Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question: What is the definition of a small class? Is there no such thing as a set which contains other sets such as {{1,2},{1},{2}, 1, 2} (ie, is this really called a "class")? What are some prototypical examples of small classes? What are some prototypical examples of large classes?

Motivation: A few things I've been paging through (Rotman's homological algebra, a few papers which talk about categories, etc.) talk about small classes using definitions which are slightly confusing to me. Most of them note that a "class" is a "collection" of "sets", and that a small class is a class which is a set. I'm not exactly sure where "small" ends and "large" begins in this respect. See the specific questions below for what I mean by this.

Related Specific Questions: If we use the notation {a,b,c} for a set with the elements 'a', 'b', and 'c', then is {a,b,c} a small class because it IS a set? Is {{a,b,c}} a large class because it is a "set which contains a set"? Is the set of real numbers a small class? Is the set of natural numbers a small class? Is the set of all functions from the real numbers to the real numbers a small class? Is the class of all small classes a large class?

share|improve this question
    
[Wikipedia](en.wikipedia.org/wiki/Class_(set_theory)) mentions that every set is a class. –  user2468 Dec 26 '10 at 0:33

3 Answers 3

up vote 2 down vote accepted

The sets that we consider are actually hereditary sets meaning that all of their members are themselves hereditary sets. What this informally means is that a hereditary set is a set of sets of sets of sets, etc.

For example, the empty set ($\emptyset$) is a hereditary set because it has no members so all of its members are vacuously hereditary sets. Then $\{\emptyset\}$ is a hereditary set because its only member is a hereditary set. Then $\{\emptyset, \{\emptyset\}\}$ is a hereditary set because both of its members are hereditary sets, and we can continue this process of tacking on the previous hereditary set as a member ad infinitum to get an infinite collection of hereditary sets. This process describes how the Natural numbers are constructed:

$0 = \emptyset$

$1 = \{\emptyset\} = \{0\}$

$2 = \{\emptyset, \{\emptyset\}\} = \{0, 1\}$

$3 = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} = \{0, 1, 2\}$

$\vdots$

$n + 1 = \{0, 1, 2, \ldots, n\}$

By the axiom of infinity, there exists an inductive set (set that contains $0$ and the successor of any of its members) so the class of Natural numbers is actually a hereditary set. The integers can then be constructed from the Natural numbers by considering equivalence classes of ordered pairs of Natural numbers where $(a, b)$ is equivalent to $(c, d)$ exactly when $a + d = b + c$ so $(a, b)$ will correspond to the integer $a - b$. Ordered pairs of hereditary sets are also hereditary sets because $(a, b)$ can be formalized as $\{\{a\}, \{a, b\}\}$. Consequently, the class of integers will also be a hereditary set. The set of Rational numbers can then be defined as equivalence classes of ordered pairs of integers so it too will be a hereditary set. The Real numbers, which can be constructed by considering select subsets of Rational numbers (or elements from the powerset of the Rational numbers), is also a hereditary set. Once we formalize the Real numbers as a hereditary set, we can show that a function from the Reals into the Reals, which is a collection of ordered pairs of Reals, is a hereditary set. Then we can show that the collection of all functions from the Reals into the Reals is a hereditary set. In short, all of the natural objects would be "small classes" because they can be formed from the axioms of set theory (ZFC).

The point here is that what determines whether a class is "small" or not is not whether it contains elements that look like sets such as $\{1, 2, 3\}$ or $\{a, b, c\}$ because we mostly restrict ourselves to hereditary sets anyway. Instead, we have as a minimal requirement for smallness that the objects can either be provably constructible from the axioms of ZFC or the assumption of their existence as sets does not lead to a contradiction in ZFC. I include the first possibility here because we cannot prove (within ZFC) that the axioms of ZFC won't themselves lead to a contradiction.

As an example of what we would consider a large class is the class of all small classes or the class of all sets. This is because such a class cannot be constructed from the axioms of ZFC, and it provably leads to a contradiction from ZFC (see Russell's Paradox from Brad's post). In a nutshell, if we called the class of all sets $X$ and $X$ were a set, then $A = \{x \in X| x \notin X\}$ would have to be a set by the comprehension axiom. But then $X \in X$ since $X$ is the class of all sets so $X \in A$ if and only if $X \notin A$, an immediate contradiction. Other examples of "large classes" include the class of ordinals or as Ricky mentioned any object that does not fall into the $V$ hierarchy.

