Equivalent formulations of Global Choice in NBG (without Foundation)

Recently I was looking for various formulations of strong Global Choice in NBG (not assuming Foundation, necessarily), apart from the existence of a well-ordering on the universe or the guarantee of a class of representatives of a given relational class (under Foundation these are well-known to be equivalent to weak Global Choice, which demands the existence of a universal choice or selection functional class). In the same vein, are there "class" versions of "Zorn"'s Lemma, the Knaster-Tarski and Bourbaki-Kneser Theorems ? Any help or insightful comments would be appreciated !

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Rubin H., Rubin, J.E. Equivalents of the Axiom of Choice, II North-Holland, 1985 pp.271-278 "Class Forms"

Some examples from the list:

1. (CAC1) If $S$ is a class of non-empty sets then there is a function $F$ that for every $x\in S$, $F(x)\in x$.

2. (CM1) If $R$ is a partial ordering relation on a non-empty class $X$ and if every subclass of $X$ which is linearly ordered by $R$ has an $R$-upper bound, then $X$ has an $R$-maximal element.

3. (CWO3) Every class can be well ordered

4. (CWO1) There exists a function $F$ such that for every $x$, $F(x)$ well orders $x$.

It goes on, of course, and has many generalizations of known choice equivalences to class form.

Axiom E: There is a function $F$ such that $F(x) \in x$ for every non-empty set $x$. (Taken from Jech, T. Set Theory Springer, 2003 p.70)

• Obviously CAC1 implies E, since $V\setminus\{\varnothing\}$ is a non-empty class. In the other direction, simply take $F\cap (S\times V)$, which is a function from $S$ into the elements of $S$.

• As well CWO1 implies E in a pretty clear manner. In the other direction, note that E implies AC for sets, therefore we can well order every set in $V$. We can define a function $x\mapsto\{f\colon x\to|x|\Big| f\ \text{ bijection}\}$, and the choice would choose one well ordering from the range of this class. Therefore if we can choose, we can choose a well ordering.

(Due to lack of time at the moment I will add the proof for CAC1 equivalent to CM1 (Zorn's lemma for classes) later.)

• Assume CWO1, and let $(A,R)$ be a partially ordered class, that every $R$-chain in $A$ has an upper bound. By CWO1 the class $A$ can be well ordered, denote such well ordering as $<$, without the loss of generality the order type is $\operatorname{Ord}$. Let $x\in A$, then $C_0=\{x\}$ is a chain in $R$, take the $<$-least element which is an upper bound for $C_0$. By transfinite induction, carry on in making the chains longer and longer.
Suppose by contradiction that there is no maximal element, then $R$ has an unbounded chain of length $\operatorname{Ord}$, in contradiction to the fact every chain is bounded.

• In the other direction, suppose CM1. Let $A$ be the class of choice functions, and $f\le g$ if $f\subseteq g$. Let $C$ be a chain in $<$, note that $C$ defines a function. If $\operatorname{Dom}(C)\neq V\setminus\{\varnothing\}$, then there is some non-empty $x\in V\setminus C$. Then $\bigcup C\cup\{\langle x,y\rangle\}$, for some $y\in x$ is an upper bound for $C$. (while $\bigcup C$ is not literally defined, it can be easily written as the $\lbrace y\mid\exists z\in C(y\in z)\rbrace$). Therefore there exists a maximal element in $A$, which is a choice function on $V\setminus\{\varnothing\}$.

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Thank you already, Asaf Karagila: (CAC1) is, of course, the weak form (Gödel's axiom E, I believe), and only known to be equivalent to the strong in case one has regularity (as mentioned); (CWO1) it seems to me, would fall into the same category (weak form), and I believe I mentioned the well-ordering principle (CWO3) (which is indeed equivalent to the strong version), which only leaves (CM1), i.e. a class version of Zorn (et al.). As I don't have access to a copy of Rubin/Rubin's book at present, could you enlighten me as to how Zorn is proven in this case ? Kind regards ! Stephan F. Kroneck. – bonnbaki Sep 8 '11 at 9:43
P.S. My problem is that in all the derivations of Zorn's lemma in set form that I know of, one cannot get around forming power sets in one way or other (be it using the Bourbaki-Kneser fixed point theorem for completely ordered posets, or Kneser's proof), clearly not an option for classes. In truth, what is neede is the ability to form the union (of a self-indexed) class of classes. Or is there some trick in finding the correct predicative description of the class required in the respective proofs ? Kind regards ! Stephan F. Kroneck. – bonnbaki Sep 8 '11 at 9:49
I have added some proofs to show that regularity is not needed. – Asaf Karagila Sep 8 '11 at 10:19
Thank you again, Asaf Karagila ! As mentioned in my original question and the follow-up, I am aware of the relationships between conditions (CAC1), (CWO1) and (CWO3) (no problem proving that); new to me is only (CM1) (which I've been trying to prove on my own, but see aforementioned snags; which, by the way, I also encounter when trying to generalize the Knaster-Tarski fixed point theorem to classes, in hope providing a simple proof of Schröder-Bernstein for classes, despite the moot nature of this result). (ctd.) – bonnbaki Sep 8 '11 at 11:45
How does the condition that every surjective functional class should have a right inverse fit in, for instance ? Kind regards - Stephan F. Kroneck. – bonnbaki Sep 8 '11 at 11:49