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.

For some formulae $\phi(x)$ it can be proved from the axioms of ZFC, that there is no set $X$ with $(x)x\in X \equiv \phi(x)$. Thus the collection $\lbrace x\ |\ \phi(x)\rbrace$ is a proper class.

Furthermore, there is no set $Y$ that is in bijection with such a collection, right? Thus, every proper class must be in some sense larger than any set, right? ("too large to be a set")

For two formulae $\phi(x)$, $\psi(x)$, one of them defining a proper class, it may be possible two show that $(x) \phi(x) \equiv \psi(x)$. This means a) that also the second collection is a proper class, and b) that the two have equal size in some sense.

Can the notion of size of proper classes (or equipollency) be made precise? Is it possible that two proper classes do not have equal size in this sense? Would it thus follow, that there are smaller and larger proper classes? Or do all proper classes have the same size (of $V$)?

And - finally - can the size of $V$ be characterized? To talk at large, maybe like this (if sense of $P(V)$ could be made): $V$ is so large such that $|V| = |P(V)|$ (which never holds for sets)? (This would admittedly be counter-intuitive, since the discrepancy between $|X|$ and $|P(X)|$ grows with $|X|$.)

share|improve this question
    
The descrepancy between $|X|$ and $|P(X)|$ only provably grows with $|X|$ for finite $X$. For infinite $|X|$, just about anything is possible (but not everything!). In particular, it can be the case that the generalized continuum hypothesis holds - that $|P(X)|$ is the very next size after $|X|$ for all infinite $|X|$. –  Jason DeVito Feb 21 '11 at 16:06
    
But at least |X| < |P(X)| holds for all sets X, doesn't it? –  Hans Stricker Feb 21 '11 at 16:08
    
You can say a bit more than that: you can say that |X|<|Y| implies |P(X)| <= |P(Y)| and you can add one minor technical condition about the cofinality of the power set, but (roughly) other than that, anything goes. –  Steven Stadnicki Feb 22 '11 at 2:54
add comment

2 Answers

up vote 6 down vote accepted

This question was asked at MO a while ago. Below I quote the relevant part of my answer. But let me add some comments.

In ZFC (proper) classes are not actual objects, we only treat them formally, and they are really just short-cuts for formulas (with parameters), i.e., we identify formulas $\phi(x)$ with the collection of sets that satisfy $\phi$.

This makes talking of relations between classes somewhat awkward. Arturo's nice answer, for example, indicates how one can make sense of, say, a bijection between two classes. A bit more formally, if $X$ is the class of $x$ such that $\phi(x)$ and $Y$ is the class of $y$ such that $\psi(y)$, then a bijection between $X$ and $Y$ is a class $Z$ given by a formula $\rho(x,y)$ such that:

  1. For any $x,y,z$, if $\rho(x,y)$ and $\rho(x,z)$ then $y=z$.
  2. Similarly, for any $x,y,z$, if$ \rho(x,y)$ and $\rho(z,y)$, then $x=z$.
  3. For any $x,y$, $\rho(x,y)$ implies that $\phi(x)$ and $\psi(y)$ holds.
  4. For any $x$ such that $\phi(x)$ there is a $y$ such that $\rho(x,y)$.
  5. And, for any $y$ such that $\psi(y)$, there is an $x$ such that $\rho(x,y)$.

This awkwardness also makes it impossible to develop an appropriate theory of classes. For example, the natural question: "Are $X$ and $Y$ pf the same size?" cannot even be asked in ZFC. Of, course, if there is a $\rho$ as above, then we can say that they have the same size, as witnessed by $Z$, and if there is no such $\rho$, outside of ZFC we can say that thy don't have the same size, but all we can do in ZFC is to say, of any given formula $\rho$, that $\rho$ does not define a bijection between $X$ and $Y$.

Of course, equipotency is only an example of what we cannot discuss freely about classes. Thus when interested in proper classes, we move from ZFC to other theories. There are (at least) two natural extensions of ZFC where classes can be treated as formal objects. One is NBG (Von Neumann-Bernays-Goedel). Here, the objects are classes, and sets are classes that belong to other classes. $Z$ is a bijection between $X$ and $Y$ iff $A$ is a function, its domain is $X$ and its range is $Y$ (just as with sets). NBG is usually preferred because it is a conservative extension of ZFC; in a sense, all we have done is to allow references to classes without changing what we mean by "set" in any way. Formally, any theorem of NBG that only mentions sets is a theorem of ZFC. And our interpretation of classes as given by formulas gives us a way of extending any model of ZFC to one of NBG.

The fact that NBG is conservative over ZFC is in a sense a serious limitation, as we limit the notion of class somewhat artificially. When discussing elementary embedding formulations of large cardinal axioms (something very common in modern set theory), for example, the discussion can be carried out in NBG, but not in the smoothest possible fashion. In technical terms, the comprehension axiom in NBG is predicative, but it is hard to justify this when we are allowing proper classes at all.

The most natural extension of ZFC to treat proper classes is Morse-Kelley (MK). Here, comprehension is unrestricted, so it is more natural to combine classes into new ones by usual operations. The cost of this is that MK is not a conservative extension of ZFC. In fact, MK can prove the consistency of NBG (and therefore of ZFC). That being said, to discuss equipotency of classes, MK seems the appropriate framework.

Here is what I said in the MO post mentioned above:

In extensions of set theory where classes are allowed (not just formally as in ZFC, but as actual objects as in MK or GB), sometimes it is suggested to add an axiom (due to Von Neumann, I believe) stating that any two classes are in bijection with one another. Under this axiom, the "cardinality" of a proper class would be ORD, the class of all ordinals. (By the way, by class forcing, given any proper class, one can add a bijection between the class and ORD without adding sets, so this assumption bears no implications for set theory proper.)

