Equivalent formulations of Global Choice in NBG (without Foundation)

Question

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 !

Answer 1

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 $V\setminus\{\varnothing\}$.