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Kunen's Set Theory uses the following definitions of Axiom of Pairing and Axiom of Union:

Pairing $$\forall x \forall y \exists z (x \in z \wedge y \in z)$$ Union $$\forall \mathcal{F} \exists A \forall Y \forall x (x \in Y \wedge Y \in \mathcal{F} \rightarrow x \in A)$$

Jech's approach to 'weaker' means less specific union/intersection sets, whereas I am using it here to mean pairwise (e.g. Kunen's Pairing, 'weaker') vs. arbitrary (e.g. Kunen's Union, 'stronger') formula. This question is motivated by an (analogous?) asymmetry between unions and intersections in topology.

I understand the restriction of intersections to pairwise formula in topology (as to avoid deriving singleton open sets from nets). However, I don't understand the limitation in ZFC. Is it simply because a 'strong axiom of pairing' is also derivable from ZFC (or is it not, or only in certain models)? Or rather, are there some metalogical reasons to restrict the intersection axiom to a pairwise formula?

In short, why is the axiom of pairing not: $$\exists A \forall x \forall \mathcal{F} [(\forall y (y \in \mathcal{F} \rightarrow x \in y)) \rightarrow x \in A] $$

silly-little-guy
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  • (Not 100% about my attempted formulation of a strong axiom of pairing...) – silly-little-guy Dec 14 '24 at 16:14
  • Pairing form pair, means two. – Mauro ALLEGRANZA Dec 14 '24 at 16:32
  • What is $A$ in the last formula? Did you mean $\mathcal{F}$? – Jean Abou Samra Dec 14 '24 at 17:04
  • @JeanAbouSamra forgot the $\exists$, sorry. same $A$ as in the original axiom (i.e., the actual resulting union). – silly-little-guy Dec 14 '24 at 17:05
  • What does a strong axiom of pairing mean? – DRF Dec 14 '24 at 17:09
  • @DRF as hinted by Mauro's discomfort, perhaps i would be better rephrasing by replacing 'Axiom of Pairing' with 'Axiom of Intersection'. for me, a 'weak' axiom of intersection means the statement uses only pairs - as in Kunen's formulation. a 'strong' axiom of intersection uses a family/collection of sets (as opposed to a pair), analogous to the Axiom of Union's $\mathcal{F}$. – silly-little-guy Dec 14 '24 at 17:13
  • Oh I think I see, what you mean assuming you're taking Jexh's definition. But in that case both the pairing and union axiom's in Kunen's text are weak versions. The thing you wrote at the end has many flaws (quantifier ordering being the main) but eve assuming you fix that you don't get a "strong" version. You also don't get a pairing axiom. I'll write up an answer later based on what I think you might be asking. – DRF Dec 14 '24 at 17:14
  • @DRF my understanding is bad but from my reading, Jech's definition of 'weak' vs. 'strong' is whether or not the elements of the union/intersection set are the only elements. (e.g., for Jech, a weak axiom of pairing means that the set $A$ formed from the pairing axiom might also contain arbitrary $z \in A$ with $z \neq x \wedge z \neq y$.) (i.e., not the same as my definition!) – silly-little-guy Dec 14 '24 at 17:22
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    Really quickly. The reason there is no intersection axiom weak or strong is because the set of axiom's in Kunen (and I assume Jech) is in some sense minimal and intersection is provable. Pairing is not intersection. Also surprisingly pairing creates "bigger" sets whereas union creates "sets of the same size" which is sort of anti intuitive. But both are weak as written in Kunen. The reason I remember him giving for it (I was lucky enough to actually take set theory from him) was that it was cleaner in his view. – DRF Dec 14 '24 at 17:26
  • @DRF i think the size issue you've mentioned may be where my discomfort lies, and ties in with Jean's answer. to lay out my thoughts using your addition: $\mathcal{F}$ is not necessarily a set in the Axiom of Union - the only reason this is possible/allowed is because regardless of $\mathcal{F}$'s 'size', the Axiom of Union is bounded by $\mathcal{F}$'s elements which are sets. however, the 'Axiom of Association' (as i'm redubbing it, per comments below) doesn't bound its constructed set, so a strong definition would be perverse (allowing construction of nonsets). is this gibberish? – silly-little-guy Dec 14 '24 at 18:18
  • @DRF (also that is extremely cool you got to study under Kunen! i'm a big fan of just the little work i've read so far.) – silly-little-guy Dec 14 '24 at 18:21
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    @silly-little-guy In the axiom of union, $\mathcal{F}$ definitely is a set. In set theory, you can only quantify over sets. – Jean Abou Samra Dec 14 '24 at 18:22

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I think there are several misunderstandings at play.

