Talk:Axiom of regularity

message: Is it not the case that a colon means 'such that' and a right-pointing arrow 'implies'? So that the experssion of the axiom on this article reads "For all A such that A is non the empty set, implies there exists B..."
That doesn't make any sense, does it? 'such that' OR 'implies' would be fine, but putting both together seems to me to render it meaningless. |Olaf Davis 17:12, 23 Sep 2006 (UTC)
replies: text: No, the colon doesn't mean anything.  Here it's just syntactic sugar to separate the quantifier and bound variable from the rest of the formula.  In other words, this notation assumes that whenver \phi is a well-formed formula, so is \forall x: \phi.  The formula can be read in many ways, but not the way you had in mind.  For example: "for all, A, colon, A, not equals, open brace, close brace, right arrow,..." or, in the only slightly more informative (but potentially misleading) "for all A, if A is not equal to the empty set, then...".  I've changed the notation and added explicit parentheses to mitigate the confusion. --MarkSweep (call me collect) 02:31, 24 September 2006 (UTC)
replies: text: I felt that the formula was still hard to read. So I rewrote it again. I hope that you agree with me that this is clearer. JRSpriggs 06:52, 24 September 2006 (UTC)
thread_title:  First-order logic formation 

message: I probably have some math wrong in this, but I feel there must exsist a set that has itself as a member. For example, take the universal set, which is defined as EV,Ax(x=x <-> x<V) Now, its possible, though difficult, to prove that Ax(x=x), so using a version of modus ponens, we get EV,Ax(x<V). Now, by the axiom of power sets, there is a set that contains all of the subsets of the universal set. Since all members of sets are sets, and the universal set is a subset of the universal set, the universal set is a set, and therefore
:Ax,EV(x<V)
:EV(x<V)
:EV(V<V)
Now, the lemma states
:~(V<V)
Generalize
:AV~(V<V)
Take the earlier expression
:EV(V<V)
Expand the E
:~AV~(V<V)
and we also have
:AV~(V<V)
I use the principle of explosion to say that I did something wrong, the lemma is incorectly justified, or that there is a big hole in set theory.
Does anyone know which of the three it is? Thanks!
Indeed123 02:21, 26 May 2007 (UTC)
replies: text: There is no all-purpose "universal set" in ZF. The claim that \exists V \ni \forall x \left ( x = x \leftrightarrow x \subset V \right ) requires unrestricted comprehension — that is, it requires that you be able to define a set based only on a predicate (in this case, being an element of another set), rather than based on a predicate plus a superset. (If you've no clue what I'm talking about, see Axiom schema of specification#Unrestricted comprehension.) So, all you've shown is that the axiom of regularity is incompatible with some axioms, which is fine; the only necessary thing is that it be compatible with the other axioms of ZF (though there are other sets of axioms it's also compatible with). —RuakhTALK 06:01, 26 May 2007 (UTC)
thread_title:  A strange lemma 

message: The article includes the following lines:
:The axiom of regularity enables defining the ordered pair (a,b) as {a,{a,b}}
:This definition eliminates one pair of braces from Kuratowski's canonical definition (a,b) = {{a},{a,b}}.
Why is the axiom of regularity necessary for this? If we define the ordered pair (a,b) as {a,{a,b}}, then the only way we couldn't extract the order is if {a,{a,b}}={b,{a,b}}, which can only be the case if a=b or a={a,b} and b={a,b}, in which case again a=b, and so (a,b)=(b,a) and there's no problem. skeptical scientist (talk) 13:41, 8 October 2007 (UTC)
replies: text: If a = {a,b}, then { a, {a,b} } = { {a,b} } is a singleton. — Carl (CBM · talk) 14:50, 8 October 2007 (UTC)
replies: text:  And more to the point, if a = {a', b'} and a' = {a, b}, where b and b' are distinct, then (a, b) = { a, {a, b} } = { {a', b'}, a' } = (a', b'), but a is not a' and b is not b'. So the definition doesn't work. 70.135.20.81 06:33, 4 December 2007 (UTC)
thread_title:  Defining the orrdered pair 

message: I thought the ordinal ω was included in ZF.  If so, isn't it true that ω intersects \omega-1?  It seems then, that since ω is in ZF, we can create the sequence \omega\ni\omega-1\ni\omega-2\ni\cdots.  What's my error? 129.107.240.1 (talk) 12:25, 23 December 2008 (UTC)
replies: text: There is no \omega-1 in the universe of sets. JRSpriggs (talk) 01:53, 24 December 2008 (UTC)
thread_title:  infinite decreasing sequences 

message: "Let A be a set such that A is an element of itself and define B = {A}, which is a set by the axiom of pairing. Applying the axiom of regularity to B, we see that the only element of B, namely, A, must be disjoint from B. But A is both an element of itself and an element of B. Thus B does not satisfy the axiom of regularity and we have a contradiction, proving that A cannot exist."
This appears to prove merely that no singleton is an element of itself, which does not necessarily generalize to sets with more than one element.  Consider the set A = {A, {∅} }.  In this case, {∅} is both an element of A and disjoint from A, which appears to satisfy the axiom of regularity even though A is an element of itself.  Can anyone explain why this doesn't work?
Jesin (talk) 17:02, 2 August 2010 (UTC)
replies: text: The article is saying that if A is any set that is an element of itself, then the axiom of regularity as stated here fails for the corresponding set {A}. The fact that the axiom of regularity would work for the other set {A, ∅} is neither here nor there. It just takes one failure of the axiom of regularity to show that the set A could not be an element of itself. — Carl (CBM · talk) 18:18, 2 August 2010 (UTC)
replies: text: I don't think that the proof as written is sufficient--it's incomplete and has poor pedagogy because of this. A more complete proof might note one of the various other axioms that require that if A is an element, then there must be a set that contains A and only A. If we have such a set and call it B, then the axiom of regularity fails because both A and B contain A.Guyminuslife (talk) 00:57, 22 July 2012 (UTC)
replies: text: The article clearly states that B={A}={A,A} is a set due to the axiom of pairing. JRSpriggs (talk) 12:16, 22 July 2012 (UTC)
thread_title:  Clarify "No set is an element of itself"? 

message: In the proof that no infinitely decreasing membership sequence of sets exists, it claims range(f) is a set due to the Axiom Schema of Replacement.  In fact, the Axioms of Union and Restricted Comprehension are sufficient given the definition of a function, meaning Zermelo Set Theory is sufficient -- we don't need to use ZF.
--Achur (talk) 19:38, 14 March 2011 (UTC)
replies: text: I see what you mean, and I'm not certain why it is phrased that way. perhaps the reasoning was that the current proof shows something stronger, that there is not even a definable function that enumerates an infinite sequence of sets each of which is an element of the previous one. — Carl (CBM · talk) 20:20, 14 March 2011 (UTC)
thread_title:  Axiom Schema of Replacement Not Necessary 

message: Substituted at 01:47, 5 May 2016 (UTC)
thread_title: Assessment comment

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replies: text: I reverted since there was no problem with the original site, and the substitute had problems. JRSpriggs (talk) 14:58, 23 October 2016 (UTC)
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