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I recently started a course in set theory and it was said that a model of set theory consists of a nonempty collection $U$ of elements and a nonempty collection $E$ of ordered pairs $(u,v)$, the components of which belong to $U$. Then the elements of $U$ are sets in the model and a set $u$ is interpreted as an element of $v$ if $(u,v) \in E$. It was also said that $U$ can also be a set and then $E$ is a relation in the set $U$ so that the ordered pair $(U,E)$ is a directed graph and reversely, any ordered graph $(U,E)$ can be used as a model of set theory.

There have been examples of different models now where some of the axioms of ZFC do not hold and some do, but the axiom of extensionality has always held and I for some reason don't seem to comprehend enough of that axiom and its usage. Can someone tell an example of some collections $E$ and $U$ where the axiom of extensionality wouldn't hold?

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Terminology: Any ordered graph can be an interpretation of the language of set theory. It is only a model if every axiom is true under the interpretation. –  Henning Makholm Sep 12 '11 at 15:24
    
Though they may not be models of ZFC without the axiom of extensionality, one does on occasion consider set theories with so-called "urelements", i.e. "objects" that are not "the" empty set, but rather do not contain any elements, yet differ from each other and are well distinguishable. (In many ways such set theories come closer to the everyday notion of what a set is.) I'm just wondering how you would formulate some of the axioms in a system of ZFC without extensionality (and why) ? Kind regards - Stephan F. Kroneck. –  bonnbaki Sep 12 '11 at 15:30

4 Answers 4

The axiom of extensionality says:

$$\forall x\forall y\left(x=y\leftrightarrow \forall z\left(z\in x\leftrightarrow z\in y\right)\right)$$

Obviously, if two sets are the same then they have the same elements. So in order to violate this axiom we need to have different sets which the model would think have the same elements.

If you just want a model of sets in which the axiom of extensionality does not hold, consider for $a\neq b$ the following: $$\left(U=\Big\{\{a,b\},\{a\},a\Big\}, \in\right)$$

We have that $a\in\{a\}$, and $a\in\{a,b\}$. Since $a\neq b$ we have that $\{a\}\neq\{a,b\}$, however for all $x\in U$ we have $x\in\{a\}\leftrightarrow x\in\{a,b\}$.

This is because $U$ does not know about $b$. It just knows that $\{a,b\}$ and $\{a\}$ are two distinct beings. It is unable to say why, in terms of $\in$ relation.

The problems begins when you want more axioms. The more axioms you would want to have, the more complicated your universe will have to get.

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The point of the axiom of extensionality is to avoid the situation where the way in which the set is defined, not just its members, affects which sets contain the set. So to make extensionality fail you will just need a model in which there are two sets $A,B$ which have the same elements but are not equal.

The easiest way to do this is to take any set $A$, make a copy of it, color one copy red, and color the other copy blue. Declare that any set that was a member of $A$ is a member of both red-$A$ and blue-$A$, and that any set which contained $A$ contains both red-$A$ and blue-$A$. Then the resulting thing will still be a model of ZFC in the language with just $\in$ (without $=$ yet).

The point of the axiom of extensionality is then to say that this situation does not occur: there are not two sets which are somehow different despite having all the same elements.

In fact, a modification of the axiom of extensionality lets us define $=$ in terms of $\in$, by declaring that two sets will be regarded as equal if they have all the same members. To make this work, we rewrite the axiom of extensionality without $=$ as:

$$ (\forall x)(x \in A \Leftrightarrow x \in B) \Rightarrow (\forall y)(A \in y \Leftrightarrow B \in y) $$ Then we can define $$ A = B \Leftrightarrow (\forall x)(x \in A \Leftrightarrow x \in B) $$ and we can prove the substitution axioms for equality in terms of the equality-free axiom of extensionality and the definition of $=$.

This can be done in any model of ZFC in the language with just $\in$ (satisfying the modified version of extensionality) to obtain a model of ZFC in the language ($\in$, $=$), including the usual axiom of extensionality. But here $=$ may not be interpreted as the true equality relation, for example if the original model had red sets and blue sets. If we want the $=$ symbol to be interpreted as actual equality, we have to mod out by the equivalence relation induced by the interpretation of $=$ in our model. In the setting where $=$ is treated as a logical symbol which must be interpreted as true equality, the point of the axiom of extensionality is to ensure that this modding-out has already been performed: if we would define to sets to be "equal" in this indirect way, then they are already equal.

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See Theorem 2 (p. 157) in the following paper, which is freely available on the internet:

Alexander Abian and Samuel LaMacchia, "On the consistency and independence of some set-theoretical axioms", Notre Dame Journal of Formal Logic 19 (1976), 155-158.

http://projecteuclid.org/euclid.ndjfl/1093888220

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I changed your link into a persistent one. Most article pages contain a link to themselves that is guaranteed to be stable, either using an internal linking system (like projecteuclid), or using the DOI system. –  t.b. Sep 12 '11 at 16:46

I haven't checked all the details, but I expect it would work simply to take a model of ZFC and then remove the empty set from the universe. All other sets and their membership relations stay unchanged. The new model still has an empty set (namely the one that was formerly $\{ \emptyset\}$), and I think all of the other axioms except extensionality would still hold, though verifying this could be tedious.

However, the two individuals that used to be $\{\{\emptyset\}\}$ and $\{\emptyset,\{\emptyset\}\}$ now have the same elements -- but the set that used to be $\{\{\{\emptyset\}\}\}$ contains one but not the other, violating extensionality.

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