One more thing I should mention here is that you should always be aware of the context when considering whether an object is a large class (also known as proper class) or not. Specifically, most of the natural objects that you consider will be small classes with respect to the set-theoretic universe. However, sometimes classes are considered with respect to individual models. A Grothendieck Universe for example, is a proper class if there are no inaccessible cardinals and we cannot prove that such a large cardinal is even possible from the ZFC axioms of set theory. As another example, we sometimes do not allow ourselves use of all of the ZFC axioms, and the set of all even numbers for example would be a large class if we explicitly assumed that all sets were finite.

share|improve this answer
1  
Thank you, this was clearly written! You've also sparked my interest in set theory again, which is no small feat. –  james Dec 26 '10 at 17:25

As I understand it, the terms "set" and "small class" are equivalent. Certainly a set may have other sets as elements, so {{1,2},{1},{2},1,2} is a set.

Informally, a class is a collection of sets which can be "unambiguously defined by a property that all its members share," to quote Wikipedia.

To formalize the notion of a large class -- that is, a class which is not small -- requires a choice of set theory that allows the formulation of such a concept. The standard framework of the Zermelo-Fraenkel axioms (ZF) does not have a formal notion of a large class: everything in consideration in that theory is a set.

Some standard examples of large classes are:

  • the class of all sets
  • the class of all groups
  • the class of all topological spaces

etc.

These are large classes because they are not small. For example, the class of all sets is not itself a set: Russell's paradox can be thought of as a proof of this statement.

To answer your specific questions in light of the above:

  • {a,b,c} is a small class, because it is a set.
  • {{a,b,c}} is a small class, because it is a set.
  • Each of the set of real numbers and the set of natural numbers is a small class.
  • The class of all small classes (i.e. the class of all sets) is a large class, as above.
share|improve this answer
    
+1. But, what about the class of all classes (large and small)? By having large classes (like itself perhaps), shouldn't it be affected by the Russel Paradox? –  JMCF125 Jul 13 '13 at 22:15

You already know the definition of a "small class", it is what you gave.


Definition:

Let X be a class.

Then X is a set if and only there exists a class Y such that X is an element of Y.


Theorem:

Let X be a class.

Then X is a set if and only if either/both of the following equivalent conditions hold:

  1. There exists an ordinal alpha such that X is an element of V_alpha.
  2. There exists an ordinal alpha such that X is a subclass of V_alpha.

(ordinal is defined at en.wikipedia.org/wiki/Ordinal_number,
V_alpha is defined at en.wikipedia.org/wiki/Von_neumann_universe)

share|improve this answer
    
So, for example, using this V_alpha definition, it notes in the wiki article that showing the set of all sets is not a set comes down to the fact that the union of these V_alpha type things is a proper class. How do they jump to this conclusion? –  james Dec 26 '10 at 3:13
    
For all ordinals alpha, alpha+1 is also an ordinal, and alpha is an element of V_(alpha+1); so the class of ordinals is a subclass of the union of the V_alpha. By en.wikipedia.org/wiki/Burali-Forti_paradox, the class of all ordinals is not a set. Since a subclass of a set is also a set, the union of the V_alpha is not a set either. –  Ricky Demer Dec 26 '10 at 16:33
    
Thank you, this was an interesting way to look at this! –  james Dec 26 '10 at 17:24
    
Point of clarification: A definable subclass of a set is also a set by the axiom of comprehension. There is a philosophical question of whether there "should" be an absolute universe for set theory, and we often work with a variety of models of set theory and consider what are called forcing extensions. In a forcing extension, we are adding a subset of a set in the model that's not already in the model (e.g., we can add a subset of the Natural numbers not in the model by forcing to add one). –  Jason Jan 7 '11 at 0:40
    
Also as an obvious cautionary flag, care should always be taken when appealing to a theorem to prove another one so that you don't wind up with one using circular logic. For example, you can prove the Burali-Forti paradox by using the fact that there is no set of all sets. Specifically, if the set ORD of all ordinals were a set, then $V = \bigcup_{\alpha \in ORD} V_{\alpha}$ would be also be a set, contradicting the fact that there is no set of all sets. But fortunately, this is a nonessential use of the theorem that there is no set of all sets in the proof of Burali-Forti so you're fine. –  Jason Jan 7 '11 at 0:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.