Without assuming Von Neumann's axiom, or the axiom of choice, I know of no sensible way of making sense of this notion, as now we could have some proper classes that are "thinner" than others, or even incomparable. Of course, we could study models where this happens (for example, work in ZF, assume there is a strong inaccessible $\kappa$, and consider $V_\kappa$ as the universe of sets, and $Def(V_\kappa)$ in Gödel's sense (or even $V_{\kappa+1}$) as the collection of classes).

Let me expand on this a bit. I argue in ZFC as this is the best known framework of the three mentioned above:

First, class forcing allows us to add a bijection between $V$ and $ORD$ without adding sets. Essentially, what we do is to "thread" through the class of bijections between sets and ordinals. In the resulting extension, we have added no sets, but we have a new class $G$. The resulting structure $(V,G,\in)$ is a model of the strong version of ZFC where we allow $G$ to appear as a predicate in instances of the replacement axiom. Also, any proper class $A$ here (in the sense of "definable by a formula") is also in bijection with the ordinals, and such a bijection is easily definable from $A$ and $G$.

This shows that the assumption that all classes have the same size is completely harmless: No new theorems of ZFC can be proved by adding this assumption.

The model we obtain is not a "natural" model of NBG, though, since G is not definable.

Arturo's suggestion ($V=L$) gives us models where bijections between classes and the ordinals are definable. It has the disadvantage that V=L is quite limiting. As I mentioned in the comments to his answer, there is an alternative: We can assume $G$ definable. This is because, over any model of ZFC, we can force to obtain a new model where $V=HOD$ (on the other hand, once $V\ne L$ we cannot force to have the equality back).

$HOD$ is the class of hereditarily ordinal definable sets. This means that $x\in HOD$ implies that $x\subset HOD$, and that to be in HOD, $x$ must be definable from ordinal parameters. Of course $V=L$ implies $V=HOD$, but $HOD$ is compatible with all known large cardinals, while $L$ is not. Moreover, $HOD$ carries a definable well-ordering (in order type ORD), so if $V=HOD$, then any proper class is in bijection with the ordinals. Moreover, $V=HOD$ is equivalent to the statement that there is a (definable) bijection between $V$ and $ORD$. So, at least in ZFC, this characterizes in a natural way when all proper classes have the same size (and it is really the only way that "sizes of proper classes" can be freely discussed in ZFC).

Finally, it is consistent that $V\ne HOD$, in which case $V$ and $ORD$ have different cardinality in the ZFC sense defined above, and in fact one can arrange that there are incomparable "sizes" of proper classes.

share|improve this answer
    
Thanks for the extended (and better than mine) answer. –  Arturo Magidin Feb 21 '11 at 19:14
    
Definitely! If I had seen it, I would have marked it as a favorite question, but I didn't. Thus, I didn't have seen it. –  Hans Stricker Feb 21 '11 at 19:44
add comment

The answer seems to depend on your set theory.

Specifically, in an August 2006 post in sci.math, Herman Rubin said that if $\mathbf{V}=\mathbf{L}$ (Goedel's Constructible Universe), then every proper class is equipollent to (bijectable with) the class of all ordinals, hence equipollent with one another, so you only have a single "size" of proper classes.

However, he also said that there are models of ZFC with different versions of the Axiom of Choice for proper classes and in some of these not all proper classes are equipollent (even if you assume the Axiom of Choice for sets).

For completeness: what do we mean by equipollency of proper classes? Goedel defined equivalence of proper classes as follows: For proper classes $X$ and $Y$, $X\sim Y$ if and only if there exists $Z$ such that:

  1. $Z$ is a relation (contained in $\mathbf{V}^2$);
  2. $Z$ is twice unary ($Z$ is single-valued, and so is its inverse);
  3. The domain of $Z$ is $X$; and
  4. The domain of the inverse of $Z$ is $Y$.

One model where all proper classes are equipollent is when we assume the "Axiom of Global Choice", which states that there exists a unary relation $A$ such that for all nonempty sets $X$, there exists $y\in X$ such that $(X,y)\in A$. This is given by Goedel in his paper on the consistency of AC and GCH with ZF. It implies that any class is equipollent to an initial segment of the ordinals, and since any proper initial segment of the ordinals is a set, a proper class must be equipollent to the proper class of all ordinals, giving the result that all proper classes are equipollent.

share|improve this answer
    
Do I understand correctly: equivalence of proper classes depends on the existence of a set Z? –  Hans Stricker Feb 21 '11 at 16:26
    
@Hans: The referred $Z$ is a class. –  Asaf Karagila Feb 21 '11 at 16:36
    
@Hans: No, $Z$ is a proper class. Joel Hamkins has addressed this issue elsewhere on this site. More generally than V=L, what you need is precisely that V=HOD, the class of hereditarily ordinal definable sets. –  Andres Caicedo Feb 21 '11 at 16:37
    
@Andres: Do you mean this response? math.stackexchange.com/questions/9037/… If so, I'll add a link to the answer. –  Arturo Magidin Feb 21 '11 at 16:41
1  
Hi Arturo, I think... (Joel may have said something else in comments to another question as well). In any case: The point is: V=HOD gives a global (definable) well-ordering of the universe (in order type ORD). From this it follows that any proper class is in bijection with the ordinals. And, in ZF, V=HOD is equivalent to the existence of such bijection. (Part of the issue here is technical: In ZF, proper classes are given by formulas (with parameters). In other set theories like MK where proper classes need not be definable, a well-ordering of V does not imply V=HOD.) –  Andres Caicedo Feb 21 '11 at 16:50
show 3 more comments

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.