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I know every set is a class and not the other way around, but can one consider the set of, say, two classes? Is this "well-defined"?

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@Ross thanks for the suggestion. Done. –  klanth Jan 2 '13 at 4:04
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In standard set theory, sets come equipped with a rank (an ordinal), the rank of a set is completely determined by the ranks of its elements, if $x\in y$ then the rank of $x$ exists and is strictly below the rank of $y$, and proper classes have no rank. This precludes proper classes from being elements of any set. –  Andres Caicedo Jan 2 '13 at 4:07
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In standard set theories where one can talk about classes, one can actually define sets as those classes that belong to some class. So, if $x,y$ are classes, and there is a class $Z=\{x,y\}$, then $x$ and $y$ are actually sets (and so is $Z$). –  Andres Caicedo Jan 2 '13 at 4:09
    
(Of course, there are variants of standard set theory where not every set has a rank, so we need another tool to tell the difference between sets and proper classes. In Cantorian set theory, it is the notion of size --cardinality-- that plays this role, and our axioms ensure that sets are the objects whose size is not "too large" and that this notion is hereditary, so elements of sets are also sets.) –  Andres Caicedo Jan 2 '13 at 4:15

1 Answer 1

up vote 8 down vote accepted

If von Neumann-Bernays-Gödel Set Theory (NBG Set Theory) is presented as a one-sorted theory, then a set is defined as a class that is an element of another class (possibly itself, but this is ruled out by the Axiom of Regularity for NBG Set Theory). A class that is not a set (i.e., one that is not an element of another class) is called a proper class.

Let $ A $ and $ B $ be classes. If there exists a set $ C $ whose elements are precisely $ A $ and $ B $, then by definition, $ A $ and $ B $ are sets. Conversely, by the Axiom of Pairing for NBG Set Theory, if $ A $ and $ B $ are sets, then there exists a set $ C $ whose elements are precisely $ A $ and $ B $. Therefore, a set consisting of precisely two classes is ‘well-defined’ if and only if the two classes are sets.

In order to deal with the case when the two classes are proper, we must go beyond sets and classes by creating a third type of object, called a conglomerate. We require that conglomerates at least satisfy Extensionality, Pairing, Union, Power and the Axiom Schema of Specification. By default, all classes and sets are conglomerates. However, there exist conglomerates that are neither a set nor a class, such as the object $ \{ V \} $, where $ V $ is the class of all sets. We already know that $ V $ is a proper class, so in NBG Set Theory, the object $ \{ V \} $ does not exist. In the theory of conglomerates, however, $ \{ V \} $ is a well-defined conglomerate.

We can therefore view conglomerates as generally being one level higher than classes and two levels higher than sets. This is reminiscent of Russell's theory of types.

By Pairing, one can thus form a conglomerate of two proper classes, which is one way of resolving the problem of ‘pairing’ two proper classes. You may wish to refer to David Murfet's Foundations for Category Theory for a discussion of conglomerates.

One way of avoiding conglomerates is to use inaccessible cardinals. I believe that Murfet talks about this in the same essay, albeit briefly. I have also provided below two other references where conglomerates are mentioned.

References

  • Osborne, M. Scott. Basic Homological Algebra (Graduate Texts in Mathematics), Springer, 2000.

  • Xu, Kongshi. Advances in Chinese Computer Science, Volume 3, World Scientific, 1991, pp. 155-170.

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I should've mentioned it, I meant proper classes. I'd never heard of these things. I'm going to take a look at the link. Thanks! –  klanth Jan 2 '13 at 4:03
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I much prefer the terminology of $2$-classes to conglomerates. Sure, the latter sounds fancy, but how would you describe a collection of conglomerates? A cartel? How about the collection of those? With $\alpha$-classes it's easier. –  Asaf Karagila Jan 2 '13 at 8:28

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