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
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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 2:47

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.

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