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I don't think I could ever say I understand ZFC if I don't get to the bottom of this question. It all started when we had extensionality and comprehension. Then Russell finds a paradox and here comes ZFC to the rescue. We keep extensionality and we introduce a series of axioms that build the sets we need and whatever we haven't built is out. Here's what puzzles me. I don't think there is any ZFC axiom that would let us build a set that is an element of itself. So why do I need the axiom of regularity? I guess I cannot prove $x \notin x$ without it, but what does it matter if with the axioms I have I will never be able to build such set? If ZFC with regularity is consistent, doesn't that mean that it will always be impossible to use the other axioms to provide a set that is an element of itself? I am puzzled because, apart from extensionality which existed before ZFC, all axioms build sets that would not exist otherwise, while regularity seems to prevent me from having sets I can't build in the first place.

  • The axiom of regularity gives "structure" to the universe of sets. If you accept regularity then every set can be built up starting from the empty set, i.e, every set appears in $V_{\alpha}$ for some ordinal $\alpha$. Without regularity you loose this structure. – Vivaan Daga Feb 20 '24 at 15:01
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    The way regularity was first explained to me is that it (sort of, in an "internal" sense...) says that every set in the universe is genuinely "built using" and "reachable from" the other axioms of ZF. This matches up with how I first learned naïve set theory and how I use it "in real life" - sets are always "built" by taking unions, or powersets, or subsets. In Conway's terminology, it says "every set has a birthday". – Izaak van Dongen Feb 20 '24 at 17:12

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We don't really need the Axiom of Regularity in ZFC. Having it is a matter of convenience.

We've known since the 1930s that if the variant of Zermelo-Fraenkel set theory that lacks the axiom of regularity is consistent, then so is Zermelo-Fraenkel set theory itself. The argument is essentially what you allude to in your post: if we have a structure that satisfies all the axioms of ZF set theory without Regularity, then the well-founded sets of the structure in fact satisfy all the axioms of ZF, including Regularity. And we don't need anything else apart from these well-founded sets to formalize ordinary mathematical notions such as $\mathbb{R}$ and function spaces.

So why do we take Regularity anyway?

  • It makes certain foundational activities a lot easier. For example, it lets us define the ordered pair $(a,b)$ as $\{a,\{a,b\}\}$, and an ordinal number simply as a transitive set of transitive sets. In turn, this simplifies the definition of cardinal numbers (which are just specific kinds of ordinals), and combined with the well-ordering principle lets us show that every set has a cardinality, and cardinalities have a linear order (the latter question is common enough even in "ordinary" mathematics: undergraduates I taught predictably asked about it when we discussed the uncountability of $\mathbb{R}$).

  • In general, Regularity comes in handy when set-theoretic and "ordinary" mathematical constructions interact. For example, you might want to study which ordinals embed into the Dedekind-complete ordered field, and then you really don't want to worry about whether the answer depends on choosing one of the isomorphic copies of $\mathbb{R}$ that happens to contain a non-well-founded $S = \{S\}$ or not. Thanks to Regularity, you can just assume it doesn't. Similarly, the Mostowski collapse theorem is occasionally useful, and wouldn't work without Regularity.

  • By itself, ZFC minus Regularity is not a convenient setting to study non-well-founded sets anyway. Consider the simple case, where we have two non-well-founded sets $S = \{S\}$ and $T = \{T\}$, and we'd like to know whether $S = T$ holds. To use Extensionality, we'd want to exhibit a member of $S$ that $T$ does not contain, or the other way around. But to do that, we'd have to know whether $S=T$ holds in the first place. So if we do get rid of Regularity with the aim of studying non-well-founded sets, it makes sense to adopt some strong anti-foundation axiom in its stead. Symmetrically, if we don't intend to study non-well-founded sets, it does no harm, and makes a lot of sense, to just take Regularity, instead of working around its absence (even though most of the working around would not happen in "ordinary mathematics").

  • Finally, when we study restricted, less powerful subsystems of ZFC, certain principles which do imply Regularity become very important. One can use the axiom of epsilon-induction to construct transitive closures of sets: ordinarily, this requires Replacement.

