8
$\begingroup$

I want to know if there is a standard terminology for this among set theorists working with element-based set theories like ZFC. I will follow the convention that a class in any given set theory is a definable class--it is all the sets that have some property $\phi(x)$ expressible in the theory. So it is generally not a legitimate entity in the theory, but we can talk about it in the theory.

Then I call a class a proper class if it is not coextensive with any set. In ZFC that is well expressed by saying a proper class is too big to be a set. But that already fails in Zermelo set theory. Zermelo set theory has a definable class of all finitely iterated power sets of the natural numbers, which is not provably a set but is provably countable. In Zermelo set theory we can still say a proper class is a class not contained in any set (since the axiom scheme of separation says the intersection of any class with a set is a set).

The matter is yet more complex in Bounded Zermelo set theory, where the separation axiom scheme only holds for defining conditions wilh all quantifiers bounded. In this theory a definable subclass of a set need not be a set. For example Bounded Zermelo set theory can define the class of all natural numbers $n$ such that there exists an $n$-th iterated power set of the natural numbers. But it cannot prove this class is a set. (See: A.R.D. Mathias, The Strength of Mac Lane Set Theory, Annals of Pure and Applied Logic 110 (2001) 107–234 doi:10.1016/S0168-0072(00)00031-2)

So definable class in Bounded Zermelo can be too complex to be a set (in the sense of quantifier complexity) even when it is contained in some set. But do people in the field use some other term than "too complex"?

$\endgroup$
  • $\begingroup$ And $Z$ (which contains the separation schema) cannot prove that set exists.. In particular $V_{\omega+\omega}$ models $Z$ and does not contain the set $\{ \mathcal{P}^n(\omega): n\in \omega\}$ $\endgroup$ – Not Mike Jan 6 at 23:54
  • 5
    $\begingroup$ @NotMike The language of $Z$ certainly can express: "$X$ is the last entry in some length $n$ sequence of sets where the first entry is $\omega$ and each successive entry is the power set of the one before." And $Z$ proves there are such sequences for every $n\in \mathbb{N}$. Again, of course, it does not prove there is one set of all of them. $\endgroup$ – Colin McLarty Jan 7 at 4:03
  • 1
    $\begingroup$ @NotMike Good point. Luckily i am not asking for a formula describing the union of the class, just describing the class. $\endgroup$ – Colin McLarty Jan 7 at 5:23
  • 2
    $\begingroup$ @NotMike The class $\{\mathcal{P}^n(\omega)\mid n\in \omega\}$ is defined by the following formula $\varphi(x)$: There exists $f$ such that $f$ is a function, $\text{dom}(f) \in \omega\setminus \{0\}$, $f(0) = \omega$, $f(\text{dom}(f)-1) = x$, and for all $n< \text{dom}(f)-1$, $f(n+1) = \mathcal{P}(f(n))$. $\endgroup$ – Alex Kruckman Jan 7 at 5:30
  • 4
    $\begingroup$ I don't know if it is used by bounded Zermelo people, but in some circles, subclasses of sets are called semisets. $\endgroup$ – Emil Jeřábek Jan 7 at 12:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.