18
$\begingroup$

In every textbook on ZF set theory I come across the Axiom of extensionality, which basically says that if two sets have the same elements, then those two sets are equal: $$\forall x \forall y (\forall z(z\in x \iff z\in y)\Rightarrow x=y)$$ But why is such statement made to be an axiom? It surely looks like a straightforward definition, because it only relies on the primitive notion of membership, logical connectives and quantifiers. What am I missing here?

$\endgroup$
2
  • 3
    $\begingroup$ Doesn't every formula of ZFC just rely on membership, logical connectives, and quantifiers? $\endgroup$ Jun 30, 2014 at 1:09
  • 1
    $\begingroup$ I thought that was the point, because equality is a new predicate, so it appears on first glance that extentionality defines a predicate in terms of primitives rather than assuming a new axiom about how sets behave in terms of the primitives themselves. $\endgroup$
    – dezakin
    Jun 30, 2014 at 23:11

3 Answers 3

18
$\begingroup$

It is true that different authors have suggested that extensionality feels closer to a logical axiom than something else, and can be seen as defining equality in terms of membership. The question is then whether this is indeed innocuous.

This was considered seriously by Dana Scott, in

Dana S. Scott. More on the axiom of extensionality, in Essays on the foundations of mathematics, dedicated to Prof. A. H. Fraenkel on his 70th birthday, Magnes Press, The Hebrew University, Jerusalem, 1961, pp. 115–131. MR0163838 (29 #1137).

Scott considers Zermelo's set theory $\mathsf{Z}$, the standard system of Zermelo-Fraenkel $\mathsf{ZF}$, and their variants $\mathsf{Z}^{\ne}$ and $\mathsf{ZF}^{\ne}$ where extensionality is not assumed. Recall that in $\mathsf{Z}$ we do not assume replacement. (The precise formulation of these theories has some minor technical caveats, see section 1 of the paper.)

Scott shows:

1. There is a relative interpretation of $\mathsf{Z}$ inside $\mathsf{Z}^{\ne}$.

2. There is a relative interpretation of $\mathsf{ZF}^{\ne}$ inside $\mathsf{Z}$.

The significance of these results is that of course there is no such interpretation of $\mathsf{ZF}$ within $\mathsf{Z}$, since in fact $\mathsf{ZF}$ proves the consistency of $\mathsf{Z}$. On the other hand, 1 and 2 show that the theories $\mathsf{Z}$, $\mathsf{Z}^{\ne}$, and $\mathsf{ZF}^{\ne}$ are equiconsistent. This proves that the removal of extensionality actually significantly weakens the reach of set theory, and leaves us with a theory of considerably lower consistency strength.

A different issue is whether this would affect mathematical practice, since most of "classical" mathematics does not require replacement anyway (so it can be carried out in $\mathsf{Z}$ and therefore in $\mathsf{Z}^{\ne}$). However, I feel this argument no longer holds water (if it ever did), since big portions of "modern" mathematics do use and need replacement routinely (and so the weaknesses of $\mathsf{ZF}^{\ne}$ with respect to $\mathsf{ZF}$ become now matters of serious concern).

$\endgroup$
13
$\begingroup$

We want to connect the two basic relation symbols, $\in$ and $=$. The axiom of extensionality tells us that if two sets are distinct then there is some witness of this in the universe. Consider the very contrived example $\{\varnothing,\{a\}\}$, with the usual $\in$. The element $a$ is not in the universe of this example, so it cannot internally discern between $\varnothing$ and $\{a\}$.

$\endgroup$
4
  • $\begingroup$ Late-late reply, apologies. Would it mean that, by your example, the set $\{ a \}$ will be concluded as a null set, cause object $a$ is ambiguous? Thus the name suggests, extension of membership to equality? $\{a\} = \varnothing$ $\endgroup$ Feb 20 at 14:46
  • $\begingroup$ No, the idea is that in the structure $(M,\in)$ where $M=\{\varnothing,\{a\}\}$ where $a\neq\varnothing$, it is the case that no object is an element of any other object. In particular, the axiom of extensionality fails there, since $\varnothing$ and $\{a\}$ are distinct objects. The idea, in a nutshell, of the axiom of extensionality is that if two objects are different, then the $\in$ relationship "can see that". Namely, there is something which is an element of the one and not the other. Whereas in our $M$ this isn't the case. The two objects are distinct, but $\in$ fails to see that. $\endgroup$
    – Asaf Karagila
    Feb 20 at 14:49
  • $\begingroup$ To recap, the two sets $\{\},\{a\}$ are "seen" distinct by $\in$; even when we are not "aware" of object $a$. Could you please explain what you mean by set membership $\in$, how do you comprehend it? $\endgroup$ Feb 20 at 14:59
  • 1
    $\begingroup$ The $\in$ membership is the one you know and love. It just means that $x\in y$ if $x$ is an element of $y$. In this case, $\varnothing$ has no elements, and $\{a\}$ has exactly one, $a$. $\endgroup$
    – Asaf Karagila
    Feb 20 at 15:50
3
$\begingroup$

Suppose the set $S$ contains the element $a$ and no other elements.

Suppose the set $T$ contains the element $a$ and no other elements.

Should these sets be equal? If you think so, then you'll have to use extensionality to argue as such: it is the only axiom of ZFC that allows you to infer that two sets are equal.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .