4

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.

HDatta
  • 99
  • 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." – Mauro ALLEGRANZA May 14 '18 at 05:58
  • @MauroALLEGRANZA so are you trying to say that the main motivation was never to prevent sets containing themselves? – HDatta May 14 '18 at 06:00
  • See the para about History. See also Well-founded relation and Von Neumann universe. – Mauro ALLEGRANZA May 14 '18 at 06:03
  • @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? – HDatta May 14 '18 at 06:11
  • 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. – Mauro ALLEGRANZA May 14 '18 at 06:18
  • Au contraire. Because a naive approach to set theory avoids technical axioms like Regularity, and "constructing" sets is generally something that should happen in a naive setting (not to be confused with naive set theory), and because ZFC implies this naive settings is consistent, then Regularity is consistent with the naive approach to set theory. Therefore it is impossible to construct a set which is an element of itself. You need to assume its existence (or some other axioms which imply it, e.g. stratified comprehension à la Quine's New Foundations). – Asaf Karagila May 14 '18 at 06:21

1 Answers1

4

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.

Asaf Karagila
  • 393,674
  • 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? – HDatta May 14 '18 at 06:27
  • How did you make those conclusions from what I wrote? – Asaf Karagila May 14 '18 at 06:28
  • 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. – HDatta May 14 '18 at 06:36
  • 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 $\sqrt2$? – Asaf Karagila May 14 '18 at 06:38
  • 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. – HDatta May 14 '18 at 06:41
  • In that context, "constructing" means defining an object which will function as $\sqrt2$. Just like "constructing the reals" means we define a set (an object in the universe of set theory) which functions like we would expect from the real numbers. If you look at it closely, ZFC only defines the properties of $\in$. – Asaf Karagila May 14 '18 at 06:41
  • So I guess we could define reals using the axioms of a complete ordered field, which would be equivalent to constructing $\sqrt2$ (as well as all the other reals). – HDatta May 14 '18 at 06:47
  • Yes, but what about $\Bbb Q$? Isn't this a model of the axioms of a field? Can you construct $\sqrt2$ there? – Asaf Karagila May 14 '18 at 06:48
  • Surely not, considering only the field axioms. – HDatta May 14 '18 at 06:49
  • Good. So going back to ZFC, look at Regularity as some sort of analog for the completeness axiom (this is wildly incorrect, but bear with me on this). It guarantees that $\sqrt2$ exists, but it doesn't mean that omitting it "lets you construct counterexamples to the existence of $\sqrt2$". We include Regularity because it "makes sense" on the paper, and because it has surprisingly important consequences (the von Neumann hierarchy and $\in$-induction to name two). But omitting it won't let you construct a set $x={x}$ or a set of all the sets not containing themselves. – Asaf Karagila May 14 '18 at 06:53
  • But Regularity does not guarantee anything new, we already knew that all sets are well-founded or did we not know that? – HDatta May 14 '18 at 06:58
  • It's time to grow out of Platonism, or at least grow out of that point of view that we know in advance everything, and mathematics is just about trying to write it down into the books of history. We only know what we proved to follow from what we already assumed. Since people believed that concentrating of regular sets (yes, that was a term), is enough—and they were right—they added Regularity. – Asaf Karagila May 14 '18 at 07:01
  • You should probably also look at https://math.stackexchange.com/questions/1046863/how-can-a-set-contain-itself and https://math.stackexchange.com/questions/253818/example-of-set-which-contains-itself and https://math.stackexchange.com/questions/2720915/real-life-meaning-of-a-set-containing-itself-as-an-element and https://math.stackexchange.com/questions/761681/how-can-i-prove-that-there-is-no-set-containing-itself-without-using-axiom-of-fo – Asaf Karagila May 14 '18 at 07:04
  • Okay that makes a lot of sense now. So regularity is a way of mentioning that we are only going to study well-founded sets. If we do not include it in ZFC, then we do not say anything about what sets are being studied. We cannot prove or disprove the existence of non-well-founded sets. Moreover it is still possible to study non-well-founded sets, just not in ZFC. – HDatta May 14 '18 at 07:10
  • Correct. There is a family of axioms called Anti-Foundation Axioms which posits the existence of various non-well founded sets in various ways. – Asaf Karagila May 14 '18 at 07:11
  • Ok, thank you very much. – HDatta May 14 '18 at 07:12