0
$\begingroup$

$\DeclareMathOperator\Cls{Cls}\newcommand{\ZFC}{\mathrm{ZFC}}$Let ZF be extended with class and set predicates, and extend the theory with a class abstraction schema so that for all formulas $\phi$:

$$\forall \vec{z}\;\exists Y\;(\Cls(Y)\wedge \forall x(x\in Y\leftrightarrow Set(x)\wedge\phi(\vec{z},x)))$$

Let us call the theory $\ZFC^{\Cls}$.

We know that many classes in $\ZFC^{\Cls}$ are proper classes, such as Russell's class $R=\{x|x\notin x\}$.

What happens if we instead of the usual axiom of choice have axioms which for any set s, of nonempty non-overlapping sets, just postulate a class Y, which has precisely one member from each member of s?

For example, it seems clear that the Vitali set is not definable in $\ZFC^{\Cls}$.

Does $\ZFC^{\Cls}$ extend $\mathrm{ZF}$ in any interesting way?

Improve this question
$\endgroup$
11
  • $\begingroup$ @ZuhairAl-Johar You are right. Thanks! $\endgroup$
    – Frode Alfson Bjørdal
    Commented Oct 22, 2021 at 19:08
  • $\begingroup$ I presume you mean class variables, not class predicates, since you quantify over $Y$. Isn't the theory just equivalent to ZFC? (You haven't really spelled out the axioms of your theory regarding classes. But I expect the axioms to be arranged so that you can use Separation to separate a choice set from the choice class for $s$, and in the other direction, any choice set for $s$ should give you a choice class for $s$.) $\endgroup$
    – Farmer S
    Commented Oct 23, 2021 at 5:49
  • $\begingroup$ @FarmerS, the exposition is incomplete, for example Extensinaity is not an axiom and also all axioms of ZF are relativized to the predicate $Set$, I think by using class and set predicates he actuay means a mono-sorted FIRST order language extended with two primitive one place predicate symbols $Cls; Set$ denoting "is a class" and "is a set" respectively. The question actually needs to be re-written by the author to clarify his intentions. $\endgroup$
    – Zuhair Al-Johar
    Commented Oct 23, 2021 at 7:51
  • 1
    $\begingroup$ I think (not sure) that Randall Holmes's purported proof of Con(NF) uses a trick where choice is in some sense external. Notice that there is no known inconsistency with Quines ML + Class choice over sets, but of course it's inconsistent with Set choice $\endgroup$
    – Zuhair Al-Johar
    Commented Oct 23, 2021 at 8:45
  • $\begingroup$ @Zuhair My exposition perhaps simplifed too much. As is usual, Set(x) is short for Ey( x\in y). Cls(x) is short for not-Set(x). The axiom (x)(Cls(x)) is presupposed. $\endgroup$
    – Frode Alfson Bjørdal
    Commented Oct 23, 2021 at 13:06

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .