A I can't understand axiom of regularity...

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1. Feb 7, 2016

ankit.jain

It states that no set can be an element of itself..what does it exactly mean?why can't. We consider the whole set as a single element?

2. Feb 7, 2016

Staff: Mentor

You have to exclude the possibility that a set can contain itself as an element because otherwise you will run into paradoxa: "The barber in this village shaves everyone except those who shave themselves."
So does the barber shave himself or not?

Of course you can consider sets which contain other sets as elements.
E.g. consider a set $S=\{1,2,3\}$. Then the set of all subsets of $S$ is $P(S) = \{∅,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$.
Please regard that although $S$ is an element of $P(S)$ neither of both contains itself as an element. $P(S)$ is called power set of $S$.

3. Feb 7, 2016

micromass

This is a very popular misconception. The axiom of regularity does not avoid any paradoxes. The barber paradox cannot be avoided by the axiom of regularity. It is the strengthening of the comprehension axiom that avoids the barber paradox.

Indeed, it has been shown that there are many nice models of ZFC- that are perfectly consistent (given ZFC- is consistent) and that do not satisfy the axiom of regularity. For an example of such models, see the nice book by Hrbacek and Jech.

As for the OP: it's just an axiom, and it's not an essential axiom. Almost all of math goes through perfectly fine if we choose to accept the axiom, or if we do not choose to accept the axiom. To 99% of mathematics, the axiom is irrelevant (and that can be given an informal argument). I myself choose not to accept the axiom of regularity because it is irrelevant. I never work with sets which are an element of itself anyway, even though I'm perfectly fine with them existing.

4. Feb 7, 2016

Staff: Mentor

Thank you for clarification.

At this point I got stuck. If the axiom of regularity reads ∀x (x ≠ ∅ ⇒ ∃y ∈ x (y∩x = ∅) ) so why isn't the construction S = {S} forbidden by it, since x = y = S are the only elements and therefore in contradiction to the axiom. At which point am I wrong?

5. Feb 7, 2016

micromass

You are right that constructions of the form $S = \{S\}$ are forbidden by the regularity axiom. But that was not my point. My point is that it has been proven that the axiom of regularity is relative consistent with ZFC, but so is its negation. This means that if ZFC has a contradiction somewhere, then so will ZFC + regularity. The contradiction will just be slightly different, but it is there.

So if some kind of barber's paradox shows up without regularity, then some paradox will show up with regularity too.

6. Feb 7, 2016

ankit.jain

Still not clear.

7. Feb 8, 2016

Samy_A

What exactly is still not clear?
Yes, the whole set is considered as a single element. The axiom of regularity implies that for any set S, S∈S is not allowed.

As micromass explained, this axiom isn't terribly important for most of Mathematics.

8. Feb 21, 2016

mbs

Yes. It isn't that important because development of the integers and arithmetic do not depend on it at all. It simply forces the notion of a "set" to agree with what our intuition says a set should represent. Infinite or circular inclusion chains do not make a whole lot of practical sense.

9. Feb 21, 2016

FactChecker

How can you ever say that you have precisely defined a set that contains itself? I am curious if there is a well-defined example of such a thing. If not, I think it is irrelevant to mathematics.

10. Feb 21, 2016

micromass

There are axioms of anti-regularity that deal with such things. So there are quite precise way of dealing with this. It's still true it's irrelevant to most mathematics though.

11. Mar 3, 2016

mbs

From my understanding, you don't have to define them. You simply omit the axiom of regularity/foundation and study the hypothetical "sets" that would violate it. $\in$ is just a binary predicate relating two objects we happen to call "sets" in set theory. Every object in the domain of the first-order system of ZFC is called a "set". It is the axioms themselves that demand "sets" operate as we would envision collections of "things" in the real world. Of course, we can have other things in the universe for which the extensionality axiom doesn't apply. You then have to add an "is a set" predicate and preface the extensionality axiom with it

12. Mar 3, 2016

micromass

Of course we can just leave out the regularity axiom, but that doesn't get us far. All that negating the foundation axiom does, is saying there is one set for which the foundation axiom fails. You don't even have any information regarding the set. And even if you do. For example, take a set $x$ for which $x=\{x\}$. Is such a set $x$ unique? How do we check equality with such sets? The extensionality axiom is useless here.
So just leaving out the regularity axiom is not useful, to have something useful, we should replace it with a better axiom for which we can actually prove things with. For example https://en.wikipedia.org/wiki/Aczel's_anti-foundation_axiom

13. Mar 3, 2016

FactChecker

I think I understand. So leaving it out does not necessarily mean that there is, in fact, a set that contains itself. It just means that you can not use that axiom (no set contains itself) in proofs.

14. Mar 3, 2016

micromass

You're correct of course, but I interpreted his post as assuming the negation of regularity.

15. Mar 3, 2016

FactChecker

I guess you could hypothetically define an example, just by saying that the set X has itself as an element. Although I would like to say that X is not well-defined, I would need to use the axiom to say that.

16. Mar 3, 2016

mbs

Okay. I don't think I said anything that contradicts anything you're saying. A set for which the axiom of foundation fails is simply a (non-empty) set S in which every member of S contains another member of S. Simply removing the foundation axiom keeps such sets purely hypothetical. You would need addition axioms to formally construct the existence of all such sets (much like the axiom of infinity constructs the integers). Simply having one non-regular set isn't enough to prove anything.

17. Mar 4, 2016

MrAnchovy

What a fascinating area of study - I couldn't understand why I had not seen this before, then realised that I finished full-time academic study two years before Aczel's paper. Can you point me to a good, relatively discursive book? I was never very fond of books containing long formal proofs of theorem after theorem with no apparent motivation and 30 years out of study has not made me any more so.

18. Mar 4, 2016

micromass

The book by Hrbacek and Jech contains a nice chapter on this anti-foundation thing. It's a really great book to read. https://www.amazon.com/Introduction-Edition-Revised-Expanded-Mathematics/dp/0824779150

19. Mar 4, 2016