3
$\begingroup$

The set of all sets that do not contain themselves, the set of all ordinal numbers, and the set of all sets represent proper classes that would clearly be paradoxical if they were admitted as 'sets' within ZFC--they are the kinds of collections that ZFC intended to make proper classes. My question is, are there known to be proper classes which do not induce paradoxes within ZFC--more clearly: does there exist a proper class $A$ whose existence could be added as an axiom to ZFC and still not induce any paradoxes within ZFC?

$\endgroup$
  • 2
    $\begingroup$ I think you need to be much more precise with what you call a paradox here. However, one thing all classes $C$ (sometimes called 'virtual classes') have in common in $\mathrm{ZFC}$ is that they are unbounded, i.e. there is no $\alpha \in \mathrm{Ord}$ such that $C \subseteq V_{\alpha}$. This is guaranteed by the comprehension scheme (and not true for weaker axiomatic systems like $\mathrm{KP}$). $\endgroup$ – Stefan Mesken Aug 1 '18 at 18:48
6
$\begingroup$

Note that the following are equivalent, for $C\subseteq V$:

Here I'm taking $V$ to be the true universe of sets. If we want to work instead in the context of an arbitrary model of ZFC - e.g., if we are skeptical of the existence of a true universe of sets in the first place - then we need to add the additional rule that $C$ is definable inside the model we're working in.

  • $C$ is a proper class, rather than a set.

  • $C$ contains elements of arbitrarily high rank: for each ordinal $\alpha$ there is an element of $C$ not in $V_\alpha$.

The point is that ZFC (indeed, much less) proves that every set contains only elements of bounded rank, and so in this sense the set-hood of any proper class is inconsistent with ZFC.


EDIT: At the same time, there are plenty of classes (= formulas defining collections of sets) about whose set-hood ZFC is ambivalent. Besides the silly examples (like "$(x=x)\wedge$ CH," which defines a proper class (the universe) if CH holds and a set (the emptyset) if CH fails), there are reasonably natural candidates such as:

  • The class of inaccessible cardinals (or any other type of large cardinal) is consistently empty, but (under appropriate consistency assumptions) could be a proper class.

  • The class of regular cardinals $\kappa$ satisfying $2^\kappa=\kappa^+$ can be basically anything, by Easton's theorem. Note that unlike the previous example, no consistency strength considerations are needed.

  • Over ZF rather than ZFC, the class $\{\alpha\in Ord: $ Every set in $V_\alpha$ is well-orderable$\}$ is the set of ranks below which we don't have a counterexample to the axiom of choice; this is an initial segment of $Ord$ which is a set iff choice fails.

So another important point is that a formula could fail to define a set because of a "coincidence" about the set-theoretic universe.

$\endgroup$
2
$\begingroup$

No (to the question in the title), or at least not in the way I suspect you imagine it.

There is no rule in ZFC that every collection must be a set unless having it be a set would lead to a paradox.

Not only is there no explicit rule saying this: It is not actually true about arbitrary models of ZFC that they will contain all collections that wouldn't contradict the rest of the model.

To see this, we need to assume that ZFC has any models at all, of course -- that is, that it is consistent. Then, thanks to the Löwenheim-Skolem theorem, it will have a countable model. Looking at that model from the outside we can see that it doesn't contain all possible subsets of its $\mathbb N$, because we know uncountably many of them and only countably many (from our perspective) are sets in the model. Pick one of the subsets that don't exist in the model; we can then, by compactness construct an elementary extension of the model where that subset is actually a set. This shows that this subset simply happened to be absent in the original model -- it wasn't missing because having it would have led to a contradiction.

This is not what is usually called a "proper class", though -- because proper classes are usually supposed to be definable, possible with parameters. If you're using that concept, Noah Schweber's answer applies.

$\endgroup$
  • $\begingroup$ i could be wrong here but I’m not sure the picture you paint of adding one of the reals that isn’t in the model works unless the model is standard. If it’s standard I know what the missing reals look like and exactly where to find one. I’m not sure otherwise. $\endgroup$ – spaceisdarkgreen Aug 1 '18 at 20:03
  • 1
    $\begingroup$ @space There is no issue, although we may need some rephrasing. Henning explains how to proceed (via compactness). To be more precise, given any (standard) subset $S$ of $\omega$ absent from the model (meaning, for any $T$ that the model believes is a subset of $\mathbb N$, $S$ is different from the intersection of $T$ with the (external) set of standard integers), by compactness we can form an elementary extension of the model with a set $T$ in it such that the intersection of $T$ and the standard integers is precisely $S$. $\endgroup$ – Andrés E. Caicedo Aug 1 '18 at 20:27
  • $\begingroup$ @Andres Thank you for clarifying. $\endgroup$ – spaceisdarkgreen Aug 1 '18 at 20:36
  • $\begingroup$ Hmm, I was actually thinking of an $S$ that's a subset of the model's $\omega$. However, @Andrés's clarification made me realize that I may have been too optimistic when I (tacitly) assumed we can adjoin it to the model without also getting more integers in the process. Intuitively I would expect we can, but just compactness would not suffice to prove that. $\endgroup$ – Henning Makholm Aug 1 '18 at 20:54
1
$\begingroup$

All proper classes are paradoxical. By definition. Proper classes are not sets, and classes are sets if and only if they are subclasses of a set, this is the axiom schema of separation.

However, for the most part, classes are defined by a formula. So you might want to ask, if $\varphi(x)$ defines a class, as any formula does, if it is not provable that $\varphi$ defines a set, does it necessarily define a proper class?

The answer to that is negative. One example, as given by Noah, is the class of all cardinals $\kappa$ such that $2^\kappa\neq\kappa^+$. You can convince yourself that this is a class indeed. Whereas $\sf GCH$ implies this class is empty, and thus a set, it is also consistent that this class contains all successor cardinals, and therefore a proper class.

But an even simpler example can be given. Let $\psi$ be a sentence which is independent of $\sf ZFC$. For example the Continuum Hypothesis. Define $\varphi(x)$ to be $\psi\rightarrow x\neq x$. This means that assuming $\psi$ holds, $\varphi$ defines the empty class, which is a set by all means; but if $\psi$ is false, then $\varphi$ is always true and thus $\varphi$ defines the class of all sets, which is certainly a proper class.


So what is the real issue here?

Well the real issue is exactly that subtle difference between theory and meta-theory. Classes, or rather formulas which define classes, are object of the meta-theory, this is where we argue and prove things about $\sf ZFC$ (e.g. $\sf ZFC$ can neither prove nor disprove $\sf CH$). Classes are objects of the meta-theory. While the theory, $\sf ZFC$, can deal with classes on a limited basis, it is easy to trick it. Especially because $\sf ZFC$ is not a complete theory: there are statements which are independent of it. This lets us play a lot of tricks, and produce a lot of formulas which in some models of $\sf ZFC$ would define a set, and in other would define a proper class.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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