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In ZFC, the axiom of regularity states that $$\forall x \neq \emptyset . \exists y \in x . y \cap x = \emptyset $$

The axiom of global choice on the other hand asserts the existence of a global choice operator $\tau$ such that $$ \forall x \neq \emptyset . \tau x \in x$$

These look very similar - in particular, both assert that every nonempty set contains an element, which may(in case of regularity) or may not (in case of choice) have interesting properties.

The question then is - why is it not possible to derive a choice operator by setting $\tau x = y$, (where y is the set whose existence is asserted by regularity) or more strictly $$\tau x = \\\{ z \in y : z \nsubseteq x \\\} $$

This looks almost identical to the structure for the axioms of pairing, union, and powerset, where we first assert the existence of a set using the respective axiom, and then use the separation axiom to exclude extraneous members. The only difference I can see is that regularity/choice are not available for all sets but I'm not sure why that would matter?

TLDR; What prevents us from simply referring to the set whose existence is guaranteed by the axiom of regularity as our global choice "function"/operator

Edit: I understand that "existence" doesn't imply "unique existence" , but I don't really understand how other operators like pair, union or powerset manage to go from "there exists a unique set with these properties" to specifying that a certain bit of syntax allows you to access that set.

Christoph Sachse
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    Regularity tells you that if $x$ is nonempty, then ${y\in x\mid y\cap x=\varnothing}$ is nonempty. You cannot talk about "the set $y$ whose existence is asserted by regularity", you just know such a set exists. You still have to pick one such set for each $x$. – Arturo Magidin Sep 10 '24 at 21:24
  • @ArturoMagidin How do other operators do it? What are the actual mechanics from knowing that a unique such set exists somewhere, to actually "selecting" it without secretly piling up existential quantifiers somewhere? – Christoph Sachse Sep 10 '24 at 22:34
  • What "others"? There is no "operator" here. – Arturo Magidin Sep 10 '24 at 23:03
  • @ArturoMagidin I might be mixing up terminology, sorry. Let's say powerset for instance - given a set X, I know that there exists a unique set of all of its subsets. How does one formally go from there to saying that this set is "called" P(X), and what property of uniqueness (up to exetnsionality) makes this possible? – Christoph Sachse Sep 10 '24 at 23:15
  • Depends on the wording of your axioms. You can state the axiom of the power set as "$\forall x\exists y(z\in y\leftrightarrow (\forall w(w\in z\to w\in x))$, and then extensionality tells you that any two sets satisfying this condition are equal. Then "$P(x)$" is shorthand for that set. Or your axiom may be stated with an implication, then you use separation to describe a set with the biconditional, prove any two such sets are equal by extensionality, and proceed. – Arturo Magidin Sep 10 '24 at 23:23
  • @ArturoMagidin I got the first part, but when we have ∃(∈↔(∀(∈→∈)), how do we "extract" the y from the existential quantifier, and why can I only "extract" it if it's unique (in this case, followed by the biconditional)

    I'm realizing this is more of a first order logic question than set theory at this point :/

     – Christoph Sachse Sep 10 '24 at 23:32
  • It's called existential instantiation. Bit the point is that because you know that any existential instantiation of that formula will always produce the same object, you can use that object to define other objects without ambiguity. An existential instantiation of regularity does not always produce the same object, so you cannot use them to try to then define specific objects. You can do it for the duration of an argument with finitely many steps, but you cannot existentially instantiate a definite $y$ for infinitely many sets at once. – Arturo Magidin Sep 10 '24 at 23:38
  • @ArturoMagidin Thank you so much, that helps explain what I was getting hung up on! – Christoph Sachse Sep 10 '24 at 23:44

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No, of course not. $\sf ZF$ includes the Axiom of Regularity and does not even imply the Axiom of Choice, let alone Global Choice.

The mistake, of course, is that guarantee of existence does not guarantee uniqueness, or that some element is specified. This is exactly where choice is usually needed.

It is, true, however, that with Global Choice we can choose a minimal element from every nonempty set, in the presence of Regularity, of course.

Asaf Karagila
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  • I guess that leads into a more fundamental question - if you do have a proof of uniqueness, ie $\exists x . \forall x' . P(x') \rightarrow x = x' $, how does that lead to an operator (say, unions or powersets) . Do you have to define some structure at the class level over all of V, or...? – Christoph Sachse Sep 10 '24 at 21:43