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Suppose you in a more comprehensive framework has the result that U is the least set such that all axioms of NBG-- (NBG, minus extensionality and regularity), hold; replacement is here, as expected, obtained relative to the least class containing U and (membership class) E, and as well closed under complement, intersection, domain, Cartesian product with U, circular permutation and transposition.

It seems obvious that regularity holds for U.

Is it obvious that extensionality holds for U?

Edit:

The question(s) presuppose that NBG is consistent.

Edit 2: I changed the original V to U, so as to avoid confusion with the cumulative hierarchy.

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    $\begingroup$ @FrodeAlfsonBjørdal Are you sure that that's what's shown in More on the Axiom of Extensionality? I'm looking at a copy of it right now and it looks like Scott is proving that ZF minus extensionality and foundation is strictly weaker than ZF minus foundations. $\endgroup$
    – James E Hanson
    Commented Oct 17, 2023 at 3:32
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    $\begingroup$ @FrodeAlfsonBjørdal Okay well then the statement depends in a strong way on the precise formulation of replacement, so I feel like you should specify that when making this kind of statement, especially when the commonly cited result in this area (given here for instance) is more or less the opposite. $\endgroup$
    – James E Hanson
    Commented Oct 17, 2023 at 4:22
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    $\begingroup$ @FrodeAlfsonBjørdal I don't think you can really claim that your statement is correct in an unqualified sense when the paper you're citing uses a materially different definition of 'ZF minus extensionality and foundation' and literally states in the second paragraph that ZF minus extensionality and foundation is strictly weaker than ZF minus foundation. $\endgroup$
    – James E Hanson
    Commented Oct 17, 2023 at 5:05
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    $\begingroup$ @FrodeAlfsonBjørdal Frankly, your original question is also just poorly written. You don't specify which version of replacement you are using (which as we have just established, can significantly change the character of the theory in question), and you expected the reader to just know what 'closure under E' and 'circular permutation' mean. In general, you seem to be expecting others to do the lion's share of the work of interpreting your question, rather than putting the effort in yourself to communicate an unambiguous question. $\endgroup$
    – James E Hanson
    Commented Oct 17, 2023 at 7:21
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    $\begingroup$ @FrodeAlfsonBjørdal regardless of the other parts of the discussion between yourself and James, I share the opinion that pasting two raw BibTeX entries is not helping the reader. Most preferable is something that looks like a proper bibliographic entry as you'd see in a paper, and, if possible, a link (and best of all a stable link). This book does not seem easy to find, so obfuscating its information inside BibTeX code makes matters even harder for people trying to answer and help here. $\endgroup$
    – David Roberts ♦
    Commented Oct 18, 2023 at 0:13

1 Answer 1

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$U$ can be shown to only contain wellfounded sets by an inductive argument, given the assumption stated that replacement is obtained relative to the least class containing $U$ and $E$, and as well closed complement, intersection, domain, Cartesian product with $U$, circular permutation and transposition. There is no special assumption upon replacement here.

Given Gandy's result in Gandy, R. On the Axiom of Extensionality II, Journal of Symbolic Logic, 24.4, 287–300, 1959, one cannot prove the existence of non-extensional classes in $U$.

So the more comprehensive theory may safely postulate that $$(1) \ \ \forall x\in U\colon \mathrm{Wellfounded}(x),$$ and $$(2) \ \ \forall (y,z)\in U^2\colon .\forall x\in U\colon(x\in y\leftrightarrow x\in z)\to \forall x\in U\colon(y\in x\to z\in x).$$

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