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Some posts [1] [2] suggest that the axiom of foundation (AoF) is essentially pointless, except that it allows us to prove that every set has a rank. I mostly deal with math built atop set theory, and I'm not familiar with any results that actually require (or are more easily proven by) the fact that every set has a rank.

Are any "important" theorems lost (or made harder to prove) by removing AoF? By important, I mean a result that is used in practice in mathematical theories built atop ZFC. For example, Zorn's lemma is important because it helps us prove that every vector space has a basis and every field has an algebraic closure. I'd like to know about similar applications of AoF, if they exist.

WillG
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  • Doesn't the Axiom of Foundation make forcing proofs much easier specifically because you know that sets all have ranks? – Robert Shore Jul 15 '21 at 20:06
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    A particularly nifty application of Foundation (and for which it is necessary) is called Scott's trick. For example, when you have an equivalence relation on a proper class, you can work as though each equivalence class is a set by considering, for each equicalence class, the subclass consisting of sets of lowest rank in that class. Foundation then ensures that such subclasses are sets. This is useful in constructing ultrapowers and also in defining cardinality within choiceless contexts. – Jason Zesheng Chen Jul 15 '21 at 20:13
  • This question and its answers may be of interest. – spaceisdarkgreen Jul 16 '21 at 00:24

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Consider the axiom scheme of collection, which says roughly $\forall x \exists y P(x, y) \implies \exists c \forall x \in B \exists y \in c P(x, y)$ for any predicate $P$.

I don't know of any way to prove collection from replacement without using foundation. The proof with foundation is:

For each $x$, let $\alpha_x$ be the smallest ordinal such that there is some $y$ of rank $\alpha$ such that $P(x, y)$. Then define $g(x) = \{y$ of rank $\alpha_x$ | $P(x, y)\}$. Then $\{g(x) | x \in B\}$ forms a set. Take its union $c$. Then $\forall x \in B$, $g(x)$ is nonempty and hence there is some $y \in g(x) \subseteq c$, and this $y$ must satisfy $P(x, y)$.

As mentioned in the comments, another application of foundation is showing that proper classes can be quotiented by equivalence relations.

Edit: according to this answer, collection does not follow from ZFC without foundation. So it's actually a necessary component of the proof.

Mark Saving
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