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A set $a$ can be called extensional if it has the following propery: $$\forall b\left[\forall x\left[x\in b\iff x\in a\right]\Rightarrow a=b\right]$$ Based on this the axiom of extensionality can be formulated as: $$\forall a\left[a\text{ is extensional}\right]$$

How exactly should this axiom be interpreted?

I see two options:

1) The statement that every set is extensional.

2) The statement that sets that are not extensional are to be neglected.

The first looks like a statement of a dictator. The second more like an admittance that there is more, but that we agree to keep that out of sight. If statement 2) could be practicized then I would choose for it, but I have my doubts about that, which is the reason for asking this.

addendum

Inspired by comments and answers (thank you very much!) I have decided to share something about my motivation and also my doubts about option 2).

A set $a$ is regular if it has the following property: $$\forall b\left[a\in b\Rightarrow\exists c\in b\; c\cap b=\emptyset\right]$$ Based on that the axiom of regularity can be formulated as: $$\forall b\left[b\text{ is regular}\right]$$ This axiom can be accepted in the awareness that you just restrict your scope and focus on studying the regular ones. You could say that option 2) 'works' here. If $\mathbf{V}$ denotes the class of all extensional sets then there is a subclass $\mathbf{G}$ of regular extensionals. Option 1) applied on regulars says: $\mathbf{V}=\mathbf{G}$ while option 2) says: focus on $\mathbf{G}$. Here I prefer the second option. This is facilitated by the fact that the elements of regular sets are regular, which means that by focusing on regulars their elements do not get out of sight. It should be remarked here that you need the axioms PAIR and SUM to prove this. This facilitation lacks (or seems to lack; uptil now I was looking for it in vain) when it comes to extensionals, which was the reason for my doubt concerning the second option applied on extensionals.

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It is 1). It says a set is determined by its elements. That's something most people would agree with. The set of primes less than $4$ is the same as the set of integers that are strictly between $1$ and $4$. Different descriptions, same set. –  André Nicolas Feb 8 at 9:22
    
@AndréNicolas. The question arose when I was thinking about of the axiom of regularity. That can be formulated as 'every set is regular' and if it is accepted on the way of 2) then you keep sight on sets that are not regular. You can say things like 'if $\mathbf{V}=\mathbf{G}$ then...if not then....' Here $\mathbf{V}$ denotes the class of extensionals and $\mathbf{G}$ the class of regular extensionals. –  drhab Feb 8 at 9:34
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The Axiom of Regularity is not a consequence of our basic intuitive notion of set. It is convenient for developing the theory of ordinals, and harmless, but I do not think many people in set theory consider it necessary. –  André Nicolas Feb 8 at 9:39
    
@AndréNicolas I have added something, so if you are interested... –  drhab Feb 8 at 11:06
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Axioms are statements of dictatorship. Have you got some philosophical problem with that? –  wim Feb 8 at 12:13

6 Answers 6

up vote 2 down vote accepted

I think that can be useful to start from the debate about extensionality in modern mathematics and logic, that dates back to Cantor and Frege and their researches about classes and extensions.

We can start with the (very simplified) picture of a concept or “universal”, we can symbolize it as $\phi(x)$, and its extension, defined as the set of objects such that $\phi$ holds of them, we symbolize it as $\{ x : \phi(x) \}$.

Following the discovery of paradoxes of Cantor’s set theory (with the Comprehension axiom : for every “condition” $\phi(x)$, there exists the corresponding extension $\{ x : \phi(x) \}$) and of Frege’s logic (with Basic Law V), several attempts was explored to solve them.

One of the most relevant attempt was due to Russell, in his Principles and then in the Principia.

The basic idea was to avoid the “language of classes (or set)” and adopt instead the language of “propositional functions”; think of them as “attributes”, and assume for simplicity that they correspond to open formula (i.e. to $\phi(x)$).

According to Russell, propositional functions are primitive, and classes or sets are reduced to suitable “abbreviations” .

The following comment by the logician and philosopher Quine (Whitehead and the Rise of Modern Logic, 1941, reprinted in Selected Logic Papers, new edition 1995, page 22) is a good clarification of the issue :

To have reduced classes to attributes is of little philosophical consequence, for attributes are no less universal, abstract, intangible, than classes themselves.

[…] classes are identified when they coincide in point of members, whereas it may be held that attributes sometimes differ though they are attributes of just the same things. It is precisely this difference, in fact, and nothing more that Russell’s contextual definition of classes accommodates; his is a technical construction enabling us to speak ostensibly of identical classes by way of shorthand for discourse about coincident but perhaps non-identical attributes.

Such definition rests the clearer on the obscurer, and the more economical on the less. Classes are more economical than attributes because they are scarcer: they coalesce when their members are the same. [emphasis added] Classes are clearer then attributes because they have a relatively definite principle of individuation: they differ from one another just in case their members differ, whereas attributes (if they diverge from classes at all) differ from one another also under additional circumstances whose nature is left, in Principia, quite unspecified. […]

In any case there are no specific attributes that can be proved in Principia to be true of just the same things and yet to differ from one another.

Due also to the immediate success of Zermelo’s axiomatic set theory as a mathematical theory, compared to the solution developed into the Principia (the ramified theory of types), the “more clear and economical” conception of set as “extensional” has become, since the beginning of last century, the “mainstream view” about classes or set.

In conclusion, referring to Malice Vidrine’s answer, I think that is perfectly possible to think of “abandoning” our “intuitions about sets”: from one side, we have a lot of interesting results based on extensional set theory; form the other side, we need support for “a case for deviation from it”.

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1) is correct. There are things which are not extensional, but in a theory which accepts the axiom of extensionality, those things are not called "sets".

