Talk:Axiom of regularity

for every non-empty set S there is an element a in it which is disjoint from S

I really don't understand what this is saying (but this could well just be me being thick). Are we saying that the element a is both 'in the set S' and 'disjoint from S'? That's what the sentence says to me - does the word 'it' refer to the set S? How can element a be both within the set and disjoint from it? --Stuart Presnell 29/11/2002

One has to keep in mind that elements are themselves sets. So while A ∈ S, we can also consider A ∩ S. S = { A , B , C } and A = { B }, for example. Then A ∩ S = { B }

Ok, now it's my turn being thick :P saying that a is an element of S but they are disjoint sounds to me like a paradox. What subset of S = {A, B, C} is disjoint to S? I mean the only subsets are

a = {}

a = {A}

a = {B}

a = {C}

...

a = {A, B, C}

and so on. None of these are disjoint from S (not counting {} obviously), they all share atleast one element! Can some one please explain :S Gkhan 16:59, Apr 8, 2005 (UTC)

The axiom is about elements of S being disjoint from S, not about subsets. --MarkSweep 19:07, 8 Apr 2005 (UTC)

---

Surely sets and elements are different! a is not the same as {a} or {{a}}.

This needs some care. -- User:David Martland

You're confusing several things here. First, sets and elements are not ontologically different, since there are sets of sets, i.e., sets whose elements are themselves sets. If you want to, you can introduce a unary predicate is-a-set into most axiomatic set theories, but that doesn't buy you a whole lot. In any case, this issue is somewhat orthogonal to the question of whether generally a is distinct from {a} (they are of course notationally distinct, but the real issue is under what conditions they might be equal).

Second, regarding a not being the same as {a}, this is a good illustration of why axiomatic set theory exists in the first place: people don't have reliable intuitions about what sets are and under what conditions two sets, intuitively defined, are the same. Naive set theory is based on intuitions about well-founded finite sets that one may encounter in the physical world; for those sets <math>a\neq\{a\}<math> holds without exception. But what if you were to allow the set {{{{...}}}} (call it <math>\Omega<math> and define it as <math>\Omega=\{\Omega\}<math>, i.e. the singleton set that is a member of itself) -- then you would finde that <math>a\neq\{a\}<math> does not hold universally, but has exceptions such as <math>a=\Omega<math>. Fact is that people have no reliable intuitions about non-well-founded sets. Zermelo had a problem with them and that's why the Axiom of Foundation rules them out. Aczel and others find non-well-founded sets useful, and consequently there are non-standard set theories that scrap the foundation axiom altogether and indeed postulate the existence of non-well-founded sets. In either case, the axiomatic approach forces people to spell out precisely what they think sets are and how they behave. The Axiom of Foundation is one way of resolving this issue. It is a bit cryptic (or make that "elegant") and its full power only becomes clear when one considers its interaction with the other axioms of Zermelo(-Fraenkel) set theory.

--MarkSweep 08:56, 7 Sep 2004 (UTC)

---

Under the axiom of choice, this axiom is equivalent to saying there is no infinite sequence {a_n} such that a_i+1 is a member of a_i

I'm changing this, because the axiom of choice is not required to prove the result. Onebyone 16:08, 25 Oct 2003 (UTC)

... in one of the directions, I meant to say. Onebyone 21:01, 10 Nov 2003 (UTC)