I am trying to self study Set Theory. While studying the ZFC axioms, I am introduced to Russell's Paradox and why the Universal Set does not exist. With the Axiom of Restricted Comprehension, Russell's Paradox ceases to exist and it can be shown that the Universal Set does not exist.

My confusion here is why include the Axiom of Regularity if possible cases of sets containing themselves have been eliminated by Restricted Comprehension. There are obviously many more implications of the axiom, but surely the main motivation was to eliminate such sets.

Are their sets that can be constructed using the axioms of ZFC excluding Regularity. Because this to me is the only potential reason why someone would include the Axiom of Regularity.

  • $\begingroup$ See Axiom of regularity : "Virtually all results in the branches of mathematics based on set theory hold even in the absence of regularity. However, regularity makes some properties of ordinals easier to prove. Given the other axioms of Zermelo–Fraenkel set theory, the axiom of regularity is equivalent to the axiom of induction." $\endgroup$ May 14 '18 at 5:58
  • $\begingroup$ @MauroALLEGRANZA so are you trying to say that the main motivation was never to prevent sets containing themselves? $\endgroup$
    – HDatta
    May 14 '18 at 6:00
  • $\begingroup$ See the para about History. See also Well-founded relation and Von Neumann universe. $\endgroup$ May 14 '18 at 6:03
  • $\begingroup$ @MauroALLEGRANZA so well-founded sets are those that do not have an infinite descending chain similar to ${{{{{...}}}}}$ (please correct me if I am wrong). Regularity lets ZFC only be limited to well-founded sets but there are many systems that do not do so. My question still remains the same: if regularity is needed to ensure that non-well founded sets do not exist in ZFC, then there must be a way to construct such sets. Or am I still missing something? $\endgroup$
    – HDatta
    May 14 '18 at 6:11
  • $\begingroup$ Yes, there are Non-well-founded set theories where the axiom is rejected and that allow sets to contain themselves, violating the rule of well-foundedness. $\endgroup$ May 14 '18 at 6:18

There are a few things here:

  1. Russell's paradox predates the Axiom of Regularity. It comes to show how unrestricted Comprehension is inconsistent.

  2. The Axiom of Regularity has nothing to do with the paradox. If $\sf ZF$ is consistent, then $\sf ZF-Reg+\lnot Reg$ is consistent. So the paradox shouldn't be affected from this. So it is also not true that the Axiom of Regularity was formulated to "avoid Russell's paradox" (an unfortunate mistake you can find all over the place).

  3. In the usual proof of Russell's paradox, the Axiom of Regularity makes it one step shorter. $R=\{x\mid x\notin x\}$, then $R$ is the class of all sets in our case, and since $R\notin R$ by the Axiom of Regularity, then by definition $R\in R$. Oops... then it's not a set.

  • $\begingroup$ Thank you for answering. Anyways, so constructing sets that contain themselves must be impossible in ZFC when regularity is excluded. But that would mean all sets in ZFC are well-founded and there would be no reason to include Regularity. If the thing that we need regularity for already been established then all results related to Regularity must also be provable without Regularity. Is there something more to regularity? $\endgroup$
    – HDatta
    May 14 '18 at 6:27
  • $\begingroup$ How did you make those conclusions from what I wrote? $\endgroup$
    – Asaf Karagila
    May 14 '18 at 6:28
  • $\begingroup$ What you are saying is that ZFC without Regularity is completely usable. But Regularity is still included in ZFC by mathematicians. As regularity implies that all sets are well-founded, then there must be sets that can be constructed from the other axioms that are non-well founded and regularity was included to stop that. But if ZFC without Regularity is completely usable then sets containing themselves must not be an issue in ZFC. This where I drew my conclusion from. $\endgroup$
    – HDatta
    May 14 '18 at 6:36
  • $\begingroup$ Okay, let's digress for a second. I'm assuming that you are familiar with the axioms of a field, right? Are these consistent? Do the axioms of a field allow you to construct $\sqrt2$? $\endgroup$
    – Asaf Karagila
    May 14 '18 at 6:38
  • $\begingroup$ I am familiar with the field axioms (though have no experience with abstract algebra). No I do not think so. The axioms seem to have nothing to do with construction of any kind of object and just refer to properties of a field. $\endgroup$
    – HDatta
    May 14 '18 at 6:41

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