  • The axiom of pairing is not an axiom of intersection restricted to an intersection of two sets. Intersection for two sets $x, y$ means you can form the set $x ∩ y$, pairing lets you form the set $\{x, y\}$ which is entirely unrelated.

  • The asymmetry between having an axiom of union and no axiom for intersection is only apparent. It is provable in ZFC that you can take the intersection of all sets in a given set:

    $$∀ F, ∃ A, ∀ x, x ∈ A ⇔ (∀ f ∈ F, x ∈ f)$$

    Compare with union:

    $$∀ F, ∃ A, ∀ x, x ∈ A ⇔ (∃ f ∈ F, x ∈ f)$$

    The way to derive this form of intersection is to take the union thanks to the axiom of union and then select only those elements which belong to all elements of $F$, rather than any element of $F$. The reason you first need to apply union is that specification will not let you build the set of elements having any property (see Russell's famous paradox), but only the elements from an already constructed set having the property.

  • There is no reasonable "generalized" version of pairing where you "pair" any number of elements from a given set… it would basically say that given a set, one can construct the set of all its elements, but we already have the set itself. The reason we need pairing is that otherwise there is no way we can write sets constructed explicitly from a finite list of elements.

Jean Abou Samra
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  • i've mucked up this whole discussion by incorrectly introducing 'Intersection' to try and generalize 'Pairing'. I should have said 'Association' or something like that. – silly-little-guy Dec 14 '24 at 17:35
  • @silly-little-guy I'm still not sure what a generalized “axiom of association” would be. Your proposal $∃A, ∀x, ∀F, (∀y ∈ F, x∈y) ⇒ x∈A$ is disprovable, because $A$ would need to contain all sets. – Jean Abou Samra Dec 14 '24 at 17:38
  • am i correct in noting that the essential difference between your formulation and mine (other than the use of $f$ in place of $y$, and the biconditional) is the quantifier ordering $\forall F \exists A \forall x$ in place of $\exists A \forall x \forall \mathcal{F}$? (apologies and thanks for your answer. trying to work my way down my confusions here...) – silly-little-guy Dec 14 '24 at 17:44
  • @JeanAbouSamra Ignore the messed up quantification order and put the exists at the end. You'll see the idea. But that's just singleton pairing followed by union. – DRF Dec 14 '24 at 17:51
  • the metalogical status of 'constructing explicitly from a finite list of elements' is why I am interested in a generalization, i think. if the Axiom of Association were instead phrased to (strongly, like the Axiom of Union) use a collection (not necessarily a set, right?) $\mathcal{F}$ rather than (weakly) any two elements of the universe, we would be allowing ourselves to construct sets explicitly from (potentially) infinite lists of elements. Whereas, with the pairwise axiom only permits finite lists. I'm not sure how to prove the strong version (or if its even true). – silly-little-guy Dec 14 '24 at 18:13
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    If your list of elements is not finite and explicit, then you cannot write it down in your proof, so you have to find some other way to specify it. That might be an already constructed set… in which case why try to construct the set of these things if you already have it? And if it's a proper class (what you call a collection), then in general you simply cannot make it a set. For example, you can consider the class of all sets, but there is no set of all sets. – Jean Abou Samra Dec 14 '24 at 18:25
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    F most definitely is a set. Union is a proper axiom not an axiom schema. Since you're quantifiing ovef F it better be a set. – DRF Dec 14 '24 at 18:25
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    There are so-called infinitary logics where proofs can be infinite. There this might make sense, but not in traditional set theory. – Jean Abou Samra Dec 14 '24 at 18:26
  • i was clearly confused on a lot here, but i think the status of $\mathcal{F}$ was my primary hurdle. thanks everyone for helping me detangle! – silly-little-guy Dec 14 '24 at 18:35
  • @si I really recommend playing around with small bits of ZFC to see what you create/what satisfies them. It gives you a good idea what they actually do. – DRF Dec 14 '24 at 19:45
  • @DRF i've benefitted immensely from Kunen's approach in Foundations of Mathematics where we test each axiom against a graph-theoretic representation of the $\epsilon$-membership relation. seeing extensionality that way finally helped me grasp its specificity. – silly-little-guy Dec 14 '24 at 19:57