Z. A. K.
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    (+1) on a side note, Transitive closures also require axiom of infinity (in the same sense that they require replacement, i.e. ZF-inf doesn’t prove TC unless we have $\in$-induction rather than just set regularity) – spaceisdarkgreen Feb 20 '24 at 16:44
  • Isn't the ability to represent the property '$\alpha$ is an ordinal' using a $\Lambda_0$ formula - thereby making it absolute in transitive classes - a key pillar of the field of consistency and independence proofs? Without it, transitive classes destined to be models can have new ordinals popping in them out of thin air. The field would collapse, or at least require a thorough review of what's still valid. – Chad K Feb 20 '24 at 17:36
  • @ChadK That's an argument that we need regularity (or well-founded sets) as an idea, not that we need every set to be well-founded, foundationally speaking. I'd consider the axiom as a powerful tool in establishing consistency results, though not necessarily "true" of naive set theory (much like the axiom of constructibility). One can use antifoundation axioms in consistency arguments as well. – spaceisdarkgreen Feb 20 '24 at 18:19
  • Hi, why did you state that "it lets us define the ordered pair (a,b) as ${a,{a,b}}$" ? I don't see why axiom of regularity needed for this definition of ordered pair. Can you please explain me why ? Many thanks! – InTheSearchForKnowledge Aug 04 '24 at 21:18
  • @InTheSearchForKnowledge: Try proving that $(a_1, b_1)=(a_2,b_2)$ implies $a_1=a_2$ and $b_1=b_2$ without invoking Regularity. – Z. A. K. Aug 05 '24 at 01:52
  • Hello, for example the proof in the page 23, section 6 (ordered pairs) of Naive set theory by Halmos, there is no mention of Regularity. Can you please look at it ? – InTheSearchForKnowledge Aug 05 '24 at 09:28
  • @InTheSearchForKnowledge: That's because it's not ${a,{a,b}}$ that Halmos uses, but rather ${{a},{a,b}}$. Different set. NB The general point here is that it's harder to characterize the behavior of set-based constructions in even some very simple cases when we don't have Regularity, not "which definition of ordered pair is best". – Z. A. K. Aug 05 '24 at 09:46
  • @InTheSearchForKnowledge: On one hand I'm happy to answer questions. OTOH I can't help but think that you'd have noticed the difference had you tried proving that $(a_1,b_1)=(a_2,b_2)$ implies $a_1=a_2$ and $b_1=b_2$ when I suggested :/ – Z. A. K. Aug 05 '24 at 09:53
  • My bad, actually I tried to prove it but failed to notice the subtle difference between the definitions. Many thanks for your responses! And please forgive me for my stupidity :) – InTheSearchForKnowledge Aug 05 '24 at 11:26
  • Hi, can you explain why in order to use extensionality, we should know whether $S = T$ in the first place ? I can't see any reason why because without extensionality, there is no definition of "equal". And how regularity helps us with this ? Many thanks! – InTheSearchForKnowledge Aug 05 '24 at 20:45
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    @InTheSearchForKnowledge: Extensionality is not a definition of equality, it's an axiom that states that two sets are equal if they have the exact same elements. Let's say you want to prove $S=T$ using Extensionality. The only element of $S$ is $S$, and similarly for $T$: to prove that $S$ and $T$ have the same elements, you'd first have to prove that their elements are equal, i.e.... $S=T$ again. Given Regularity, sets like $S$ and $T$ don't even exist: problem solved. – Z. A. K. Aug 06 '24 at 00:24
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Most of the axioms of $\mathsf{ZFC}$ are "weak" in the sense that they only tell you what sets definitely exist, not that those are the only sets or that some sets don't exist. The exceptions are extensionality, foundation, and choice. In the case of foundation, it explicitly concerns set that don't exist, so indeed you don't "need" it. If you omit foundation you get what's called "neutral non-well-founded set theory," but it doesn't not have non-well-founded sets: there are models with and without non-well-founded sets; whether it has them is independent of the theory.

Leaving the question of foundation open doesn't really make much sense though. After all, as you note, in neutral non-well-founded set theory all you can talk about is the well-founded part: you still have to hypothesize that the non-well-founded part actually exists to say anything about it. Thus you could argue that you should either stick to just the well-founded part, taking foundation and the convenient assumption that all sets are well-founded, or be explicit that the non-well-founded part exists, adding an "anti-foundation" axiom that allows you to construct non-well-founded sets.

So why do we tend to take foundation rather than an anti-foundation principle? Besides aesthetic and convenience reasons, probably mostly historical reasons. Not all early set theories were well-founded, but well-founded reasoning has been present throughout, and in particular non-well-founded set theories tend to refute choice, which adds another explicit hypothesis required to develop analysis. There's a rich body of work in the set theory of well-founded sets and ordinals and comparatively little in non-well-founded set theory, although it does see its uses.

Soundwave
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