This is not "a statement of a dictator", whatever you mean by that, this is how language works: every term in every language has a definition, which describes what the term means and what it does not mean. In the same way that an elephant is not an insect, something which is not extensional is not a set.

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I have added something, so if you are interested. –  drhab Feb 8 at 11:09

You should interpret it the first way. Axioms of formal systems are to be satisfied without explanation. A model which instantiates a formal system must satisfy the axioms of the formal system.

Thus any model of ZF does not have a set that is not extensional, they are not merely 'not considered'. I mean that they cannot be neglected because non-extensional sets do not exist in a model of ZF.

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I have added something, so if you are interested. –  drhab Feb 8 at 11:09
    
I personally do not accept regularity; that is, I always try to make assertions that I know are still true in its absence. Why do I deny it? You can imagine sets as 'trees' where elements are represented by nodes and set membership by lines connecting nodes. The axiom of infinity can then be thought of as 'there are infinitely wide trees' while regularity can be thought of as 'there are no infinitely tall trees'. I sort of find the claims contradictory. Not much changes without regularity. But the AMC need not be equivalent to AC. So I can't claim "every vector space has a basis" implies AC. –  Bryan Feb 8 at 11:27
    
Bryan, those are very weird interpretation of the axioms. Moreover regularity important for recursion, while not requiring further consistency strength. –  Asaf Karagila Feb 8 at 12:04

To add to the other answers:

(1) does say that every set the quantifier ranges over is extensional, full stop. There's no room in any model of extensionality for non-extensional objects. But at the same time there are other theories of which ZF is a subtheory, or in which ZF is interpretable, including some with very different properties (some strong extensions of NFU, Esser's positive set theory).

The question, and it's a philosophical one, is whether to think of some of these other theories as actually axiomatizing a theory of sets. NFU+Choice doesn't satisfy extensionality, but it's usually interpreted as a theory of sets and other stuff; the things that are sets in the theory are characterised precisely by extensionality. As extensionality is a strong part of many people's intuitions about sets, you would have to make a case for deviation from it.

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I have added something, so if you are interested. –  drhab Feb 8 at 11:08

Something nice I learned from Wikipedia is that, even in logical systems that don't have an equality predicate built in, it's possible to express the axiom of extensionality as

$$\forall a,b: (\forall x: x \in a \iff x \in b) \iff (\forall y: a \in y \iff b \in y), \tag1\label{ik-eqn:1}$$

or, in words: "if $a$ and $b$ have all the same members, then they belong in all the same sets."

(In ZF set theory, at least, only the forward implication actually needs to be given as an axiom, since the converse follows from the other axioms. However, if you don't mind a slight redundancy in your axioms, I prefer to include both implications in the formula to highlight its symmetry.)

The nice thing about the statement $\eqref{ik-eqn:1}$ is that the two sides of it can be seen as two halves of a definition of equality. That is to say, we may rewrite it as

$$\forall a,b: a \overset R\sim b \iff a \overset L\sim b \tag2\label{ik-eqn:2}$$

where we define the predicates $\overset L\sim$ and $\overset R\sim$ as shorthand for

  • $a \overset L\sim b \overset{def}\iff (\forall y: a \in y \iff b \in y)$, i.e. "$a$ and $b$ are substitutable to the left of $\in$," and

  • $a \overset R\sim b \overset{def}\iff (\forall x: x \in a \iff x \in b)$, i.e. "$a$ and $b$ are substitutable to the right of $\in$."

It's easy to see that both of these are equivalence relations, and that any two sets $a$, $b$ that satisfy both $a \overset L\sim b$ and $a \overset R\sim b$ in fact fully satisfy the substitution property of equality with respect to $\in$, and are therefore indistinguishable by any set-theoretic formula (i.e. a formula that can be expressed using only $\in$ and logical symbols).

The full axiom $\eqref{ik-eqn:2}$ then simply asserts that either of these definitions of "half-equality" is, in fact, equivalent to the other (and thus to full equality, in the sense of indistinguishability by set-theoretic formulae): if $a$ and $b$ can be freely interchanged on the right-hand side of $\in$, then they can also be freely interchanged on the left-hand side of it.


So, how does this relate to your original question, then?

In simplest terms, the interpretation of extensionality I've given above falls under your interpretation #1: asserting that $\overset L\sim$ and $\overset R\sim$ are equivalent is essentially the same as asserting that all sets are extensional.

What I find nice about this interpretation, though, is that, by not requiring any fundamental equality predicate, it sidesteps any potential "metaphysical" issues related to it. We simply have a definition of an equivalence relation $a \sim b \iff a \overset L\sim b \land a \overset R\sim b$, and an axiom further stating that, in fact, $a \overset L\sim b \iff a \overset R\sim b \iff a \sim b$.

It may be useful to contrast this view with user18921's "interpretation #3", which claims that the axiom of extensionality is itself a definition of set equivalence / equality. In a sense, these views are the same: if we take $\overset R\sim$ as the definition of set equality, and assert that sets equal under this definition are freely substitutable, as in Malice Vidrine's comment, then we obtain an extensional set theory.

On the other hand, what I'm really saying is that there are two natural definitions of set equivalence with respect to $\in$, $\overset L\sim$ and $\overset R\sim$, which together imply indistinguishability using set-theoretical formulae. Viewed this way, the axiom of extensionality does make a substantive, non-definitional assertion: it asserts that these two relations are, in fact, the same, and thus that any sets equivalent under either of them are in fact indistinguishable.

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Suppose I want to focus on G.

Among the things that are true about G is

Every set is regular extensional

so, in the course of focusing on G, I will invoke this "dictatorial" fact which is true about the objects I'm studying.

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