Talk:Axiom schema of replacement

message: Occasionally the axiom is quoted without the uniqueness requirement. In other words, 
 allowing any predicate, instead of only functional predicates. 
To me this means just take the statement at the top of the page and drop the exclamation mark, i.e.:
:(\forall X, \exist\, Y: P(X, Y)) \rightarrow \forall A, \exist B, \forall C: C \in B \iff \exist D: D \in A \land P(D, C)
which can't be right. I think we want something like
:(\forall X, \exist\, Y: P(X, Y)) \rightarrow \forall A, \exist B, \forall C: C \in A \rightarrow \exist D: D \in B \land P(C, D)
right? Also:
 Given a predicate P, define a functional predicate F by F(x) = 
 { y is of rank α | P(x,y)}.  
 Using replacement, one defines F[X], the union of which satisfies the 
 requirements of boundedness.
Can't there be two elements in x with the same rank? How would F be functional then? -Dan 21:40, 30 April 2006 (UTC)
replies: text: Got it. x is actually the element, and f(x) maps to the class { y | P(x,y) }, collapsed to minimum rank. Errr, this leaves out the bit which says that we can collapse a class to a set like this, the article on rank doesn't help, and the intuition for "collapse" is much weaker than the intuition for "replacement" in the first place! -Dan 16:32, 1 May 2006 (UTC)

text: All boundedness/collection guarantees is that there exists a B which contains an R-image of x for every x in A. It doesn't give us a set containing all the R-images. Thus in this case all it says is that there is a set which has an element. Algebraist 21:16, 30 March 2008 (UTC)
thread_title:  Axiom of boundedness? 

message: How can the axiom of boundedness be equivalent to the axiom of replacement?  Take the relation x R y to mean 0=0 (ie always true), then the mapping under R of {x} is U, the universal class, certainly not a set.  —Preceding unsigned comment added by 81.210.250.222 (talk) 08:11, 30 March 2008 (UTC)
replies: text: All boundedness/collection guarantees is that there exists a B which contains an R-image of x for every x in A. It doesn't give us a set containing all the R-images. Thus in this case all it says is that there is a set which has an element. Algebraist 21:16, 30 March 2008 (UTC)
thread_title:  Axiom of Boundedness (part 2) 

message: A function is a relation. A relation is a subset of cartesian product.  A cartesian product is a set of ordered pairs. An ordered pair is a set. But a set is something that satisfies axioms that is defined by a function. 
lol, wat are u talking about? A is a B , B is a A.
:The issue here is that the "function" being described in the axiom of replacement is not actually a set function; it's really just a formula φ(x,y) that happens associate a single y with each x. — Carl (CBM · talk) 21:24, 30 March 2008 (UTC)
thread_title:  Function 

message: The section Axiom schema of replacement#History and philosophy seems to imply that Zermelo set theory is adequate to prove the existence of each set in Vω·2 (provided that it is a definable subclass of Vω·2). However, I do not see how one could prove that Vω is a set without using replacement. So this section may be misleading in that respect. JRSpriggs (talk) 05:42, 25 February 2010 (UTC)
replies: text: It may depend on exactly how the axiom of infinity is stated, but the axiom of infinity will usually give you Vω and so it's Vω + ω that's problematical. If it doesn't, I would take that as a sign that the axiom of infinity wasn't stated correctly for use in Z set theory. — Carl (CBM · talk) 12:24, 25 February 2010 (UTC)
replies: text: Yes. If, instead of the usual axiom of infinity aimed at producing ω, you said that there is a set which contains the empty set and is either closed under power set and separation or closed under pairing and union, then you could establish the existence of Vω. But I have not seen that version. JRSpriggs (talk) 01:50, 27 February 2010 (UTC)
replies: text: Better, use \exists I \, ( \{\} \in I \land \forall x \in I \, \forall y \in I \, ( x \cup \{ y \} \in I ) ) \,. The only difference between this and the usual axiom of infinity is that x and y may be distinct here and are forced to be the same there. JRSpriggs (talk) 18:32, 28 February 2010 (UTC)
thread_title:  Models of Zermelo's set theory 

message: "defining formulas" may identify surjective functions that are not bijections. Example: Union is definable and there exists only one for each set. The image of the set including only the empty set and the set that includes only the empty set under the definable union function is the set including only the empty set.
Note: This comment may be deleted when the issue is deemed resolved by an expert in the theory of sets.  — Preceding unsigned comment added by 85.97.49.95 (talk) 23:29, 26 October 2012 (UTC)
replies: text: Well caught. Everyone else had overlooked this egregious error until you fixed it. Thank you. JRSpriggs (talk) 05:07, 27 October 2012 (UTC)
thread_title:  Bijection or Surjection 

message: I think the axiom schema of collection, which is often more suited to constructive settings and other settings (see e.g. https://arxiv.org/abs/1110.2430, constructive set theory, and strong collection) should receive its own article. It seems to have quite different properties from replacement when other axioms are changed. Would anyone be interested in helping create and write such an article? To start, I would move the current content under the "Collection" section to such an article, and link to it from this article instead. In general, I think it's best that conceptually distinct axiom schemas receive their own articles (see e.g. axiom of empty set, axiom of adjunction, axiom of regularity vs. epsilon-induction, etc.), since it helps keep things organized and mitigates duplication in other articles (which can just directly link to the article in question instead). 2601:547:501:8F90:B99F:CF8E:E353:38A5 (talk) 06:03, 9 September 2022 (UTC)
thread_title:  Axiom schema of collection deserves its own article